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I have a general question in partial differential equations.

Can we say that when an even function is expressed as a Fourier series, the Fourier cosine series is also the Fourier series?

My thinking is that a Fourier series has the form,

$$f(x) = \frac{a_0}{2}+\sum^{\infty}_{n=1} a_n cos(nx) + \sum^\infty_{n=1}b_nsin(nx)$$

where $$a_0 = \frac{1}{\pi}\int ^\infty _{-\infty}f(x)dx$$, $$a_n = \frac{1}{\pi}\int ^\infty _{\infty}f(x)cos(nx)dx$$, $$b_n = \frac{1}{\pi}f(x)sin(nx)dx$$

where $cos$ is a even function and $sin$ is a odd function. Then if $f(x)$ is even, multiplying a even and odd function together gives a odd function which is $0$ which would eliminate the $b_n$ term, leaving just $a_0$ and $a_n$ which is the fourier cosine series. Therefore yes this is true

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    $\begingroup$ Yes, exactly correct. $\endgroup$ – ziggurism Nov 20 '17 at 2:32
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    $\begingroup$ More generally, you can say that any function has an even part and an odd part, by writing $f(x) = (f(x)+f(-x))/2 + (f(x)-f(-x))/2.$ The fourier series for the even part is the cosine terms. The fourier series for the odd part is the sine terms. $\endgroup$ – ziggurism Nov 20 '17 at 5:29
  • $\begingroup$ @ziggurism I have made another post if you had a second to look at it, thanks for that though! $\endgroup$ – user104 Nov 20 '17 at 5:32

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