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Find the fourier series representation of $$f(x)=\frac12(\pi-x), 0 \lt x \lt 2\pi$$ to calculate $a_0, a_n \text{ and } b_n$ I have to use the integrals $\int_{-L}^L$ but is $-L=0 \text{ and } L=2\pi \text{ or is } -L=-\pi \text{ and } L=\pi$ for this problem? I think the period is $2\pi$ I am not sure if I use the bounds of the interval as the bounds of the integral or the fact that the period should be 2L?

Martin Sleziak
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idknuttin
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  • yes it is a similar question but can you just explain why they introduced g(x) by splitting f(x) into two functions? because I stilll don't understand where they got the bounds for the integrals of $a_n$ and $b_n$? – idknuttin Dec 04 '15 at 16:53
  • Do not think of the function $f$ as defined on just $[0,2\pi]$. It should be thought of as defined everywhere so as to be $2\pi$-periodic. Then you can compute the Fourier coefficients by integrating on any interval of length $2\pi$. It is a lot easier to integrate on $[-\pi,\pi]$ than on $[0,2\pi]$. But to do so you will have to figure out what the values of the function would have to be on that interval. We already know the values on $[0,\pi]$ so just try to determine the values on $[-\pi,0]$. That is where the "split" function $g$ in the other answer comes in. – B. S. Thomson Dec 04 '15 at 17:51
  • i did found the fourier series twice, once with the bound $[-\pi, \pi]$ with two functions $g(x)$ and then I did it with the bounds $[0, 2\pi]$ and it was so much quicker to do it without splitting it into two function because you have half the number of ingrals to solve and plugging in $0$ and $2\pi$ after integrating was very simple. splitting the function so you can sole it over $[-\pi, \pi]$ makes the problem twice as long. – idknuttin Dec 06 '15 at 18:20
  • Some people find it easier to use the fact the function is odd and then just integrate over $[0,\pi]$ to get a sin series. But the point of the exercise was just to figure what it all meant and you have succeeded. Nobody really has much fun computing Fourier coefficients--the theory is fun though. – B. S. Thomson Dec 07 '15 at 20:55

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