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Consider the functions

$(1-a)x,\quad 0 ≤ x ≤ a,$

$a(1-x),\quad a ≤ x ≤ 1,$

over the interval $[0, 1]$ and $0 < a < 1$.

My question is: since this is not over a symmetric interval, how would I go about representing this function as a Fourier series? Any help would be appreciated!

Mars Plastic
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  • Ti is not very clear what you meant...if at least you'd use some commas. Besides this, it seems to be they want you to extend somehow the function's definition domain and then do Fourier. Or maby they meant you'll find out about what verical line is that function symmetric and... – DonAntonio Mar 13 '20 at 15:07
  • You could could consider its even or odd (periodic) extension to the interval [-1, 1] and find the Fourier series of that. – Paul Mar 13 '20 at 15:39
  • I tried but it was not correct when I tried the extension to the interval [-1, 1] – 123 Mar 13 '20 at 16:01
  • But Fourier series can also be computed on the interval $[0, 1]$, I don't think you need to do it on a symmetric interval about 0. – Fede Poncio Mar 13 '20 at 17:59
  • But it does have an odd/even extension which has a fourier series, so what do you mean by "it was not correct"? You computed it incorrectly or it was not in a form you recognized? – Paul Mar 13 '20 at 18:11
  • I only have the solution and my solution wasn't the same as the solution, so I computed it incorrectly or it doesn't have an odd/even extension. – 123 Mar 13 '20 at 18:31
  • @FedePoncio Do you know how to compute it if it isn't a symmetric interval? – 123 Mar 13 '20 at 18:32

1 Answers1

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If you want to expand with respect to the orthonormal system $U,C_1,S_1,C_2,S_2,\cdots$ with $U(x)=1$, $C_n(x)=\frac{\sqrt2}2\cos(2\pi nx)$ and $S_n(x)=\frac{\sqrt2}2\sin(2\pi nx)$, then recall $\int x\cos(bx)=\frac xb\sin(bx)+\frac1{b^2}\cos(bx)$ and $\int x\sin(bx)=-\frac xb\cos(bx)+\frac1{b^2}\sin(bx)$. Thus \begin{align}\int_0^1f(x)U(x)\,dx&=(1-a)\frac{a^2}2+a\left(1-\frac12-a+\frac{a^2}2\right)=-\frac{a^2+a}2\\ \int_0^1f(x)C_n(x)&=2^{-1/2}(1-a)\int_0^a(1-x)\cos(2\pi nx)\,dx+2^{-1/2}a\int_a^1x\cos(2\pi nx)\,dx=\\&=2^{-1/2}(1-a)\left(\frac{1}{2\pi n}\sin(2\pi na)-\frac a{2\pi n}\sin(2\pi na)-\frac1{4\pi^2n^2}\cos(2\pi na)+\frac1{4\pi^2n^2}\right)+\\&+2^{-1/2}a\left(\frac1{4\pi^2n^2}-\frac{a}{2\pi n}\sin(2\pi na)-\frac1{4\pi^2n^2}\cos(2\pi na)\right)=\\&=2^{-1/2}\left(\frac{1-\cos(2\pi na)}{4\pi^2n^2}+\frac{(1-2a)}{2\pi n}\sin(2\pi na)\right)\\\int_0^1f(x)S_n(x)&=2^{-1/2}(1-a)\int_0^a(1-x)\sin(2\pi nx)\,dx+2^{-1/2}a\int_a^1x\sin(2\pi nx)\,dx=\\&=2^{-1/2}(1-a)\left(\frac{1-\cos(2\pi na)}{2\pi n}+\frac{a}{2\pi n}\cos(2\pi na)-\frac1{4\pi^2n^2}\sin(2\pi na)\right)+\\&+2^{-1/2}a\left(-\frac{1}{2\pi n}+\frac a{2\pi n}\cos(2\pi na)-\frac1{4\pi^2n^2}\sin(2\pi na)\right)=\\&=2^{-1/2}\left(\frac{(1-2a)(1+\cos(2\pi na))}{2\pi n}-\frac{\sin(2\pi na)}{4\pi^2n^2}\right)\end{align}