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# Sum to infinity of recipricol of squares watch

1. Can anyone show me how to get the sum to infinty of: 1/1² + 1/2² + 1/3²...

My friend says its π²/6, which looks about right but I have no idea how to get there.

Thanks
2. Your friend is correct, see picture.But sorry my Mac worked out the answer.
Attached Images

3. I'm thinking you need to use the Taylor series but we never really covered that in class...
Anyone else?
4. It's an extremely interesting result.
http://en.wikipedia.org/wiki/Basel_problem

The taylor series can be used. Very briefly it states:
5. Cheers Gaz, he must have brought it up because I said I was about to start reading the music of primes (and therefore the Riemann Hypothesis).
6. (Original post by Jump)
Can anyone show me how to get the sum to infinty of: 1/1² + 1/2² + 1/3²...

My friend says its ?²/6, which looks about right but I have no idea how to get there.

Thanks
I have found this solution: but I don't understand it. looks very complicated.
Attached Images

7. (Original post by Gaz031)
It's an extremely interesting result.
http://en.wikipedia.org/wiki/Basel_problem

The taylor series can be used. Very briefly it states:
That's Euler's argument and wouldn't really count as a proof these days - you'd need to be more rigorous regarding convergence etc.
8. (Original post by RichE)
That's Euler's argument and wouldn't really count as a proof these days - you'd need to be more rigorous regarding convergence etc.
I'm reading some introductory analysis [eg and can only imagine how hard it must be to produce a proper rigorous proof. I think i'll just admire the result for now
9. (Original post by Gaz031)
I'm reading some introductory analysis [eg and can only imagine how hard it must be to produce a proper rigorous proof.
Well some of the claims in the proof are wrong. It is true that sinx/x equals the given infinite product but one cannot claim generally that a function's power series equals an infinite product of factors containing the function's roots - for example look at the exponential series

e^x = 1 + x + x^2/2! + ...

yet the function e^x has no roots, not even counting complex ones. Yet by Euler's reckoning these "roots" should add to -1.
10. (Original post by RichE)
Well some of the claims in the proof are wrong. It is true that sinx/x equals the given infinite product but one cannot claim generally that a function's power series equals an infinite product of factors containing the function's roots - for example look at the exponential series

e^x = 1 + x + x^2/2! + ...

yet the function e^x has no roots, not even counting complex ones. Yet by Euler's reckoning these "roots" should add to -1.
Interesting. Using infinite series to prove properties seems to require delicate work. Is there a way to prove other than using power series?
11. (Original post by RichE)
Well some of the claims in the proof are wrong. It is true that sinx/x equals the given infinite product but one cannot claim generally that a function's power series equals an infinite product of factors containing the function's roots - for example look at the exponential series

e^x = 1 + x + x^2/2! + ...

yet the function e^x has no roots, not even counting complex ones. Yet by Euler's reckoning these "roots" should add to -1.
You can't actually critise Euler's reasoning and belief, as they were the earlier years of mathematics, rather admire and correct him.

It seems that the broadening and discovering of more math has come to a halt.
12. (Original post by Vijay1)
You can't actually critise Euler's reasoning and belief, as they were the earlier years of mathematics, rather admire and correct him.

It seems that the broadening and discovering of more math has come to a halt.
I acknowledge that Euler was the giant of eighteenth century maths.

But it is fair for me to say some of the proof is wrong - because it is - it is relying on facts that are wrong. You can't simply cavalierly presume that facts about polynomials carry over to power series. But that doesn't stop the fact a young Euler got the right answer to the unsolved problem.

I have no idea what your last sentence (in bold) means as maths research is in an undeniable golden age.
13. (Original post by Gaz031)
Interesting. Using infinite series to prove properties seems to require delicate work. Is there a way to prove other than using power series?
More often than not sine, cosine and e^z are defined as power series to start with and the result follows from the definition.

Also you can define sine and cosine as complex exponential functions anyway, and the result is again obvious. It is all dependant on the definition.
14. (Original post by AMM)
More often than not sine, cosine and e^z are defined as power series to start with and the result follows from the definition.

Also you can define sine and cosine as complex exponential functions anyway, and the result is again obvious. It is all dependant on the definition.
The definitions of sine and cosine in complex exp. functions:

cosx = ½(eix+e-ix)
sinx = ½i(eix-e-ix)

These don't seem to be used often though.
15. (Original post by Jump)
1/1² + 1/2² + 1/3²...
=ζ(2)=(π²/6)

Newton.
16. (Original post by Newton)
=ζ(2)=(π²/6)

Newton.
People who are doing / have just finished A-Levels (including myself) won't really know much about the riemann zeta function (sp?)
17. But it wouldn't do them any harm to learn!

Anyone who can find the value of ζ(3) will get their own little piece of fame
18. (Original post by davros)
But it wouldn't do them any harm to learn!

Anyone who can find the value of ζ(3) will get their own little piece of fame
It's known to be irrational and that is all I think. A result due to Apery and so zeta(3) is often called Apery's constant.

I guess most students see a proof of zeta(2) = pi^2/6 through Fourier Series.
19. (Original post by RichE)
It's known to be irrational and that is all I think. A result due to Apery and so zeta(3) is often called Apery's constant.

I guess most students see a proof of zeta(2) = pi^2/6 through Fourier Series.
Yes, and you can evaluate a close approximation to zeta(3) using Bernouli numbers.

The same method applies for all zeta(2k+1), k E N.
20. I've seen zeta(2k) expressed in terms of Bernoulli numbers, but I thought all the odd Bernoulli numbers were zero, so the same technique didn't work for zeta(2k+1)

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