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I am trying to solve the following,

Find the Fourier series of $h(x) = \text{e}^x, x \in [0,\pi)$.

I'm not sure how to approach it since the question does not specify whether to use an even or an odd extension.

Any help would be much appreciated!

ThisIsNotAnId
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Tom
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  • Maybe it's because it is only mean to be $\pi$-periodic. So I'm afraid you have to do three cases, unless you obtain more information: odd extension $2\pi$-periodic, even extension $2\pi$-periodic, $\pi$-periodic. – Julien May 07 '13 at 19:37
  • I think it is supposed to be pi-periodic as the question does make a point of saying h(x) has period pi. How would I calculate the coefficients in this case? – Tom May 07 '13 at 19:40
  • Here is a formula in the most general case. – Julien May 07 '13 at 19:42
  • I've never seen that type of Fourier coefficient before. Could you help me with the integrals I need to calculate in terms of cos and sin? – Tom May 07 '13 at 19:45
  • Compute the coefficients with respect to $e^{2in x}$ first. It will be easier to calculate the integrals. Then you can deduce the $\cos, \sin $ form. – Julien May 07 '13 at 19:47
  • ok thanks for your help! – Tom May 07 '13 at 19:50
  • The coefficient of $e^{2inx}$ is $c_n=\frac{1}{\pi}\int_0^\pi e^xe^{-2inx}dx=\frac{1}{\pi}\int_0^\pi e^{(1-2in)x}dx$. That's easy to find an antiderivative. Then each $c_ne^{2inx}+c_{-n}e^{-2inx}$ will give you $a_n \cos (2nx) +b_n\sin (2nx)$ after a little bit of complex algebra. – Julien May 07 '13 at 19:50

1 Answers1

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You may use an even extension or an odd extension. Since:

$$ \int_{0}^{\pi}e^{x}\cos(nx)\,dx = \frac{1}{n^2+1}\left(-1+(-1)^n e^{\pi}\right), $$ $$ \int_{0}^{\pi}e^{x}\sin(nx)\,dx = \frac{n}{n^2+1}\left(1-(-1)^n e^{\pi}\right) $$ that are both consequences (take the real or imaginary part) of $$ \int_{0}^{\pi} e^{x} e^{nix}\,dx = \left.\frac{e^{(1+ni)x}}{1+ni}\right|_{0}^{\pi},$$ by considering the Fourier cosine series of $e^{|x|}$ over $(-\pi,\pi)$ we have: $$ e^x = \frac{e^\pi-1}{\pi}+\frac{2}{\pi}\sum_{n\geq 1}\left((-1)^n e^\pi-1\right)\frac{\cos(nx)}{n^2+1}$$ and by considering the Fourier sine series of $\text{sign}(x)\,e^{|x|}$ over $(-\pi,\pi)$ we have: $$ e^x = \frac{2}{\pi}\sum_{n\geq 1}\left(1-(-1)^n e^{\pi}\right)\frac{n\sin(nx)}{n^2+1}.$$ However, the last series converges to zero for $x=0$, so, if we want a pointwise convergent representation over $[0,\pi)$, it is best to take the even one (also because it converges faster, since $e^{|x|}$ is a way more regular function than $\text{sign}(x)\,e^{|x|}$).

Jack D'Aurizio
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