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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 $(1)$.

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  • $\begingroup$ Set $x=0$ then $x=1$ then subtract. $\endgroup$ – Alexey Burdin Jun 18 '15 at 5:18
  • $\begingroup$ but in x=0 and 1 the function is not defined know...then how can i do this??@AlexeyBurdin $\endgroup$ – David Jun 18 '15 at 5:57
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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 $$ \sum\limits_{n=1}^\infty\dfrac1 {(2n-1)^2}=\dfrac{\pi^2}{8} $$

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  • $\begingroup$ how it hold even the fuction is not defined on 1 and 2. $\endgroup$ – David Jun 18 '15 at 6:39
  • $\begingroup$ $f$ becomes a periodic function of $1$ so it repeats on $[1,2]$ from $[0,1]$, and that is why you get $f(x)=\frac 1 3+\sum\limits_{n=1}^\infty \frac 4 {n^2\pi^2} \cos n\pi x$. $\endgroup$ – hermes Jun 18 '15 at 6:44
  • $\begingroup$ ya,,, thanks for the comment $\endgroup$ – David Jun 18 '15 at 7:02

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