In the next few days, we will be looking at a topic called "Sequences and Pattern Recognition," which is basically finding the next number in a pattern. In first grade, you make the little bumps with a plus two on top or something to get a feel for simple patterns. We are going to take it to the next level, and look at a quadratic pattern which will use systems of linear equations to find. These are one of my favorite parts of Algebra.
Easy Problem: Since Tuesday, we have been solving equations, we will keep that going. Tomorrow, we will look at a sequence, but today, we'll just keep it simple. In our equation, we will have two steps after you plug in b and simplify. First, you will have to get rid of some addition at the end by subtracting from both sides. Then, you will divide on both sides to reach your answer. Good luck!
Plug in b and solve for z: 100 = 6bz + 4
z = ___
Hard Problem: When given a sequence, it is essential that you create an equation that has you put in the spot that you are looking for and have it equal to the number in that spot. For instance, the sequence 3, 12, 21, 30, ... will have the equation n = 9x - 6 with x being the spot you put it in and n being the value in that spot. In order to figure out the equation, you need to look for common differences (for arithmetic sequences). In that one, you would have:
3 12 21 30
\/ \/ \/
9 9 9
In that case, we had the same differences at our first step. That means our equation is in the form y = mx+b. Hence, we create a system where we are solving for m and b by plugging in 1 for x and 3 for y in the first one. 2 for x and 12 for y is the second, and so on. You'll notice these equations are very easy to do elimination in because terms are already isolated.
If you don't get common differences, try finding the common differences of the common differences and see if those are the same. If they are, plug values into y = ax^2 + bx + c and solve for a, b, and c. This will require three equations.
1) Find the value for p and q.
s + 0.9/6 = p
t - 3.9 = q
2) Find the common differences in this sequence: p, 20, q, 48, ...
3) Create a system that could be used to find the equation for this sequence.
If you want to put yourself one step ahead, try to solve the system and determine what the equation is. Since we went over systems last month, you should be able to figure it out. Just remember to eliminate variables with no coefficient. Tomorrow, everybody is doing sequences! It's going to be fun!!
Tuesday, July 19, 2011
Monday, July 18, 2011
The Problem of the Week Day 1: Week of 7/17 - 7/23
The problem of the week is back for its second time! Make sure you did last month's problem because the answers are going up this Saturday. Don't forget to round to the nearest tenth unless told otherwise.
Easy Problem: Last month, we learned about the Pythagorean Theorem and how to figure out the missing side of a right triangle. Just to refresh your memory, the square of the longest side: c, equals the sum of the squares of the two shorter sides, a and b. So, a^2 + b^2 = c^2.
If you were missing a and b, it is just another algebra equation, but the inverse operation to squaring is square rooting.
If you have a right triangle with a =12 and c = 20, what does b equal?
b = ___
Hard Problem: Last month, we learned about the sine function. Now, we will look at another function that you might not have on your average calculator, but if you hit the 2nd button on a scientific calculator or iPhone calculator, you will get a button where the sine function has a little -1 above it, in the place of an exponent. That button takes the sine of an angle, and turns it into the angle. So, you could divide the side opposite to an angle by c and get the sine, and then hit that button to retrieve your angle.
If a right triangle had a = 6 inches, b = 8 inches and c = 10 inches, what would the two missing angles be? The angle opposite of a will be called t and the angle opposite of b will be called s.
s = ____
t = ____
Tip: Every triangle, right or not, is guaranteed to have its angles sum up to 180° if on a flat surface. Therefore, you only need to use trigonometry for one angle, and use arithmetic for the other.
Easy Problem: Last month, we learned about the Pythagorean Theorem and how to figure out the missing side of a right triangle. Just to refresh your memory, the square of the longest side: c, equals the sum of the squares of the two shorter sides, a and b. So, a^2 + b^2 = c^2.
If you were missing a and b, it is just another algebra equation, but the inverse operation to squaring is square rooting.
If you have a right triangle with a =12 and c = 20, what does b equal?
b = ___
Hard Problem: Last month, we learned about the sine function. Now, we will look at another function that you might not have on your average calculator, but if you hit the 2nd button on a scientific calculator or iPhone calculator, you will get a button where the sine function has a little -1 above it, in the place of an exponent. That button takes the sine of an angle, and turns it into the angle. So, you could divide the side opposite to an angle by c and get the sine, and then hit that button to retrieve your angle.
If a right triangle had a = 6 inches, b = 8 inches and c = 10 inches, what would the two missing angles be? The angle opposite of a will be called t and the angle opposite of b will be called s.
s = ____
t = ____
Tip: Every triangle, right or not, is guaranteed to have its angles sum up to 180° if on a flat surface. Therefore, you only need to use trigonometry for one angle, and use arithmetic for the other.
