Check out my 2011 Greatest Hits Video and learn a pattern as well.

A couple of posts ago, I presented my greatest hits video for my Mathemagics performance. If you haven’t seen it before, it is what it says, a combination of math and magic. Now that 2011 has just twelve hours left, it might be a good time to present my 2011 greatest hits video.

For some upcoming shows, I will be at Gathering for Gardner early spring in Atlanta, Curious1729 Con in St. Louis, and many other exciting performances, listed on www.EthanMath.com.

As you probably noticed in the Mathemagics performance, squaring numbers is a big part of my show. I’ve already shown how to square 2-digit numbers that end in five, as well as the close-together method (which works for squaring as well), but there is a certain type of 2-digit number, or any digit number for that matter, that follows a specific pattern. Let me show you:

3^2 = 09 (or just 9)

33^2 = 1089

333^2 = 110889

3333^2 = 11108889

33333^2 = 1111088889

6^2 = 36

66^2 = 4356

666^2 = 443556

6666^2 = 44435556

66666^2 = 4444355556

9^2 = 81

99^2 = 9801

999^2 = 998001

9999^2 = 99980001

99999^2 = 9999800001

See the pattern?  This only works for numbers that are all 3’s, 6’s, or 9’s, but it is pretty cool! It’s not a very impressive effect to do for someone, but a really cool pattern, and trick.

Algebra + Geometry + Trigonometry + Arithmetic = A Perfect Cool Math Stuff Post...

A few weeks ago, we saw the second and third cube roots of one, which used those complex numbers, with i in it. While proving that their cubes were one, we learned how to do some operations with them using algebra. However, there is a really cool geometric way to do the math as well.

First off, take the Cartesian Plane. We'll label the x-axis with the integers and the y-axis with the imaginary numbers.

To plot a point, just move the constant to the left/right and the coefficient up/down. So, for 2 + 3i, you would move two to the right and three up. For 4 - 8i, you would go four right and eight down.

Let's say you had to add these two numbers together. First off, you could do it algebraically, which isn't that cool, and you get 6 - 5i. However, you could also add them on the imaginary plane.

Let's plot the points:

How about we just see what happens when we plot 6 - 5i. We'll connect all of the points to make a little quadrilateral.

What do you notice? The points have formed a parallelogram, with the opposite sides equal in length and parallel. At this point, we should go over how to figure out the lengths.

To figure out the lengths, we use the Pythagorean Theorem. Remember back to the problems of the week? We saw that a^2 + b^2 = c^2? We will be proving that in a later post, believe it or not. Anyways, we will be doing exactly that, literally! The form for a complex number is a + bi, so we can just plug those in and solve for the length c.

In the case of 2 + 3i, we will do 2^2 + 3^2 = c^2.

2^2 + 3^2 = c^2

4 + 9 = c^2

13 = c^2

3.606 ≈ c

To keep things more simple, we will use two different points; 8 + 6i and -3 + 4i. In this case, the lengths are 10 and 5.

Of course, we could add them together pretty easily with algebra or geometry to get 5 + 10i. However, let's figure out the angles of the lines. To do it, we use trigonometry, which we also used for the problem of the weeks.

To briefly review/explain, we will take the b term (which is the opposite side) and divide it by the c term, or hypotenuse, to get the sine of the angle. To retrieve the angle, press the sin^-1 button on your calculator.

In this case, here are our angles:

8 + 6i: 37° (approx.)

-3 + 4i: 127° (approx.)

Now, we will learn how to multiply them together. All you have to do is two easy steps: multiply the lengths and add the angles. For this one, we multiply 5 and 10 and add 37 and 127.

5 x 10 = 50

37 + 127 = 164

If you do some trigonometry, you will get an approximate answer of 48 + 14i, which happens to be the correct answer.

You can also easily multiply complex numbers by real numbers; just multiply the length by that number. If you think about it, this corresponds to the original method. These are just a handful of the many cool things you can do with the imaginary plane. To be honest, I really love algebra and arithmetic, but I'm not one of those people who is all over geometry (I'm not Archimedes, the famous Greek mathematician who died doing geometry in the sand), but this little method is just so cool.

To infinity and beyond!! With numbers I mean...

