tag:blogger.com,1999:blog-22077897416937892962024-03-17T23:03:08.657-04:00Cool Math StuffEthan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.comBlogger189125tag:blogger.com,1999:blog-2207789741693789296.post-70294988871902959712014-04-12T12:00:00.000-04:002014-04-12T12:00:03.883-04:00Math Awareness Month Part 2: Infinite SeriesToday's page for Math Awareness Month is about a recent video that caused some huge debate. I saw the video a month or two ago, and was very intrigued. I showed it to some of my friends, and we were arguing about the content for quite a while. It also spread rapidly around the math department at Andover, with some teachers bringing up in their classes.<br />
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Take a look at the page and try some of the exercises. You will find the outcomes very interesting and mind-boggling. The concept of infinity is difficult for any human being to grasp, making it tons of fun to think about.<br />
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<a href="http://www.mathaware.org/mam/2014/calendar/infinity.html">http://www.mathaware.org/mam/2014/calendar/infinity.html</a><br />
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Comment below what you think of the video. Do you think it is accurate? What do you think the fallacies are? How could this be a part of string theory if it is mathematically flawed?<br />
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In math class last week, we were given the following problem:<br />
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I then did the math and determined that the limit would be -1/12. I then called my teacher over, and pointed to that answer. Recalling the video, I asked him if I could rewrite that -1/12 as 1+2+3+4+5+6+7+... as my final answer. Thankfully, he got the reference. In addition to being a funny anecdote, the fact that people got the joke shows how wide of an audience this information has reached and captivated, which is amazing to see.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-83832012534025648732014-04-05T12:00:00.000-04:002014-04-05T12:00:05.820-04:00Math Awareness Month Part 1: Magic SquaresI explained a bit in my last post that April is Math Awareness Month, as well as linked to the poster on www.mathaware.org. In honor of this occasion, I plan to make my posts this month relevant to the pages on the website and the mathematicians hosting them.<br />
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April 1st was a day on magic squares, and I am honored to have been the host of that page. There is a recent performance of me doing it, tutorials on how to make various magic squares, and different activities and questions that can further your magic square experience. Click <a href="http://www.mathaware.org/mam/2014/calendar/magicsquares.html" target="_blank">here</a> to see the page.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-88426149036011424502014-03-29T12:00:00.000-04:002014-03-29T12:00:02.815-04:00Conclusion to Half-Tau MonthThough pi day passed a few weeks ago, my brother made the interesting observation that this month is "Pi Month." It is March of 2014, or 3/14. Since I spent the month focused on trigonometry, I never had a chance to honor this joyous occasion until today.<br />
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Interestingly, pi does play a huge role in trigonometry. From wrapping functions to sine curves to angle measurements, pi is always popping up. Though this is kind of interesting, trigonometry is also one of the areas where tau really shines. Having just finished topics such as trigonometry, polar coordinates, and wrapping functions, I have found that it is a real struggle to use pi. I found myself converting most of my problems to tau before solving them just because pi made it too confusion.<br />
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No video describes these sorts of issues better than Vi Hart's "Pi is (still) Wrong" video, which gets into some of the issues involved with using pi, one of these being trigonometry.<br />
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I would also like to make you all aware that next month is Math Awareness Month. The theme this year is Mathematics, Magic, and Mystery, in part to honor the centennial of <a href="http://en.wikipedia.org/wiki/Martin_Gardner" target="_blank">Martin Gardner</a>'s birth. At<a href="http://www.mathaware.org/" target="_blank"> www.mathaware.org</a>, there is a poster with 30 squares on it to represent the 30 days of April. On each day of the month, the next square becomes active. I will try to keep an eye on these webpages, as I will probably use April to comment on the topics posted there. Also, April 1st is a page on magic squares, and I am extremely honored to be hosting that day.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-25897703700909154602014-03-22T12:00:00.000-04:002014-03-22T12:00:02.204-04:00Rediscovering Trigonometry Part 4: More Useful Formulas<span style="font-family: inherit;"><span style="background-color: white; color: #444444; line-height: 18px;">Click <a href="http://coolmathstuff123.blogspot.com/2014/03/rediscovering-trigonometry-part-1.html" style="color: #4d469c; text-decoration: none;" target="_blank">here</a> to see part one of this four week series.</span><br style="background-color: white; color: #444444; line-height: 18px;" /><span style="background-color: white; color: #444444; line-height: 18px;">Click <a href="http://coolmathstuff123.blogspot.com/2014/03/rediscovering-trigonometry-part-2.html" style="color: #4d469c; text-decoration: none;" target="_blank">here</a> to see part two of this four week series.</span></span><br />
<span style="font-family: inherit;"><span style="background-color: white; color: #444444; line-height: 18px;">Click <a href="http://coolmathstuff123.blogspot.com/2014/03/rediscovering-trigonometry-part-3-half.html" target="_blank">here</a> to see part three of this four week series.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">Now that we have discovered some useful trigonometric identities, we can continue to build on them and create many more. There are an infinite number of trigonometric identities out there (not all of them have been created of course), but we will stick to two in this post: the product-to-sum formulas and the sum-to-product formulas.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">Take the four angle addition/subtraction formulas we discovered in our first week. I will use <i>A</i> and <i>B</i> as our letters rather than alpha and beta.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">1. sin(<i>A</i> + <i>B</i>) = sin<i>A</i>cos<i>B</i> + cos<i>A</i>sin<i>B</i></span></span><br />
<span style="color: #444444;"><span style="line-height: 18px;">2. sin(<i>A</i> – <i>B</i>) = sin<i>A</i>cos<i>B</i> – cos<i>A</i>sin<i>B</i></span></span><br />
<span style="color: #444444;"><span style="line-height: 18px;">3. cos(<i>A</i> + <i>B</i>) = cos<i>A</i>cos<i>B</i> – sin<i>A</i>sin<i>B</i></span></span><br />
<span style="color: #444444;"><span style="line-height: 18px;">4. cos(<i>A</i> – <i>B</i>) = cos<i>A</i>cos<i>B</i> + sin<i>A</i>sin<i>B</i></span></span><br />
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These can be messed with very easily to create some new formulas. For instance, adding together the first two formulas gives:<br />
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<span style="color: #444444; line-height: 18px;">sin(</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;"> + </span><i style="color: #444444; line-height: 18px;">B</i><span style="color: #444444; line-height: 18px;">) + </span><span style="color: #444444; line-height: 18px;">sin(</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;"> – </span><i style="color: #444444; line-height: 18px;">B</i><span style="color: #444444; line-height: 18px;">)</span><span style="color: #444444; line-height: 18px;"> = sin</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;">cos</span><i style="color: #444444; line-height: 18px;">B</i><span style="color: #444444; line-height: 18px;"> + cos</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;">sin</span><i style="color: #444444; line-height: 18px;">B + </i><span style="color: #444444; line-height: 18px;">sin</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;">cos</span><i style="color: #444444; line-height: 18px;">B</i><span style="color: #444444; line-height: 18px;"> – cos</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;">sin</span><i style="color: #444444; line-height: 18px;">B</i><br />
<span style="color: #444444;"><span style="line-height: 18px;">2sin<i>A</i>cos<i>B</i> = </span></span><span style="color: #444444; line-height: 18px;">sin(</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;"> + </span><i style="color: #444444; line-height: 18px;">B</i><span style="color: #444444; line-height: 18px;">) + </span><span style="color: #444444; line-height: 18px;">sin(</span><i style="color: #444444; line-height: 18px;">A</i><span style="color: #444444; line-height: 18px;"> – </span><i style="color: #444444; line-height: 18px;">B</i><span style="color: #444444; line-height: 18px;">)</span><br />
<span style="color: #444444; line-height: 18px;">sin<i>A</i>cos<i>B</i> = [sin(<i>A</i> + <i>B</i>) + sin(<i>A</i> – <i>B</i>)]/2</span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">We now have a new identity. This can be now be used to solve a whole new range of problems and generate a whole new range of identities. The same steps can be done by subtracting the second equation from the first, adding the third and fourth together, and subtracting the fourth from the third. This creates the four product-to-sum identities.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">These four formulas can be rewritten in a way that converts the sum into a product. Let's rewrite the variables as the following:</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;"><i>a</i> + <i>b</i> = <i>A</i></span></span><br />
<span style="color: #444444;"><span style="line-height: 18px;"><i>a</i> – <i>b</i> = <i>B</i></span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">Making this change, we can then perform some operations to get a whole new set of formulas. These are called the sum-to-product identities.</span></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbexJjzxjUC9Suz_xyh-F4gwRmveia6oKxAv7qRnAgB7jbEqlGwCpRnMUZbB0u8dHFFAhXU5vPL5vkID5223fnfBb000lwQnWTQd5nnHfJ3Bix8jgH-BnFqcHkHTpLJP_zTnlWDhVK8mwQ/s1600/img6.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbexJjzxjUC9Suz_xyh-F4gwRmveia6oKxAv7qRnAgB7jbEqlGwCpRnMUZbB0u8dHFFAhXU5vPL5vkID5223fnfBb000lwQnWTQd5nnHfJ3Bix8jgH-BnFqcHkHTpLJP_zTnlWDhVK8mwQ/s1600/img6.gif" height="240" width="400" /></a></div>
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<span style="color: #444444;"><span style="line-height: 18px;">These can then be built upon to generate whole new sets of formulas as well. Though the actual mathematics here might be a bit complicated, the idea is simple. Mathematics is always continuing to be developed, and this can be done through building upon previous ideas to form new ideas that help solve new problems. Trigonometry is a great place to see this sort of thing happen.</span></span>Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-71476422470064481192014-03-15T12:00:00.000-04:002014-03-15T12:00:01.576-04:00Rediscovering Trigonometry Part 3: Half Angle Formulas<span style="font-family: inherit;"><span style="background-color: white; color: #444444; line-height: 18px;">Click </span><a href="http://coolmathstuff123.blogspot.com/2014/03/rediscovering-trigonometry-part-1.html" style="background-color: white; color: #4d469c; line-height: 18px; text-decoration: none;" target="_blank">here</a><span style="background-color: white; color: #444444; line-height: 18px;"> to see part one of this four week series.</span></span><br />
<span style="font-family: inherit;"><span style="background-color: white; color: #444444; line-height: 18px;">Click </span><a href="http://coolmathstuff123.blogspot.com/2014/03/rediscovering-trigonometry-part-2.html" style="background-color: white; color: #4d469c; line-height: 18px; text-decoration: none;" target="_blank">here</a><span style="background-color: white; color: #444444; line-height: 18px;"> to see part two of this four week series.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">Last week, we figured out a way to figure out trigonometric functions for twice a given angle. This week, we will do the same, but for determining the trigonometric functions for half a given angle. First, let's discuss what half of an angle means.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">Remember that there are 360° in a circle, or 360° in a full revolution. This means that a number like 370° can also be expressed as 10°. Though they are different measurements, plugging 370° into a trigonometric function will yield the same answer as 10°. It is also equivalent in most other situations in mathematics.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">For double angle formulas, we did not need to discuss this. This is because when doing double angle formula calculations with these measurements, there would be no issue.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">sin(2 • 10°) = sin(20°)</span></span><br />
<span style="color: #444444;"><span style="line-height: 18px;">sin(2 • 370°) = sin(740°) = sin(740° - 720°) = sin(20°)</span></span><br />
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<span style="color: #444444; font-size: x-small;"><span style="line-height: 18px;">Note that 720° is two full revolutions around a circle, and thus, it can be subtracted off when performing a trigonometric operation.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">But performing a half angle calculation will create more of an issue. Let's use 10° and 370° again.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">sin(1/2 • 10°) = sin(5°)</span></span><br />
<span style="color: #444444;"><span style="line-height: 18px;">sin(1/2 • 370°) = sin(185°)</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">These answers are not the same. Since they are both in the 0° - 360° interval, we cannot make any assumptions. We do know that the sine and cosine of 185° are the negative sine and negative cosine of 5° respectively, but this proves that they are not equal. If one were go up to 730°, they would be back to normal, however.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">sin(1/2 • 730°) = sin(365°) = sin(365° - 360°) = sin(5°)</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">This means that for every angle, the half sine and half cosine function should yield two answers. The two answers should have the same absolute value, but different signs (they are the same number, but one is negative and one is positive). You may already have a function in your head that can create this type of situation, but we will be able to derive it as well.</span></span><br />
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<span style="color: #444444;"><span style="line-height: 18px;">Take a variation of the cosine double angle formula that we derived last week:</span></span><br />
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<span style="background-color: white; color: #444444; line-height: 18px;"><span style="font-family: inherit;"><span style="line-height: 19px; text-align: center;">cos(2</span><span style="line-height: 19px; text-align: center;">α) = 1</span><span style="line-height: 19px; text-align: center;"> – 2sin</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="line-height: 19px; text-align: center;">α</span></span></span><br />
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<span style="background-color: white; color: #444444; line-height: 18px;"><span style="font-family: inherit;"><span style="line-height: 19px; text-align: center;">Let's try to isolate sin</span></span></span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α. We would first add that to the left hand side and subtract the cosine of 2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α over to the right hand side.</span><br />
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<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">2sin</span><sup style="background-color: white; color: #444444; line-height: 19px; text-align: center;">2</sup><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α = 1 – cos(2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α)</span><br />
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<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">Divide through by 2 to get:</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">sin</span><sup style="background-color: white; color: #444444; line-height: 19px; text-align: center;">2</sup><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α = (1 – cos(2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α))/2</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">And square root both sides to get:</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">sin</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α = ±√((1 - cos(2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α))/2)</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; text-align: center;"><span style="color: #444444;"><span style="line-height: 19px;">Notice how there is a ± sign in front of the square root. This is because when one squares a positive or negative value, it becomes positive. For instance, the equation <i>x</i><sup>2</sup> = 25 would be solved as <i>x</i> = ±5 because (5)(5) = 25 and (–5)(–5) = 25. The same thing happened here. But also remember what we found before. We proved through logic that the half sine and half cosine of an angle has two answers, one negative and one positive. With that in mind, we can see that this is the accurate way to write the square root (some derivations call for just a positive answer such as the Distance Formula).</span></span></span><br />
<span style="background-color: white; text-align: center;"><span style="color: #444444;"><span style="line-height: 19px;"><br /></span></span></span>
<span style="background-color: white; text-align: center;"><span style="color: #444444;"><span style="line-height: 19px;">Let's replace angle </span></span></span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α with </span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α/2 to keep the half angle definition. This gives a formula of:</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">sin(</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α/2) = </span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">±√((1 - cos</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α)/2)</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">Let's derive a cosine half angle formula. We can take another variation on the cosine double angle formula and go forward.</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 18px;"><span style="font-family: inherit;"><span style="line-height: 19px; text-align: center;">cos(2</span><span style="line-height: 19px; text-align: center;">α) = 2</span><span style="line-height: 19px; text-align: center;">cos</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="line-height: 19px; text-align: center;">α – 1</span></span></span><br />
<span style="background-color: white; color: #444444; line-height: 18px;"><span style="font-family: inherit;"><span style="line-height: 19px; text-align: center;"><br /></span></span></span>
<span style="background-color: white; color: #444444; line-height: 18px;"><span style="font-family: inherit;"><span style="line-height: 19px; text-align: center;">This time, we will only need to add one to both sides to isolate the cos</span></span></span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α term. Let's also flip the equation around to make it simpler.</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">cos</span><sup style="background-color: white; color: #444444; line-height: 19px; text-align: center;">2</sup><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α = 1 + cos(2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α)</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">Divide through by 2 to get:</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">cos</span><sup style="background-color: white; color: #444444; line-height: 19px; text-align: center;">2</sup><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α = (1 + cos(2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α))/2</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">And square rooting both sides yields:</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">cos</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α = ±√((1 + cos(2</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α))/2)</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">Again, we end up with a ± sign in the formula. This means that we probably did everything correctly, as logic shows we will need this sort of sign to create two answers. Rewriting </span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α as </span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α/2 gives a final formula of:</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">cos(</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α/2) = ±√((1 + cos</span><span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">α)/2)</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;">It is tough to see what these formulas actually look like in this formatting, so I have written them out in LaTeX so you can see what is going on more conveniently.</span><br />
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNWKRunrfdzy_W57tfElTuhPHU3VxhvK7zdky-tMCb_Ch7E5Qk349HqvhNGWYVOYooNZWFvXrtjoQT6_0EokiRBfAzFF-ziXOhFNNM5KUM01R8jfBMbjRNwYraKB-QswTNGJNju5drjkw_/s1600/Screen+Shot+2014-03-10+at+10.54.52+AM.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNWKRunrfdzy_W57tfElTuhPHU3VxhvK7zdky-tMCb_Ch7E5Qk349HqvhNGWYVOYooNZWFvXrtjoQT6_0EokiRBfAzFF-ziXOhFNNM5KUM01R8jfBMbjRNwYraKB-QswTNGJNju5drjkw_/s1600/Screen+Shot+2014-03-10+at+10.54.52+AM.png" /></a></div>
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These formulas themselves are pretty cool, but the logic involved in finding them is also very interesting. The fact that we could predict the nature of the function before we even found it is really helpful. This can be huge in trying to figure out the right way to go about solving a problem.</div>
<span style="background-color: white; color: #444444; line-height: 19px; text-align: center;"><br /></span>Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-70156982761225717752014-03-08T12:00:00.000-05:002014-03-08T22:03:17.488-05:00Rediscovering Trigonometry Part 2: Double Angle FormulasClick <a href="http://coolmathstuff123.blogspot.com/2014/03/rediscovering-trigonometry-part-1.html" target="_blank">here</a> to see part one of this four week series.<br />
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Last week, we developed some formulas that could calculate the sine, cosine, and tangent of the sum and difference of two angles. This proves useful when trying to calculate the exact sines/cosines/tangents of angles that are not in special right triangles (45-45-90, 30-60-90, 18-72-90).<br />
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What if we wanted to find the exact values for double a certain angle. Let's say we know the following:<br />
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sin18 = (√(5) - 1)/4<br />
cos18 = (√(10 + 2√(5)))/4<br />
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How could we calculate the sine of 36°? We know the sine angle addition formula from last week, but it would be much easier to have a generalized version of this. Let's take a look at it.<br />
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sin(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> + β) = sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ + cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ<br />
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What if we were finding the sine of 2<span style="background-color: white; line-height: 19px; text-align: center;">α? Let's see what the formula would tell us:</span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; line-height: 19px; text-align: center;">sin(2</span><span style="background-color: white; line-height: 19px; text-align: center;">α) = sin(</span><span style="background-color: white; line-height: 19px; text-align: center;">α + </span><span style="background-color: white; line-height: 19px; text-align: center;">α) = sin</span><span style="background-color: white; line-height: 19px; text-align: center;">α cos</span><span style="background-color: white; line-height: 19px; text-align: center;">α + cos</span><span style="background-color: white; line-height: 19px; text-align: center;">α sin</span><span style="background-color: white; line-height: 19px; text-align: center;">α = 2sin</span><span style="background-color: white; line-height: 19px; text-align: center;">α cos</span><span style="background-color: white; line-height: 19px; text-align: center;">α</span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; line-height: 19px; text-align: center;">So with some pretty simple computations, we get the formula:</span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; line-height: 19px; text-align: center;"><b>sin(2</b></span><span style="background-color: white; line-height: 19px; text-align: center;"><b>α) = 2sin</b></span><span style="background-color: white; line-height: 19px; text-align: center;"><b>α cos</b></span><span style="background-color: white; line-height: 19px; text-align: center;"><b>α</b></span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><b><br /></b></span>
<span style="background-color: white; line-height: 19px; text-align: center;">That's a pretty simple formula. To find the sine of 36°, we could just do some easy manipulation with the values from up top.</span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; line-height: 19px; text-align: center;">sin36 = sin(2 • 18) = 2 • (</span>(√(5) - 1)/4) • ((√(10 + 2√(5)))/4) = (√(10 - 2√(5)))/4<br />
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This sort of computation is clearly very useful when you are studying trigonometry. It wouldn't be practical in the sense that we use it in our day-to-day lives, but I think it is clear how useful this is to mathematicians and astronomers. I also find it really cool that these seemingly random irrational values can be derived exactly using just some basic mathematics.<br />
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Let's create a cosine formula. We know from last week that:<br />
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cos(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> + β) = cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ – sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ<br />
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Substituting <span style="background-color: white; line-height: 19px; text-align: center;">α in for</span> β gives:<br />
