## Saturday, April 28, 2012

### A Neat and Simple Formula for the Golden Ratio

This week, I thought I would talk a little bit about the golden ratio. The golden ratio is a topic in Fibonacci numbers, but this post won’t be involving them, so I did not bother saving it for a Fibonacci day.
There is a formula for the golden ratio, as follows:
(1 + √5)/2
However, there is another formula that has a little more beauty in it.
1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/...
I’d say that is a little cooler than the previous formula. However, this is a pretty difficult statement to prove what this is equal to. However, we can figure it out.
First, let’s set this formula equal to x.
x = 1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/...
Ignore the 1 + 1 at the beginning. What do we have?
1 + 1/(1 + 1/(1 + 1/(1 + 1/...
If you’ll notice, this is the same exact thing as before, since they are both an infinite series. Therefore, the equation can be rewritten as:
x = 1 + 1/x
Now, we can multiply through by x.
x(x) = x(1 + 1/x)
x^2 = x + 1
Now, we will subtract x + 1 from both sides to get:
x^2 - x - 1 = 0
We have just turned that infinite series into a solvable equation! A little while ago, we talked about a formula called the quadratic formula, which is used to solve equations just like this one. The formula is:
(-b ± √(b^2 - 4ac))/2a
With a, b, and c being the first, second, and third coefficient. So, let’s plug them in and see what we get.
(-(-1) ± √((-1)^2 - 4(1)(-1)))/2(1)
(1 ± √(1 + 4))/2
(1 ± √5)/2
We are in home stretch. Now, here is where logic will come in. One answer here is correct, and one is called an extraneous solution (based on logic, it is not a valid solution). Since we are adding a positive quantity to one, we know the solution must be positive. Therefore, (1 - √5)/2 would not be a valid answer. So, we are left with:
x = (1 + √5)/2
So, this proves the golden ratio equal to that top number.
The golden ratio’s formula is very cool, but there are more continued fractions such as this one. For instance, pi (or as I now call it: half-tau) has a formula as well. Check it out:
π = 3 + 1/(6 + 9/(6 + 25/(6 + 49/(6 + 81/...
This one isn’t quite as cool, but still very interesting.
Bonus: A month or so ago, I posted a puzzle involving light bulbs. If you figured it out, great. If you haven’t, here is the answer.
Turn on two of the light switches and wait for a little while. After some time, turn off one of them, and leave the other one on. Then, go upstairs.
You will see the on light bulb, which you can immediately match up. For the two off bulbs, feel them and see which has more heat. The one with more heat is the one you turned on and then turned off (it would still have heat from it being lit up), and the one with less heat was never turned on.

## Saturday, April 21, 2012

### Fibonacci Day: Divide in all the Primes. Each and every one

I don't know if you noticed, but today is a Fibonacci day. It is April 21, and 21 is the eighth Fibonacci number.  Let's keep on the topic of prime numbers, and talk about a really cool correlation between the two sequences.

Take a prime number, like 11. Is eleven a factor of a Fibonacci number? Well, yes, it is. It divides the number 55.

What about 23? It divides 46368, the twenty-fourth Fibonacci number.

Take 101, a big prime. It actually divides the 100th Fibonacci number, which is 354224848179261915075.

In fact, every single prime number divides a Fibonacci number. Unfortunately, I do not know the proof for this, but if you are familiar with it, please comment below and tell us. This statement can actually be made cooler, believe it or not.

Every prime number (that we will call P) ending in 1 or 9 is a factor of the Fibonacci number in position P-1. Furthermore, any prime ending in 3 or 7 is a factor of the Fibonacci number in position P+1.  I thought that even though I don't know the proof, it is pretty cool that this correlation exists nonetheless.

## Saturday, April 14, 2012

### Probability, The Number e, and Magic all in one

Since I have not done any mathematical magic posts recently, I wanted to make sure we got one in.
Here is a fun trick you can do for your friends that is all based on simple math. Ask them what the odds would be that if you give them a card in the deck, they could guess the value of the card. They should say one in thirteen, or about eight percent.
What about guessing the value wrong. That would be 92%.
Challenge them to get through the whole entire deck and get every single card’s value wrong. Pretty easy right, you only have an eight percent chance of getting it right. Watch their stunned reaction when they get halfway through and get one correct.
The Method: As I said, it is all based on mathematics. To calculate the odds, you must multiply together the odds of getting it wrong every single time. So, you would have to multiply the .92 by itself fifty-two times. This gives you:
(12/13)^52 = .0183...
Basically, there is a 1.83% chance that you will get through the deck getting it wrong every time. I thought that was pretty cool.
If you’re in a mathematical mood, it gets even cooler. Say you did it with the full card rather than just the value. This would give you:
(51/52)^52 = .3643...
Still good odds. But check this out. Remember back when we studied the number e? Well, what is the reciprocal of e, or 1/e?
1/e = .3678...
They aren’t exact, but that is an extremely close estimate, right! This is true because of our formula for e^x. Let’s go back to it.
e^x ≈ (1 + x/n)^n
In this case, 51/52 would be written as:
(1 + -1/52)^52
With this rule, this should be about e^-1, which means the same thing as the reciprocal of e.
You could use this same logic to figure it out for just the value. (12/13)^52 is the same as:
(1 + -4/52)^52
This means that the odds are approximately e^-4, which is 1/e^4. I thought that this trick is cool, but the reasoning behind it is even cooler.

## Saturday, April 7, 2012

### A Weekend at Gathering 4 Gardner

I just spent last weekend at a conference in Atlanta, Georgia called Gathering 4 Gardner. This is a conference based on the interests of Martin Gardner, who wrote a math column in Scientific American as well as dozens of math, magic, and puzzle books. Though he did not do difficult mental calculations of any sort, he was the first person to hold the title of “Mathemagician.”
This conference had so many cool things, and I would like to share a few of these cool things. First off, I would like to mention a Fibonacci conversion that I learned. I know, it’s not Fibonacci day, but it is pretty cool.
Let’s say you want to convert from miles to kilometers. If you want to approximate 5 miles in kilometers, we would just go up to the next Fibonacci number. Since it goes 1, 1, 2, 3, 5, 8, five miles is eight kilometers.
What about a half-marathon. Well, minus .1 miles. After 13 comes 21 meaning that there are 21 kilometers in 13 miles.
What about 10 miles? Well, this isn’t a Fibonacci number, but you can still do it just fine. How can you create 10 as the sum of two Fibonacci numbers?

8 + 2 = 10
Now, just bring the 8 and the 2 up to the next Fibonacci number.
13 + 3 = 16
There are about 16 kilometers in 10 miles. This is all because the ratio of miles to kilometers is extremely close to the golden ratio; somewhere around 1.61. The golden ratio is the ratio of Fibonacci numbers, which proves this estimation system.
There were also a bunch of cool puzzles that I saw. Here is one of them:

For this puzzle, you have to try to move just one line to make the giraffe a different giraffe. Good luck! As usual, I will post the answer in around a month.
I also heard a pretty funny math joke at the conference that I thought I would share. Here it is:
A few people were inside of a cave and wondering if the cave makes an echo. So, they shout out, “We are lost.” A little later, they hear, “You are lost.” They then realize it was a mathematician and there are three reasons why. One, it took so long. Two, it was accurate. Three, it was completely useless.