We have done a lot with infinite series already, we used it to supposedly prove the Communitive Property wrong, to practice systems of equations in the problem of the week, and even prove that the prime numbers fall into this category. However, we haven't really looked at infinity. Is it a number? Can we define it? What is greater than infinity?

Let's look at an infinite series. Take the natural numbers; 1, 2, 3, 4, 5, 6, 7, ... This sequence takes us out to infinity. How about 10, 20, 30, 40, 50, 60, 70, ... This takes us to infinity also. However, which sequence has more terms in it?

One side of you is saying, for every term in the multiple of ten series, you need ten terms from the natural number series to get that high. Therefore, the first series must be ten times bigger. The other side of you is saying, for both series, we are going out forever. This means they must be equal. When I first saw this, I was leaning greatly towards the first side. However, this isn't quite right.

For each number in the first series, we can pair it with a number in the second. For example, we can pair 1 with 10. Then 2 with 20. Then 3 with 30, 4 with 40, 5 with 50, and so on. If the first side is greater, then you must run out of terms on the second side. However, the second side is infinite also. If the second side were greater (which I cannot make an argument for that), then we would run out of terms on the first side. This means that the sides must be equal.

In fact, any series is the same size as the set of natural numbers if you can write it out with no infinite gaps in between. For instance, the integers:

... -3, -2, -1, 0, 1, 2, 3 ...

You can rewrite this to get:

0, 1, -1, 2, -2, 3, -3, 4, -4...

Since there are no infinite gaps here, that must mean that it is equal to the set of natural numbers.

What about the fractions (you could say the rational numbers, but rational numbers are simplified and fractions can or cannot be)? It's a big statement, but we can try it.

1/1	1/2	1/3	1/4	1/5
2/1	2/2	2/3	2/4	2/5
3/1	3/2	3/3	3/4	3/5
4/1	4/2	4/3	4/4	4/5
5/1	5/2	5/3	5/4	5/5

This is a table of the fractions. Is there a way to write this table without any infinite gaps? Turns out there is. If we draw lines diagonally, and put it all together, it will have no infinite gaps. It would look like this:

1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3, 1/4, ...

This means the amount of fractions is the same! This quantity is denoted with the hebrew letter alef, also called alef zero, or alef naught. However, do you think there is another type of infinity, or is every set equal? I will come back to this in a couple months, and we will find the answer and why.

A Few Awesome Properties of Quadratics

In the problem of the week, I would occasionally throw in some quadratics, or problems with squaring involved. If you’ve seen my performances, you might have noticed I like things that have to do with squaring, and quadratics is definitely one of them. Quadratics is normally done in the form f(x) = ax^2 + bx + c, which looks nice, but isn’t so useful. However, there is a form that is pretty commonly used that is almost magical in a way.

This form is the form f(x) = a(x - h)^2 + k. First, I will show you how to get into this form, then we will look at its properties.

The technique we will use is called “completing the square.” The first thing you do is factor the a out of the equation. Let’s use 1/2x^2 + 3x + 5 as an example.

1/2x^2 + 3x + 5

1/2(x^2 + 6x + 10)

Next, plug that new b term into b^2/4. What I find easier is to divide the b term by two and then square it, so the division doesn’t get too messy.

6/2 = 3

3^2 = 9

What this means is that the equation (x^2 + 6x + 9) is square, or a “perfect square trinomial.” It is in fact (x + 3)(x + 3). To figure out that three, all you have to do is divide that b term by two.

There is only one problem though. We have x^2 + 6x + 10, not x^2 + 6x + 9. However, we can put the (x + 3)^2 there, but we must add one to the (x + 3)^2, which ends up getting multiplied by the 1/2 to get a constant of one-half at the end. So, we have:

1/2(x + 3)^2 + 1/2

And that is vertex form. Let’s look at what is cool about it. 

First off, both a’s happen to be equal. This is not coincidence, as we factored out the a in order to switch forms.

What I find really cool is those two terms we couldn’t control, h and k. In this case, they are -3 and 1/2 (the equation is a(x - h)^2 + k, not a(x + h)^2 + k, so h is -3). What does that have to do with anything? Take a look at the graph of this equation.

