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Let $f$ be a $2\pi$-periodic and Riemann integrable function on $\mathbb R$, where the Fourier series of the function f can be written as $$f(\theta)\thicksim \hat f(0)+\sum_{n\geq 1}[\hat f(n)+\hat f(-n)]\cos(n\theta)+i[\hat f(n)-\hat f(-n)]\sin(n\theta)$$

And, question is :
Prove that if $f$ is even, then $\hat f(n) = \hat f(-n)$, and we get a cosine series.
Prove that if $f$ is odd, then $\hat f(n) = -\hat f(-n)$, and we get a sine series.
Also note that $$A_n=\hat f(n)+\hat f(-n)=\frac 1 {\pi}\int_0^{2\pi}f(s)\cos(ns)ds \\ B_n=i[\hat f(n)-\hat f(-n)]=\frac 1 {\pi}\int_0^{2\pi}f(s)\sin(ns)ds$$

If $f$ is odd, then $A_n$ must be odd, i.e $f(s)\cos(ns)$ must be odd, which is indeed odd. Now in the solution manual they conclude saying $A_n=0$, how do they find that?

Secondly, if $f$ is even, then $B_n$ must be odd, i.e $f(s)\sin(ns)$ must be odd, which is also true. And again in the solution manual it states that $B_n=0$ How do they come up with these?

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Hint: For even or odd $f$ use

$$f(-x)=f(x)\qquad \text{or} \qquad f(-x)=-f(x)$$

and plug in the fourier series and see what happens. Use that sine is an odd function and cosine is an even function.

If you want to show it like in the manual then you should shift the integration bounds for $0$ to $2\pi$ to the interval $-\pi$ to $\pi$, you can do this because the integrand is $2\pi$ periodic. Then use

$$\int_{-a}^{a}F(u)du=0$$

if $F(u)$ is odd.

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  • $\begingroup$ Most probably they came up with that.. Okay thanks! $\endgroup$ – Leyla Alkan Mar 15 '18 at 20:18

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