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I have to find the Fourier series of $coshx$ on $(-l,l)$.What I did was I found the Fourier series of $e^{x}=\sum _{n=-\infty}^{\infty }{(-1)^n (\ell^2+in\pi)\over{l^2+n^2\pi^2}}\sinh(\ell)e^{{in\pi x}\over\ell}$

Since $\cosh x={e^{x}+e^{-x} \over 2}$ ,to find $e^{-x}$ I substituted x=-x for $e^{x}$.Can I do that substitution?
I am using the complex form of fourier series $f(x)=\sum_{n=-\infty}^{\infty } C_n e^{in\pi x \over l}$ where $C_n={1\over 2l}\int_{-l}^l f(x)e^{-in\pi x \over l} dx$

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Yes, you can. Let's do it more carefully: introduce a new variable $t$, assigning to it the value $t=-x$. Then $$e^{-t}=\sum _{n=-\infty}^{\infty }{(-1)^n (\ell^2+in\pi)\over{l^2+n^2\pi^2}}\sinh(\ell)e^{{-in\pi t}\over\ell} \tag1$$ But what's in a name? The formula (1) is valid, no matter what the variable is called. In particular, if could be called $x$: $$e^{-x}=\sum _{n=-\infty}^{\infty }{(-1)^n (\ell^2+in\pi)\over{l^2+n^2\pi^2}}\sinh(\ell)e^{{-in\pi x}\over\ell} \tag2$$

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  • $\begingroup$ :Thanks.I wanted to verify if I would get the same answer to $e^{-x}$ from this method and when I do it from scratch using formula of complex Fourier $\endgroup$ – clarkson Jun 9 '14 at 15:00
  • $\begingroup$ @clarkson There is just one complex Fourier series on a given interval. You can't possibly get a different one, no matter what method you use. Unless you mess up the calculations. $\endgroup$ – user147263 Jun 9 '14 at 15:04

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