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Continuing from an earlier question of mine: Fourier-Series of a part-wise defined function?

I now got a fourier series which I believe is the correct one: $$\frac{\pi(b-a)}{2} + \sum\limits_{n=1}^{\infty} \frac{(a-b)(1-(-1)^n)}{n^2\pi}\cos(nx) + \frac{(-1)^n(b-a)}{n}\sin(nx)$$

Now my next task confused me a little - "What is the sum of this series?". By definition, this should just be $f(x)$ ($\frac{ax+bx}{2}$ if $x$ is a whole multiple of $\pi$) right? So I figured I am probably supposed to find a closed form for this - although given the definition of the function, I find it hard to imagine that it even exists. Am I wrong? If so, what is the closed form of this series?

(This wasn't the correct fourier series after all - the right one is

$$\frac{\pi(b-a)}{2} + \sum\limits_{n=1}^{\infty} \frac{(a-b)(1-(-1)^n)}{n^2\pi}\cos(nx) + \frac{(-1)^{n+1}(a+b)}{n}\sin(nx)$$

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I think that's all you need to do: to state the answer and quote a theorem on pointwise convergence to justify it. –  user31373 Jun 18 '12 at 22:00
    
They are pretty close to for example the sum over $\sin(nx)/n$ which is periodic in the interval $(0,2\pi)$ and given by $(\pi-x)/2$. More generally the sums over $\sin(nx)/n^k$ and $\cos(nx)/n^k$ are related to periodic Bernoulli polynomials. But maybe the alternating sign messes that up a bit. –  Peter Sheldrick Jun 18 '12 at 22:23
    
This isn't part of the question title - but I now noticed that this doesn't work with a=b - the sum of the series would become 0. But f(x) is not constant 0 for a=b... But I don't recall having made assumptions about a and b not being equal in my calculations... –  Cubic Jun 18 '12 at 22:46
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1 Answer

up vote 2 down vote accepted

Hint1: prove following sums for $x \in (-\pi,\pi)$ and conclude using the appropriate linear combination. $$\tag{1} \sum_{n=1}^\infty \frac {\cos(nx)}{n^2}=\frac{(\pi-|x|)^2}4-\frac{\pi^2}{12}$$ $$\tag{2} \sum_{n=1}^\infty \frac {(-1)^n\cos(nx)}{n^2}=\frac{x^2}4-\frac{\pi^2}{12}$$ $$\tag{3} \sum_{n=1}^\infty \frac {(-1)^n\sin(nx)}n=-\frac x2$$

Hint2: To prove these identities start by computing the Fourier series of $f(x)=\frac x2$ : you should get minus the last identity (it is the classical 'Sawtooth wave').
Setting $x:=y+\pi$ you may get a fourth identity ('sign' is the 'Sign function' that is $\pm$) :
$$\tag{4} \sum_{n=1}^\infty \frac {\sin(nx)}n=\operatorname{sign}(x)\frac {\pi-x}2$$ The integral of $(4)$ will give you minus the first identity (the constant of integration is obtained by considering $x$ at $0$).
The integral of $(3)$ will give you minus the second identity.

Hint3: The result is rather simple (and was probably the starting point!) and it should be easier to separate the cases $x\in (-\pi,0)$ and $x\in (0,\pi)$.

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I guess that makes sense, except I have no idea how to prove these sums either. Do you happen to know what causes my problem with (a=b)? –  Cubic Jun 19 '12 at 5:40
    
@Cubic: I gave more details concerning their evaluation. Concerning $b-a$ note that $(b-a)$ may be taken as an overall factor so that $b=a$ implies indeed that the result is $0$. –  Raymond Manzoni Jun 19 '12 at 7:52
    
Yep, as it turns out my fourier series was wrong (again). It should have been $\frac{(a+b)(-1)^{n+1}}{n}$ where it says b-a in the series (I missed a '-' in the scalar product). Yeah, but I don't think I'm gonna include a closed form for the sum in my homework after all - what you just said is a little over my head, we just briefly touched fourier series in 2 or 3 classes and then went on to the next topic. –  Cubic Jun 19 '12 at 8:34
    
@Cubic: in this case perhaps that (as suggested by Leonid) you just got the Fourier terms that you deduced earlier. The result should then just be the function you had at the start except at the points of discontinuity where you'll get half (the limit of the value at the left plus the limit at the right). –  Raymond Manzoni Jun 19 '12 at 9:44
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