Saturday, July 16, 2011
The Monty Hall Paradox: What are the Odds?
One subject in mathematics always confuses people. No, not Calculus. This is the one some kids call "easy." Probability and Statistics is so difficult for us to understand. Let's look at one of my favorite problems that is famous in the world of game theory: The Monty Hall Paradox.
Suppose you are a contestant on the old TV show Let’s Make a Deal®, hosted by Monty Hall. Monty shows you three doors. Behind two of the doors is trash; behind one of them is a new car. You choose a door, and Monty then opens one of the other doors, revealing trash (he can always do this). You are then given a chance to switch your choice to the other door. What do you do?
The average person will stay because they don't like to be wrong. People also have a strong tendency to go with their gut instinct. Mathematically, the average person also stays because well, it's 50-50 right?
Turns out this is not the case. Switching doors doubles your chance of getting the car, bringing the odds from 1/3 to 2/3. This fact was impossible for me to understand, but I eventually figured out a reason, and understood why humans don't understand statistics.
Let's bring it to the basics, experimental probability. You used these principles to solve June's problem of the week (I hope you did it!!). We will look at all the possible scenarios.
1)You pick the first door and the second door has the car (the second door will always have the car for us). Monty reveals that the third door has trash. Since we said it's better to switch, we switch to door two and get the car.
2) You choose the second door, and Monty reveals that the third one has trash. As a math enthusiast, you know it is best to switch, so you move to door one and see a big pile of junk. Plain, dirty junk.
3) You go to the third door and Monty reveals the first. You then switch to the second door and drive home in a brand new car.
There were three different possibilities, and two got you the car. This got me satisfied. If you have any other reasonings for the people who aren't convinced, please post them! Isn't that cool?!
Suppose you are a contestant on the old TV show Let’s Make a Deal®, hosted by Monty Hall. Monty shows you three doors. Behind two of the doors is trash; behind one of them is a new car. You choose a door, and Monty then opens one of the other doors, revealing trash (he can always do this). You are then given a chance to switch your choice to the other door. What do you do?
The average person will stay because they don't like to be wrong. People also have a strong tendency to go with their gut instinct. Mathematically, the average person also stays because well, it's 50-50 right?
Turns out this is not the case. Switching doors doubles your chance of getting the car, bringing the odds from 1/3 to 2/3. This fact was impossible for me to understand, but I eventually figured out a reason, and understood why humans don't understand statistics.
Let's bring it to the basics, experimental probability. You used these principles to solve June's problem of the week (I hope you did it!!). We will look at all the possible scenarios.
1)You pick the first door and the second door has the car (the second door will always have the car for us). Monty reveals that the third door has trash. Since we said it's better to switch, we switch to door two and get the car.
2) You choose the second door, and Monty reveals that the third one has trash. As a math enthusiast, you know it is best to switch, so you move to door one and see a big pile of junk. Plain, dirty junk.
3) You go to the third door and Monty reveals the first. You then switch to the second door and drive home in a brand new car.
There were three different possibilities, and two got you the car. This got me satisfied. If you have any other reasonings for the people who aren't convinced, please post them! Isn't that cool?!
Saturday, July 9, 2011
Divide Almost Any Odd Number into a Number Consisting of all Nines
Using Discrete Mathematics, we can do so many things. My all-time favorite of them is definitely the amazing proof that states that any odd number that isn't a multiple of five can divide a number consisting of all nines. Sounds hard to believe, but 713 does go into some number consisting of all nines. I'll prove it.
If you think about it, if you divide a number by 713, there are only 713 possible remainders because 713 isn't a remainder. With that in mind, since we are dealing with the numbers consisting of all nines, let's list some out:
9, 99, 999, 9999, ... 9999999...999 (with 714 nines)
Since we have 714 numbers here, two of them must have the same remainder when divided by 713 because there are only 713 possible remainders. Let's take those two numbers, the bigger one is x and the other one is y. Each one of those is equal to 713 times a quotient, two different quotients (say q and p), plus the same remainder (r).
999...999 (with x nines) = 713q + r
999...999 (with y nines) = 713p + r
Technically, we can subtract these two equations from each other. The x nines - the y nines would end up with x - y nines in the front and y zeros at the end because the nines would cancel. According to the Distributive Law, 713q - 713p = 713(q - p) because you can factor a 713 out of both terms. However, the r's at the end cancel each other out, leaving us with no remainder!!! Since q and p are both integers, subtracting them will also be an integer.