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cos(2<span style="background-color: white; line-height: 19px; text-align: center;">α) = cos(</span><span style="background-color: white; line-height: 19px; text-align: center;">α + </span><span style="background-color: white; line-height: 19px; text-align: center;">α) = cos</span><span style="background-color: white; line-height: 19px; text-align: center;">α cos</span><span style="background-color: white; line-height: 19px; text-align: center;">α – sin</span><span style="background-color: white; line-height: 19px; text-align: center;">α sin</span><span style="background-color: white; line-height: 19px; text-align: center;">α = cos<sup>2</sup>α – sin<sup>2</sup>α</span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; line-height: 19px; text-align: center;">This is the formula that naturally comes out:</span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><br /></span>
<span style="background-color: white; line-height: 19px; text-align: center;">cos(2</span><span style="background-color: white; line-height: 19px; text-align: center;">α) = </span><span style="background-color: white; line-height: 19px; text-align: center;">cos</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α – sin</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α</span><br />
<b><span style="background-color: white; line-height: 19px; text-align: center;"><br /></span></b>
<span style="background-color: white; line-height: 19px; text-align: center;">Knowing that </span><span style="background-color: white; line-height: 19px; text-align: center;">sin</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α + </span><span style="background-color: white; line-height: 19px; text-align: center;">cos</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α = 1, this can be rewritten in a few different ways:</span><br />
<span style="background-color: white; line-height: 19px; text-align: center;"><br /></span>
<b><span style="background-color: white; line-height: 19px; text-align: center;">cos(2</span><span style="background-color: white; line-height: 19px; text-align: center;">α) = </span><span style="background-color: white; line-height: 19px; text-align: center;">cos</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α – sin</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α</span></b><br />
<b><span style="background-color: white; line-height: 19px; text-align: center;">cos(2</span><span style="background-color: white; line-height: 19px; text-align: center;">α) = 2</span><span style="background-color: white; line-height: 19px; text-align: center;">cos</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α – 1</span></b><br />
<b><span style="background-color: white; line-height: 19px; text-align: center;">cos(2</span><span style="background-color: white; line-height: 19px; text-align: center;">α) = 1</span><span style="background-color: white; line-height: 19px; text-align: center;"> – 2sin</span><sup style="line-height: 19px; text-align: center;">2</sup><span style="background-color: white; line-height: 19px; text-align: center;">α</span></b><br />
<b><span style="background-color: white; line-height: 19px; text-align: center;"><br /></span></b>
<span style="background-color: white; text-align: center;"><span style="line-height: 19px;">This is extremely convenient, as the problem can be made much easier depending on what information you have. If you only know the sine of the angle, the third formula will work. If you only know the cosine, the second formula will work. There are also times where the first formula might be most convenient.</span></span><br />
<span style="background-color: white; text-align: center;"><span style="line-height: 19px;"><br /></span></span>
<span style="background-color: white; text-align: center;"><span style="line-height: 19px;">As you can see, it is pretty simple to do the work to come up with the double angle formulas, probably even easier than applying them in many cases. I encourage you to do the same process as we have done for sines and cosines to generate one for tangents, using the tangent formula we found last week. You will get a very pretty result.</span></span><br />
<span style="background-color: white; text-align: center;"><span style="line-height: 19px;"><br /></span></span>
<span style="background-color: white; text-align: center;"><span style="line-height: 19px;">Also, it can be fun to play around with these and find the exact sines and cosines of various angles. You will also see that there are many ways to write these different values. This is because they are irrational numbers. You could find the sine of 105° by doing sin(45 + 60), sin(90 + 15), sin(180 - 75), or many other variations. These could very well give different looking answers, but if the math was correct, the actual results will be equal.</span></span>Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-7260711353441846482014-03-01T12:00:00.000-05:002014-03-01T12:00:09.329-05:00Rediscovering Trigonometry Part 1: Addition and Subtraction FormulasIn January, I did a series with some heavier mathematics, involving lots of number theory, combinatorics, and some fairly difficult manipulation. Last month, we took a little rest from this by talking a bit about casino games and optimal strategies. The mathematics we used was more probability and game theory. This month, I thought we could explore something completely different that is also extremely interesting. It is taught in school (much of the material in this post actually comes straight from my math class), but maybe not in a fun, thought-provoking way. This month, we will dive into trigonometry.<br />
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If you want to brush up on your basic foundational trigonometry, please click <a href="http://coolmathstuff123.blogspot.com/2013/07/the-law-of-cosines.html" target="_blank">here</a>. It is my post on the Law of Cosines, but also explains at the beginning what the sine, cosine, and tangent functions are. I will make sure to prove any other trigonometric identities we use in this series (or link to a proof), so all you need to know is what a sine, cosine, and tangent is.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZDiwsxQk26RlhEaQxrclNVLzjM5CBOZS82sqk-vj13Ug8sPbN4b9RfmiKBGnwWqCFOy7xgb-vhWQAPzf6S1PZRkYJPo2zG_n3tJ0xwRRtpfejyGXvFUwtcsJZ6el5DbRn2lRhvgSdpGuO/s1600/funny-puns-fruity-trig.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZDiwsxQk26RlhEaQxrclNVLzjM5CBOZS82sqk-vj13Ug8sPbN4b9RfmiKBGnwWqCFOy7xgb-vhWQAPzf6S1PZRkYJPo2zG_n3tJ0xwRRtpfejyGXvFUwtcsJZ6el5DbRn2lRhvgSdpGuO/s1600/funny-puns-fruity-trig.jpg" height="177" width="400" /></a></div>
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Now we are ready to derive our first identity. We will try to figure out a simpler way of writing sin(<span style="background-color: white; line-height: 19px; text-align: center;">α</span>+β). There are many proofs of this, but I am using one that I found at www.themathpage.com. Let's start with a diagram to help explain this idea.</div>
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Create the line AB.</div>
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Rotate it to point C, creating the angle <span style="background-color: white; line-height: 19px;"><span style="font-family: inherit;">α</span></span>.</div>
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Rotate it to point D, creating the angle β.</div>
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Draw DE perpendicular to AB.</div>
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Draw DF perpendicular to AC.</div>
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Draw FG perpendicular to AB.</div>
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Draw FH perpendicular to ED.</div>
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We are trying to figure out the sine of <span style="background-color: white; line-height: 19px; text-align: center;">α</span>+β, which can be written as (ED/DA). First, let's determine the value of the angle HDF (which has already been written in for us).</div>
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Note that the lines HF and AG are parallel, with a transversal (a line that intersects both lines) of AF. In geometry, there is a theorem called the Alternate Interior Angle Theorem, that essentially tells us that in this circumstance, the measure of angle GAF is equal to the measure of angle AFH. In other words, angle AFH has a measure of <span style="background-color: white; line-height: 19px; text-align: center;">α</span>. Noting that angle AFD has a measure of 90 and AFH has a measure of <span style="background-color: white; line-height: 19px; text-align: center;">α</span>, that means angle HFD would have a measure of 90–<span style="background-color: white; line-height: 19px; text-align: center;">α</span>. Since a triangle has 180 degrees, having two angle measures in a triangle is enough to get the third. Triangle HFD has an angle of 90, an angle of 90–<span style="background-color: white; line-height: 19px; text-align: center;">α</span>, and one more angle that creates a sum of 180 degrees. Do the math and you will find that the angle must be <span style="background-color: white; line-height: 19px; text-align: center;">α</span>, and thus, angle HDF has a measure of <span style="background-color: white; line-height: 19px; text-align: center;">α</span> as depicted in the diagram.</div>
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Now we can try to find the formula. Let's start with the following equality:</div>
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ED = GF + HD</div>
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Note that we are trying to find what (ED/DA) equals. Because of this, let's divide both sides by DA.</div>
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(ED/DA) = (GF/DA) + (HD/DA)</div>
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Let's complicate some terms on the right hand side. Multiply the fraction on the left by (AF/AF) and the one on the right by (FD/FD)</div>
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(ED/DA) = (GF/AF)(AF/DA) + (HD/FD)(FD/DA)</div>
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(ED/DA) is simply the sine of <span style="background-color: white; line-height: 19px; text-align: center;">α</span>+β. Looking back at the diagram, (GF/AF) would be the sine of <span style="background-color: white; line-height: 19px; text-align: center;">α</span>. (AF/DA) would be the cosine of β. (HD/FD) would be the cosine of <span style="background-color: white; line-height: 19px; text-align: center;">α</span>. (FD/DA) would be the sine of β. This gives us the following formula:</div>
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<b>sin(</b><span style="background-color: white; line-height: 19px; text-align: center;"><b>α</b></span><b> + β) = sin</b><span style="background-color: white; line-height: 19px; text-align: center;"><b>α</b></span><b> cosβ + cos</b><span style="background-color: white; line-height: 19px; text-align: center;"><b>α</b></span><b> sinβ</b></div>
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This is a beautiful result. The sine of the sum of two quantities can just be pulled apart so easily. This is really cool on its own, and furthermore, it ends up being the foundation of tons more identities. Let's say we wanted to find the sine of <span style="background-color: white; line-height: 19px; text-align: center;">α</span>–β. We could just plug (-β) in for β and see what we get.</div>
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sin(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> – β) = sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cos(-β) + cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sin(-β)</div>
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These negative terms can be analyzed in a number of ways, but the easiest way to do it is to look at the graphs of the sine and cosine functions.</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilmkSZf-G1xFTZyfusnMjdOX7rgskI5tkXyJpLukW3FdjdXsxQZNR0u4oLuyh0zd7LLQ7LUQg4i4n6ZSaDktGrTKrGs48ZQkf87jMZQP35tZIPQ7uvnYf6A96OMK15tkb_1mx_9ulta9GJ/s1600/render.php.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilmkSZf-G1xFTZyfusnMjdOX7rgskI5tkXyJpLukW3FdjdXsxQZNR0u4oLuyh0zd7LLQ7LUQg4i4n6ZSaDktGrTKrGs48ZQkf87jMZQP35tZIPQ7uvnYf6A96OMK15tkb_1mx_9ulta9GJ/s1600/render.php.png" height="246" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Graph of y = sin(x)</td></tr>
</tbody></table>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhU7ivzcP1Zelzzv_sWmz4RgzPYmjUA4V5Fqrn6XkjbOFVuyJ4GTSrVptustnUA2Niubzt4j75lTRmWB11QKpuQ8HxsOXqVavkHi2FgMqGRadKjGVSXqLw_GqjY77R-ErE9a2riHd1hN5bf/s1600/render-1.php.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhU7ivzcP1Zelzzv_sWmz4RgzPYmjUA4V5Fqrn6XkjbOFVuyJ4GTSrVptustnUA2Niubzt4j75lTRmWB11QKpuQ8HxsOXqVavkHi2FgMqGRadKjGVSXqLw_GqjY77R-ErE9a2riHd1hN5bf/s1600/render-1.php.png" height="246" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Graph of y = cos(x)</td></tr>
</tbody></table>
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Notice how the sine graph is symmetric over the origin. Any sin(x) will equal -sin(-x) and any sin(-x) will equal -sin(x). On the other hand, the cosine graph is symmetric over the y-axis. Any cos(x) will equal cos(-x) and -cos(x) will equal -cos(-x). This enables us to simplify the formula much more.<br />
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sin(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> – β) = sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cos(-β) + cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sin(-β)</div>
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sin(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> – β) = sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ + cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> (-sinβ)</div>
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<b>sin(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> – β) = sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ – cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ</b></div>
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Fantastic. What if we wanted to find the cosine of the sum of two angles? Going back to our original diagram, we can start with the following expression:</div>
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EA = GA - FH</div>
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Dividing through by AD creates the desired left hand side.</div>
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(EA/AD) = (GA/AD) - (FH/AD)</div>
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Multiply the term on the left of the right hand side by (AF/AF) and the right term by (DF/DF) to get:</div>
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(EA/AD) = (GA/AF)(AF/AD) - (FH/DF)(DF/AD)</div>
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And turning this into the appropriate functions gives:</div>
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<b>cos(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> + β) = cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ – sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ</b></div>
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Doing the same thing as before gives a subtraction formula of:</div>
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<b>cos(<span style="background-color: white; line-height: 19px; text-align: center;">α</span> – β) = cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ + sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ</b><br />
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Finally, let's try to create a tangent formula. Knowing that dividing the sine by the cosine will give the tangent, let's try dividing the two formulas by each other.</div>
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<u>sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ + cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ</u></div>
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cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ – sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ</div>
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Divide each term through by cos<img src="http://www.themathpage.com/atrig/trig_IMG/alpha.gif" /> cosβ.</div>
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<u>(sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ)/(cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ) + (cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ)/(cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ)</u></div>
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(cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ)/(cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ) – (sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span> sinβ)/(cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span> cosβ)</div>
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Many terms will now cancel.</div>
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<u>(sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span>)/(cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span>) + (sinβ)/(cosβ)</u></div>
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1 – ((sin<span style="background-color: white; line-height: 19px; text-align: center;">α</span>)/(cos<span style="background-color: white; line-height: 19px; text-align: center;">α</span>))((sinβ)/(cosβ))</div>
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Turning the sine and cosine ratios into tangents gives a formula of:</div>
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<u><b><br /></b></u></div>
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<u><b>tan<span style="background-color: white; line-height: 19px; text-align: center;">α</span> + tanβ</b></u></div>
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<b>1 – tan<span style="background-color: white; line-height: 19px; text-align: center;">α</span> tanβ</b></div>
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And there's the formula! If you wanted to create a subtraction formula, you would just note that the tangent function is symmetric over the origin (or it is an odd function), and plug in the necessary values to get:</div>
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<u><b>tan<span style="background-color: white; line-height: 19px; text-align: center;">α</span> – tanβ</b></u></div>
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<b>1 + tan<span style="background-color: white; line-height: 19px; text-align: center;">α</span> tanβ</b></div>
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I think that these are really cool formulas. But they are also very practical. For instance, let's say you want to find the exact sine of 75°. We know that sin(45) = (√2)/2, sin(30) = 1/2, cos(45) = (√2)/2, and cos(30) = (√3)/2. So, this can be determined with the sine angle addition formula.</div>
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sin(75) = sin(30 + 45) = sin30cos45 + cos30sin45 = (1/2)((√2)/2) + ((√3)/2)((√2)/2) = (√(2) + √(6))/4</div>
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Knowing that the Greeks did some manipulation with the regular pentagon to find an exact sine of 72°, we can plug all of this information into the sine angle subtraction formula to find the exact sine of 3°. We can then eventually make our way down to 1°, which gives us the ability to generate the sines of all of the angles on the trig tables that mathematicians now use today.</div>
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Next week, we will use these formulas to create more generalizations that will make many of these trig table computations much much simpler. And give us some really pretty formulas in the process.</div>
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Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-62400844709483390572014-02-22T12:00:00.000-05:002014-02-22T12:00:04.270-05:00Gambler's Ruin Problem Part 4: Odds in Video Poker<div class="separator" style="clear: both; text-align: left;">
<span style="font-family: inherit;"><span style="background-color: white; line-height: 18px;">Click <a href="http://coolmathstuff123.blogspot.com/2014/02/gamblers-ruin-problem-part-1-odds-in.html" style="text-decoration: none;" target="_blank">here</a> to see part 1 of this four week series.</span><br style="background-color: white; line-height: 18px;" /><span style="background-color: white; line-height: 18px;">Click <a href="http://coolmathstuff123.blogspot.com/2014/02/gamblers-ruin-problem-part-2-odds-in.html" style="text-decoration: none;" target="_blank">here</a> to see part 2 of this four week series.</span></span></div>
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<span style="font-family: inherit;"><span style="background-color: white; line-height: 18px;">Click <a href="http://coolmathstuff123.blogspot.com/2014/02/gamblers-ruin-problem-part-3-odds-in.html" target="_blank">here</a> to see part 3 of this four week series.</span></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiIzFjsGJfokWjcZ9jPkOVHld6l347j_uoLXrqdfyKBuWKs5GkzVfKI5cKqadXfx2-_eSfG51QyURd7FqF1mQdihEQovTKmoxPM0Hb_vYBYfwv_KkrZsGjo4_AKK8tQiihlnw0k5O8UqhcF/s1600/FunnyPart-com-casino.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiIzFjsGJfokWjcZ9jPkOVHld6l347j_uoLXrqdfyKBuWKs5GkzVfKI5cKqadXfx2-_eSfG51QyURd7FqF1mQdihEQovTKmoxPM0Hb_vYBYfwv_KkrZsGjo4_AKK8tQiihlnw0k5O8UqhcF/s400/FunnyPart-com-casino.jpg" height="250" width="400" /></a></div>
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<span style="font-weight: bold;">Suppose that you find yourself in a city with a casino and you have $60 in your pocket. There is a concert in town that you really want to see, but the tickets cost $100. You decide that you will place $1 bets until you either reach $100 or go broke. Which casino game should you play? How likely are you to reach your goal of $100?</span><br />
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This week is the final week in my series on the Gambler's Ruin Problem, which I have explained again in the paragraph above. We have analyzed roulette, craps, and blackjack, finding out various odds, expected values, and success probabilities. Here is what we have so far:<br />
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<table border="1" style="text-align: center;"><tbody>
<tr><th>Game</th><th>Avg Gain Per $1 Bet</th><th>P of Reaching $100</th></tr>
<tr><th>Favorable Game (51-49 odds)</th><td>2¢</td><td>92.6%</td></tr>
<tr><th>Fair Game (50-50 odds)</th><td>0¢</td><td>60%</td></tr>
<tr><th>Blackjack</th><td>-0.5¢</td><td>47.8%</td></tr>
<tr><th>Craps</th><td>-1.4¢</td><td>28%</td></tr>
<tr><th>Roulette</th><td>-5.3¢</td><td>1.3%</td></tr>
</tbody></table>
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Let's see how video poker fairs. I chose not to do Texas Hold'em or another popular version of real poker because there is so much strategy involved, and determining a universal statistic is near to impossible. If you are also considering reading players, pot odds, and the occasional bluff, it becomes extremely complicated. However, video poker is another game that is you against the house and can be played with a more concrete strategy like blackjack. Let's go over the rules.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_ZL6nrpOGsXZyA-2r0FuH2DJ5_30MXVl4ZSZ4iiXs8FRsGtEccOSSK-InsOA3qjwlDik0Mvtmv8aflUgQCa8K4GoS9iJuNIK8pcwbkcs2vqrwN1upoVZjxt-AnxW7r4MBMwXoXFYdJrAk/s1600/video-poker.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_ZL6nrpOGsXZyA-2r0FuH2DJ5_30MXVl4ZSZ4iiXs8FRsGtEccOSSK-InsOA3qjwlDik0Mvtmv8aflUgQCa8K4GoS9iJuNIK8pcwbkcs2vqrwN1upoVZjxt-AnxW7r4MBMwXoXFYdJrAk/s400/video-poker.jpg" height="236" width="400" /></a></div>
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As you can see in the picture, video poker is pretty much a slot machine, except it gives much better odds. The game of poker played on it is five card draw. The player is given five cards to start. They then are given the opportunity to either keep them all, get rid of one, get rid of two, get rid of three, get rid of four, or get rid of all of them. The cards they got rid of are then replaced, and the hand they have left is identified and the proper money exchange is made. If you are unfamiliar with the poker hands, click <a href="http://en.wikipedia.org/wiki/List_of_poker_hands#Hand_categories" target="_blank">here</a>. This is the list of odds the casino gives for each hand:<br />
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Royal Flush - 250:1<br />
Straight Flush - 50:1<br />
Four of a Kind - 25:1<br />
Full House - 9:1<br />
Flush - 6:1<br />
Straight - 4:1<br />
Three of a Kind - 3:1<br />
Two Pair - 2:1<br />
Pair of Jacks or Better - 1:1<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGiz-rXoLY_5TWXJnejGwb4jyjwH6aFPwnE3bF1YIN3Tj1705cXXsS5Q6_kdw37XuQQMsVqFcpRizCWZ37PR7Icsp8fURiYSkbasmCuh_mrmjPSENUOduexE1ORllZxpii5F5BEE8DbI93/s1600/JacksOrBetter-Single.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGiz-rXoLY_5TWXJnejGwb4jyjwH6aFPwnE3bF1YIN3Tj1705cXXsS5Q6_kdw37XuQQMsVqFcpRizCWZ37PR7Icsp8fURiYSkbasmCuh_mrmjPSENUOduexE1ORllZxpii5F5BEE8DbI93/s400/JacksOrBetter-Single.jpg" height="300" width="400" /></a></div>
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As you can see, the rarer the hand, the bigger the payoff is. This creates the difficulty in choosing which cards to hold and which to discard in any given turn.<br />
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I will start by just going through the basic strategy. Then, we will look at some of the exceptions and why they are encouraged.<br />
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1. With two pair or higher, keep the cards needed for the hand and discard the rest (for instance, draw one card in a two pair, don't draw any in a flush)<br />
2. With one pair, keep the pair and discard the rest<br />
3. With no high cards, discard all five<br />
4. With one high card, keep it and discard the rest<br />
5. With two high cards, keep them and discard the rest<br />
6. With three or four high cards, keep two. If there are two of the same suit, keep them. Otherwise, take the two with the <i>lowest</i> value.<br />
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Pretty simple when it comes to poker strategy. That one won't take up an entire book. Now let's look at some exceptions. The first one is the following:<br />
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If four cards are in a straight flush, discard the fifth one (but don't break a straight or flush unless it can become a royal flush)<br />
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For instance, take the following example:<br />
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7C 8C 9C JC JD<br />
<br />
In this hand, the basic strategy would say to discard the seven eight and nine, but this exception tells us to discard the jack of diamonds. Let's see why. There are 52 cards in a deck, and five have been used in this hand. So, the remaining possibilities to take the place of the jack of diamonds should total 47.<br />
<br />
Here are the possible hands and their probabilities:<br />
<br />
P(Straight Flush) = 1/47<br />
P(Flush) = 8/47<br />
P(Straight) = 3/47<br />
P(2 Jacks) = 2/47<br />
P(Nothing) = 33/47<br />
<br />
EV(Discard JD) = (1/47)(50) + (8/47)(6) + (3/47)(4) + (2/47)(1) + (33/47)(-1) = 79/47 ≈ $1.68<br />
