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Find the half range cosine fourier series expansion for $$f(x)=(x-1)^2,\quad 0<x<1$$
and hence deduce that $$\pi^2=8\left(\frac 1 {1^2}+\frac 1 {3^2}+\frac 1 {5^2}+\ldots\right)\tag{1}$$
My work
I have derived that the expansion is $$f(x)=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi x$$ But I could not deduce that $(1)$.
Define $f$ as a periodic function of $1$ on $\mathbb{R}$ as $$ f(x)=(x−1)^2, \hspace{5 mm} 0\le x<1 $$ Set $x=0$, then $$ f(0)=1=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} $$
So we have $$ \dfrac{2}{3}=\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \hspace{4 mm} \text{and} \hspace{4 mm} \dfrac{\pi^2}{6}=\sum\limits_{n=1}^\infty \dfrac1{n^2} $$
And set $x\to1^{-}$, then $$ \lim\limits_{x\to1^{-}}f(x)=0=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi $$
Thus \begin{align} \dfrac{\pi^2}{12}&=\sum\limits_{n=1}^\infty \dfrac1 {(2n-1)^2}-\sum\limits_{n=1}^\infty \dfrac1 {(2n)^2} \\ &=\sum\limits_{n=1}^\infty \dfrac1 {(2n-1)^2}-\frac1{4}\sum\limits_{n=1}^\infty \dfrac1 {n^2} \\ &=\sum\limits_{n=1}^\infty\dfrac1 {(2n-1)^2}-\dfrac{\pi^2}{24} \end{align}
So $$ \sum\limits_{n=1}^\infty\dfrac1 {(2n-1)^2}=\dfrac{\pi^2}{8} $$