If you’ll notice, the vertex of this parabola (graph of a quadratic function) is in fact (-3, 1/2). In fact, you can actually graph a quadratic equation using only vertex form, just with this fact and a. I think that is cool that there is a format out there that can do that.

Bonus: Say that instead of doing a(x - h)^2 + k, you did (bx - h)^2 + k. b is unfortunately not the same as the other b, but much more useful. This form is messier, but there is a cool thing about it. In this case, the vertex is actually (bh, k), which is interesting.

What the a factor did was it told us the factor for the “vertical stretch/compression” of the  parabola, or the number you multiply every single y value by. In the case before, every y value of plain (x + 3)^2 + 1 was multiplied by 1/2 to create the graph.

For this form, there is actually a “horizontal stretch/compression,” which isn’t common in quadratics. This form gives you a stretch/compression of 1/b. You might see it as just b, but it is usually written like this:

((1/b)x - h)^2 + k)

Either way, it is pretty cool.

Fibonacci Day: Fibonacci Magic Trick

I don’t know if you noticed, but today is a Fibonacci Day! It is December third, and three is a Fibonacci number. We’ve looked at some cool patterns in Fibonacci numbers, but it’s time we learn how to do some magic with them! Let’s look at the Fibonacci numbers, but this time in lines.

Line 1: 1

Line 2: 1

Line 3: 2

Line 4: 3

Line 5: 5

Line 6: 8

Line 7: 13

Line 8: 21

Line 9: 34

Line 10: 55

Line 11: 89

Line 12: 144

Line 13: 233

Line 14: 377

Line 15: 610

Line 16: 987

Line 17: 1597

Line 18: 2584

Line 19: 4181

Line 20: 6765

What do you think the sum is of all of the numbers up until line, say thirteen? I can tell you immediately that it is 609. How?

Wait, we know this! Remember when we were adding Fibonacci numbers? The answer was always two ahead minus one! But we can take it one step further.

Let’s make our own Fibonacci sequence this time, starting with any two numbers we want. If you are doing it on pencil and paper, you might want to stick 1 – 10, or you can make an excel or numbers spreadsheet and do it as high or little as you want. In this case, we’ll start with 4 and 7.

Line 1: 4

Line 2: 7

Line 3: 11

Line 4: 18

Line 5: 29

Line 6: 47

Line 7: 76

Line 8: 123

Line 9: 199

Line 10: 322

Line 11: 521

Line 12: 843

Line 13: 1364

Line 14: 2207

Line 15: 3571

Line 16: 5778

Line 17: 9349

Line 18: 15127

Line 19: 24476

Line 20: 39603

What is the grand total up to line eight, you can do any line. The answer is 315. It’s still the same exact principle, except for one little thing.

We move up two lines, and then subtract line two. It is the easiest of all things to do! No matter how gigantic the numbers are, you can still pull it off. What’s even cooler is that there is no specific line you are adding up to, unlike other methods that only go up to line ten.

Bonus Trick: Make one of these sequences yourself, and make sure you have at least ten lines. Now, divide the last line by the one before it. In this case, we would be doing 39603 ÷ 24476. You should have 1.61, right?

This is the same thing as the golden ratio appearing in the Fibonacci sequence. To prove it, we will actually do something a little different than usual. We will add fractions “badly.” If you were a young kid, how would you guess adding fractions works?

I’d say add the numerators, then add the denominators. Like ½ + ¼ should be ²/₆, or ¹/₃. This doesn’t give you the right answer, but it does assure that the answer is in between the two fractions. In this case, we are using line ten and line nine.

Line nine has its own formula: 13x + 21y, assuming that line 1 is x and line 2 is y. Line ten has formula 21x + 34y. So, we have:

^{21x + 34y}/_{13x + 21y}

This is the same as adding fractions badly. This says that this ratio is between ^21x/_13x and ^34y/_21y.

^21x/_13x = ²¹/₁₃ = 1.61538…

^34y/_21y = ³⁴/₂₁ = 1.61904…

Both of these numbers begin with 1.61, meaning any number in between them will begin with 1.61. This proves that line ten over line nine is always 1.61, a great bonus prediction effect to the trick.