<br />
And here are the possible hands and their probabilities for keeping the jacks:<br />
<br />
P(4 of a Kind) = (3 ways of choosing)(2/47 third jack options)(1/46 fourth jack options) = 3/1081<br />
P(Full House) = (3 ways of choosing placement of third jack)(2/47 third jack options)(46/46 • 3/45 pair options) + (12 value choices of three of a kind)(4/47 first cards)(3/46 second cards)(2/45 third cards) = 62/5405<br />
P(3 Jacks) = (3 ways of choosing placement of third jack)(2/47 third jack options) = 6/47<br />
P(2 Pair) = (3 ways of choosing)(47/47 • 3/46 pair options) = 9/46<br />
P(2 Jacks) = 1 - P(4 of a Kind) - P(Full House) - P(3 Jacks) - P(2 Pair) = 7161/10810<br />
<br />
EV(Keep Jacks) = (3/1081)(25) + (62/5405)(9) + (6/47)(3) + (9/46)(2) + (7161/10810)(1) = 17397/10810 ≈ $1.61<br />
<br />
As you can see, keeping the jacks has less of a payoff than this exception. It isn't a huge difference, but still one to take note of.<br />
<br />
Here are some other exceptions:<br />
<br />
If you have four cards in a flush, go for the flush.<br />
If you have three cards in a royal flush, go for the royal flush.<br />
If you have an open-ended straight, go for the straight.<br />
If you have three straight flush cards, go for the straight flush.<br />
If you have J 10 suited, Q 10 suited or K 10 suited, keep the 10 in addition to the high card.<br />
<br />
All of those exceptions could be proven with the probability techniques. Using this strategy, you will have very similar odds to blackjack with about a 0.5¢ loss per turn. Most players will lose 4¢ per turn, so this puts us way ahead of the game.<br />
<br />
<table border="1" style="text-align: center;"><tbody>
<tr><th>Game</th><th>Avg Gain Per $1 Bet</th><th>P of Reaching $100</th></tr>
<tr><th>Favorable Game (51-49 odds)</th><td>2¢</td><td>92.6%</td></tr>
<tr><th>Fair Game (50-50 odds)</th><td>0¢</td><td>60%</td></tr>
<tr><th>Blackjack</th><td>-0.5¢</td><td>47.8%</td></tr>
<tr><th>Video Poker</th><td>-0.5¢</td><td>47.8%</td></tr>
<tr><th>Craps</th><td>-1.4¢</td><td>28%</td></tr>
<tr><th>Roulette</th><td>-5.3¢</td><td>1.3%</td></tr>
<tr><th>Average Vegas Slot Machine</th><td>-6.6¢</td><td>0.5%</td></tr>
</tbody></table>
<br />
I added an average for Vegas slot machines onto the bottom of our table just to give a taste as to how bad their odds are. Of course, each slot machine varies and I'm not sure how accurate the information I found was for them. But the odds for the games we discussed are all completely based on mathematics and are a lot of fun to play around with. If you play other games in the casino, I'm sure you can determine all of the same information for the Gambler's Ruin Problem and looking for optimal strategy.<br />
<br />
So what's the answer to the Gambler's Ruin Problem? I don't know. We took a close look at many different games, but we can never have enough information to have a solid answer. There are dozens of different games and bets out there, many of whom can't be completely analyzed with probability and game theory. But this problem is a gateway to having fun with numbers and probability to make yourself a better player at the casino and learn a bit about mathematics. I hope some of the things we've learned in the last month become useful to you next time you have sixty bucks in a city with a casino.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-67068138238870796072014-02-15T12:00:00.000-05:002014-02-15T12:00:01.485-05:00Gambler's Ruin Problem Part 3: Odds in Blackjack<span style="font-family: inherit;"><span style="background-color: white; color: #444444; line-height: 18px;">Click </span><a href="http://coolmathstuff123.blogspot.com/2014/02/gamblers-ruin-problem-part-1-odds-in.html" style="background-color: white; color: #4d469c; line-height: 18px; text-decoration: none;" target="_blank">here</a><span style="background-color: white; color: #444444; line-height: 18px;"> to see part 1 of this four week series.</span></span><br />
<span style="font-family: inherit;"><span style="background-color: white; color: #444444; line-height: 18px;">Click <a href="http://coolmathstuff123.blogspot.com/2014/02/gamblers-ruin-problem-part-2-odds-in.html" target="_blank">here</a> to see part 2 of </span><span style="color: #444444;"><span style="line-height: 18px;">this four week series.</span></span></span><br />
<span style="font-family: inherit;"><br /></span>
<span style="font-family: inherit;"><br /></span>
<span style="font-family: inherit;">We spent the last two weeks talking about the Gambler's Ruin Problem and how it is a gateway to analyze differe</span>nt games like craps and roulette. Here is a refresher of what we have discovered so far, as well as what the Gambler's Ruin Problem is:<br />
<br />
<span style="font-weight: bold;">Suppose that you find yourself in a city with a casino and you have $60 in your pocket. There is a concert in town that you really want to see, but the tickets cost $100. You decide that you will place $1 bets until you either reach $100 or go broke. Which casino game should you play? How likely are you to reach your goal of $100?</span><br />
<span style="font-weight: bold;"><br /></span>
And here were the probabilities we found for all bets in roulette, the pass line bet in craps, and the other two control statistics.<br />
<br />
<table border="1" style="text-align: center;"><tbody>
<tr><th>Game</th><th>Avg Gain Per $1 Bet</th><th>P of Reaching $100</th></tr>
<tr><th>Favorable Game (51-49 odds)</th><td>2¢</td><td>92.6%</td></tr>
<tr><th>Fair Game (50-50 odds)</th><td>0¢</td><td>60%</td></tr>
<tr><th>Craps</th><td>-1.4¢</td><td>28%</td></tr>
<tr><th>Roulette</th><td>-5.3¢</td><td>1.3%</td></tr>
</tbody></table>
<br />
We found that roulette is a game of minimal strategy; each bet had the same expected value. Craps is a more difficult game because of the many different bet options each with different perks and odds, but players still don't have to worry about utilizing strategies on the fly. Blackjack starts to get into some more complicated strategy building.<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9xGiH8esAP6Ew1uCEkZfDcncjsS-tPbFS8kFNxnuwqidsiSf6TWw5aPgK5ZNjRHRdAY-aF11-J_uh9Pp_xxynVGGJ9ZuJw5qjlObZs-tePokv8-IB9FbUYX71zpwuInT8XPB7PUNY9ia8/s1600/black-jack-tables-for-rent.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9xGiH8esAP6Ew1uCEkZfDcncjsS-tPbFS8kFNxnuwqidsiSf6TWw5aPgK5ZNjRHRdAY-aF11-J_uh9Pp_xxynVGGJ9ZuJw5qjlObZs-tePokv8-IB9FbUYX71zpwuInT8XPB7PUNY9ia8/s400/black-jack-tables-for-rent.jpg" height="300" width="400" /></a></div>
<br />
Though there are other players at the table, like craps and roulette, blackjack is a game of you against the house (which is represented by the dealer). You will receive two face up cards and the dealer will receive one face up and one face down card. The sum of the numbers on the card is what your total is (J = 10, Q = 10, K = 10, A = 1 <b>or</b> 11). For instance, K 2 would be 12, 9 7 would be 16, A 5 would be 6 or 16. Hands with an ace counting for eleven are called soft hands because they are easier to work with; they can be lowered by ten at any time if need be.<br />
<br />
Once you get your two hands, the dealer will ask if you want to "hit" or "stand." Hitting is when you take another card and add it to your total. Standing is when you do not take any more cards and your hand becomes final. If you hit and your total goes over 21, you "bust" and lose your bet.<br />
<br />
Once everyone at the table stands or busts, the dealer then turns over his facedown card. He then must hit until his total is 17 or above. Once he gets to there, he must stand. You then lose your bet if his/her total beats yours and you get even money (1:1) if you beat his/her total.<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEicWDragfckzTshvhDfonnUej570zJvphTKpt3JdG_gByDGx7lrRk5EaYTGK3SifbFDT_g3osz4dqM63kBelkBlqYe-mZwSkCeyI2XORuON2T23e-FuLLBNm26q1K-8M7L6cZKY1SdwrlZO/s1600/bjtable.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEicWDragfckzTshvhDfonnUej570zJvphTKpt3JdG_gByDGx7lrRk5EaYTGK3SifbFDT_g3osz4dqM63kBelkBlqYe-mZwSkCeyI2XORuON2T23e-FuLLBNm26q1K-8M7L6cZKY1SdwrlZO/s400/bjtable.jpg" height="281" width="400" /></a></div>
When you get 21 with just two cards (A K, A J, etc), your hand is called blackjack. As you can see on the table above, blackjack pays 3:2 instead of just 1:1. This is one of the perks that you have as a player; the dealer cannot request that you pay him extra because he got blackjack.<br />
<br />
As a player, you also have a few additional betting options that I will explain below.<br />
<br />
<b>Doubling Down</b> is when you double your bet on the table and agree to only take one more card. With certain types of starting hands, you can increase your payoff in this way.<br />
<br />
<b>Splitting</b> is only allowed when you have two cards of the same number. You can "split" the cards, put a second equal bet on the other card, and play both hands simultaneously against the dealer. Most casinos let you re-split cards, but rarely do they let you double down after splitting.<br />
<br />
<b>Taking Insurance</b> is only allowed when the dealer's face up card is an ace. You can bet up to half of your original bet that the dealer's facedown card is a 10, J, Q, or K. This bet pays 2:1, as depicted on the table above.<br />
<br />
Blackjack doesn't have to deal with as many weird ratios and random bets as craps does, but developing strategy becomes more difficult because of this. Aside from taking insurance, it is all just inflicting different probabilities into a lone 1:1 bet. But how can we optimize our odds on this 1:1 bet?<br />
<br />
When you think about it, it seems like you have a pretty big advantage over the dealer. You can stand before 17, hit after 17, split, double down, take insurance, and get paid 3:2 on blackjack. How do the casinos make money? Well, there is one thing that the dealer has an advantage on, which doesn't even cross our minds. Every time you bust, you lose. Even if the dealer busts. This one advantage is what makes blackjack worth it for the casinos.<br />
<br />
This doesn't mean that we can't try our best though. A beginner might just make random bets based on intuition, but more experienced players are able to get their odds much closer to 1:1. Let's start with looking at when to hit or stand. First, hard hands:<br />
<br />
<table border="1" style="text-align: center;">
<tbody>
<tr>
<th>Dealer's Up Card</th>
<th>Hit Until You Reach</th>
</tr>
<tr>
<td>7, 8, 9, 10, A</td>
<td>17</td>
</tr>
<tr>
<td>4, 5, 6</td>
<td>12</td>
</tr>
<tr>
<td>2, 3</td>
<td>13</td>
</tr>
</tbody></table>
<br />
So let's say your starting hand is 7 2 and the dealer is showing a 3.<br />
<br />
7 2<br />
? 3<br />
<br />
Your total is nine, and you hit until you reach 13. So you hit.<br />
<br />
7 2 2<br />
? 3<br />
<br />
Now your total is eleven, so you must hit again.<br />
<br />
7 2 2 K<br />
? 3<br />
<br />
Now your total is 21, and you obviously would stand here.<br />
<br />
7 2 2 K<br />
5 3 J<br />
<br />
You win! Great job. That guideline is probably the most basic part of the strategy, and will come into play in almost every round. If you memorize any of the strategies I discuss here, that is the one to remember.<br />
<br />
That was for hard hands (without an ace acting as an eleven). What about soft hands?<br />
<br />
<table border="1" style="text-align: center;"><tbody>
<tr><th>Dealer's Up Card</th><th>Hit Until You Reach</th></tr>
<tr><td>9, 10, A</td><td>Soft 19</td></tr>
<tr><td>8 or below</td><td>Soft 18</td></tr>
</tbody></table>
<br />
Let's try it.<br />
<br />
A 5<br />
? 9<br />
<br />
Our strategy says we should hit.<br />
<br />
A 5 7<br />
? 9<br />
<br />
Now we have a hard hand of thirteen, so we must switch back to the hard hand strategy and hit until we reach seventeen.<br />
<br />
A 5 7 4<br />
? 9<br />
<br />
This is one of the most debated parts of blackjack. We have a hand totaling sixteen and we need to hit or stand. Many players are conservative and choose to stand here. However, we know that probability tells us to hit. You could map out all of the possible outcomes of this hand for yourself and the dealer and compare them, and you would find that hitting actually does end up giving you more success.<br />
<br />
A 5 7 4 Q<br />
? 9<br />
<br />
And you busted.<br />
<br />
A 5 7 4 Q<br />
K 9<br />
<br />
But you would have lost anyways, so hitting was worth a try on that one. Now let's look at double down and splitting strategies.<br />
<br />
<table border="1" style="text-align: center;"><tbody>
<tr><th>Your First Two Cards</th><th>Double if Dealer Has</th></tr>
<tr><td>Total 11</td><td>10 or below</td></tr>
<tr><td>Total 10</td><td>9 or below</td></tr>
<tr><td>Total 9</td><td>4, 5, 6</td></tr>
<tr><td>A2 through A7</td><td>4, 5, 6</td></tr>
</tbody></table>
<br />
<table border="1" style="text-align: center;"><tbody>
<tr><th>When do you split</th><th>If Dealer Has</th></tr>
<tr><td>A, 8</td><td>any card</td></tr>
<tr><td>4, 5, 10</td><td>never</td></tr>
<tr><td>2, 3, 6, 7</td><td>2, 3, 4, 5, 6</td></tr>
<tr><td>9</td><td>2, 3, 4, 5, 6, 8, 9</td></tr>
</tbody></table>
<br />
Those are for doubling down and splitting. What about taking insurance? This is the easiest rule of them all.<br />
<br />
<b>Don't take insurance</b><br />
<b><br /></b>
Ever. Let's look at why. The dealer's face down card could be any of the following:<br />
<br />
A 2 3 4 5 6 7 8 9 10 J Q K<br />
<br />
There are nine cards that don't win you the bet and four cards that do. So, the fair odds for the casino to offer would be 9:4. But, they only offer 2:1 or 8:4, meaning that insurance is not worth it (unless you are a card counter).<br />
<br />
If you figure out the expected value of your $1 bet incorporating all of these strategies, you end up getting an average loss of 0.5¢. This is much better than craps and roulette! These are the best odds we have seen so far!<br />
<br />
<table border="1" style="text-align: center;"><tbody>
<tr><th>Game</th><th>Avg Gain Per $1 Bet</th><th>P of Reaching $100</th></tr>
<tr><th>Favorable Game (51-49 odds)</th><td>2¢</td><td>92.6%</td></tr>
<tr><th>Fair Game (50-50 odds)</th><td>0¢</td><td>60%</td></tr>
<tr><th>Blackjack</th><td>-0.5¢</td><td>47.8%</td></tr>
<tr><th>Craps</th><td>-1.4¢</td><td>28%</td></tr>
<tr><th>Roulette</th><td>-5.3¢</td><td>1.3%</td></tr>
</tbody></table>
<br />
These odds are not bad. In fact, we are not far from having a 50-50 shot of turning our $60 into $100!<br />
<br />
This strategy is great. But what I always wonder is why it works. Why do we hit on sixteen if the dealer has a seven showing? Let's figure it out.<br />
<br />
First, we determine the dealer's odds of busting with a seven. To do this, we start with the odds of busting with sixteen (the dealer won't bust if they get to seventeen because they are required to stand). This is a sum of their odds of getting a 6, 7, 8, 9, or 10 as their next card value (remember that 10 can be achieved four ways).<br />
<br />
P(bust with sixteen) = P(6) + P(7) + P(8) + P(9) + P(10)<br />
P(bust with sixteen) = 1/13 + 1/13 + 1/13 + 1/13 + 4/13<br />
P(bust with sixteen) = 8/13 ≈ 0.615<br />
<br />
Now, we can determine the odds of busting with a fifteen. They could either hit and get a 7, 8, 9, 10, or an ace followed by a bust.<br />
<br />
P(bust with fifteen) = P(7) + P(8) + P(9) + P(10) + [P(A) • P(bust with sixteen)]<br />
P(bust with fifteen) = 1/13 + 1/13 + 1/13 + 4/13 + [1/13 • 8/13]<br />
P(bust with fifteen) = 99/169 ≈ 0.586<br />
<br />
We could then figure it out for busting with a fourteen. They could either hit and get a 8, 9, 10, ace-bust, or two-bust.<br />
<br />
P(bust with fourteen) = P(8) + P(9) + P(10) + [P(A) • P(bust with fifteen)] + [P(2) • P(bust with sixteen)]<br />
P(bust with fourteen) = 1/13 + 1/13 + 4/13 + [1/13 • 99/169] + [1/13 • 8/13]<br />
P(bust with fourteen) = 1217/2197 ≈ 0.554<br />
<br />
This process can be continued until you get to the probability of busting with a seven, which ends up being around 0.262. All of the probabilities of the dealer's outcomes with certain face up cards are on the table below:<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh7pzxV_2Thu4ywqsqqp-t9t3GIbVF_nO3rCse1bkC5l9QTtzpuqax7Y_d71sL8SJik-mQ6KAWPJsOul6mgf-2n4IkjJ-uWNdaaCEdfc1pe3zjsVXyIMjBlgBMTFp7uAPMSoBRI_155rws7/s1600/Screen+Shot+2014-01-03+at+1.59.34+PM.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh7pzxV_2Thu4ywqsqqp-t9t3GIbVF_nO3rCse1bkC5l9QTtzpuqax7Y_d71sL8SJik-mQ6KAWPJsOul6mgf-2n4IkjJ-uWNdaaCEdfc1pe3zjsVXyIMjBlgBMTFp7uAPMSoBRI_155rws7/s400/Screen+Shot+2014-01-03+at+1.59.34+PM.png" height="286" width="400" /></a></div>
<br />
So the chances of our sixteen beating the dealer's seven if we stand is 0.262; the only way we would win is if the dealer busts. If we hit, then we have a good chance of busting but also a chance of getting a number between 17 and 21, each with their own chance of winning.<br />
<br />
P(win with an A) = P(dealer busts) + <span style="letter-spacing: 0px;"><span style="font-family: inherit;">½P(dealer gets 17)</span></span><br />
<span style="letter-spacing: 0px;"><span style="font-family: inherit;">P(win with an A) = 0.262 + </span></span>½(0.369)<br />
P(win with an A) = 0.447<br />
<br />
P(win with a 2) = P(dealer busts) + P(dealer gets 17) + ½P(dealer gets 18)<br />
P(win with a 2) = 0.262 + 0.369 + ½(0.138)<br />
P(win with a 2) = 0.730<br />
<br />
P(win with a 3) = P(bust) + P(17) + P(18) + ½P(19)<br />
P(win with a 3) = 0.262 + 0.369 + 0.138 + ½(0.079)<br />
P(win with a 3) = 0.809<br />
<br />
P(win with a 4) = P(bust) + P(17) + P(18) + P(19) + ½P(20)<br />
P(win with a 4) = 0.262 + 0.369 + 0.138 + 0.079 + ½(0.079)<br />
P(win with a 4) = 0.886<br />
<br />
P(win with a 5) = P(bust) + P(17) + P(18) + P(19) + P(20) + ½P(21)<br />
P(win with a 5) = 0.262 + 0.369 + 0.138 + 0.079 + 0.079 + ½(0.074)<br />
P(win with a 5) = 0.963<br />
<br />
P(win with a 6, 7, 8, 9, 10, J, Q, K) = 0<br />
<br />
Now that we have all of this, let's do an expected value equation to see what our average odds are.<br />
<br />
EV(hit on 16) = (1/13)(0.447) + (1/13)(0.730) + (1/13)(0.809) + (1/13)(0.886) + (1/13)(0.963) + (8/13)(0) = 0.295<br />
<br />
So hitting on 16 gives 0.295 odds while standing gives 0.262 odds. And surprisingly enough, the hitting odds do end up better. Yes you are likely to bust when you hit, but it is better than banking on the worse odds of the dealer busting. The strategies for all of the other things can be derived in similar ways.<br />
<br />
Next week, we will complete our series on the Gambler's Ruin Problem and see if video poker can offer odds as good as blackjack.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-13187430912072113372014-02-08T12:00:00.000-05:002014-02-08T12:00:04.635-05:00Gambler's Ruin Problem Part 2: Odds in CrapsClick <a href="http://coolmathstuff123.blogspot.com/2014/02/gamblers-ruin-problem-part-1-odds-in.html" target="_blank">here</a> to see part 1 of this four week series.<br />
<br />
<br />
<br />
This week is week two in a four week series on the Gambler's Ruin Problem. Unlike my series on the proof to Bertrand's Postulate, this one can be started in the middle if you have an understanding of some elementary probability concepts like expected value. It will take away some of the initial surprise when faced with the probability of success in previous games, but it is not like Bertrand's Postulate where I am recalling or building upon information that was developed in an earlier post. They are all very separate pieces of information that can be understood on their own or together as an attempt at the Gambler's Ruin Problem.<br />
<br />
The idea of the Gambler's Ruin Problem is simple. Here is the problem if you have forgotten it:<br />
<br />
<span style="font-weight: bold;">Suppose that you find yourself in a city with a casino and you have $60 in your pocket. There is a concert in town that you really want to see, but the tickets cost $100. You decide that you will place $1 bets until you either reach $100 or go broke. Which casino game should you play? How likely are you to reach your goal of $100?</span><br />
<span style="font-weight: bold;"><br /></span>
Last week, we discovered that roulette loses on average 5.3¢ per dollar bet, which gives you a shockingly low 1.3% chance of reaching your goal of $100.<br />
<br />
<table border="1" style="text-align: center;">
<tbody>
<tr>
<th>Game</th>
<th>Avg Gain Per $1 Bet</th>
<th>P of Reaching $100</th>
</tr>
<tr>
<th>Favorable Game (51-49 odds)</th>
<td>2¢</td>
<td>92.6%</td>
</tr>
<tr>
<th>Fair Game (50-50 odds)</th>
<td><div style="text-align: center;">
0¢</div>
</td>
<td><div style="text-align: center;">
60%</div>
</td>
</tr>
<tr>
<th>Roulette</th>
<td><div style="text-align: center;">
-5.3¢</div>
</td>
<td><div style="text-align: center;">
1.3%</div>
</td>
</tr>
</tbody></table>
<div style="text-align: center;">
<span style="font-size: x-small;">Note: I added a 51-49 "favorable game" and its statistics. This was not discussed last week, but it is more informative for our needs than a 100% probability game like we discussed last week. It has a 2¢ average gain and a 92.6% success probability.</span></div>
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We have found all of the necessary data with roulette, and odds didn't look great. So let's try another game. This week we will look at craps, which is one of the most popular casino games involving dice.</div>
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<span style="background-color: white;"><span style="font-family: inherit;"><span style="line-height: 20px;">The rules of </span></span><span style="line-height: 20px;">craps take a couple minutes to remember in its entirety, but are pretty easy to understand. This explanation comes from bigmcasino.com, which laid them out very well.</span></span></div>
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<span style="background-color: white;"><span style="font-family: inherit;">The players take turn rolling two dice. The player that is rolling the dice is considered the shooter. The shooter <span class="caps">MUST</span> bet at least the table minimum on either the <em>pass line</em> or the <em>don’t pass line</em>.</span></span></div>
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<span style="background-color: white;"><span style="font-family: inherit;">The game is played in rounds consisting of two phases: come out and point.</span></span></div>
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<span style="background-color: white;"><span style="font-family: inherit;"><strong>Come Out</strong> – to start a round, the shooter makes a “come out” roll</span></span></div>
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<span style="background-color: white;"><span style="font-family: inherit;">If the come out roll is a 2, 3, or 12, then the round ends. The rules of craps state that the shooter is said to “crap out” and players lose their pass line bets.</span></span></div>
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<span style="background-color: white;"><span style="font-family: inherit;">If the come out roll is a 7 or 11, this results in a win for pass line bets.</span></span></div>
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<span style="background-color: white;"><span style="font-family: inherit;">The shooter continues to make come out rolls until he rolls 4, 5, 6, 8, 9, or 10. This number becomes the point and in turn the point phase begins.</span></span></div>
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<span style="background-color: white;"><span style="font-family: inherit;"><strong>Point</strong> – during this phase, if the shooter rolls a point number then it’s a win for the pass line bets. If the shooter rolls a seven, it’s a loss for the pass line bets and the round is over.</span></span></div>
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There are many different bets that can be made in craps, but each can be analyzed through probability. I don't have nearly enough time and energy to go through them all, but I will discuss a few.</div>
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The most common craps bet is the pass line bet. This is the one discussed in the rules that wins on a seven, eleven, or the roll of the point number. To figure out its odds, we can just create a table of percentages for each of the eleven possible dice totals, determine the probability of winning the pass line bet, and perform an expected value equation like we mastered last week.</div>
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<table border="1" style="text-align: center;">
<tbody>
<tr>
<th>Come Out Roll</th>
<th>Odds for Pass Line</th>
</tr>
<tr>
<td>2</td>
<td>0</td>
</tr>
<tr>
<td>3</td>
<td>0</td>
</tr>
<tr>
<td>4</td>
<td></td>
</tr>
<tr>
<td>5</td>
<td></td>
</tr>
<tr>
<td>6</td>
<td></td>
</tr>
<tr>
<td>7</td>
<td>1</td>
</tr>
<tr>
<td>8</td>
<td></td>
</tr>
<tr>
<td>9</td>
<td></td>
</tr>
<tr>
<td>10</td>
<td></td>
</tr>
<tr>
<td>11</td>
<td>1</td>
</tr>
<tr>
<td>12</td>
<td>0</td>
</tr>
</tbody></table>
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I have already put in the ones that we know from the craps rules. The other six become point rolls and need to appear before a seven. Here are the possible roll totals from two dice:</div>
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Let's start with determining the odds for four. There are three ways to roll a four, six ways to roll a seven, and twenty-seven ways to roll something else. To determine the odds of success with a four, there are a few different methods. We will use a more logical method the first time, and then discover a quicker method to figure out the rest.</div>
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To roll a four before a seven, you must add up the probabilities of rolling it on the first try, rolling something other than a four/seven on the first and a four on the second, rolling something else on the first two and a four on the third, and so on. </div>
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Rolling a four on the first try would be simply 3/36. Rolling something else on the first try (27 possibilities) and a four on the second try, the odds would be (27/36)(3/36). Rolling something else the first two times and a four on the third would be (27/36)(27/36)(3/36), or (27/36)<sup>2</sup>(3/36). Rolling something else on the first three times and a four on the fourth would be (27/36)(27/36)(27/36)(3/36), or (27/36)<sup>3</sup>(3/36). This pattern will continue forever.<br />
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Adding this infinite series up will give the following:<br />
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3/36 + (27/36)(3/36) + (27/36)<sup>2</sup>(3/36) + (27/36)<sup>3</sup>(3/36) + (27/36)<sup>4</sup>(3/36) + ...<br />
(3/36)(1 + 27/36 + (27/36)<sup>2 </sup>+ (27/36)<sup>3 </sup>+ (27/36)<sup>4 </sup>+ ...)<br />
(3/36)(36/9)<br />
1/3<br />
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So your probability when rolling a four is 1/3. That way was the most direct way to calculate it, but also a little complicated and requiring some basic knowledge of infinite series. A much easier approach would be to realize that the remaining 27 options have no significance as to your success. There are 3 good rolls and 6 bad rolls and that is all that matters.<br />
