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Find the half range sine series for f(x) = ?x - x2 in [ 0, ? ]. Hence find \[\sum_{n=1}^{\infty{}}\frac{1}{{(\ 2n-1)}^6}\]
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$For \ sine \ series \\[2ex] \displaystyle a_o=0\ \ \ \ \ a_n=0 \\[2ex] \begin{equation} \displaystyle Let\ \ f(x)\ =\ \pi{}x\ -\ x^2=\ \sum_{n=1}^{\infty{}}b_n\sin{nx}\ \ \ .....(1)\\[2ex] \displaystyle b_n=\ \frac{2}{\pi{}}\int_0^{\pi{}}f\left(x\right)\sin nx\ dx\ %eq1 \end{equation}\\ \displaystyle =\ \frac{2}{\pi{}}\int_0^{\pi{}}\left(\pi{}x\ -\ x^2\right)\sin nx\ dx\ \\[2ex] \displaystyle =\ \frac{2}{\pi{}}\ {\left[\left(\pi{}x\ -\ x^2\right)\left(-\frac{\cos{nx}}{n}\right)-\ \left(\pi{}-2x\right)\left(-\frac{\sin{nx}}{n^2}\right)+\ (-2)\left(\frac{\cos{nx}}{n^3}\right)\right]}_0^{\pi{}} \\[2ex] \displaystyle =\ \frac{2}{\pi{}}\ \left[\left(0-0-\frac{2}{n^3}\cos{n\pi{}}\right)-\left(0-0-\frac{2}{n^3}\right)\right] \\[2ex] \displaystyle =\ \frac{4}{\pi{}}\left[\frac{1-\cos{n\pi{}}}{n^3}\right] \\[2ex] \displaystyle =\ \frac{4}{n^3\pi{}}\left[1-{\left(-1\right)}^n\right] \\[6ex] Substituting \ in \ (1) $

$\begin{align*} \\ \displaystyle \therefore{}\ \pi{}x\ -\ x^2&=\ \sum_{n=1}^{\infty{}}\frac{4}{n^3\pi{}}\left[1-{\left(-1\right)}^n\right]\sin{nx} \\[2ex] \displaystyle &=\ \frac{8}{\pi{}}\left[\frac{1}{1^3}\sin{\pi{}x+\ \frac{1}{3^3}\sin{3\pi{}x+\ \frac{1}{5^3}\sin{5\pi{}x+\ …….\ \ }}}\right] \end{align*}\\[6ex] For \ half \ range \ sine \ series \ Parseval's \ Tdentity \ is \\[2ex] $

$ \begin{equation} \displaystyle \frac{1}{\pi{}}\int_0^{\pi{}}{\left[f\left(x\right)\ \right]}^2\ dx=\ \frac{1}{2}\ \left[b_1^2+\ b_2^2+\ b_3^2+\ …..\right]\ \ \ \ \ \ \ \ \ .....(2) %eq2 \end{equation}\\[2ex] \displaystyle \therefore{}\ \frac{1}{\pi{}}\int_0^{\pi{}}{\left[\pi{}x-\ x^2\ \right]}^2\ dx=\ \frac{1}{\pi{}}\ \int_0^{\pi{}}\begin{array}{l}\left({\pi{}}^2x^2-\ 2\pi{}x^3+\ x^4\right)\ dx \\ \ \end{array} \\[2ex] \displaystyle =\ \frac{1}{\pi{}}\ \left[{\pi{}}^2{\left(\frac{x^3}{3}\right)}_0^{\pi{}}-\ 2\pi{}{\left(\frac{x^4}{4}\right)}_0^{\pi{}}+\ {\left(\frac{x^5}{5}\right)}_0^{\pi{}}\right] \\[2ex] \displaystyle =\ \frac{1}{\pi{}}\left[\frac{{\pi{}}^5}{3}-\frac{{\pi{}}^5}{2}+\frac{{\pi{}}^5}{5}\right] \\[2ex] \displaystyle =\ \frac{{\pi{}}^4}{30} \\[4ex] From (2)\\[2ex] \displaystyle \frac{{\pi{}}^4}{30}=\frac{1}{2}\ .\ \frac{8^2}{{\pi{}}^2}\left[\frac{1}{1^6}+\ \frac{1}{3^6}+\ \frac{1}{5^6}+\ …..\right]\ \\[2ex] \displaystyle \frac{{\pi{}}^4}{30}=\ \frac{1}{2}\ .\ \frac{64}{{\pi{}}^2}\ \sum_{n=1}^{\infty{}}\frac{1}{{(\ 2n-1)}^6} \\[2ex] \displaystyle \frac{{\pi{}}^6}{30\times{}32}\ =\ \sum_{n=1}^{\infty{}}\frac{1}{{(\ 2n-1)}^6} \\[2ex] \displaystyle \therefore{}\ \sum_{n=1}^{\infty{}}\frac{1}{{(\ 2n-1)}^6}=\ \frac{{\pi{}}^6}{960} \\[2ex] $

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