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enter image description here Only when the function is odd its Fourier coefficient of cosnx is equal to 0?

park ning
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1 Answers1

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First, your function is the sum of a constant and an odd function. Because of the constant, $a_0\ne0$, so not all $a_n$ are zero. But the ones for $n\ge1$ do vanish, because $a_n$ for $n\ge1$ vanish both in the case of a constant and for odd functions.

It is indeed true that if $a_n=0$ for all $n$ including $n=0$ then the function is odd. The easiest way to see this is to consider $f(x)+f(-x)$. It's Fourier cosine coefficients will be $2a_n=0$, and its sine coefficients vanish by a direct calculation. Hence all Fourier coefficients of $f(x)+f(-x)$, so this function is zero by the uniqueness theorem for Fourier series.

Harald Hanche-Olsen
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  • To emphasize, in case the OP missed it: The derivation of $a_0$ in the question is wrong. The derivation of $a_n$ for $n > 0$ is correct, though. – fgp May 04 '13 at 11:59
  • @fgp To me, it looked more like the calculation of $a_0$ was missing. Don't they teach that you have to treat it separately these days? But you're right that I should have been louder on this point. – Harald Hanche-Olsen May 04 '13 at 12:03