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$$ \text { Decompose a function } f(x)=e^{a x}, 0<x<\pi \text { into Fourier series (a) of cosines; (b) of sines. } $$ I don't understend, what's ment by "series of sine/cosine". I tried to use standart formulas of Fourier decomposition, but the unswer, obviously, is single., no "cosine" or "sine" series. What is my mistake?

Voizz
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  • Standard formulas give the coefficients of the series $f(x)=a_0+\sum\limits_{n=1}^\infty (a_ncos(nx)+b_nsin(nx))$. What are you doing? – herb steinberg Apr 03 '21 at 20:56

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The Fourier series of function consist only of cosines (or sines) then and only then function is even (or odd). So you should extend your function on $-\pi <x< +\pi$ to be even (or odd) and build a fourier series of a new function. This fourier series will be consist only of cosines (or sines) and will be converge uniformly to old function on the $0 <x< +\pi$

In more details. Consider new function $ g(x) = \begin{cases} e^{ax} &\text{ $x\in [0,\pi)$}\\ e^{-ax} &\text{ $x \in (-\pi,0)$} \end{cases} $

Calculate its Fourier series $g(x)=a_0+\sum\limits_{i=1}^n a_n \cos nx$

$a_0 = \dfrac{1}{2\pi} \int\limits_{-\pi}^{\pi} g(x)dx = \dfrac{(e^{a \pi}-1)}{\pi a}$

$a_n = \dfrac{1}{\pi} \int\limits_{-\pi}^{\pi} g(x) \cos nx dx= \dfrac{2 a ((-1)^{n} e^{a \pi}-1)}{\pi (a^2+n^2)}$

And $f(x)=g(x)$ on $0 <x< +\pi$

Gregory
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