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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\qquad\text{\(f(x)\) as a sine series}\] \begin{equation*} f(x)=\cos(x),\qquad0<x<\pi \end{equation*} \begin{equation*} % g(x) g(x)= \begin{cases} \cos(x),&0<x<\pi\\ \cos(x),&\pi<x<0 \end{cases}\qquad\text{odd extension} \end{equation*}

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} % a_0,n a_0&=a_n=0\qquad\text{(odd function)}\\ \end{align*}\] \begin{align*} % b_n b_n&=\frac1\pi\int\limits_{\pi}^\pi g(x)\sin(nx)\,\text dx\\ &=\frac2\pi\int\limits_{0}^\pi \cos(x)\sin(nx)\,\text dx\qquad\text{(even integrand)}\\ &=\frac1\pi\int\limits_{0}^\pi \sin\big((1+n)x\big)\sin\big((1n)x\big)\,\text dx\\ &=\frac1\pi\left[\frac{\cos\big((1+n)x\big)}{1+n}+\frac{\cos\big((1n)x\big)}{1n}\Big_0^\pi\right]\\ &=\frac1\pi\left[\frac{1\cos\big((1+n)\pi\big)}{1+n}+\frac{\cos\big((1n)\pi\big)1}{1n}\right]\\ &=\frac1\pi\left[\frac{1(1)^{n+1}}{1+n}+\frac{(1)^{n+1}1}{1n}\right]\\ &=\frac1\pi\left[\frac{1+(1)^{n}}{1+n}+\frac{(1)^{n+1}1}{1n}\right]\\ &=\frac1\pi\left[\frac{(1n)\big(1+(1)^n\big)(1+n)\big((1)^{n}+1\big)}{1n^2}\right]\\ &=\frac1\pi\left[\frac{2n\big((1)^{n}+1\big)}{1n^2}\right]\\ &=\frac2\pi\left[\frac{n\big((1)^{n}+1\big)}{n^21}\right]\\ \end{align*}

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align*} % S(x) S(x)&=\frac2\pi\sum\limits_{n=1}^\infty\frac{n\big((1)^{n}+1\big)}{n^21}\sin(nx)\\ &=\frac4\pi\sum\limits_{n=2,4,6,\dots}^\infty\frac{n\sin(nx)}{n^21}\\ &=\frac8\pi\sum\limits_{r=1}^\infty\frac{r\sin(2rx)}{4r^21}\\ \end{align*}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0for some reason in the back of my book it has \[\frac8\pi\sum\limits_{r=1}^\infty\frac{r\sin(2rx)}{4r^2+1}\] as the answer, which seams like a mistake to me , but maybe i have overlooked something

NewOne
 one year ago
Best ResponseYou've already chosen the best response.1He used the last trig identity on this page convert it. http://www.sosmath.com/trig/Trig5/trig5/trig5.html

NewOne
 one year ago
Best ResponseYou've already chosen the best response.1Were you expecting bn as 0?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0the only qualm i have on this question is the plus sign in the denominator of answer in the back of the book, i got a minus sign , which seams to graph the correct thing so im not sure .

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i guess the back of my book is wrong then ,

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0yeah, even i get minus doing it another way also...

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0Thanks to both of you
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