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$ f(x) = e^{|x|} $ with x $\in (-1,1)$ and $f(x+2) = f(x)$ for all $x$ $ T = 2 = 2L $ => $ L = 1$ $f(x) = \frac{A_0}{2} + \sum_{n=1}^{\infty} A_n \cos(n \pi x) $ Where $A_n = $ $ \int_{-1}^1 f(x) \cos(n \pi x)\,dx \ = $ $ 2\int_{0}^1 f(x) \cos(n \pi x)\,dx \ $ $ f(x) = e^{|x|} = e^x $ for all $x \in [0,1]$ So: $ A_n = 2\int_{0}^1 e^x \cos(n \pi x)\,dx \ $ By parts: $ \int u\,dv = uv - \int v\,du$ Let u = $ e^x$, du = $ e^x $, dv = $\cos(n \pi x)$ v = $ \frac{\sin(n \pi x)}{n \pi} $ .. therefor: $A_n = $$ 2\int_0^1 e^x \cos(n \pi x)\,dx \ = $ $ \frac{2}{n \pi} \left[e^x\sin(n \pi x) \right]_0^1 -$ $\frac{2}{{n \pi}} \int_0^1 e^x \sin( n \pi x )\,dx$ Let $u = e^x$, $du = e^x$, $dv = \sin(n \pi x),$ $v = -\frac{\cos(n \pi x)}{n \pi} $ $A_n = $$ 2\int_0^1 e^x \cos(n \pi x)\,dx \ = $ $ \frac{2}{n \pi} \left[e^x\sin(n \pi x) \right]_0^1 -$ $ \frac{2}{n^2 \pi^2} \left[e^x\cos(n \pi x) \right]_0^1 -$ $ \frac{2}{n^2 \pi^2} \int_0^1 e^x \cos(n \pi x)$
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Thu, 22 Nov 2012 20:32 GMT