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The Fourier coefficients for a function with period 2a over a symmetric interval is obtained by integrating over $(-a,a)$. If the interval however, is not symmetric, say $(0,2a)$, one can still integrate over $(-a,a)$ if the period is 2a and get the same Fourier Series.

I will attach pictures with full solutions, but here's a summary of my conundrum:

I thought this was pretty straightforward and I calculated the F-series for $f(x)=x~~$ on $(0,1)$ in part a) by integrating the coefficients on $(-\frac{1}{2},\frac{1}{2})$. This will be a sine series, so I shifted the sine function by using $sin(x-\frac{1}{2})$ in the series representation and it gave the correct series.

In part b) for the function $f(x) = ax^2+bx+c$ on $(0,2\pi)$ I used the same method and it did not work. The solution is close but the correct solution has an extra term of $2a\pi ~sin(nx)$ which I can't explain. (see top of third pic) What am I missing?

Showing what I did in LaTex will take a lot of time to write, hence the pictures.

TheMercury79
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  • The link wants to advertise to me so I don't want to look at it. Why shift at all though, simply do your integration over one time period so that 0 to 1 will do in your first example and 0 to $2\pi$ in your second. – Paul Feb 26 '20 at 09:29
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    Your second function is periodic and has a description $ax^2+bx+c$ in the interval 0 to $2\pi$. It does not have the same description in the interval $-\pi$ to 0. If your quadratic is $(x-1)^2$ then its description in $-2\pi$ to 0 is $(x+2\pi-1)^2$. – Paul Feb 26 '20 at 09:47

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