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So out of the nine total significant rolls, three of them are winners. Using this logic, the odds for success would be 3/9, or 1/3 as discovered before. Either method generates a 1/3 chance of success.<br />
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Using the approach we just discovered, we can very quickly determine the odds for the other totals and complete the table from before.<br />
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<table border="1" style="text-align: center;"><tbody>
<tr><th>Come Out Roll</th><th>Odds for Pass Line</th></tr>
<tr><td>2</td><td>0</td></tr>
<tr><td>3</td><td>0</td></tr>
<tr><td>4</td><td>1/3</td></tr>
<tr><td>5</td><td>2/5</td></tr>
<tr><td>6</td><td>5/11</td></tr>
<tr><td>7</td><td>1</td></tr>
<tr><td>8</td><td>5/11</td></tr>
<tr><td>9</td><td>2/5</td></tr>
<tr><td>10</td><td>1/3</td></tr>
<tr><td>11</td><td>1</td></tr>
<tr><td>12</td><td>0</td></tr>
</tbody></table>
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This could then be used to determine the probability of winning the pass line bet. Keep in mind that it is not averaging the eleven totals, but multiplying their odds by roll frequency.<br />
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(1/36)(0) + (2/36)(0) + (3/36)(1/3) + (4/36)(2/5) + (5/36)(5/11) + (6/36)(1) + (5/36)(5/11) + (4/36)(2/5) + (3/36)(1/3) + (2/36)(1) + (1/36)(0) ≈ 49.293%<br />
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There is about a 49.293% chance of winning the pass line bet. Let's see what our expected value is when we make a dollar bet.<br />
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EV($1 bet on pass line) = 0.49293(1) + 0.50707(-1) = -1.4¢<br />
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A pass line bet in craps has an average loss of 1.4¢, which is much better than the 5.3¢ loss we continued to see in roulette. But are there other bets in craps that beat that?<br />
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Another bet in craps is called the don't pass line. This is pretty much betting that the shooter will lose rather than the shooter winning. Seems fair, right? It actually seems pretty favorable. But before you bet your mortgage payment on it, keep in mind one little rule. If the shooter rolls a twelve on his come out roll, then it is a push. In other words, a roll of twelve will not be included in the table or equations.<br />
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<table border="1" style="text-align: center;"><tbody>
<tr><th>Come Out Roll</th><th>Odds for Don't Pass Line</th></tr>
<tr><td>2</td><td>1</td></tr>
<tr><td>3</td><td>1</td></tr>
<tr><td>4</td><td>2/3</td></tr>
<tr><td>5</td><td>3/5</td></tr>
<tr><td>6</td><td>6/11</td></tr>
<tr><td>7</td><td>0</td></tr>
<tr><td>8</td><td>6/11</td></tr>
<tr><td>9</td><td>3/5</td></tr>
<tr><td>10</td><td>2/3</td></tr>
<tr><td>11</td><td>0</td></tr>
</tbody></table>
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(1/36)(1) + (2/36)(1) + (3/36)(2/3) + (4/36)(3/5) + (5/36)(6/11) + (6/36)(0) + (5/36)(6/11) + (4/36)(3/5) + (3/36)(2/3) + (2/36)(0) ≈ 47.929%</div>
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EV($1 bet on don't pass line) = 0.47929(1) + 0.52071(-1) = -4.1¢</div>
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With this minuscule adjustment of pushing a twelve, this bet actually becomes less favorable than the pass line bet. Rather than losing 1.4¢ on average, you would lose 4.1¢. Still better odds than roulette, but why bet against the shooter when you are benefiting more by betting for him?</div>
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There are many more bets in craps that I don't have the time to go through, but I would encourage you to perform the calculations with some of them. If you play craps at the casino, try taking one of your favorite bets and see if it measures up to the pass line bet or other options either within craps or in other casino games. Here are a few different options:</div>
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<b>Place Bets</b> are bets on a specific number. After the come out roll, if your number appears before a seven, you win the bet. Place bets on a six or eight pay 7:6, bets on a five or nine pay 7:5, and bets on a four or ten pay 9:5.<br />
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<b>Place to Lose Bets</b> are bets against a specific number. After the come out roll, if a seven appears before your number, you win the bet. Place to lose bets on a six or eight pay 4:5, bets on a five or nine pay 5:8, and bets on a four or ten pay 5:11.<br />
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<b>Buy Bets</b> are like place bets, except the casino gives you better odds in exchange for a 5% commission fee for making the bet. For instance, a $1 bet would be split where about 4.76¢ go to the casino and 95.24¢ go towards your bet. Buy bets on a six or eight pay 6:5, bets on a five or nine pay 3:2, and bets on a four or ten pay 2:1.<br />
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<b>Lay Bets</b> are like place to lose bets, except the casino gives you better odds in exchange for a 5% commission fee for making the bet (see buy bets). Lay bets on a six or eight pay 5:6, bets on a five or nine pay 2:3, and bets on a four or ten pay 1:2.<br />
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<b>Hard Way Bets</b> are bets on a specific pair (either two 2's, two 3's, two 4's, or two 5's). After the come out roll, if your pair appears before a seven, you win the bet. In the United States (and most other countries) hard way bets on a pair of twos or fives pay 9:1 and bets on a pair of threes or fours pay 7:1.<br />
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<b>Hop Bets</b> are bets on a specific combination of numbers (1&4, 3&5, 2&2, etc.), but they only last for one roll. During the point rolls, you can make a hop bet for that turn. If you get your combination, you win. If you don't, you lose that money. Hop bets on a combination with two different numbers (such as 1&4 and 3&5) pay 15:1 and bets on a pair (such as 2&2 and 4&4) pay 30:1. This is different from a hard way bet because the hop bet lasts for one turn while the hard way bet lasts until the shooter gets a seven.<br />
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There are still even more bets such as odds bets and proposition bets, but these should be a good start. The most common craps bet is the pass line bet, so we will use this when comparing casino games. Our table of games now looks like:<br />
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<table border="1" style="text-align: center;"><tbody>
<tr><th>Game</th><th>Avg Gain Per $1 Bet</th><th>P of Reaching $100</th></tr>
<tr><th>Favorable Game (51-49 odds)</th><td>2¢</td><td>92.6%</td></tr>
<tr><th>Fair Game (50-50 odds)</th><td>0¢</td><td>60%</td></tr>
<tr><th>Craps</th><td>-1.4¢</td><td>28%</td></tr>
<tr><th>Roulette</th><td>-5.3¢</td><td>1.3%</td></tr>
</tbody></table>
</div>
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If you find some craps bets or combinations of bets (ex: pass line/odds - odds bets cannot be placed on their own) that have better odds than the ones we've discussed, please comment your insights. Probability in casino games is always a fun area of mathematics to discuss and debate about.</div>
Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-29376223999216387882014-02-01T12:00:00.000-05:002014-02-01T12:00:01.205-05:00Gambler's Ruin Problem Part 1: Odds in RouletteWhen I was at MAAthFest (a mathematics conference run by the Mathematical Association of America) last summer, I got to spend the first two days in a short course on the Mathematics of Games and Puzzles put on by Dr. Arthur Benjamin, who is actually my mentor in learning mental math. He recently came out with a video course through the Great Courses on the Mathematics of Games and Puzzles, which you can click <a href="http://www.thegreatcourses.com/tgc/Courses/course_detail.aspx?cid=1401" target="_blank">here</a> to see or purchase.<br />
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Last month, we did a four week series on Bertrand's Postulate. It was a fun and interesting proof, but it did require some hard concentration as well as relatively heavy algebra, number theory, and combinatorics. So, I thought looking at a lighter and more practical problem might be a good idea. It is certainly still deep and thought-provoking, but it won't require as much higher-level mathematics. This problem is called the "Gambler's Ruin Problem," and we spent a lot of time discussing it in that short course. This problem goes as follows:<br />
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<b>Suppose that you find yourself in a city with a casino and you have $60 in your pocket. There is a concert in town that you really want to see, but the tickets cost $100. You decide that you will place $1 bets until you either reach $100 or go broke. Which casino game should you play? How likely are you to reach your goal of $100?</b><br />
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This problem can go on forever because there are so many casino games, each one with different betting options and strategies and variations. But it is also a really fun problem, because throughout the process of figuring it out, you will learn a lot about how to improve your skills at the casino. Though probability shows that you likely won't make money, you will learn ways to play for a longer period of time without going broke. And if you use these strategies enough times, you might just get lucky. As a side note, I find it ironic that I as a fourteen-year-old am choosing to write blog posts about casino games.<br />
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There are too many casino games to fully analyze in just a few posts, but we do have time to take a look at some of the most popular. This week, we will focus on roulette. Next week will be craps. February 15th will be blackjack. February 22nd will be video poker (I would have probably preferred to do Texas Hold'em, but there is so much strategy, bluffing, reading players, and chance involved that it would be way above my head to analyze odds in that game). Notice that I am choosing to ignore slot machines. This is because they are one of the worst bets one can make in the casino. Just look at the rearrangement of the letters!<br />
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<div style="text-align: center;">
SLOT MACHINES</div>
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CASH LOST IN 'EM</div>
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So now let's start our basic analysis of roulette.</div>
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I'm sure many of you know the rules of roulette already, but I will quickly try to explain them to refresh everyone's memory. Though roulette can be played by many people at once, the game itself is just you against the house. Each spin of the wheel (at the bottom right corner of the picture above) is a game. The wheel is numbered from 1 to 36 with alternating red and black numbers. There is also a green zero as well as a green double-zero. The objective is to correctly guess a characteristic of the slot that the ball falls in to, whether it be the color, number, size, or parity (odd/even). Bets are placed on the green felt table, as shown more clearly in the picture below.</div>
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One of the more popular roulette bets is to bet on a color, say betting red. This bet pays evenly; you risk one dollar to gain one dollar (or whatever money amount you choose). When you bet red, there are 18 red numbers out of 38 total numbers, giving an 18/38 or 47.3% chance of winning.</div>
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To figure out how worthwhile this bet is, we do something called finding the expected value, which I discussed in my post about the <a href="http://coolmathstuff123.blogspot.com/2012/06/st-petersburg-paradox-at-cty.html" target="_blank">Saint Petersburg Paradox</a>. Expected value is essentially a weighted average; you average together your winning payoff and your losing payoff with the odds of achieving each one taken into account. For this example, winning a $1 bet would earn you 1 dollar and losing a $1 bet would cost you 1 dollar, or earn you -1 dollars. So, the calculation would be set up as follows:</div>
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EV($1 bet on red) = (18/38)(1) + (20/38)(–1) = -2/38 ≈ -0.0526316</div>
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In other words, each $1 bet on red will lose you 5.3¢ on average. Let's look at another bet, say betting on the first twelve numbers. This bet pays 2 to 1; you risk one dollar to gain two dollars. But, the odds of landing on one of those twelve are now 12/38 instead of 18/38.</div>
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Setting up the expected value equation gives us:</div>
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<br /></div>
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EV($1 bet on 1st 12) = (12/38)(2) + (26/38)(–1) = -2/38 ≈ -0.0526316</div>
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Again, the answer comes out to -2/38, or an average 5.3¢ loss. What if we were to place a bet on a single number. Let's say we bet on 26, the only natural number that is directly between a square (25 = 5<sup>2</sup>) and a cube (27 = 3<sup>3</sup>). Casinos pay 35 to 1 on this bet; you risk one dollar to gain thirty-five dollars. The odds of landing on the number 26 would be 1/38. This expected value calculation would be:<br />
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EV($1 bet on 26) = (1/38)(35) + (37/38)(–1) = -2/38 ≈ -0.0526316<br />
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The answer is the same again: an average loss of 5.3¢. In fact, casinos choose the ratios for roulette such that every bet is the same expected value. Because of this, it does not matter to them what the players choose to bet on. They will always be making the same amount of money on average.<br />
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Let's return to the Gambler's Ruin Problem, and see how likely these odds are to achieve the $100. We mentioned before that the odds for winning with a bet on red is 47.3%. Though the odds differ with other bets, the properties of expected value end up making the game of roulette come out to a 47.3% chance of success. In other words, really close to 50-50, but just a hair below.<br />
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With the Gambler's Ruin Problem, a game with 0% odds would give 0% chance of success. A game with 100% odds would give 100% chance of success. Since you are starting with $60, a game of 50% odds actually gives a 60% chance of success. With a fair game, your amount of starting money determines your success interestingly enough.<br />
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What about a game of 47.3% odds? Would it be above 50% or below 50%? Sounds like it would be around that neck of the woods, considering that a 50-50 game gives a 60% chance. Turns out that there is a formula for determining the probability of turning $60 into $100.<br />
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Let <i>p</i> = probability of winning the game (in decimal form)<br />
Let <i>q</i> = probability of losing the game (in decimal form)<br />
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg22SS1Yn3rZr8zGuxyV5a-yppiAJp8yT6_eQ80R0MAVWOyqlWwuHCbYuAYO8-4kUne-ps8MpS5gDi9GrkzgZ5wOO66MLWQKJqvzhLPRtGRVvT-qy5NmuGU1maXAU1PcyQmC4XIEqHUjIPI/s1600/Screen+Shot+2013-12-31+at+4.23.34+PM.png" imageanchor="1" style="clear: left; display: inline !important; margin-bottom: 1em; margin-right: 1em; text-align: center;"><img border="0" height="123" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg22SS1Yn3rZr8zGuxyV5a-yppiAJp8yT6_eQ80R0MAVWOyqlWwuHCbYuAYO8-4kUne-ps8MpS5gDi9GrkzgZ5wOO66MLWQKJqvzhLPRtGRVvT-qy5NmuGU1maXAU1PcyQmC4XIEqHUjIPI/s200/Screen+Shot+2013-12-31+at+4.23.34+PM.png" width="200" /></a><br />
Plugging 0.473 in for <i>p</i> and 0.527 in for <i>q</i> predicts the probability of success to be about 1.3%. This was really surprising to me at first. Just that 2.7% difference between fair and unfavorable costs you so much when it comes to succeeding in the Gambler's Ruin Problem. But probability is full of surprises, as you can see in my past probability posts as well as the posts in the rest of the month.</div>
Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-71852508850877313012014-01-25T12:00:00.000-05:002014-01-25T12:00:00.534-05:00Bertrand's Postulate Part 4: The Proof<div class="separator" style="background-color: white; clear: both; line-height: 18px;">
<span style="font-family: inherit;">Click <a href="http://coolmathstuff123.blogspot.com/2014/01/bertrands-postulate-part-1-definitions.html" style="text-decoration: none;" target="_blank">here</a> to see part 1 of this four week series.</span></div>
<div class="separator" style="background-color: white; clear: both; line-height: 18px;">
<span style="font-family: inherit;">Click <a href="http://coolmathstuff123.blogspot.com/2014/01/bertrands-postulate-part-2-restricting.html" style="text-decoration: none;" target="_blank">here</a> to see part 2 of this four week series.</span></div>
<div class="separator" style="background-color: white; clear: both; line-height: 18px;">
<span style="font-family: inherit;">Click <a href="http://coolmathstuff123.blogspot.com/2014/01/bertrands-postulate-part-3-bounding.html" target="_blank">here</a> to see part 3 of this four week series.</span></div>
<div class="separator" style="background-color: white; clear: both; color: #444444; line-height: 18px;">
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This is the final post in my four week series on Bertrand's Postulate. Using the information we have discussed in the last three weeks, we will officially prove that there is always a prime between a number and its double.<br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This final step five is the step that will complete the upper bound and provide the proof. Since we have restricted the values of the prime factors so much over the last four steps, we will be able to utilize a technique called proof by contradiction. Last week, we did a proof by induction where we assumed the conclusion was true and did the math from there. With a proof by contradiction, one assumes that the conclusion is false and will later run into a problem.</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">To prove Bertrand’s Postulate, we will assume that there are no primes between <i>n</i> and 2<i>n</i>. So, this means that there are no prime factors that are:</span></span><br />
<ul>
<li style="margin: 0px;"><span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">greater than 2<i>n</i> by step 2</span></span></li>
<li style="margin: 0px;"><span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">equal to 2<i>n</i> because 2<i>n</i> is even, and therefore, not prime</span></span></li>
<li style="margin: 0px;"><span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">between <i>n</i> and 2<i>n</i> because of our assumption</span></span></li>
<li style="margin: 0px;"><span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">between ⅔<i>n</i> and <i>n</i> by step 1</span></span></li>
</ul>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This means that all prime factors of “2<i>n</i> choose <i>n</i>” are less than ⅔<i>n</i>.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Keep in mind that this does not include prime powers. We did prove that there are no prime powers greater than 2<i>n</i>, but they can exist below that. So, an upper bound on each prime power would be 2<i>n</i>.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">We know that (√(2<i>n</i>))</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> is 2<i>n</i>. So, any number greater than √(2<i>n</i>) cannot have a prime power factor. If it did, then the factor would be greater than 2<i>n</i>, which is impossible (see step two). This means that there are at most √(2<i>n</i>) values of 2<i>n</i> (the maximum of a prime power) in the prime factorization of “2<i>n</i> choose <i>n</i>.” So, we can set an upper bound on any prime factor below √(2<i>n</i>) as 2<i>n</i></span><span style="letter-spacing: 0px;"><sup>√(2</sup><i><sup>n</sup></i><sup>)</sup></span><span style="letter-spacing: 0.0px;">.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">But there can still be prime factors between √(2<i>n</i>) and ⅔<i>n</i>. They just will have an exponent of one. So, the product of all the primes between √(2<i>n</i>) and ⅔<i>n </i>will cover everything in that interval, since there are no exponents. To make the math simpler, we will just use (⅔<i>n</i>)╫. Multiplying these two quantities together will yield an upper bound on “2<i>n</i> choose <i>n</i>.”</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">“2<i>n</i> choose <i>n</i>” < 2<i>n</i></span><span style="letter-spacing: 0px;"><sup>√(2</sup><i><sup>n</sup></i><sup>)</sup></span><span style="letter-spacing: 0.0px;"> • (⅔<i>n</i>)╫</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">In step 4, we set an upper bound on that primorial. (⅔<i>n</i>)╫ is definitely less than 4</span><span style="letter-spacing: 0px;"><sup>⅔</sup><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">. So, we can plug that into the right-hand side to get:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">“2<i>n</i> choose <i>n</i>” < 2<i>n</i></span><span style="letter-spacing: 0px;"><sup>√(2</sup><i><sup>n</sup></i><sup>)</sup></span><span style="letter-spacing: 0.0px;"> • 4</span><span style="letter-spacing: 0px;"><sup>⅔</sup><i><sup>n</sup></i></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">Also, a lower bound was set on “2<i>n</i> choose <i>n</i>” in step 3. We know that it can’t be smaller than 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">/2<i>n</i>. Plug that into the left-hand side to get:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">/2<i>n</i> < 2<i>n</i></span><span style="letter-spacing: 0px;"><sup>√(2</sup><i><sup>n</sup></i><sup>)</sup></span><span style="letter-spacing: 0.0px;"> • 4</span><span style="letter-spacing: 0px;"><sup>⅔</sup><i><sup>n</sup></i></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i><sup></sup></i></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">And now, we are at an equation that can be solved with algebra! Simplifying this takes a lot of logarithms and complicated graphing, but it will end up as:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><i>n</i> < 468</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">That means that any time <i>n</i> is bigger than 468, the bounds that were created after assuming that there are no primes between <i>n</i> and 2<i>n</i> will not work. This means that there must be a prime between <i>n</i> and 2<i>n</i> every time <i>n</i> is greater than 468 because any other possibility was proven impossible.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0px;"><span style="font-family: inherit;">The only time where it is theoretically possible for an <i>n</i> value to have no prime between <i>n</i> and 2<i>n</i> is when <i>n</i> is less than 468. But, it was mentioned before that Joseph Bertrand himself found a prime for all <i>n</i> values up through three-million. So, there can’t be a number less than 468 that the postulate doesn’t work for. And thus, Bertrand’s Postulate is proven.</span></span><br />
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Personally, I do think this proof is really cool. However, it is difficult to be as engaged when it takes four fairly long posts to prove. But what I find more interesting than the actual mathematical side of it is that it is understandable to just someone with a couple years of high school mathematics. I did categorize these posts with the other advanced ones, but there is no need for calculus or even much precalculus to understand (the only precalculus is really the logarithms that I left out at the end). The real barrier in understanding them is the notation that mathematicians use. The actual paper used much more complex language as well as things like sigma notation that wouldn't make as much sense initially to a high school student. But after that language is translated into something simpler, the content itself is certainly understandable and even quite interesting. I am looking forward to making more of these series with other mathematical papers in the months to come.</div>
Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-69064033008608598902014-01-18T12:00:00.000-05:002014-01-18T12:00:05.624-05:00Bertrand's Postulate Part 3: Bounding the Central Binomial Coefficient<div class="separator" style="clear: both; text-align: left;">
Click <a href="http://coolmathstuff123.blogspot.com/2014/01/bertrands-postulate-part-1-definitions.html" target="_blank">here</a> to see part 1 of this four week series.</div>
<div class="separator" style="clear: both; text-align: left;">
Click <a href="http://coolmathstuff123.blogspot.com/2014/01/bertrands-postulate-part-2-restricting.html" target="_blank">here</a> to see part 2 of this four week series.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjnjYmr2Swf5toEInM4l7K84muDYo4LxudD6EeWWuw_lGN8eLenXjma0nHLu5ZnLAN_TkVWAVV9clUTvksk57caCnIbqQxxQ7BtX4r72kQu0dniv1k2amkouTYZr5PiiEV0QstJ7pnoL4uy/s1600/h6AECA924.jpeg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="166" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjnjYmr2Swf5toEInM4l7K84muDYo4LxudD6EeWWuw_lGN8eLenXjma0nHLu5ZnLAN_TkVWAVV9clUTvksk57caCnIbqQxxQ7BtX4r72kQu0dniv1k2amkouTYZr5PiiEV0QstJ7pnoL4uy/s400/h6AECA924.jpeg" width="400" /></a></div>
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This is week three of the four week series on Bertrand's Postulate. Last week, we restricted the range of the prime factors of "2<i>n</i> choose <i>n</i>." This week, we are going to restrict the size of it. This will use many of the techniques and identities we have discussed in the last two weeks to do.<br />
<span style="font-family: Helvetica; font-size: 12px; letter-spacing: 0px;"><br /></span>
