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Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by letting $$f(x)=1-x^2 \ \ \ \ \ \text{for} \ \ 0\leq x\leq 1$$ and requiring that $$ f(x+2)=f(x) \ \ \ \ \ \text{and} \ \ \ \ f(-x)=f(x) \ \ \forall x\in\mathbb{R} $$ Find $Sf(x)$, the Fourier series of $f$.

My solution:

$$ Sf(x) = \frac{a_0}{2}+\sum_{k = 1}^{\infty} a_k\ \text{cos}(k\pi x) \ \ \ \ \ \text{as} \ b_k=0\ \ (f\ \text{is an even function}) $$ Now for $a_0$, $$ a_0=2\int_{0}^{1} 1-x^2 dx=\frac{4}{3} $$ For $a_k$, $$ a_k=2\int_{0}^{1} (1-x^2)\text{cos}(k\pi x) dx $$ $$ a_k=2\int_{0}^{1} \text{cos}(k\pi x) dx-2\int_{0}^{1} x^2\text{cos}(k\pi x) dx $$ $$ a_k=-\frac{4}{k\pi}\int_{0}^{1} x\ \text{sin}(k\pi x) dx dx $$ $$ a_k=\frac{4}{k\pi}\Big(-\frac{\text{cos}(k\pi )}{k\pi}\Big) $$ $$ a_k=\frac{4(-1)^{k+1}}{(k\pi)^2} $$

Hence $$ Sf(x) = \frac{2}{3}+\frac{4}{\pi^2}\sum_{k = 1}^{\infty} \frac{(-1)^{k+1}}{k^2}\ \text{cos}(k\pi x) $$

Is my solution correct? I am unsure.

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    $\begingroup$ It looks correct to me! $\endgroup$ – Nosrati Jun 14 '18 at 15:11
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Based on feedback from user Dylan, I have to agree that the above from OP is correct, assuming that $f(-x)=f(x)$ and that $f(x+2)=f(x)$.

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  • $\begingroup$ But this isn't the assumption. The problem requires that $f(-x)=f(x)$, i.e. an even periodic extension $\endgroup$ – Dylan Jun 17 '18 at 7:49

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