## Saturday, June 22, 2013

### History of Math: Leonhard Euler

Today, I am giving a talk about Mending Mathematics Education in America at TEDxBushnellPark in Hartford, Connecticut. I did a post on this issue (click here to see it), as well as a huge school project on the issue (click here to see the research paper).

In the talk, I will be bringing up a famous math problem that originated with Leonhard Euler. I won't post about that problem today, but I will give a little story from the life of Euler.

Leonhard Euler was born on April 15, 1707 in Basel, Switzerland. He chose the path of a mathematician (if he didn't, I probably wouldn't be talking about him right now), and ended up as one of the most prolific mathematics writers of all time.

Some of his biggest contributions include standardizing the notation for the number e and π, coming up with Euler paths and circuits, and revamping most of the branches of mathematics that were known in his time. He was said to be able to entertain a child, scratch a cat, and calculate math problems simultaneously.

On this blog, I have mentioned sequences a lot. For instance, the triangular numbers are a sequence. The negative-first powers are a sequence. The powers of 1/2 are a sequence. Let's look at these three sequences:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55...
1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10...
1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128...

We haven't done it before, but a common thing to do is add up all of the numbers in these sequences. In some sequences, this seems odd to be doing, but in other ones, it seems completely understandable. Let's try it with those three.

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 +...

The individual numbers themselves are going towards infinity, so the sum of them is probably infinity as well. Generally, if the numbers in a sequence are growing, then the sum will be infinite. This type of sequence is called a divergent sequence, and questions like their sum are generally not asked.

What about the next series. This one is a little harder to tell, but if we group some things together, it will be easier to see.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...

Compare this to the following sequence, which is clearly less than this one.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...
1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + ...

Since all of the terms in the first sequence seem to be bigger than the second one, then the first one has to sum to a greater total. But, the second sequence simplifies to 1 + 1/2 + 1/2 + 1/2 +..., which will end up as infinity. So, a sequence that has a bigger sum must also be infinite as well. This makes that a diverging sequence.

What about the third one? Well, combining terms together won't do much, but let's try figuring out what the sum is by setting the sequence equal to x.

x = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...

Now, what if you were to multiply that by two? You would get:

2x = 2(1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...)
2x = 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...
2x = 2 + (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...)
2x = 2 + x
x = 2

So, the sum of this sequence is not infinity, but rather the natural number two. Since the sequence does not diverge to infinity, it is called a convergent sequence. These are the type that are generally summed to their total, since that number can then be applied.

Anyways, some of Euler's students were analyzing a much more complicated convergent sequence to figure out its sum. They had been adding numbers, and after they got to the seventeenth number in the sequence, they found their answers to be different at the fiftieth decimal place. Since they didn't want to go back through that tedious process, they argued and argued over who got it right.

Eventually, they went to their professor, Euler, to ask him for assistance. After asking for the problem, he mentally calculated out the answer to the fiftieth decimal, and was able to determine who was correct.

I found that story kind of funny, but also really impressive. To be able to calculate that difficult of a problem to that high of a degree shows some real mathematical talent.

#### 1 comment:

1. I waited all τDay for you to post about τ, but you never did. ;-(