<span style="letter-spacing: 0px;">The third step here is to determine the lower bound. The fourth step will be to find the upper bound, and we will then have an inequality to </span>work with. <span style="letter-spacing: 0px;"><span style="font-family: inherit;">To find a lower bound, we must use another property of Pascal’s Triangle.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 2 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 3 3 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 4 6 4 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 5 10 10 5 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 6 15 20 15 6 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 7 21 35 35 21 7 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 8 28 56 70 56 28 8 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 9 36 84 126 126 84 36 9 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 10 45 120 210 252 210 120 45 10 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">We will now expand a few polynomials (see definition of polynomial expansion in the first post). The Pascal’s Triangle application will become clear in a minute.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>0</sup></span><span style="letter-spacing: 0.0px;"> = 1</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>1</sup></span><span style="letter-spacing: 0.0px;"> = 1a + 1b</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> = 1a</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> + 2ab + 1b</span><span style="letter-spacing: 0px;"><sup>2</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;"> = 1a</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;"> + 3a</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">b + 3ab</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> + 1b</span><span style="letter-spacing: 0px;"><sup>3</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>4</sup></span><span style="letter-spacing: 0.0px;"> = 1a</span><span style="letter-spacing: 0px;"><sup>4</sup></span><span style="letter-spacing: 0.0px;"> + 4a</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;">b + 6a</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">b</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> + 4ab</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;"> + 1b</span><span style="letter-spacing: 0px;"><sup>4</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Do you see a pattern? Let me make it clearer:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"> (a + b)</span><span style="letter-spacing: 0px;"><sup>0</sup></span><span style="letter-spacing: 0.0px;"> = <b>1</b></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"> (a + b)</span><span style="letter-spacing: 0px;"><sup>1</sup></span><span style="letter-spacing: 0.0px;"> = <b>1</b>a + <b>1</b>b</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"> (a + b)</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> = <b>1</b>a</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> + <b>2</b>ab + <b>1</b>b</span><span style="letter-spacing: 0px;"><sup>2</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0px;"><sup> </sup></span><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;"> = <b>1</b>a</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;"> + <b>3</b>a</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">b + <b>3</b>ab</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> + <b>1</b>b</span><span style="letter-spacing: 0px;"><sup>3</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>4</sup></span><span style="letter-spacing: 0.0px;"> = <b>1</b>a</span><span style="letter-spacing: 0px;"><sup>4</sup></span><span style="letter-spacing: 0.0px;"> + <b>4</b>a</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;">b + <b>6</b>a</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">b</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> + <b>4</b>ab</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;"> + <b>1</b>b</span><span style="letter-spacing: 0px;"><sup>4</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><sup></sup></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">The coefficients in each expansion are the numbers in the corresponding row of Pascal’s Triangle. In general, by using the “<i>n</i> choose <i>k</i>” notation, this theorem can be generalized as:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>n</sup></span><span style="letter-spacing: 0.0px;"> = (“n choose 1”)a</span><span style="letter-spacing: 0px;"><sup>n</sup></span><span style="letter-spacing: 0.0px;"> + (“n choose 2”)a</span><span style="letter-spacing: 0px;"><sup>n-1</sup></span><span style="letter-spacing: 0.0px;">b + (“n choose 3”)a</span><span style="letter-spacing: 0px;"><sup>n-2</sup></span><span style="letter-spacing: 0.0px;">b</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">... + (“n choose n-1”)ab</span><span style="letter-spacing: 0px;"><sup>n-1</sup></span><span style="letter-spacing: 0.0px;"> + (“n choose n”)b</span><span style="letter-spacing: 0px;"><sup>n</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">After you let that sink in for a second, the process of creating the lower bound will be simple.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">The lower bound that is used is 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">/2<i>n</i>. To prove that this is a lower bound, we must prove the following inequality to be true:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">/2<i>n </i>< “2<i>n</i> choose <i>n</i>”</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This can be rearranged to be:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">4</span><span style="letter-spacing: 0px;"><i><sup>n </sup></i></span><span style="letter-spacing: 0.0px;">< 2<i>n</i>(“2<i>n</i> choose <i>n</i>”)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> can be rewritten using the binomial theorem. A lone four is not an expandable binomial, but here is a reconfiguration of 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> that can be expanded.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0px;">4</span><span style="letter-spacing: 0.0px;"><i><sup>n</sup></i></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(2</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">)</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0px;">2</span><span style="letter-spacing: 0.0px;"><sup>2</sup><i><sup>n</sup></i></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(1 + 1)</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> </span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Plug these values into the binomial theorem for a, b, and n to get:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(a + b)</span><span style="letter-spacing: 0px;"><sup>n</sup></span><span style="letter-spacing: 0.0px;"> = (“n choose 1”)a</span><span style="letter-spacing: 0px;"><sup>n</sup></span><span style="letter-spacing: 0.0px;"> + (“n choose 2”)a</span><span style="letter-spacing: 0px;"><sup>n-1</sup></span><span style="letter-spacing: 0.0px;">b + (“n choose 3”)a</span><span style="letter-spacing: 0px;"><sup>n-2</sup></span><span style="letter-spacing: 0.0px;">b</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">... + (“n choose n-1”)ab</span><span style="letter-spacing: 0px;"><sup>n-1</sup></span><span style="letter-spacing: 0.0px;"> + (“n choose n”)b</span><span style="letter-spacing: 0px;"><sup>n</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><sup></sup></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(1 + 1)</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> = (“2<i>n</i> choose 1”)1</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> + (“2<i>n</i> choose 2”)1</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i><sup>-1</sup></span><span style="letter-spacing: 0.0px;">1 + (“2<i>n</i> choose 3”)1</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i><sup>-2</sup></span><span style="letter-spacing: 0.0px;">1</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">... + (“2<i>n</i> choose 2<i>n</i>-1”)1•1</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i><sup>-1</sup></span><span style="letter-spacing: 0.0px;"> + (“2<i>n</i> choose 2<i>n</i>”)1</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i><sup></sup></i></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Since one raised to any power is one, all of the ones cancel out to get:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i><sup></sup></i></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(1 + 1)</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> = (“2<i>n</i> choose 1”)1 + (“2<i>n</i> choose 2”)1 + (“2<i>n</i> choose 3”)1</span><span style="letter-spacing: 0px;"><sup> </sup></span><span style="letter-spacing: 0.0px;">... + (“2<i>n</i> choose 2<i>n</i>-1”)1 + (“2<i>n</i> choose 2<i>n</i>”)1</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(1 + 1)</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> = (“2<i>n</i> choose 1”) + (“2<i>n</i> choose 2”) + (“2<i>n</i> choose 3”) ... + (“2<i>n</i> choose 2<i>n</i>-1”) + (“2<i>n</i> choose 2<i>n</i>”)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This creates just a sum of all of the values in the (2<i>n</i>)-th row of Pascal’s Triangle. Earlier, it was noted that the central binomial coefficient of a row is the largest number in that row. So, “2<i>n</i> choose <i>n</i>” is the largest number in the sum above.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">How many numbers are in that sum? Since it is the (2<i>n</i>)-th row of Pascal’s Triangle, there are 2<i>n</i> entries in that row. This means that the product of 2<i>n</i> and “2<i>n</i> choose <i>n</i>” must be greater than that sum.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">(“2<i>n</i> choose 1”) + (“2<i>n</i> choose 2”) + (“2<i>n</i> choose 3”) ... + (“2<i>n</i> choose 2<i>n</i>-1”) + (“2<i>n</i> choose 2<i>n</i>”) < 2<i>n</i>(“2<i>n</i> choose <i>n</i>”)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">What is that sum equal to? It was defined to be the expansion of 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">, meaning that 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> can be substituted in for that sum.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> < 2<i>n</i>(“2<i>n</i> choose <i>n</i>”)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Dividing both sides by 2<i>n</i> gives the inequality that the step was looking for:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">/2<i>n </i>< “2<i>n</i> choose <i>n</i>”</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<br />
<span style="letter-spacing: 0px;"><span style="font-family: inherit;">And that completes the step.</span></span><br />
<span style="letter-spacing: 0px;"><span style="font-family: inherit;"><br /></span></span>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">The fourth step is to set an upper bound. A lower bound won’t do much good without an upper bound to counter it. A fully intact upper bound for “2<i>n</i> choose <i>n</i>” cannot be found until step five is partly established, but an upper bound can be placed on a different expression which will play back into the proof later on.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This step requires something called the primorial function, which is similar to the factorial function. The number <i>n</i> factorial is the product of all of the positive integers less than or equal to <i>n</i>. Similarly, the number <i>n</i> primorial (written <i>n</i>╫) is the product of all of the prime numbers less than or equal to <i>n</i>. For example:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">4╫ = 3 • 2 = 6</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">8╫ = 7 • 5 • 3 • 2 = 210</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">16╫ = 13 • 11 • 7 • 5 • 3 • 2 = 30030</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">This step will set an upper bound on the number <i>n</i>╫. The upper bound we will try to set is 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">. So, this is the inequality that needs to be proven:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>n</i>╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> </span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This proof needs to be tackled in two parts. First, it needs to be proven for all odd values of <i>n</i>. Then, it can be proven for all even values of <i>n</i>.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">If <i>n</i> is odd, then it can be rewritten as 2<i>m</i> + 1 assuming that <i>m</i> is an integer (see definition of odd number). Throughout the proof, quantities like “2<i>n</i> choose <i>n</i>” and “2<i>m</i> choose <i>m</i>” have come up, but the row of Pascal’s Triangle it refers to is always an even numbered row (2<i>n</i> and 2<i>m</i> are even). What about odd numbered rows?</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><b><span style="font-family: inherit;">1 1</span></b></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 2 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 <b> 3 3 </b> 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 4 6 4 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 5 <b>10 10 </b> 5 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 6 15 20 15 6 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 7 21 <b> 35 35</b> 21 7 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 8 28 56 70 56 28 8 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 9 36 84 <b>126 126</b> 84 36 9 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 10 45 120 210 252 210 120 45 10 1</span></span></div>
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">An odd numbered row does not have a center position, but the two positions in the middle are always equal to each other. In other words, “2<i>m</i>+1 choose <i>m</i>” is equal to “2<i>m</i>+1 choose <i>m</i>+1.”</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">“2<i>m</i>+1 choose <i>m</i>” = “2<i>m</i>+1 choose <i>m</i>+1”</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Now, add “2<i>m</i>+1 choose <i>m</i>” to both sides of this.</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">“2<i>m</i>+1 choose <i>m</i>” = “2<i>m</i>+1 choose <i>m</i>+1”</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">“2<i>m</i>+1 choose <i>m</i>” + “2<i>m</i>+1 choose <i>m</i>” = “2<i>m</i>+1 choose <i>m</i>+1” + “2<i>m</i>+1 choose <i>m</i>”</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">2(“2<i>m</i>+1 choose <i>m</i>”) = “2<i>m</i>+1 choose <i>m</i>+1” + “2<i>m</i>+1 choose <i>m</i>”</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">The right hand side of that equation can be thought of as the sum of two of the elements of the (2<i>m</i>+1)-st row of Pascal’s Triangle. But, what if the sum of all of the elements in the (2<i>m</i>+1)-st row was found? That would definitely be a bigger number than what is currently there. So, an inequality can be formed using that sum as the right-hand side.</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">2(“2<i>m</i>+1 choose <i>m</i>”) < (“2<i>m</i>+1 choose 1”) + (“2<i>m</i>+1 choose 2”) + (“2<i>m</i>+1 choose 3”) ... + (“2<i>m</i>+1 choose 2<i>m</i>”) + (“2<i>m</i>+1 choose 2<i>m</i>+1”)</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">But in step 3, we defined this to be the polynomial expansion of (1 + 1)</span><span style="letter-spacing: 0px;"><sup>2m+1</sup></span><span style="letter-spacing: 0.0px;">, or 2</span><span style="letter-spacing: 0px;"><sup>2m+1</sup></span><span style="letter-spacing: 0.0px;">. Substitute that in for the right-hand side. Then, use the laws of exponents (see definition of law of exponents) to get:</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">2(“2<i>m</i>+1 choose <i>m</i>”) < 2</span><span style="letter-spacing: 0px;"><sup>2m+1</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(“2<i>m</i>+1 choose <i>m</i>”) < 2</span><span style="letter-spacing: 0px;"><sup>2m+1</sup></span><span style="letter-spacing: 0.0px;"> ÷ 2</span><span style="letter-spacing: 0px;"><sup>1</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(“2<i>m</i>+1 choose <i>m</i>”) < 2</span><span style="letter-spacing: 0px;"><sup>2m+1–1</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(“2<i>m</i>+1 choose <i>m</i>”) < 2</span><span style="letter-spacing: 0px;"><sup>2m</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(“2<i>m</i>+1 choose <i>m</i>”) < (2</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">)</span><span style="letter-spacing: 0px;"><sup>m</sup></span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(“2<i>m</i>+1 choose <i>m</i>”) < 4</span><span style="letter-spacing: 0px;"><sup>m</sup></span></span><br />
<div style="min-height: 10px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><sup></sup></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Now, let’s look at the prime factorization of “2<i>m</i>+1 choose <i>m</i>.” Recall the formula that was used to find exponents of primes in step 1 and step 2.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = <i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>n</i>)!) - 2<i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>n</i>!)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This can be rewritten for the quantity “2<i>m</i>+1 choose <i>m</i>” fairly easily.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>m</i>+1 choose <i>m</i>”) = <i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>m</i> + 1)!) - 2<i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>m</i>!) - <i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>m </i>+ 1)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Prime numbers less than <i>m</i>+1 are difficult to analyze, but what about between <i>m</i>+2 and 2<i>m</i>+1? By similar logic as in step 1, all of these numbers will have an exponent of zero in <i>m</i>! or (<i>m</i>+1)!, but an exponent of one in (2<i>m</i>+1)!. So, it can be guaranteed that each prime between <i>m</i>+2 and 2<i>m</i>+1 is a factor of “2<i>m</i>+1 choose <i>m</i>.” This also means that the product of all primes between <i>m</i>+2 and 2<i>m</i>+1 is less than or equal to “2<i>m</i>+1 choose <i>m</i>.”</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">The inequality we are trying to reach is n╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">. So, plugging that inequality in on a different interval (say (<i>m</i>+1)╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>m</sup></i><sup>+1</sup></span><span style="letter-spacing: 0.0px;">), is a legal maneuver. This is called proof by induction.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(<i>m</i>+1)╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>m</sup></i><sup>+1</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">The left-side of the inequality above is (<i>m</i>+1)╫. It was just determined what the upper bound of the product of all primes between <i>m</i>+2 and 2<i>m</i>+1 was as well. If we multiply the primorial of <i>m</i>+1 by the product of primes between <i>m</i>+2 and 2<i>m</i>+1, we will just get the primorial of 2<i>m</i>+1. This number will be less than the product of the two right hand sides.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(2<i>m</i>+1)╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>m</sup></i><sup>+1</sup></span><span style="letter-spacing: 0.0px;"> • “2<i>m</i>+1 choose <i>m</i>”</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">Earlier, it was proven that “2<i>m</i>+1 choose <i>m</i>” has an upper bound of 4</span><span style="letter-spacing: 0px;"><sup>m</sup></span><span style="letter-spacing: 0.0px;">. Since that will only make the right side bigger, we can plug that in without a problem. This gives:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(2<i>m</i>+1)╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>m</sup></i><sup>+1</sup></span><span style="letter-spacing: 0.0px;"> • 4</span><span style="letter-spacing: 0px;"><sup>m</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Using the law of exponents brings it to:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(2<i>m</i>+1)╫ < 4</span><span style="letter-spacing: 0px;"><sup>2</sup><i><sup>m</sup></i><sup>+1</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><sup></sup></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">And substituting <i>n</i> back in for 2<i>m</i>+1 gives:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>n</i>╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;"> </span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">We are almost done with the step. The hard part is complete; any odd values of <i>n</i> are covered. All that needs to be done is to make sure the even values are covered as well.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">If <i>n</i> is even, then its primorial will always be equal to the odd number below it. For instance:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">4╫ = 6 = 3╫</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">10╫ = 210 = 9╫</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">22╫ = 9699690 = 21╫</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This is because that even number cannot be prime. If <i>n</i> is even, its primorial must be equal to (<i>n</i>-1)╫ because <i>n</i>-1 is the highest potentially prime number less than <i>n</i>.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><i>n</i>╫ = (<i>n</i>-1)╫</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Earlier, it was proved that the upper bound works on an odd numbered primorial. Since <i>n</i>-1 is odd, we can plug that in to get:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">(<i>n</i>-1)╫ < 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i><sup>–1</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><sup></sup></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">We just determined that <i>n╫ = (n</i>-1)╫, so substituting <i>n╫ </i>in for<i> (n-</i>1)╫ gives:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>n</i>╫ < 4</span><span style="letter-spacing: 0px;"><sup>n–1</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><sup></sup></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i><sup>–1 </sup></span><span style="letter-spacing: 0.0px;">is definitely less than 4</span><span style="letter-spacing: 0px;"><i><sup>n</sup></i></span><span style="letter-spacing: 0.0px;">, so that can be put on the right-hand side to get:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>n</i>╫ < 4</span><span style="letter-spacing: 0px;"><sup>n</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><sup></sup></span><br /></span></div>
<br />
<span style="letter-spacing: 0px;"><span style="font-family: inherit;">Since the bound is good on all odd numbers and all even numbers, the step is complete.</span></span><br />
<span style="letter-spacing: 0px;"><span style="font-family: inherit;"><br /></span></span>
We have most of the groundwork done. Next week, we will officially prove Bertrand's Postulate as well as finalize our upper bound.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-69836915578385165262014-01-11T12:00:00.000-05:002014-01-11T12:00:02.173-05:00Bertrand's Postulate Part 2: Restricting the Factor DomainClick <a href="http://coolmathstuff123.blogspot.com/2014/01/bertrands-postulate-part-1-definitions.html" target="_blank">here</a> to see part one of this four week series.<br />
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This is week two of a four week series on the proof to Bertrand's Postulate. Last week, I went over some history and definitions pertaining to the problem, as well as introducing some facts about Pascal's triangle that come into play in the proof. This week, we can begin the actual process of proving that there is always a prime between <i>n</i> and 2<i>n</i>.<br />
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The proof utilizes the central binomial coefficients that I explained last week, each one being written in the form "2<i>n</i> choose <i>n</i>." The final conclusion actually states that every number of the form "2<i>n</i> choose <i>n</i>" has a prime factor between <i>n</i> and 2<i>n</i>. By determining this, we will then be able to say that there must always then be a prime between <i>n</i> and 2<i>n</i>. But how do we prove that?<br />
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This week, we will look at how to limit the range of possible prime factors of "2<i>n</i> choose <i>n</i>." If there are fewer possibilities of where these factors can exist on the number line, then it will be tougher (and eventually impossible) to find a number of this form without a prime factor between <i>n</i> and 2<i>n</i>. Next week, we will set a lower and upper bound on the size of "2<i>n</i> choose <i>n</i>," which will provide an inequality to work with. And in the last post, we will show that because of all of these restraints, there must be a prime between <i>n</i> and 2<i>n</i>. So let's get to it!<br />
<span style="font-family: inherit;"><br /></span>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">The first step is to prove that none of the prime factors of “2<i>n</i> choose <i>n</i>” are between ⅔<i>n </i>and <i>n</i>. By proving this, it will restrict the amount of spots for the prime factor to exist, and hopefully at some point restricting it to between <i>n</i> and 2<i>n</i>.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">To prove this, we first need to go over the definition of factorials.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">The number <i>n</i> factorial (written <i>n</i>!) is the product of all of the positive integers less than or equal to <i>n</i>.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">For example, 4! = 4 • 3 • 2 • 1 = 24. 7! = 7 • 6 • 5 • 4 • 3 • 2 • 1 = 5040.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">How do factorials play into this proof? Well, assuming that the prime <i>p</i> is between ⅔<i>n </i>and <i>n</i>, <i>p</i> will only appear once in the prime factorization of <i>n</i>!. This is because <i>n</i>! only has prime factors that are less than <i>n</i>. For <i>p</i> to have been multiplied in twice, that means 2<i>p</i> also must be less than <i>n</i>. But, what happens when we multiply the inequality by two?</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">⅔<i>n</i> < <i>p</i> < <i>n</i></span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">2(⅔<i>n</i>) < 2<i>p</i> < 2<i>n</i></span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1⅓<i>n</i> < 2<i>p</i> < 2<i>n</i></span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This means that 2<i>p</i> is at least 1⅓<i>n</i>, which cannot be a factor of <i>n</i>!. So, a factor of <i>p</i> can only appear once. By similar logic, there are two factors of <i>p</i> in (2<i>n</i>)!.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">A simpler way to write “number of times a prime factor appears in a number” would be to use the following notation:</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>x</i>)</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Assume that <i>p</i> is the prime number and <i>x</i> is the number that the prime is a factor of. For example:</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><sub>2</sub></span><span style="letter-spacing: 0.0px;">(24) = 3</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><sub>5</sub></span><span style="letter-spacing: 0.0px;">(50) = 2</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>n</i>)!) = 1</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>n</i>)!) = 2</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">There exists a formula that enables one to find the number of times a certain prime factor appears in the factorization of a quantity like “2<i>n</i> choose <i>n</i>.” For that example, the formula goes:</span></span><br />
