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\(f(x)\) as a sine series

Differential Equations
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\[ \qquad\text{\(f(x)\) as a sine series} \begin{equation*} % f(x) f(x)=1,\qquad0
\[\begin{equation*} % g(x) g(x)= \begin{cases} 1,&0
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just change L to pi http://mathworld.wolfram.com/FourierSeriesSquareWave.html
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you got the opposite one .. shift the period by +pi
where did i go wrong?
don't know .. don't have much time right now!! but i often use that link as reference!!
sin(nx) is not even .. get rid of that minus
i'll see later!!
why does (4) on that link have a sin^2 term?
ah i found my mistake
\[\begin{align*} % b_n b_n&=\frac1\pi\int\limits_{-\pi}^{\pi} g(x)\sin(n x)\,\text dx\\ &=\frac2\pi\int\limits_0^\pi \sin(n x)\,\text dx\qquad\text{even integrand}\\ &=\frac2\pi\left(\frac{-\cos(n x)}{n}\right)\Big|_0^\pi\\ &=\frac2\pi\left(\frac{\cos(0)-\cos(n\pi )}{n}\right)\\ &=\frac2{\pi}\left(\frac{1-(-1)^n}{n}\right)\\ \end{align*}\]
\begin{align*} S(x)&=\frac2\pi\sum\limits_{n=1}^\infty\left(\frac{1-(-1)^n}{n}\right)\sin(nx)\\ &=\frac4\pi\sum\limits_{n=1,3,5,\dots}^\infty\frac{\sin(nx)}{n}\\ &=\frac4\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1} \end{align*}
just a minus sign
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guy please help me with this. This is impossible tough for me
i havent studies solving PDEs yet sorry @Echdip

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