999...999 (with x nines) = 713q + r
If you think about it, if you divide a number by 713, there are only 713 possible remainders because 713 isn't a remainder. With that in mind, since we are dealing with the numbers consisting of all nines, let's list some out:
9, 99, 999, 9999, ... 9999999...999 (with 714 nines)
Since we have 714 numbers here, two of them must have the same remainder when divided by 713 because there are only 713 possible remainders. Let's take those two numbers, the bigger one is x and the other one is y. Each one of those is equal to 713 times a quotient, two different quotients (say q and p), plus the same remainder (r).
999...999 (with x nines) = 713q + r
999...999 (with y nines) = 713p + r
Technically, we can subtract these two equations from each other. The x nines - the y nines would end up with x - y nines in the front and y zeros at the end because the nines would cancel. According to the Distributive Law, 713q - 713p = 713(q - p) because you can factor a 713 out of both terms. However, the r's at the end cancel each other out, leaving us with no remainder!!! Since q and p are both integers, subtracting them will also be an integer.
999...999 (with x nines) = 713q + r
- 999...999 (with y nines) = 713p + r
999...(x - y nines)...999000...(y zeros)...000 = 713 (q - p)
Since 713 isn't a multiple of 2 or 5, all of the zeros don't matter. Hence, we can cross them all out leaving us with 999...999 (x - y nines) = 713(q - p). Since q - p is an integer, 713 x some integer = a number consisting of all nines. Isn't that cool!!
Saturday, July 2, 2011
Fibonacci Day: Addition of Fibonacci Numbers
As you may have noticed, today is a Fibonacci Day. If you'll notice, it is the second, and two is a Fibonacci Number.
Let's look at a simple pattern within these numbers. What would happen if you add all the Fibonacci numbers up? Infinity, because they go on forever. What if you added Fibonacci numbers, and then stopped at some point. Let's see:
1 = 1
1 + 1 = 2
1 + 1 + 2 = 3
1 + 1 + 2 + 3 = 7
1 + 1 + 2 + 3 + 5 = 12
You might not see a pattern, but there is one. Let's rewrite these sums in a different format.
1 = 1 = (2 - 1)
1 + 1 = 2 = (3 - 1)
1 + 1 + 2 = 4 = (5 - 1)
1 + 1 + 2 + 3 = 7 = (8 - 1)
1 + 1 + 2 + 3 + 5 = 12 = (13 - 1)
See it now? Every sum is one less than a Fibonacci number! This pattern will actually go on forever! In order to prove it, we will use something called proof by induction.
If we were to add on the next Fibonacci Number (21), we would also be adding that to the 13 - 1 from before. However, with 13 and 21 being consecutive Fibonacci Numbers, they join together to create the next Fibonacci Number. Since the minus one remains, you always are subtracting one from a Fibonacci Number.
For another proof, let's express each Fibonacci Number as the difference of the two after it.
(2-1) + (3-2) + (5-3) + (8-5) + (13-8) + (21-13)
You'll see that in the first two expressions, the 2 and -2 cancel out. In the next two, the 3 and -3 cancel. This keeps going until you are left with the greater number in the last expression and the -1 from the first one. That is also a very beautiful proof. Isn't that cool?!
Let's look at a simple pattern within these numbers. What would happen if you add all the Fibonacci numbers up? Infinity, because they go on forever. What if you added Fibonacci numbers, and then stopped at some point. Let's see:
1 = 1
1 + 1 = 2
1 + 1 + 2 = 3
1 + 1 + 2 + 3 = 7
1 + 1 + 2 + 3 + 5 = 12
You might not see a pattern, but there is one. Let's rewrite these sums in a different format.
1 = 1 = (2 - 1)
1 + 1 = 2 = (3 - 1)
1 + 1 + 2 = 4 = (5 - 1)
1 + 1 + 2 + 3 = 7 = (8 - 1)
1 + 1 + 2 + 3 + 5 = 12 = (13 - 1)
See it now? Every sum is one less than a Fibonacci number! This pattern will actually go on forever! In order to prove it, we will use something called proof by induction.
If we were to add on the next Fibonacci Number (21), we would also be adding that to the 13 - 1 from before. However, with 13 and 21 being consecutive Fibonacci Numbers, they join together to create the next Fibonacci Number. Since the minus one remains, you always are subtracting one from a Fibonacci Number.
For another proof, let's express each Fibonacci Number as the difference of the two after it.
(2-1) + (3-2) + (5-3) + (8-5) + (13-8) + (21-13)
You'll see that in the first two expressions, the 2 and -2 cancel out. In the next two, the 3 and -3 cancel. This keeps going until you are left with the greater number in the last expression and the -1 from the first one. That is also a very beautiful proof. Isn't that cool?!
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