<div style="min-height: 14px;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = <i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>n</i>)!) - 2<i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>n</i>!)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">The values of <i>p</i> or <i>n</i> are not defined (aside from the fact that <i>p</i> is between ⅔<i>n</i> and <i>n</i>), but we just determined the number of times <i>p</i> appeared in <i>n</i>! and (2<i>n</i>)! nonetheless. It appeared in <i>n</i>! once and (2<i>n</i>)! twice. So, those values can be substituted in to see how many <i>p</i>’s can be in “2<i>n</i> choose <i>n</i>.”</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = <i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>n</i>)!) - 2<i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>n</i>!)</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = 2 - 2(1)</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = 2 - 2</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = 0</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<br />
<span style="letter-spacing: 0px;"><span style="font-family: inherit;">So, <i>p</i> appears zero times when it is between ⅔<i>n</i> and <i>n</i>. In other words, “2<i>n</i> choose <i>n</i>” has no prime factors between ⅔<i>n</i> and <i>n</i>, thus completing this part of the proof.</span></span><br />
<span style="letter-spacing: 0px;"><span style="font-family: inherit;"><br /></span></span>
<span style="letter-spacing: 0px;"><span style="font-family: inherit;">The second step is to prove that there are no prime factors of “2<i>n</i> choose <i>n</i>” that are greater than 2<i>n</i>. Furthermore, this step will also demonstrate that there are no powers of prime factors of “2<i>n</i> choose <i>n</i>” greater than 2<i>n</i>. This will continue the process of restricting possible prime factors of step one, which will later make it impossible to have no prime factors between <i>n</i> and 2<i>n</i>.</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">This step will require another definition, but this one is extremely simple. The name of it makes it sound much more complicated than it is.</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">A number’s floor function is the greatest integer less than or equal to that number. For example:</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">floor(7.5) = 7</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">floor(18) = 18</span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">floor(π) = 3</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">These floor functions play a role in Legendre’s Theorem (created by Adrien-Marie Legendre), which is used in the proof of step two. Legendre’s Theorem states that to find the prime factorization of <i>n</i>!, one simply plugs prime numbers into the following formula, which will find their exponent.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>n</i>!) = floor(<i>n</i>/<i>p</i>) + floor(<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">) + floor(<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>3</sup></span><span style="letter-spacing: 0.0px;">) + floor(<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>4</sup></span><span style="letter-spacing: 0.0px;">) + ...</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">For example, let’s find the exponent of 2 in the number 4! = 24.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>n</i>!) = floor(<i>n</i>/<i>p</i>) + floor(<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">) + floor(<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>3</sup></span><span style="letter-spacing: 0.0px;">) + floor(<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>4</sup></span><span style="letter-spacing: 0.0px;">) + ...</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>2</sub></i></span><span style="letter-spacing: 0.0px;">(4!) = floor(4/2) + floor(4/2</span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">) + floor(4/2</span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>3</sup></span><span style="letter-spacing: 0.0px;">) + floor(4/2</span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>4</sup></span><span style="letter-spacing: 0.0px;">) + ...</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>2</sub></i></span><span style="letter-spacing: 0.0px;">(4!) = floor(4/2) + floor(4/4) + floor(4/8) + floor(4/16) + ...</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>2</sub></i></span><span style="letter-spacing: 0.0px;">(4!) = floor(2) + floor(1) + floor(0.5) + floor(0.25) + ...</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>2</sub></i></span><span style="letter-spacing: 0.0px;">(4!) = 2 + 1 + 0 + 0 + ...</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>2</sub></i></span><span style="letter-spacing: 0.0px;">(4!) = 3</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Let’s return to the formula we used in the last step to find the exponent on a prime factor of “2<i>n</i> choose <i>n</i>.”</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = <i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>n</i>)!) - 2<i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>n</i>!)</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">It is impossible to use logic to define the exponent on primes greater than 2<i>n</i> for the terms in this formula. However, we can use Legendre’s Theorem to determine the exponent on prime factors of <i>n</i>! and (2<i>n</i>)!.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = <i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">((2<i>n</i>)!) - 2<i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(<i>n</i>!)</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = [floor(2<i>n</i>/<i>p</i>) + floor(2<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">) + ...] - 2[floor(<i>n</i>/<i>p</i>) + floor(<i>n</i>/<i>p</i></span><span style="letter-spacing: 0px; text-decoration: underline;"><sup>2</sup></span><span style="letter-spacing: 0.0px;">) + ...]</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Though 2<i>n</i> cannot be plugged in for <i>p</i> (2<i>n</i> is even, and therefore is not prime), but any number larger than 2<i>n</i> can be plugged in to see if it has a non-zero exponent. But, the first term in each of the two infinite series will end up as zero (2<i>n</i> divided by a number larger than 2<i>n</i> is less than one, and thus, will have a floor function of zero). Since each term in each series is shrinking, they will all end up with a floor function of zero. So, both infinite sums will simplify to zero.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = 0 - 2(0)</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = 0 - 0</span></span><br />
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><i>v</i></span><span style="letter-spacing: 0px;"><i><sub>p</sub></i></span><span style="letter-spacing: 0.0px;">(“2<i>n</i> choose <i>n</i>”) = 0</span></span><br />
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<span style="letter-spacing: 0px;"><span style="font-family: inherit;">So, all <i>p</i> values greater than 2<i>n</i> will have an exponent of zero, meaning that there are no prime factors of “2<i>n</i> choose <i>n</i>” greater than 2<i>n</i>. This completes the second step.</span></span><br />
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Make sure to return next week so we can start to restrain the size of the number "2<i>n</i> choose <i>n</i>," and then we will be just inches away from the proof of Bertrand's Postulate.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-76182044533301923132014-01-04T12:00:00.000-05:002014-01-04T12:00:04.599-05:00Bertrand's Postulate Part 1: Definitions, History, and an Introduction<span style="font-family: inherit;">Today, I am beginning a four-post series on the proof to Bertrand's Postulate. This postulate asks the following question: is there always a prime number between a number and its double. For instance, the number 3 doubled is 6, and 5 is prime. The number 4 doubled is 8, and 7 is prime. Number theorists wondered if this would work forever.</span><br />
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<span style="font-family: inherit;">The proof for this is written up on wikipedia in a pretty short text, but it has lots of advanced notating and vocabulary. I could tell when reading it that the proof itself was not above my head, but the article was still extremely difficult to understand. So, I made it a project for myself to decode that article, and I eventually figured it out. Since it is a pretty cool proof when explained in a simpler way, I decided to share it. It is a long one, which is why I am breaking it into four parts. I am also making sure that we have most of the necessary knowledge covered in this post so that we can continue on. Some of the information discussed in each post is also interesting on its own, so don't stress out if you can only follow pieces of the proof. I can barely follow parts of it as well.</span><br />
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<span style="font-family: inherit;">First, here are some definitions to make sure you understand:</span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Integer</b></span><span style="letter-spacing: 0.0px;">: a number that can be defined as positive, negative, or zero (e.g. 3, -2, and 0 are integers; 2.4 and π are non-integers)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b></b></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Even Number</b></span><span style="letter-spacing: 0.0px;">: a number that is a multiple of two (written as 2<i>m</i>)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b></b></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Odd Number</b></span><span style="letter-spacing: 0.0px;">: one more than an even number (written as 2<i>m</i> + 1)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Sum</b></span><span style="letter-spacing: 0.0px;">: the answer to an addition problem (e.g. the sum of 2 and 7 is 9)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Product</b></span><span style="letter-spacing: 0.0px;">: the answer to a multiplication problem (e.g. the product of 3 and 5 is 15)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b></b></span><br /></span></div>
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Prime Number</b></span><span style="letter-spacing: 0.0px;">: a number that is only divisible by one and itself</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Prime Power</b></span><span style="letter-spacing: 0.0px;">: a number that is a positive integer power of a single prime number</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Factor</b></span><span style="letter-spacing: 0.0px;">: a number that evenly divides into another number (e.g. 6 is a factor of 18; 3 is a prime factor of 12)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Prime Factorization</b></span><span style="letter-spacing: 0.0px;">: the decomposition of a number into a product of its prime factors (e.g. 4680 = 2</span><span style="letter-spacing: 0px;"><sup>3</sup></span><span style="letter-spacing: 0.0px;"> • 3</span><span style="letter-spacing: 0px;"><sup>2</sup></span><span style="letter-spacing: 0.0px;"> • 5 • 13)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Polynomial Expansion</b></span><span style="letter-spacing: 0.0px;">: a polynomial’s expansion is a way to write the polynomial such that it is a sum of products (e.g. 2(x - 7) has a polynomial expansion of 2x - 14)</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px; text-decoration: underline;"><b>Law of Exponents</b></span><span style="letter-spacing: 0.0px;">: the expression a</span><span style="letter-spacing: 0px;"><sup>p</sup></span><span style="letter-spacing: 0.0px;"> • a</span><span style="letter-spacing: 0px;"><sup>q</sup></span><span style="letter-spacing: 0.0px;"> can be simplified to a</span><span style="letter-spacing: 0px;"><sup>p+q</sup></span><span style="letter-spacing: 0.0px;">, or similarly, the expression a</span><span style="letter-spacing: 0px;"><sup>p</sup></span><span style="letter-spacing: 0.0px;"> ÷ a</span><span style="letter-spacing: 0px;"><sup>q</sup></span><span style="letter-spacing: 0.0px;"> can be simplified to a</span><span style="letter-spacing: 0px;"><sup>p–q</sup></span><span style="letter-spacing: 0.0px;">, or the expression (a</span><span style="letter-spacing: 0px;"><sup>p</sup></span><span style="letter-spacing: 0.0px;">)</span><span style="letter-spacing: 0px;"><sup>q</sup></span><span style="letter-spacing: 0.0px;"> can be simplified to a</span><span style="letter-spacing: 0px;"><sup>pq</sup></span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
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<span style="font-family: inherit;"><span style="letter-spacing: 0px;">Also, for some historical context, here are some of the mathematicians that had an influence on this postulate:</span><span style="letter-spacing: 0.0px;"></span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><b>Joseph Louis François Bertrand</b> - born March 11, 1822 in Paris, France; died April 5, 1900 in Paris, France. Bertrand was a professor at the École Polytechnique and Collège de France and was known for his contributions to number theory, differential geometry, probability theory, economics, and thermodynamics. He was the first to conjecture Bertrand’s Postulate, and confirmed by hand the first three-million values of <i>n</i>.</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><b>Pafnuty Lvovich Chebyshev</b> - born May 16, 1821 in Okatovo, Russia (near Moscow); died December 8, 1894 in St. Petersburg, Russia. Chebyshev was a professor at St. Petersburg University, and was known for his contributions to probability, statistics, mechanics, and number theory. In 1850, he was the first to prove Bertrand’s Postulate, which gave the theorem nicknames such as the “Bertrand-Chebyshev Theorem” or “Chebyshev’s Theorem.”</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><b>Srinivasa Ramanujan</b> - born December 22, 1887 in Erode, India (near Chennai); died April 26, 1920 in Chennai, India. Though an autodidact, Ramanujan is said to be one of the best mathematicians of all time, making a remarkable impact in mathematical analysis, number theory, infinite series, and continued fractions. Using properties of the Gamma Function (a function commonly used in statistics and combinatorics), Ramanujan was able to provide a simpler version of Chebyshev’s proof in 1919.</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><b>Paul Erdös</b> - born March 26, 1913 in Budapest, Hungary; died September 20, 1996 in Warsaw, Poland. Erdös spent his life traveling from university to university collaborating with different professors on problems in fields such as combinatorics, graph theory, number theory, set theory, and probability theory. He was known for finding extremely elementary proofs for various conjectures, which he did with Bertrand’s Postulate in 1934.</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><b>Blaise Pascal</b> - born June 19, 1623 in Clermont-Ferrand, France; died August 19, 1662 in Paris, France. In addition to his accomplishments in number theory, projective geometry, probability theory, and economics, Pascal was a physicist, philosopher, theologist, and the inventor of the first digital calculator. He is possibly best known for popularizing Pascal’s Triangle, which plays a role in Erdös’s proof of Bertrand’s Postulate.</span></span><br />
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><b>Adrien-Marie Legendre</b> - born September 18, 1752 in Paris, France; died January 10, 1833 in Paris, France. Legendre was a professor at the École Militaire and École Normale, and was known for his contribution to number theory, elliptic functions, calculus, and applied mathematics. He created the formula to find the frequency of prime factors in factorials, which is now often called Legendre’s Theorem.</span></span><br />
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<span style="font-family: inherit;">I'm sure there are more, but those six were some of the most prominent. Now, let's begin to tell the tale of proving Bertrand's Postulate.</span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">In 1845, Joseph Bertrand proposed the idea that there will always be a prime number between a number and its double, or <i>n</i> and 2<i>n. </i>Bertrand had already checked by hand every number between two and three-million and found that they all fit this hypothesis. But, numbers are infinite, and people can’t go checking every single one. This is why a mathematical proof would be needed to verify this conjecture and turn it into a theorem.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Pafnuty Chebyshev offered a full mathematical proof of this statement in 1850; he used logic and mathematics to prove that any value of <i>n</i> would have a prime between it and its double. A simpler proof was offered in 1919 by Srinivasa Ramanujan, and an even simpler one was demonstrated by Paul Erdös in 1934. A quote commonly attributed to Erdös is:</span></span><br />
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<span style="letter-spacing: 0.0px;"><i><span style="font-family: inherit;">Chebyshev said it, but I’ll say it again</span></i></span></div>
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<span style="letter-spacing: 0.0px;"><i><span style="font-family: inherit;">There is always a prime between n and 2n.</span></i></span></div>
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">To understand Erdös’s proof, it is important to first know a few properties of Pascal’s Triangle, which was popularized by Blaise Pascal. Pascal’s triangle starts with a triangular formation of ones.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">To fill in a blank space in this triangle, one then must add the two numbers above it. For instance, the space in the third row between the two ones would be filled in by adding the 1 above and to the left and the 1 above and to the right. 1 + 1 = 2, so this space would have a 2 in it.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1</span></span></div>
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<span style="font-family: inherit;"><span style="color: #ad3e00; letter-spacing: 0.0px;"><b>1</b></span><span style="letter-spacing: 0.0px;"> </span><span style="color: #e32400; letter-spacing: 0.0px;"><b>1</b></span></span></div>
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">1 </span><span style="color: #0056d6; letter-spacing: 0.0px;"><b>2</b></span><span style="letter-spacing: 0.0px;"> 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Similarly, the square underneath the 1 and the 2 can be filled in with a 3.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
<div style="color: #b51a00; text-align: center;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"><b>1 2 </b></span><span style="color: black; letter-spacing: 0.0px;"> 1</span></span></div>
<div style="text-align: center;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;">1 </span><span style="color: #0056d6; letter-spacing: 0.0px;"><b>3</b></span><span style="letter-spacing: 0.0px;"> 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">By continuing this process, the triangle can be extended infinitely.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 2 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 3 3 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 4 6 4 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 5 10 10 5 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 6 15 20 15 6 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 7 21 35 35 21 7 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 8 28 56 70 56 28 8 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 9 36 84 126 126 84 36 9 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 10 45 120 210 252 210 120 45 10 1</span></span></div>
<div style="min-height: 14px; text-align: center;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">With this triangle, each number can be defined by saying its row number and what position it is in in that row. Since the ones are the base of the triangle, and are not necessarily part of it, the top row is defined to be row zero and the left-most element in each row is element zero. By keeping this in mind, any number in Pascal’s Triangle can be quickly found. For example, the fifth element of row seven would be 21.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">Rather than going through the process of saying “the <i>k</i>th element of row <i>n</i> in Pascal’s Triangle,” mathematicians will often use the notation “<i>n</i> choose <i>k</i>,” with <i>n</i> being the row number and k being the number’s position in the row. For instance, “6 choose 2” = 15.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">In Pascal’s Triangle, there are some numbers that stick out because they are in the center of their row. For instance:</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
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<span style="letter-spacing: 0.0px;"><b><span style="font-family: inherit;">1</span></b></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 <b>2</b> 1</span></span></div>
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<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 3 3 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 4 <b>6</b> 4 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 5 10 10 5 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 6 15 <b>20</b> 15 6 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 7 21 35 35 21 7 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 8 28 56 <b>70</b> 56 28 8 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 9 36 84 126 126 84 36 9 1</span></span></div>
<div style="text-align: center;">
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">1 10 45 120 210 <b>252</b> 210 120 45 10 1</span></span></div>
<div style="min-height: 14px; text-align: center;">
<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">These numbers have a very simple notation in the “<i>n</i> choose <i>k</i>” format because <i>k</i> is half of <i>n</i>. Since <i>n</i> and <i>k</i> can never be fractions (there is no 2.7th element in a row), the best way to write these numbers is “2<i>n</i> choose <i>n</i>.” Mathematicians call these numbers as central binomial coefficients. Note that the central binomial coefficient is the largest number in each row of the triangle.</span></span><br />
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<span style="font-family: inherit;"><span style="letter-spacing: 0.0px;"></span><br /></span></div>
<br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;">These central binomial coefficients play a huge role in Erdös’s proof. What Erdös actually proved was that there is always a prime between <i>n</i> and 2<i>n</i> that is a prime factor of the corresponding central binomial coefficient: “2<i>n</i> choose <i>n</i>.” </span></span><br />
<span style="letter-spacing: 0.0px;"><span style="font-family: inherit;"><br /></span></span>
<span style="letter-spacing: 0px;"><span style="font-family: inherit;">Next week, I will continue on and show some of the first steps of the actual proof.</span></span>Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-43223381143349063302013-12-28T12:00:00.000-05:002013-12-28T12:00:01.546-05:00Happy End ProblemAs this is the last post of the year, I thought it would be appropriate to end with a post on the "Happy End Problem." I will explain the problem, which is pretty cool in itself, and then talk a little bit about the history behind it.<br />
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The initial question was if any five points were placed on a plane with no three of them in a straight line, will four of those points always form a convex quadrilateral? For example, in the following image:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidSZiAYGo8VOPm0J4BDmMxT24ldr_mghYf9jEdC1xRpcU0hFhso5WcOwVK1yiLdXdf1ErfciFKMx1k4-I1e2CrcIINlgXhKGHz5ATXZ9aGmyP6_n1Iuu7D8jzBokKVL3oadQnrS4w0XMEI/s1600/Screen+Shot+2013-11-24+at+8.56.03+PM.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="170" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidSZiAYGo8VOPm0J4BDmMxT24ldr_mghYf9jEdC1xRpcU0hFhso5WcOwVK1yiLdXdf1ErfciFKMx1k4-I1e2CrcIINlgXhKGHz5ATXZ9aGmyP6_n1Iuu7D8jzBokKVL3oadQnrS4w0XMEI/s200/Screen+Shot+2013-11-24+at+8.56.03+PM.png" width="200" /></a></div>
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The following convex quadrilateral can be formed:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMOQdxKKKUaHBWxWSciZA_cDI1GxzlPGbr7BAP6ljCTpENH8cn1B8QAT79_JhRJJx_UsxPvxF8jBdTOdKjEtfYFZ_u_EvGs3_bNR8gnxsZsZLHDSY_AD2M06QJPL2cONELOWfwwo63CRNG/s1600/Screen+Shot+2013-11-24+at+8.58.05+PM.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="183" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMOQdxKKKUaHBWxWSciZA_cDI1GxzlPGbr7BAP6ljCTpENH8cn1B8QAT79_JhRJJx_UsxPvxF8jBdTOdKjEtfYFZ_u_EvGs3_bNR8gnxsZsZLHDSY_AD2M06QJPL2cONELOWfwwo63CRNG/s200/Screen+Shot+2013-11-24+at+8.58.05+PM.png" width="200" /></a></div>
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Will this always work? As usual, I encourage you to grab a scrap piece of paper and try out a few examples. Have fun with it. Get creative! You will end up finding that no matter how you position the five points, you cannot get a combination without a convex quadrilateral.<br />
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Why is this true? In fact, there is a very easy way to prove it. Let's analyze three cases.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgL78EjxF8C1D0akbea4HhEi7d5rxVnyiuGBlF_5dF2F92skzHrdv6qtiEJh24fOUcDtjtcFOjC_6nh4ebl_nZghGx5EOA5vwmxksxMS8wZ36L4WIOfwtq-C3qKEJTmXSQGLHEnji52MtrH/s1600/300px-Happy-End-problem.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgL78EjxF8C1D0akbea4HhEi7d5rxVnyiuGBlF_5dF2F92skzHrdv6qtiEJh24fOUcDtjtcFOjC_6nh4ebl_nZghGx5EOA5vwmxksxMS8wZ36L4WIOfwtq-C3qKEJTmXSQGLHEnji52MtrH/s1600/300px-Happy-End-problem.svg.png" /></a></div>
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The first case is the top left one in the red, where the five points form a convex pentagon. In this instance, connecting any four of the points will form a convex quadrilateral by nature.<br />
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The second case is the top right one in the blue, where one point is located in between the four outside points. The illustration shows the inside point being included in the quadrilateral, but it could have just as easily been made as just the four outside points. This will continue to work for any combination of this nature by logic.<br />
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The third case is the bottom one in the yellow, where two points are enclosed in a triangle. When you draw a line between the two center points, two of the outside points will end up on one side and one will be on the other. Using the two outside points as your third and fourth vertices will form a quadrilateral without flaw.<br />
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Now, you might be wondering if one can prove a similar case with a convex pentagon. Could it be done with six? Seven? Eight? Turns out, nine points are required for it to work every time. As you can see below, eight points is just one too few.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgfXqf0QOdnbe35XkqqcsAX1T8YnCInsKLAasZZCIMXuB-cf8306K_OH2ZlWnL8fZW89DjJjhX6TWj-TaHTjoQOW6M70vzaIm2BFwM68pVjG3T-GFNr31vip4PAW1FSkSPSgWCjy6uEzS7a/s1600/220px-8-points-no-pentagon.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="200" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgfXqf0QOdnbe35XkqqcsAX1T8YnCInsKLAasZZCIMXuB-cf8306K_OH2ZlWnL8fZW89DjJjhX6TWj-TaHTjoQOW6M70vzaIm2BFwM68pVjG3T-GFNr31vip4PAW1FSkSPSgWCjy6uEzS7a/s200/220px-8-points-no-pentagon.svg.png" width="200" /></a></div>
