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If x(t) is even, then $x(t) = a_0 + \sum_{n=1}^{\infty}a_n*\cos(2\pi nt/T)$

However, based on this formula: $x(t) = a_0 + \sum_{n=1}^{\infty}a_n*\cos(2\pi nt/T) + b_n*\sin(2\pi nt/T)$ where $a_n = 2/T \int_o^Tx(t)*\cos(2\pi nt/T)$

x(t) is an even function and cos is an odd function. An even * odd = odd function. The periodic integral of an odd function is 0. Hence, it should only be the sin term remaining. However, the Fourier Series for these even and odd functions are reversed. It should be that the fourier series of an odd function is the Fourier cosine series. I don't understand.

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  • $\begingroup$ $\cos$ is an even function. $\endgroup$ – Micah Feb 26 '17 at 21:22
  • $\begingroup$ cos is even, sin is odd, so even times cos = even, while even times sin = odd. $\endgroup$ – Ian Feb 26 '17 at 21:22
  • $\begingroup$ Ah! yes I made a dire mistake. $\endgroup$ – JobHunter69 Feb 26 '17 at 23:12
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If $x(t)$ is an even function, $x(-t) = x(t)$. If you write out $x(t)$ and $x(-t)$ out in terms of the Fourier series, you see that the only way they are equal is if $b_n = 0$ for all $n$.

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