How to Verify the Equation: pi^2/8 = Sum (1/(2n+1)^2) without Prefix?

[bomba923]

How does
infinity
Sum (n^(-2)=(pi^2)/6
n=1

Please tell me if this has been posted before (afraid )
(in that case, i'll see the other post)

 

[master_coda]

Take f(x)=x. Then the Fourier coeffcients of f are a_n=0 and b_n=\frac{2}{n}(-1)^{n+1}. Parseval's theorem says that:

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left(a_k^2+b_k^2)

Since the a_n terms are all zero, this reduces to:

\frac{1}{\pi}\int_{-\pi}^\pi x^2\:dx=4\sum_{n=1}^\infty\frac{1}{n^2}

The integral is easy enough to solve, and the left hand side reduces to 2\pi^2/3. Dividing both sides by four gives us:

\frac{\pi^2}{6}=\sum_{n=1}^\infty\frac{1}{n^2}

 

[bomba923]

I see clearly now!--thanks

(Will no one answer my "digit-factorial question" thread )

 

[dextercioby]

And let's not forget Euler's original method.Combining the series he found for \frac{\sin x}{x} and the one from Taylor expansion,he was able to prove it...

Daniel.

 

[matt grime]

And there's another way:

\int\int\frac{1}{1-xy}dxdy

Evalutate that as x and y both go from 0 to 1. Do it using a substitution, and then do it by replacing the fraction inside with its series expansion and ignore the convergence issues to rearrange sum and integral.

 

[SpaceDomain]

So this is parsevals theorem?

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left( a_k^2+b_k^2)

?

 

[Mark44]

So this is parsevals theorem?

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty\left( a_k^2+b_k^2)

?

How about this?

\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)\:dx=\frac{a_0^2}{2}+\sum_{k=1}^\infty( a_k^2+b_k^2)

 

[mstation]

hello! i am new to this forum and it looks like a very nice place!

sorry for my english, i hope everyone can understand it..

sorry for spamming in this thread but it looks like it is the most close to what i need.

i think i understood the answer master_coda gave but i don't understand why he choose f(x)=x..

for instance in my exercise i am asked to verify this equation

\frac{\pi^2}{8}=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}

what f(x) should i choose for the calculation?

 

[DonAntonio]

hello! i am new to this forum and it looks like a very nice place!

sorry for my english, i hope everyone can understand it..

sorry for spamming in this thread but it looks like it is the most close to what i need.

i think i understood the answer master_coda gave but i don't understand why he choose f(x)=x..

for instance in my exercise i am asked to verify this equation

\frac{\pi^2}{8}=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}

what f(x) should i choose for the calculation?

I'm answering this, in spite of being an intent of "kidnapping" a thread because

(1) it is, perhaps unwillingly, very close to the OP, and more important

(2) This is a newcomer so he/she doesn't know (but now you do!).

Check the following:

$$\frac{\pi^2}{8}=\sum_{n=0}^\infty\frac{1}{(2n+1)^2}\Longleftrightarrow \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$

DonAntonio