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What about convex hexagons? Or heptagons? Or octagons? Or chiliagons (1000-sided polygons)? Well, what many mathematicians will do from here is look for a formula to figure out how many points are required for a given <i>n</i>-gon. We know that for a triangle (<i>n</i> = 3), just 3 points are needed (all triangles are convex). For a quadrilateral (<i>n</i> = 4), we proved that 5 points are needed. For a pentagon (<i>n</i> = 5), I mentioned that 9 points are needed. Do you see the pattern?<br />
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3, 5, 9, ...<br />
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It is not easy to spot at first, but what if I subtract one from each of those terms:<br />
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2, 4, 8, ...<br />
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They are all now powers of two! This pattern seems to fit the formula A<sub><i>n</i></sub> = 2<sup><i>n</i>-2</sup> + 1. Plugging six in for <i>n</i> would give:<br />
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A<sub>6</sub> = 2<sup>6-2</sup> + 1<br />
A<sub>6</sub> = 16 + 1<br />
A<sub>6</sub> = 17<br />
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This formula predicts that seventeen points would be required for a hexagon. Mathematicians would then work to try to prove that this is the case. Further, they would try to prove that for any value of <i>n</i>, the A<sub><i>n</i></sub> formula holds true.<br />
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George Szekeres (1911-2005; a Hungarian-Austrailian mathematician and analytical chemist) and Esther Klein (1910-2005, another Hungarian-Austrailian mathematician) worked together to prove that all values of <i>n</i> will have a finite A<sub><i>n</i></sub> output (there will be a number of points that creates the ability for a convex <i>n</i>-gon to be formed), but they could not get this bound down to the formula above. Soon after this proof was published, Szekeres and Klein married each other, which inspired the name "Happy End Problem."<br />
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Paul Erdös (1913-1996; a Hungarian mathematician), possibly one of the most influential of the twentieth century), was able to prove successfully that with 71 points, a hexagon can always be drawn.<br />
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Sixty years later, Ronald Graham (born 1935; a Californian mathematician) and his wife, Fan Chung, decided to take a swing at the problem. While on a plane ride to a math conference in New Zealand, they were able to lower Erdös's bound to 70 points, which doesn't sound like much, but it brought the problem back into the minds of mathematicians. It was also ironic that another achievement pertaining to the Happy End Problem was from a couple.<br />
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Daniel Kleitman (born 1934; an applied mathematician at MIT) and Lior Pachter (born 1972; an Israeli mathematician and molecular biologist at Berkeley College) worked together to lower the upper bound to 65 points. The number was then lowered to 37 points, and has yet to be lowered further.<br />
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Although lots of progress has been made on this problem, the overarching proof still has not yet been found. There has been no counterexample to the A<sub><i>n</i></sub> formula, and there has certainly been no guaranteed formula to generate the future values. People often wonder what a mathematician actually does for his/her job. A big part of it is trying to figure out the answers to these unsolved problems, which can often be understood by the average person. Try playing around with it and you might make a discovery too.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-11781743678586117792013-12-21T12:00:00.000-05:002013-12-21T12:00:05.427-05:00Figure Out The Number Of Digits In Gigantic Numbers<span style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-family: inherit; font-size: 18px; line-height: 24px;">In science and math, you often run across numbers that are too big to be written in standard form. They are usually written in scientific notation, but they are also sometimes written as a number to a certain power. For instance, one might say that there are 2<sup>20</sup> outcomes of the flipping of twenty coins rather than saying 1.049x10<sup>6</sup> ways.</span><br />
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;"><br /></span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;">By using that power, your information is likely more accurate. However, this power does not tell you much about the number. Most of us would have no idea if 2<sup>20</sup> is in the thousands, millions, billions, etc. at the first glance.</span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;"><br /></span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;">First, lets ask a question. What is the common logarithm of a number, or what can you gather from it? Well, the common logarithm is the power that ten has to be raised to to obtain that number. For instance:</span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;"><br /></span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;">log(100) = 2</span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;">log(5000) = 3.69897</span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;">log(<a href="tel:6283185" x-apple-data-detectors-result="0" x-apple-data-detectors-type="telephone" x-apple-data-detectors="true">6283185</a>) = 6.79818</span></div>
<div style="-webkit-composition-fill-color: rgba(130, 98, 83, 0.0976563); -webkit-composition-frame-color: rgba(191, 107, 82, 0.496094); -webkit-tap-highlight-color: rgba(26, 26, 26, 0.292969); -webkit-text-size-adjust: auto; font-size: 18px; line-height: 24px;">
<span style="font-family: inherit;"><br /></span></div>
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<span style="font-family: inherit;">What do you notice about these numbers? It's not clear at first, but count the number of digits in each of the inputs. You will find that the common log is always just a little bit below that number. In fact, to figure out the number of digits in a number, all you have to do is take the common log and round up to the nearest integer.</span></div>
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<span style="font-family: inherit;">How can this be used to find the number of digits in a power? Interestingly enough, there is a logarithmic identity stating that the log of a number raised to the power is equal to the power times the log of the number. For example,</span></div>
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<span style="font-family: inherit;">log(27) = 3log(3)</span></div>
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<span style="font-family: inherit;">ln(32) = 5ln(2)</span></div>
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<span style="font-family: inherit;">log(2<sup>20</sup>) = 20log(2)</span></div>
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<span style="font-family: inherit;"><br /></span></div>
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<span style="font-family: inherit;">Look at the last example there. We just simplified the gigantic 2<sup>20</sup> to a reasonable looking 20log(2), which is the formula to figure out the number of digits it has. In other words, the number of digits in 2<sup>20</sup> is just 20log(2) rounded to the nearest integer. Plugging this into a calculator tells you that the log is 6.0206, meaning that there are seven digits in the number. If you multiply it out, you will find that 2<sup>20</sup> = <a href="tel:1048576" x-apple-data-detectors-result="1" x-apple-data-detectors-type="telephone" x-apple-data-detectors="true">1048576</a>, which does indeed have seven digits. </span></div>
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<span style="font-family: inherit;">So whenever a type of problem pops up with a power of this sort, try to determine how many digits it is. Chances are you will gain a much better understanding of the statistic when you perform this quick calculation.</span></div>
Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com1tag:blogger.com,1999:blog-2207789741693789296.post-82554172803718342022013-12-14T12:00:00.000-05:002013-12-14T12:00:05.370-05:00Math in the News: The Influences of PoliticsThough mathematics is normally a pretty concrete subject, people's intuition for it is not. Probability and statistics in particular is a very difficult area for us to grasp, as you've seen with the <a href="http://coolmathstuff123.blogspot.com/2013/11/the-history-of-monty-hall-problem.html" target="_blank">Monty Hall Problem </a>I talked about a few weeks ago.<br />
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Here is another example of mathematical aptitude being influenced by an outside source, but this time, it is not just a matter of lack of skill or desire to be correct. It is also influenced sometimes by political views, as Kevin Drum shows in this news article. Check it out!<br />
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http://m.motherjones.com/kevin-drum/2013/09/politics-destroys-math-abilityEthan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-74814742040049500862013-12-07T12:00:00.000-05:002013-12-07T12:00:04.682-05:00Isosceles Triangle Theorem<br />
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<span style="letter-spacing: 0.0px;">The field of geometry is comprised of many different types of questions. Some are construction based, such as “what is the area of this circle?” On the other hand, many are proof based, like “why are these two triangles congruent?” This is slightly different from the proofs I normally discuss; proofs I normally post are very generalized theorems while these questions are more similar to a specific algebra or arithmetic problem.</span></div>
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<span style="letter-spacing: 0.0px;">When writing a geometric proof, many different theorems come into play. One must use the generalized theorems, properties, and postulates to arrive at a conclusion. Let’s try a simple proof just to demonstrate the nature of these theorems. Let’s prove that triangle ABE is congruent to triangle CDE.</span></div>
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<span style="letter-spacing: 0.0px;">First, we would say that it is given that segment AC is parallel to segment BD and that segment AB is parallel to segment CD. We would then say that ABCD is a parallelogram by the definition of a parallelogram (which requires two sets of parallel sides). The definition of a parallelogram also requires AB and CD to be congruent. The vertical angle theorem can be used to say that angle AEB is congruent to angle CED. The alternate interior angle theorem says that angle ABE is congruent to angle DCE. The angle-angle-side triangle congruence postulate then concludes that triangle ABE is congruent to triangle CDE.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjD7_davoTAb8ePGhtLy2znh8yQscy8coxUzmYpa8SGGwiJikXzhEsT43xOvN2bjAyqRP6l-qTqJwhSPeoP-u_cWthPru91sJtl8IC4RveZ4QJCjCyhT_FYFtsUA153XPUAe_AM3KtXQPyd/s1600/Screen+Shot+2013-12-02+at+4.56.25+PM.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="192" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjD7_davoTAb8ePGhtLy2znh8yQscy8coxUzmYpa8SGGwiJikXzhEsT43xOvN2bjAyqRP6l-qTqJwhSPeoP-u_cWthPru91sJtl8IC4RveZ4QJCjCyhT_FYFtsUA153XPUAe_AM3KtXQPyd/s400/Screen+Shot+2013-12-02+at+4.56.25+PM.png" width="400" /></a></div>
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<span style="letter-spacing: 0px;">As you can see, there are many steps in this proof, and each one uses a different definition, theorem, or postulate. In high school geometry classes, students are told these theorems and postulates, and expected to memorize them for future examples. This makes geometry boring and pointless, when it can be quite fascinating. A way to easily spice up geometric proofs is to actually prove the theorems before they are used in class. If Euclid could do it, then we can do it.</span><span style="letter-spacing: 0.0px;"></span></div>
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<span style="letter-spacing: 0px;">A fun one to prove is the Isosceles Triangle Theorem. This theorem states that when a triangle has two congruent sides, it also has two congruent angles. This can be proven in a similar way as the congruent triangle question I posed earlier. Take an isosceles triangle:</span><span style="letter-spacing: 0.0px;"></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh10ZrzZJwO4owGsyxXv89s3KED6KodDS_MqaPxXJOlfAg_JTd0NTLf9BZiS0jc9SrpXJPZLsgb_pS4ZolnHgir7yrB8DJp3k5CM0zIvUKNbv55Pgr-n_akbQMRjKAPheW-W0suFGZekl0_/s1600/Screen+Shot+2013-12-02+at+5.00.31+PM.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="101" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh10ZrzZJwO4owGsyxXv89s3KED6KodDS_MqaPxXJOlfAg_JTd0NTLf9BZiS0jc9SrpXJPZLsgb_pS4ZolnHgir7yrB8DJp3k5CM0zIvUKNbv55Pgr-n_akbQMRjKAPheW-W0suFGZekl0_/s400/Screen+Shot+2013-12-02+at+5.00.31+PM.png" width="400" /></a></div>
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<span style="letter-spacing: 0px;">If you were to bisect that top angle, it would create two new triangles. Since the original triangle is isosceles, it is given that the top left segment is congruent to the top right segment. The definition of a bisection (cutting an angle in half) states that the left part of the top angle is congruent to the right part of the top angle. The reflexive property of congruence states that the middle segment is congruent to itself. By the side-angle-side triangle congruence postulate, the left triangle is congruent to the right triangle. And finally, by CPCTC (common parts of congruent triangles are congruent), the bottom left angle is congruent to the bottom right angle.</span><span style="letter-spacing: 0.0px;"></span></div>
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<span style="letter-spacing: 0px;">This sort of geometric proof language sounds extremely long and boring. However, finding uses for it such as proving the Isosceles Triangle Theorem can make it a little more fun. Among many things, I think that schools should teach the reasons behind these theorems to make it more logical and fun to apply them to class.</span><span style="letter-spacing: 0.0px;"></span></div>
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Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-937438394033329922013-11-30T12:00:00.000-05:002013-11-30T12:00:01.540-05:00How YOU Can Memorize 2000 Digits of a NumberSince it is two days after Thanksgiving, and many of us are probably eating leftover pie, I thought it would be appropriate to do a post somewhat relevant to the numerical pi. And I could do something very mathematical, but I just finished my first trimester at <a href="http://www.andover.edu/" target="_blank">Phillips Academy Andover</a> last week, and I needed a break after my nearly impossible MATH-380 final exam. As a result, I thought it could be fun to talk about memorizing numbers, pi and tau in particular.<div>
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When people hear about my <a href="http://www.youtube.com/watch?v=x-mPs5TuNfw" target="_blank">Tau 2000</a> event, they often ask me if I have what they call a "photographic memory." This is not at all true. I don't even think the types of photographic memories advertised in pop culture really exist (I'm not an expert on neurology, so for more on that, I'd recommend reading <a href="http://www.scientificamerican.com/article.cfm?id=i-developed-what-appears-to-be-a-ph" target="_blank">this Scientific American article</a>). The way I memorized 2012 digits of tau was all learned and practiced techniques, similar to my mental math presentations. I was not born with some gift or natural talent, it was just learning the methodology and practicing until I could do it quickly. Just like anyone can do mental math, anyone can be a memory expert as well, ranging from being able to remember 57890 digits of pi to being able to remember your car keys as you leave for work. There are techniques for it all.</div>
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First, let me introduce you to the Major System. This is a phonetic code that enables you to turn numbers into words. You store them as words, and later retrieve them as numbers. Basically, each digit is associated with a specific consonant sound.</div>
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1 is the <i>t</i> or <i>d</i> sound. It can also be either of the <i>th</i> sounds (see note below).</div>
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2 is the <i>n</i> sound.</div>
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3 is the <i>m</i> sound.</div>
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4 is the <i>r</i> sound.</div>
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5 is the <i>l</i> sound.</div>
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6 is the <i>j, ch, sh, or zh</i> sound.</div>
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7 is the <i>k</i> or <i>g</i> sound.</div>
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8 is the <i>f</i> or <i>v</i> sound.</div>
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9 is the <i>p</i> or <i>b</i> sound.</div>
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0 is the <i>z</i> or <i>s</i> sound.</div>
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<span style="font-size: x-small;">Note: <i>th</i> (both the <i>th</i> in "that" and the <i>th</i> in "thing") is normally paired with 1, but there are other variations on the system that will put it with 8 or not include it.</span></div>
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This looks hard to memorize on its own, but it is actually not that hard. Here are some mnemonics that can help you.</div>
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<li style="text-align: left;">A <i>t</i> or <i>d</i> has 1 downstroke.</li>
<li style="text-align: left;">A <i>n</i> has 2 downstrokes.</li>
<li style="text-align: left;">A <i>m</i> has 3 downstrokes.</li>
<li style="text-align: left;">The number 4 ends in the letter <i>r</i>.</li>
<li style="text-align: left;">If you hold up your hand with 4 fingers up and your thumb at a 90° angle, you will see 5 fingers shaped like an L.</li>
<li style="text-align: left;">A J looks somewhat like a backwards 6.</li>
<li style="text-align: left;">A K can be drawn with two 7s back to back.</li>
<li style="text-align: left;">A lowercase <i>f</i> in cursive looks like an 8.</li>
<li style="text-align: left;">The number 9 is a backwards <i>p</i> or an upside-down <i>b</i>.</li>
<li style="text-align: left;">The word <i>zero</i> begins with the letter <i>z</i>.</li>
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You will also notice that the consonants that were paired together sound very similar.<i> </i>Your lip movement and tongue placement are the same in any of the consonant sounds chosen for a number (except for the <i>th</i> sounds, hence the inconsistency of its use).</div>
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You might be wondering why there are no vowel sounds on the list. There is also no <i>h</i>, <i>w</i>, or <i>y</i> sound. This is because you can insert these wherever you want between consonants and they mean nothing. With all of this in mind, you can begin turning numbers into words. Let's take the number 15. What words can this become?</div>
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Well, one is the <i>t</i> or <i>d</i> sound. Five is the <i>l</i> sound. Insert vowels, and you can get doll. Or tile. Or tail. You can also insert vowels at the beginning or end of the word and make deli, or Adele. You can also insert <i>h</i>s, <i>w</i>s, and <i>y</i>s to get hotel, towel, or yodel. Here are a list of the 66 words that can be made out of the number 15 (I put the ones that I might use in a mnemonic image in bold print):</div>
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Addle, daily, dale, dally, deal, delay, dell, dial, dole, <b>doll</b>, dual, duel, dull, duly, dwell, ethyl, hastily, hostile, <b>hotel</b>, hotly, huddle, ideal, ideally, idle, idly, idol, it'll, italy, oddly, othello, outlaw, outlay, saddle, sadly, seattle, settle, societal, stale, stall, steal, steel, still, stole, <b>stool</b>, style, subtle, subtly, suicidal, sweetly, <b>tail</b>, tale, tall, tally, teal, tel, tell, they'll, <b>tile</b>, till, toil, toll, <b>tool</b>,<b> towel</b>, waddle, widely, <b>yodel</b></div>
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<span style="font-size: x-small;">Note that some of the words start with s. Since s is zero, this is referring to the number 015, which is normally still 15. These words do not work if 15 is part of a string of other digits such as in pi or tau.</span></div>
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The ones that I bolded are all nouns that you can create a mental image of in your head. As the Scientific American article that I linked to states, people naturally have a better memory for visuals (the reason why you might remember someone's face, but not be able to place the name). So, you might not be able to remember the number 15, but you can probably picture a doll, or a hotel, or a yodel (for this, I would think of the chocolate pastry, not the verb). If you are trying to remember that it is someone's address or apartment number, picture a relationship between the object and the person. Maybe the person is standing up on a stool shouting to a crowd of confused, awestricken people, or they are on the couch stuffing their face with yodels. The sillier your image, the easier it is to remember.</div>
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There are lots of memory experts who will create a list of "peg words," which are essentially 100 words that they will refer to when they are trying to remember a number between 1 and 100. It is certainly not a necessity, but it can often help if you are trying to come up with a word on the fly. Every person has a different list of words that works for them, so this is something that I would encourage you to make on your own. The website <a href="http://www.phoneticmnemonic.com/">www.phoneticmnemonic.com</a> works very well to help create this list.</div>
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To memorize shorter strings of digits (something like memorizing 100 digits of pi), the best approach in my opinion is to create sentences out of your words. For instance, take the first five digits of pi: 31415. The only word that can be formed out of this is moderately, which isn't a great start to a sentence. However, it could be turned into "my turtle" or "Madrid law" or "Mother Yodel." The first 24 digits of pi create the sentence:</div>
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<b>My turtle Pancho will, my love, pick up my new mover, Ginger.</b></div>
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Say this a few times and you will sadly have it memorized. And since you now know the code, you now have the first 24 digits of pi memorized. If you want to keep going, the next 17 digits are:</div>
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<b>My movie monkey plays in a favorite bucket.</b></div>
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The next 19 are:</div>
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<b>Ship my puppy Michael to Sullivan's backrubber.</b></div>
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If you want to take it to 100 digits, you can use:</div>
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<b>A really open music video cheers Jenny F. Jones.</b></div>
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And my personal favorite:</div>
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<b>Have a baby fish knife so Marvin will marinate the goosechick.</b></div>
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This method works great for condensing large quantities of numbers into a small amount of silly, memorable sentences. However, once you get up towards 300, 400, 500 digits, it is really tough to remember the exact prepositions and linking verbs you used, which contribute to the digits. Because of this, the method I used for memorizing 2012 digits of tau is a different variation. Rather than just memorizing plain sentences, I used a technique called the memory palace.</div>
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A memory palace is essentially a place that you can mentally visualize that you put the images that you create in. For instance, your drive from your house to work might be a memory palace. Your elementary school campus could be your memory palace. You can even create an imaginary place to be your memory palace. Let's pretend your memory palace is inside of your house. The first ten loci (places to put the images) might be:</div>
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<li>Your bed (in your bedroom)</li>
<li>Your closet</li>
<li>Bathroom</li>
<li>Hallway</li>
<li>Other Bedroom</li>
<li>Stairs</li>
<li>Living Room</li>
<li>Dining Room</li>
<li>Kitchen</li>
<li>Front Porch</li>
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And you might have a grocery list with the following items:</div>
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<li>Grapes</li>
<li>Carrots</li>
<li>Corn on the Cob</li>
<li>Yogurt</li>
<li>Cheddar Cheese</li>
<li>Marshmallows</li>
<li>Cheetos</li>
<li>Salt</li>
<li>Pepper</li>
<li>Ice</li>
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All you need to do is mentally "put" each of these items into the corresponding locus in your memory palace. For instance, the first item is grapes. You would put the grapes on your bed. But you wouldn't just put them there, you must do something to make the image stand out. First of all, you must embrace the image. Not only do you see grapes, but you smell the grapes, you taste the grapes. The more of your senses that you alert, the easier the image is to remember. The image also needs to be less dull than just a few grapes sitting on your blanket. Maybe have grapevines growing out of the back of your bed. Maybe visualize the grapes to have legs, and jumping on the bed. As long as it is a silly image that stands out in your mind, you will be able to remember it.</div>
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The next item on the list is carrots. The corresponding locus is your closet. Carrots grow out of the ground, so maybe you picture all of the mud that your sneakers have tracked into the closet has carrots growing in it. As long as you pull a carrot out of the mud, you will remember it is carrots. Or maybe there is a snowman inside with a carrot nose, or a carrot shoe-horn. The actual carrot aspect of the image can absolutely be subtle, as long as you can remember the image and this image triggers the thought of carrots in your mind.</div>
