Yamahonda450
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Find the Fourier Series for f(x)=x^2 evaluate f(0) and show that the summation
\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}
The first part of this problem asked that I find a_{n} and b_{n}
Since x^2 is an even function b_n=0 and for a_n I got \frac{4(-1)^2}{n^2}
for a_0 I got \frac{\pi^2}{3}
I was able to do a related problem that asked to evaluate f(pi) and show that...
\sum^{\infty}_{n=1}\frac{1}{(n)^2}=\frac{\pi^2}{6}
I was able to show this one easily but the other one...\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}
Is a lot more challenging. I attached my work for the related problem if someone could please show me were to start with this one I'd greatly appreciate it.
\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}
The first part of this problem asked that I find a_{n} and b_{n}
Since x^2 is an even function b_n=0 and for a_n I got \frac{4(-1)^2}{n^2}
for a_0 I got \frac{\pi^2}{3}
I was able to do a related problem that asked to evaluate f(pi) and show that...
\sum^{\infty}_{n=1}\frac{1}{(n)^2}=\frac{\pi^2}{6}
I was able to show this one easily but the other one...\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}
Is a lot more challenging. I attached my work for the related problem if someone could please show me were to start with this one I'd greatly appreciate it.