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This is a question from an old exam, I'm trying to understand the key and comments provided by the examiner.

First question is simply determine the Fourier series of the function $f$ defined as $1-x^2$ on $[-\pi,0)$ and as $1+x^2$ on $[0,\pi)$. I got $$f \sim 1+\frac{2}{\pi} \sum \frac{1}{n^3} \left((2-n^2 \pi^2)(-1)^n-2 \right) \sin{nx}$$

while the key gives $$f \sim 1+\frac{2}{\pi} \sum \frac{1}{n^3} \left((n^2 \pi^2-2)(-1)^{n+1}-2 \right) \sin{nx}$$

The only difference is style, but is it of any significance?

The key also has the following to say:

Note that $f(x) = 1 + f_1(x)$, where $f_1$ is odd.

This is the only mention of $f_1$, and I can't figure out what it could possibly refer to. Neither piecewise functions equals $1$ plus the other, so that can't be it. There has to be some significance to this comment but since I don't understand the comment in the first place said significance is lost on me.

Finally one is asked to sketch the Fourier series. Specifically:

Set $g(x) :=$ (the Fourier series of $f$) for $x$ in $\mathbb{R}$. Sketch the graph of $g$ on the interval $[-2\pi,2\pi]$. Be particularly clear at the jump points!

Well, I am dumbfounded. How does one go about finding even, say, $g(1)$, when you have to calculate an infinite number of values and sum them together?

Finally, what are the jump joints? Why isn't $g$ continuous everywhere despite $\sin{nx}$ being so?

Thank you.

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  • $\begingroup$ $f_1$ must be $f_1(x) = f(x)-1 = (\operatorname{sgn} x) x^2$. $f$ is $2 \pi$-periodic. Since $f$ is Lipschitz on $(-\pi,\pi)$ it will converge pointwise there. Hence the only points to be careful with on $[-2 \pi, 2 \pi]$ are $\pm \pi$. Since $f'$ has a limit from the right at $-\pi$ and similarly for $+\pi$, we see that the Fourier series will converge to the average of the limit of $f$ at these points (that is, $1$). $\endgroup$ – copper.hat Jan 10 '14 at 6:52
  • $\begingroup$ @copper.hat Why is $f(x)-1=x^2 \mathrm{sgn}(x)$? I don't understand that at all. $\endgroup$ – Erik Vesterlund Jan 10 '14 at 6:54
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    $\begingroup$ Because $f(x) =1+x^2$ on $[0,\pi)$ and $f(x) = 1-x^2$ on $[-\pi,0)$. $\endgroup$ – copper.hat Jan 10 '14 at 6:55
  • $\begingroup$ @copper.hat That is clever, thanks! $\endgroup$ – Erik Vesterlund Jan 10 '14 at 12:32
  • $\begingroup$ If it is that obvious in the representation of the function, it is often useful to separate the odd and even parts of a function. One may have a trivial Fourier series consisting of one or two easily identified terms, and for the other you need only compute half of the coefficients. $\endgroup$ – LutzL Jan 10 '14 at 14:00

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