# Basel problem solution

1. Jun 3, 2010

### hover

http://www.maa.org/news/howeulerdidit.html"
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I found a nice solution to the Basel problem on the internet which I am liking very much. It deals with derivatives and integrals and appears to be a much more solid solution to the Basel problem than Euler's original solution. There is just one thing I am having trouble with to understand the solution. I think it is just me being an idiot but I can't understand the end of the second page,

It sounds like garble and I don't get the logic. Can someone explain to me?

Thanks!!

Last edited by a moderator: May 4, 2017
2. Jun 3, 2010

### LCKurtz

Are you referring to this quote?:

"Any number is the product of an odd number and a power of 2. For odd numbers, the power of 2 is 2^0 . Hence, any square is the product of an odd square and a power of 4. "

All it is saying is that if n is odd, n = 20*n and n2 = 40*n2, which isn't saying much.

If n is even, factor out all factors of 2 so n has the form n = 2pk for some odd integer k. Then n2=22pk2 = 4pk2.

Last edited by a moderator: May 4, 2017
3. Jun 4, 2010

### hover

I'm sorry but I'm still having a hard time understanding. I don't see what the small quote(what I quoted in my first post) is talking about or referring to. The equation above where I quoted says pi^2/8 = ∑ 1/(1+2 k)^2 (n = 0 to infinity). I don't see how euler knows to multiply by 4/3 to get pi^2/6 and solve the Basel problem. :(

4. Sep 28, 2010

### friz82

I would try to answer to your question.
I had the same problems at that point when I read the text for the first time.
I think it can be explained as follows:
Every integer number, let's say 1,2,3,4... can be written as an odd number multiplied for some power of 2. If we take any odd number, let's say 7, we can have 7=7*2^0, 14=7*2^1 and so on.
The point is that if we start from the ensemble of just the odd numbers (1,3,5,7...) we can obtain the whole list of integer numbers by multiplying every term by the ensemble of powers of 2 (1,2,4,8....)
Therefore, to pass from the sum of the reciprocals of the odd squares to the sum of the reciprocal of all the integers we multiply the first sum by the sum of the reciprocal of the powers of 4 (2 squared). This last sum is equal to 3/4.
I hope this will be of some help