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$${Expand} \; f(x)= \begin{cases} 2{A\over L}x & 0\leq x\leq {L\over 2} \\ \\ 2{A\over L}\left(L-x\right) & {L\over 2}\leq x\leq L \end{cases} $$

I have determined $A_0$ (but omitted) to be $A_0=A$. For $\ A_n \ $ and $\ B_n$, I'm rather confused where to start. For $A_n$: $$A_n={2\over L}\left[2{A\over L}\int_{0}^{L\over 2}x\cos\left({2n\pi x}\over L\right)dx+2{A\over L}\left(L\int_{L\over 2}^L\cos\left({2n\pi x}\over L\right)dx-\int_{L\over 2}^Lx\cos\left({2n\pi x}\over L\right)dx\right)\right]$$ And a similar case to $ B_n$, I'm not quite sure if I'm on the right path.

J. Herrera
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    You are doing fine. The integration of $x\cos(\alpha x)$ can be done using integration by parts $f=x$ and $g'=\cos(\alpha x)$ – Andrei May 22 '19 at 14:38
  • Why $\cos\dfrac{2n\pi x}{L}$. Shouldn't be $\dfrac{n\pi x}{L}$? – Nosrati May 22 '19 at 14:46
  • Because for my case, my period is $L$ and so half of that would be $L\over 2$ which would be placed inside $\cos$ and $\sin$. – J. Herrera May 22 '19 at 14:49

1 Answers1

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That looks correct for $A_n$, you'll just need to evaluate some of the integrals by parts from here. For $B_n$, note that the function is even, and so $B_n = 0$ for all $n$, since they are coefficients of $\sin$ which is odd.

auscrypt
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  • Even function means $f(-x)=f(x)$. But I fon't see the function defined in the negative domain. How much is $f(-L/4)$? – Andrei May 22 '19 at 14:45
  • Are you referring to both cases of the function $f(x)$ as being even? But wouldn't $A_n$ also be 0? By integrating $\cos\left({2n\pi x}\over L\right)$, the answer would yield 0 for the upper and lower boundaries for both cases because of $n\pi$? Correct me if I'm wrong. – J. Herrera May 22 '19 at 14:46
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    Fourier series is defined on periodic functions. Assuming you're using period $L$, the function is even. Also, the answer would not yield 0; the $x\cos \left( \frac{2\pi n x}{L} \right)$ terms will not yield 0. – auscrypt May 22 '19 at 14:49