- #1
Yamahonda450
- 13
- 0
Find the Fourier Series for f(x)=x^2 evaluate f(0) and show that the summation
[tex]\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}[/tex]
The first part of this problem asked that I find [tex]a_{n}[/tex] and [tex]b_{n}[/tex]
Since x^2 is an even function b_n=0 and for a_n I got [tex]\frac{4(-1)^2}{n^2}[/tex]
for a_0 I got [tex]\frac{\pi^2}{3}[/tex]
I was able to do a related problem that asked to evaluate f(pi) and show that...
[tex]\sum^{\infty}_{n=1}\frac{1}{(n)^2}=\frac{\pi^2}{6}[/tex]
I was able to show this one easily but the other one...[tex]\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}[/tex]
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.
[tex]\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}[/tex]
The first part of this problem asked that I find [tex]a_{n}[/tex] and [tex]b_{n}[/tex]
Since x^2 is an even function b_n=0 and for a_n I got [tex]\frac{4(-1)^2}{n^2}[/tex]
for a_0 I got [tex]\frac{\pi^2}{3}[/tex]
I was able to do a related problem that asked to evaluate f(pi) and show that...
[tex]\sum^{\infty}_{n=1}\frac{1}{(n)^2}=\frac{\pi^2}{6}[/tex]
I was able to show this one easily but the other one...[tex]\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}[/tex]
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.