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Continue through the list, and you will have ten images in your head that will in fact be stuck there until you use other techniques to remove them (yes, there are techniques people use to forget things). Try this out a few times, and I'm sure you will find it very useful. If you have a list of things to do at work, you need to remember when to pick up your kids and bring them to their activities (you may even use the major system for translating times into words - if you need to bring your son to baseball practice at 4:15, you may just picture your son swinging his bat at a "hurdle" (r=4, d=1, l=5) in the appropriate locus), or anything else, the memory palace is a great way to go.</div>
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How does this help one memorize the digits of a number, like tau? Well, what the major system does is turns numbers into words, which can then be turned into images. The memory palace then acts as a place holder for those images. For instance, take the digits of tau:</div>
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6.28318530717958647692528676655900598...</div>
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The first two digits are 62. What words can this form? You can say chain, gin, maybe you know someone named Jane or John. I ended up choosing the word ocean.</div>
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The next three digits are 831. This forms the word vomit. Yes, it is disgusting, but it is a word that will create a memorable image.</div>
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The next two digits are 85. From this, we can create the word waffle. So the first image will be "an <b>ocean</b> <b>vomit</b>ing a <b>waffle</b>." It sounds very silly, but it will be memorable. The smell of the saltwater, the taste of the waffles, the sound of the ocean waves crashing. This all will go into your first locus. My memory palace for tau was my middle school campus, so I remembered this image in the back parking lot of the school.</div>
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The next image is comprised of the digits 30717958. This can be turned into "a <b>mask</b> <b>tug</b>ging on a <b>bailiff</b>." This was put inside of a staff room that the back parking lot has a door to. It is a very weird image, but still memorable. Picture the bailiff really struggling to get away from this mask, while still fearfully reciting his lines: do you solemnly swear to tell the truth, the whole truth, and nothing but the truth. Make yourself feel scared of this moving mask, and sympathize with the bailiff. The more you relate to and embrace the image, the more memorable it will be. Especially when you are memorizing 2012 digits of tau (which took me 272 images), you need each image to be extremely vivid.</div>
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To retrieve the numbers from this memory palace, all you do is go back to the image, find the subject, root verb, and object of it, and translate the consonants back to numbers with the major system. With practice, this becomes easier and easier to do. I strongly recommend practicing at least memorizing grocery lists and to-do lists with the memory palace, and if you want to take it further, learn to convert numbers to words with the major system for more advanced lists and situations. Maybe even memorize your family and friends' phone numbers with the major system and memory palace. These are all great exercises for your mind, and will definitely give you a better memory.</div>
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Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com1tag:blogger.com,1999:blog-2207789741693789296.post-70950773592699612232013-11-23T12:00:00.000-05:002013-11-23T12:00:03.236-05:00The History of the Monty Hall Problem<div style="text-align: left;">
One of my very first blog posts was about the <a href="http://coolmathstuff123.blogspot.com/2011/07/monty-hall-paradox-what-are-odds.html" target="_blank">Monty Hall Problem</a>. This is an extremely classic example of a probability paradox. Let me quickly describe the problem:</div>
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Pretend you are on a game show, and the host gives you three doors to select from. One of these doors has a car behind it, while the other two have goats. Let's say you select door number one. Then, the host (who knows where the car is) opens another door to reveal a goat. Let's say he opens door number three. You are then given the option to either stick to door one or switch to door two. Does either strategy have an advantage?</div>
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The common answer would be that it is 50-50, and there is no advantage either way. However, the correct answer is that there is only a 1/3 chance of winning by staying put, and a 2/3 chance of winning by switching. Click <a href="http://coolmathstuff123.blogspot.com/2011/07/monty-hall-paradox-what-are-odds.html" target="_blank">here</a> to learn why.</div>
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This problem was first posed by Steve Selvin, but it was popularized by Marilyn vos Savant in 1990. Vos Savant is famous for once having the highest IQ in the world, as well as her "Ask Marilyn" column in <i>Parade</i> Magazine.</div>
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One week, her column was about the Monty Hall Problem. She posed the question, and then explained her reasoning as to why there is a 2:1 advantage for switching. This created a pandemonium of angry readers who insisted that she was incorrect, and furthermore, accused her of adding to the problem of innumeracy and lack of mathematical intuition in America. Some of these complaints came from a statistician at the National Institutes of Health, the deputy director of the Center for Defense Information, and professors at George Mason University, University of Florida, University of Michigan, Millikin University, Georgetown University, Dickinson State University, Western State College, and more. Even the legendary Paul Erd<span style="background-color: white; line-height: 19px;"><span style="font-family: inherit;">ős couldn't wrap his brain around the paradox.</span></span></div>
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<span style="line-height: 19px;">This problem has continued to baffle everyone it encounters, from average people to accomplished mathematicians. In 2010, Walter Herbranson and Julia Schroeder of Whitman College performed an experiment to see if playing the game multiple times could end up refining the player's strategy. The human test subjects failed to revert to the optimal strategy and switch doors in the experiment. However, when the test was performed on pigeons, with mixed grain as the prize, they were able to pick up on the fact that switching doors gave them the best chance of success. The fact that a pigeon can do better than a human in this situation is fascinating to me.</span></div>
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<span style="line-height: 19px;">The Monty Hall Paradox is something that reminds us of how humans are not wired to understand probability and statistics. This is why people can be fooled by mathematical scams and why casinos are packed full of gamblers. If our math curricula put an equal focus on probability and statistics as it did on algebra and calculus, then our world would have much better math minds and critical thinkers in general.</span></div>
Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-75016765373497333292013-11-16T12:00:00.000-05:002013-11-16T12:00:00.352-05:00The Mathematics of GhostA couple months ago, we were talking about various mathematical games where player two can very easily force a win, such as <a href="http://coolmathstuff123.blogspot.com/2013/10/chomp-proof-in-game-theory.html" target="_blank">Chomp</a> and <a href="http://coolmathstuff123.blogspot.com/2013/09/anti-tic-tac-toe.html" target="_blank">Anti Tic-Tac-Toe</a>. Those games were pretty easy to spot the unfairness, but this is one that is actually commonly played by everyday people. You might even play it yourself. The game is called Ghost.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhn9Ahx9UV3lgzuqsw0cpoEUx_J3AAvJ8KDNrtRji5MO9ZCJnIYZBo4qvGfEi2X5SIn4FwoAu9wRsDiSsMVq9cxYVJ4IUqI2KYFyk2_Xsyge9npiDuRepcCtHLKE5DxqY9kVtR4wzLPcLaK/s1600/tumblr_ml4kd4pha81so3r4ko1_500.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhn9Ahx9UV3lgzuqsw0cpoEUx_J3AAvJ8KDNrtRji5MO9ZCJnIYZBo4qvGfEi2X5SIn4FwoAu9wRsDiSsMVq9cxYVJ4IUqI2KYFyk2_Xsyge9npiDuRepcCtHLKE5DxqY9kVtR4wzLPcLaK/s400/tumblr_ml4kd4pha81so3r4ko1_500.jpg" width="400" /></a></div>
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Ghost is definitely not a scary game, despite the name. In this game, player one starts by saying a letter. Player two must respond with another letter, and they keep taking turns saying letters and forming a word. The goal is to not create a real word. However, you can't call out letters that don't allow a word to be made either. For instance:</div>
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Player 1: T</div>
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Player 2: O</div>
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Player 1: U</div>
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Player 2: C</div>
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Player 1: A</div>
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Player 2: Umm... L</div>
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Player 1: I challenge.</div>
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Player 2: I didn't have a word in mind.</div>
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Player 1 = Winner</div>
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If your opponent challenges you, you need to say a word that can be made out of the letters to win. If you can't think of a word, then you lose the round. A better strategy for player 2 in this game might be:</div>
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Player 1: T</div>
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Player 2: W</div>
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Player 1: E</div>
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Player 2: A</div>
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Player 1: K</div>
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Player 2 = Winner</div>
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This game seems pretty fair, right? If you play this game with a friend, you will find that the wins are pretty even (unless one of you has a much better vocabulary). However, it is possible for player 2 to force a win.</div>
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When all of the words in Scrabble Player's Dictionary are in play, every letter of the alphabet can be followed with another letter that forces an odd lettered word (assuming that player 2 continues perfect play). However, this dictionary has a lot of obscure words in it that might get challenged, and the average player will not know or use.</div>
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By keeping it to words that are known to the average player, player 1 does have a chance. If they play H, J, M, or Z, they can force a win if player 2 does not have the dictionary memorized. Here are the words to remember for player 1:</div>
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<b>H</b>: Hazard, Haze, Hazily, Hazy, Heterosexual, Hiatus, Hock, Huckster, Hybrid</div>
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<b>J</b>: Jazz, Jest, Jilt, Jowl, Just</div>
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<b>M</b>: Maverick, Meow, Mizzen, Mnemonic, Mozzarella, Muzzle, Muzzling, Myth</div>
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<b>Z</b>: Zaniness, Zany, Zenith, Zigzag, Zombie, Zucchini, Zwieback, Zygote</div>
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For any other letter, player 2 can still force a win. Once you know the correct followup letter to use, it is a pretty simple game. Here is the strategy:</div>
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<table border="1">
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<th>Letter</th>
<th>Followup</th>
<th>Possible Words</th>
</tr>
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<td>A</td>
<td>O</td>
<td>Aorta</td>
</tr>
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<td>B</td>
<td>L</td>
<td>Black, Blemish, Blimp, Bloat, Blubber</td>
</tr>
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<td>C</td>
<td>R</td>
<td>Craft, Crepe, Crept, Crick, Crozier, Crucial, Cry</td>
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<td>D</td>
<td>W</td>
<td>Dwarf, Dwarves, Dweeb, Dwindle, Dwindling</td>
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<td>E</td>
<td>W</td>
<td>Ewe</td>
</tr>
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<td>F</td>
<td>J</td>
<td>Fjord</td>
</tr>
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<td>G</td>
<td>H</td>
<td>Ghastliness, Ghastly, Gherkin, <b>Ghost</b></td>
</tr>
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<td>I</td>
<td>L</td>
<td>Ilk, Ill</td>
</tr>
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<td>K</td>
<td>H</td>
<td>Khaki</td>
</tr>
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<td>L</td>
<td>L</td>
<td>Llama</td>
</tr>
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<td>N</td>
<td>Y</td>
<td>Nylon, Nymph</td>
</tr>
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<td>O</td>
<td>Z</td>
<td>Ozone</td>
</tr>
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<td>P</td>
<td>N</td>
<td>Pneumonia</td>
</tr>
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<td>Q</td>
<td>U</td>
<td>Quaff, Quest, Quibble, Quibbling, Quondam</td>
</tr>
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<td>R</td>
<td>Y</td>
<td>Rye</td>
</tr>
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<td>S</td>
<td>Q</td>
<td>Squeamish, Squeeze, Squeezing, Squelch</td>
</tr>
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<td>T</td>
<td>W</td>
<td>Twang, Tweak, Twice, Two</td>
</tr>
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<td>U</td>
<td>V</td>
<td>Uvula</td>
</tr>
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<td>V</td>
<td>U</td>
<td>Vulva</td>
</tr>
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<td>W</td>
<td>H</td>
<td>Whack, Where, Whiff, Who, Why</td>
</tr>
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<td>X</td>
<td>Y</td>
<td>Xylem</td>
</tr>
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<td>Y</td>
<td>I</td>
<td>Yield, Yip</td>
</tr>
</tbody>
</table>
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By memorizing this list, you can become a perfect Ghost player, and win every time, despite the seemingly fair nature of this game. The other two games were both fun, but with this one, you can really start to impress people with your ability to win.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-59548160647505508642013-11-09T12:00:00.000-05:002013-11-09T12:00:01.308-05:00Math in the News: Teacher SalariesOne of the biggest issues in math education is teacher quality. I have discussed this in both of my TEDx talks about this topic.<br />
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We explained it the best we could in our Capstone Research Paper (link is at the top of the page), but I think this New York Times article that just came out describes it fantastically. So, I couldn't resist posting it here.<br />
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http://www.nytimes.com/2011/05/01/opinion/01eggers.html?_r=0<br />
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Enjoy!Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-87681127104821212732013-11-02T12:00:00.000-04:002013-11-02T12:00:03.743-04:00Logarithmic Proofs and IdentitiesI have talked about logarithms quite a bit on this blog, but they were always being applied to something else, whether it be <a href="http://coolmathstuff123.blogspot.com/2013/03/benfords-law_16.html" target="_blank">Benford's Law</a>, the <a href="http://coolmathstuff123.blogspot.com/2013/08/time-to-double-your-money.html" target="_blank">Law of 72</a>, or some other practical use. This week, I would like to show that logarithms are very much a part of pure mathematics as well. Of course they are in mathematical equations just as much as exponents and radicals, but they have some pretty cool features of their own.<br />
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First off, let me review what a logarithm is. I have explained it before, but once you understand the notation, you shouldn't need to have done Precalculus to understand this post. They are very easy to understand.<br />
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For instance, we know that 10<sup>2</sup> is 100.<br />
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10<sup>2</sup> = 100<br />
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If we take the logarithm of both sides, we are essentially bringing the two out of the exponent. Rather than doing an operation on the 10, we do an operation on the 100 to determine what the exponent is.<br />
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log(10<sup>2</sup>) = log(100)<br />
2 = log(100)<br />
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In most situations, it is clear what type of logarithm you are using, especially in this one because ten is a common logarithm to use. However, many people will write a subscript to clarify. For instance:<br />
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log<sub>10</sub>(100) = 2<br />
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Logarithms become extremely useful when you need to solve an algebraic equation where the variable is in the exponent. For example:<br />
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2<sup><i>x</i></sup> = 64<br />
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As you know, algebra is about doing the inverse operation. If there is addition going on, you subtract. If there is multiplication going on, you divide. Similarly, if there is exponentiation going on, you use a logarithm. In this instance, it would be taking the log<sub>2</sub> of both sides.<br />
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c2<sup><i>x</i></sup>) = log<sub>2</sub>(64)<br />
<i>x</i> = log<sub>2</sub>(64)<br />
<i>x</i> = 6<br />
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In practice, there are three bases that are extremely popular to use in a logarithm. We just used two of them: log<sub>10 </sub>and log<sub>2</sub>, which are also known as the common logarithm and the binary logarithm. The third one is log<sub><i>e</i></sub> (using the number <i>e</i> that is described <a href="http://coolmathstuff123.blogspot.com/2012/03/very-interesting-post-literally.html" target="_blank">here</a>), which is called the natural logarithm, or the natural log. This one is found on most calculators, usually next to the common logarithm.<br />
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Though logarithms are not a part of most people's day-to-day life, they do have lots of practical applications. The common logarithm is the basis of the pH system which describes the acidity of water. The natural log is a huge aspect of finance and compound interest (as we saw with the Law of 72). And of course, they are all over nature.<br />
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Let's look at an identity of logarithms. Take the following problem:<br />
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log<sub>6</sub>(24) + log<sub>6</sub>(9) =<br />
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If you just used a calculator to do this, you would get:<br />
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log<sub>6</sub>(24) + log<sub>6</sub>(9)<br />
1.773705614 + 1.226294386<br />
3<br />
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That's odd. Two random numbers happened to have logarithms that summed to three. Let's look closer at this and see if we can figure out why. What number could you find the log<sub>6</sub> of and get 3?<br />
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log<sub>6</sub>(<i>n</i>) = 3<br />
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First of all, let me point out that we just asked an algebraic question. We suddenly got curious about why something happened, so we asked a "what" question, which calls for an unknown quantity, which later becomes an algebraic variable. So when algebra seems like a drag, remember that it is all techniques for answering that "what" question. And "what" is a question asked in all branches of mathematics, science, and engineering.<br />
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Anyways, for this equation, we would want to do the inverse operation. We turn both sides into the exponent, and create a base of 6. This gives:<br />
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6<sup>log<sub>6</sub>(<i>n</i>)</sup> = 6<sup>3</sup><br />
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The left hand side cancels, leaving just <i>n</i>. The right hand side is six cubed, which is 216. So, we end up with:<br />
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<i>n</i> = 216<br />
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So, this means that the log<sub>6</sub> of 216 makes you end up with 3, or the sum of log<sub>6</sub>(24) and log<sub>6</sub>(9).<br />
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log<sub>6</sub>(24) + log<sub>6</sub>(9) = log<sub>6</sub>(216)<br />
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What is the relationship between these three numbers? Well, it shouldn't take to long to determine that 24 x 9 = 216, or:<br />
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log<sub>6</sub>(24) + log<sub>6</sub>(9) = log<sub>6</sub>(24 • 9)<br />
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In other words, the sum of the logarithms is the logarithm of the product. Wow! That's pretty cool! Is that always the case? Well, let's try to prove that it is for all logarithms.<br />
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Let's set a few terms equal to each other and see what happens. Since there are logarithms, we will need a lot of variables.<br />
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<i>x </i>= log<sub><i>a</i></sub>(<i>p</i>)<br />
<i>y</i> = log<sub><i>a</i></sub>(<i>q</i>)<br />
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In other words:<br />
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<i>p</i> = <i>a</i><sup><i>x</i></sup><br />
<i>q</i> = <i>a</i><sup><i>y</i></sup><br />
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Let's multiply those two equations together. Since they are both equal, multiplying the terms on each side by each other won't make a difference.<br />
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<i>p</i> • <i>q</i> = <i>a</i><sup><i>x </i></sup>• <i>a</i><sup><i>y</i></sup><br />
<i>pq</i> = <i>a</i><sup><i>x</i>+<i>y</i></sup><br />
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<span style="font-size: x-small;">The right hand side was simplified using the Law of Exponents, which is explained very well <a href="http://www.mathsisfun.com/algebra/exponent-laws.html" target="_blank">here</a>.</span><br />
<span style="font-size: x-small;"><br /></span>
Now, we must take the logarithm of both sides, or specifically, the log<sub><i>a </i></sub>of both sides.<br />
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log<sub><i>a</i></sub>(<i>pq</i>) = log<sub><i>a</i></sub>(<i>a</i><sup><i>x</i>+<i>y</i></sup>)<br />
log<sub><i>a</i></sub>(<i>pq</i>) = <i>x </i>+<i> y</i><br />
<i><br /></i>
But what were <i>x</i> and <i>y</i>? We defined them in terms of <i>a</i>, <i>p</i>, and <i>q </i>earlier. So, let's substitute those values in and see what we get.<br />
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log<sub><i>a</i></sub>(<i>pq</i>) = log<sub><i>a</i></sub>(<i>p</i>)<i> </i>+<i> </i>log<sub><i>a</i></sub>(<i>q</i>)<br />
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And this creates the identity that we were trying to prove: the sum of the logarithms is the logarithm of the product, and thus, completes our proof. There are other logarithmic identities like this one, but I will save that for another post.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0tag:blogger.com,1999:blog-2207789741693789296.post-50258136254321076662013-10-26T12:00:00.000-04:002013-10-26T12:00:08.286-04:00History of Math: Isaac Newton and the Schoolyard BullyWhen talking about great mathematicians of the past, many will rank the top three as <a href="http://coolmathstuff123.blogspot.com/2013/02/the-origins-of-eureka.html" target="_blank">Archimedes</a>, <a href="http://coolmathstuff123.blogspot.com/2013/09/carl-friedrich-gauss-child-prodigy.html" target="_blank">Gauss</a>, and Newton. I have posted stories about the others, but none about Isaac Newton. So, I think now is a good time.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6dYJv6lRTULVKCXQ44_lMJSlFB4iJgdEM1jNx9y4DV4GEGXO5EbYhAT3sY1_qnpxryfne-JK8LAiQ9DwBg5YQJD9eK55e4Dw06cWmrHvdGf4x2GrWFCpa_l2BUmCSIAvgpCtHCn5qIvfb/s1600/isaac-newton.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6dYJv6lRTULVKCXQ44_lMJSlFB4iJgdEM1jNx9y4DV4GEGXO5EbYhAT3sY1_qnpxryfne-JK8LAiQ9DwBg5YQJD9eK55e4Dw06cWmrHvdGf4x2GrWFCpa_l2BUmCSIAvgpCtHCn5qIvfb/s320/isaac-newton.jpg" width="301" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Isaac Newton</td></tr>
</tbody></table>
Newton is best known for his work in physics, but he also made huge contributions to calculus, algebra, geometry, and infinite series. Many mathematicians expand their expertise to different diverse branches of math, but Newton stuck to the things that applied the most to his physics and are currently ruling the American school system.<br />
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Let me tell you an interesting story about Newton. When he was a young student, he was very shy and not at all the genius that he is known as today. One day at recess, a bully came up to him and punched him in the stomach. Newton chose to fight back, and proceeded to shove his face in the mud. All of his classmates, who did not like this kid, cheered him on as he proved his superiority to the bully.<br />
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After this incident, he decided that physical prestige wasn't enough for him, and he wanted mental prestige as well. So, he started working much harder at his schoolwork, and soon after became top of the class, proving to everyone that he was smarter than the bully as well. This motivation could have been what turned him into one of the best scientists and mathematicians of all time.<br />
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I think this story shows that anyone who has drive and dedication can become a genius, and it also is a story themed around the negativity of bullying. I also like it because it is an interesting aspect about a mathematician's childhood, which help people get to know who is behind what they are learning and practicing.Ethan Brownhttp://www.blogger.com/profile/09611695185154134251noreply@blogger.com0