- #1

bznm

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"[itex]f(x)= \pi -x[/itex] in [itex][0, \pi][/itex]

Let's consider the even extension of f(x) in [itex][-\pi, \pi][/itex]

and write the Fourier Series using this set ( [itex]\frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt {\pi}} \cos nx[/itex] )

and use the Parseval identity to prove:

[itex]\sum \frac{1}{(2k+1)^2}= \frac{\pi ^2}{8}[/itex]"

My attempt:

[itex]a_0= \frac{1}{\sqrt {2 \pi}} \int_{-\pi}^{\pi} f(x)=\frac{1}{\sqrt {2 \pi}} \int_{-\pi}^{\pi} (\pi -x) dx= \frac{2 \pi^2}{\sqrt {2\pi}}[/itex]

[itex]a_n= \frac{2}{\sqrt \pi}\int_{0}^{\pi} f(x)\cos nx dx=\frac{2}{\sqrt \pi} \frac{1-cos \pi n}{n^2}= \frac{2[1+(-1)^n]}{ \sqrt {\pi n^2}}[/itex]

So,

[itex]f(x)=\frac{2 \pi^2}{\sqrt {2\pi}}+\sum_{n=1}^{\infty}\frac{2[1-(-1)^n]}{ \sqrt {\pi n^2}} \frac {\cos nx}{\sqrt \pi}[/itex]

Could you tell me if these steps are correct? Are the formulas for Fourier coefficients correct?

For the second part of the exercise, I have and enormous doubt:

It was told me that in similar cases, Parseval Identity is:

[itex]\int_{-\pi}^{\pi} |f(x)|^2 dx= a_o^2+ \sum |a_n|^2[/itex]

I don't know how to do [itex]|a_n|[/itex], because there is the term [itex](-1)^n[/itex]..

And I have a lot of doubts to obtain the result [itex]{\pi}^2/8[/itex] using the results that I have obtained...

Many thanks for your precious help!