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I'm doing exercise 15 on page 255 in Kreyszig:

  1. To illustrate that a Fourier series of a function $f$ may converge even at a point where $f$ is discontinuous, find the Fourier series of

$$ f(x) = \begin{cases} 0 & x \in [-\pi, 0) \\ 1 & x \in [0, \pi) \end{cases}$$


My solution:

(i) For the $n$-th character, $n \in \mathbb N$, we compute the $n$-th coefficient as follows: $$ \hat{f}(e^{inx}) = \langle f, e^{inx} \rangle = \int_0^\pi e^{-inx} dx = \frac{i}{n}(e^{in \pi} - 1)$$

(ii) For the $-n$-th character we compute $$\hat{f}(e^{inx}) = \langle f, e^{-inx} \rangle = \frac{-i}{n}(e^{in \pi}-1)$$

(iii) For the $0$-character $e^{i0x} = 1$ we compute $$\hat{f}(e^{i0x}) = \langle f, e^{-i0x} \rangle = \int_0^\pi 1 dx = \pi$$

So that the Fourier series of $f$ is $$ F(f(x)) = \pi , \hspace{1cm} x \in [-\pi, \pi)$$

Which is clearly wrong. What did I do wrong? Thanks for your help.

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    Doesn't Kreyszig normalize the integral with a factor?2012-07-30
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    @t.b. Isn't choosing the normalising constants up to me?2012-07-30
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    I think you need to normalize the coefficients by the period2012-07-30
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    Yes, they are up to you up to the point where you need to make them compatible with the inversion theorem.2012-07-30
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    Does it matter that you have $\int_0^\pi{e^{-inx} dx} = {i\over n}e^{in\pi}$ instead of ${i\over n}\left(e^{in\pi}-1\right)$ ?2012-07-30
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    @MarkDominus That is of course a mistake in my sums. Thank you very much for spotting that!2012-07-30
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    @MarkDominus It makes not difference though because they cancel out. My problem is at the zero-character.2012-07-30
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    You need to divide by the period $2 \pi$ somewhere along the line. I would expect the series to converge to the average of limits from either side, so at $x=0$, I would expect $\frac{1}{2}$. It is not immediately obvious to me how you conclude with $f(x) = \pi$. I would expect $f(0) = \frac{1}{2}$.2012-07-30
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    @t.b. But no matter what normalisation I choose: the zero coefficient is non-zero. Didn't we discuss yesterday that the Fourier series should be zero? Although that wouldn't make sense either because the Fourier should equal the function.2012-07-30
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    I certainly didn't discuss *that* (I said I couldn't follow) I only said that you can reduce to a pure sine wave by subtracting $1/2$. See also: [Gibbs phenomenon](http://en.wikipedia.org/wiki/Gibbs_phenomenon)2012-07-30
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    @t.b. Ok : ) Didn't mean to accuse you of it. Just misremembered. Going to read link now.2012-07-30
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    @MattN.: The zero coefficient is the average of the function and is definitely not zero here...2012-07-30
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    @copper.hat Ooh, thank you! I knew that actually, but didn't remember! But: if all my other coefficients cancel out, then the Fourier series equals $\hat{f}(e^{i0})$. But shouldn't the Fourier series equal the function we take the Fourier series of? Or maybe not since we're in $L^2(G)$ and we only have convergence in norm...2012-07-30
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    Well, in this case the series will converge to the function everywhere except at $x=0,\pm \pi$.2012-07-30
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    @t.b. Wait, but: where does Gibbs come in here? My function is $1$ at zero and my $n$-th partial Fourier series is $\frac12$ at zero. So there is no overshooting in this example. Well at least not the way they draw it in the example pictures on Wikipedia. Perhaps being constant $\frac12$ on $[-\pi,0)$ counts as overshooting, too.2012-07-30
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    There is **definitely** overshooting in this example! The Gibbs phenomenon applies to finite sums, not the limit.2012-07-30
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    Note the reconstructed function is $0$ on $(-\pi,0)$ and $1$ on $(0,\pi)$, not $\frac{1}{2}$.2012-07-30
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    All that @copper.hat says is correct. The page on the Gibbs phenomenon has the example you ask about completely worked out along with pictures of how the partial sums approximate the function. This *is* the easiest instance of the Gibbs phenomenon.2012-07-30
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    @t.b. Yes I know. But I wouldn't trust Wikipedia. Examples there might be wrong, including the pictures. (Although it's not very likely.)2012-07-30
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    Yeah right :) They also perpetuate such blatant falsehoods as the uncountability of the reals... Come on: a page on a phenomenon that is *specifically* about this effect might have some $\varepsilon$'s off but won't be completely wrong.2012-07-30
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    @t.b. Probably. But I'm just very insecure about my mathematical thoughts in general and particularly tired today. Both together is a bad combination. Anyway. Things are looking up slightly.2012-07-30
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    @t.b. Like for example the [stub article about lifts](http://en.wikipedia.org/wiki/Lift_%28mathematics%29): If you have two morphisms $f: X \to Y$ and $g: Z \to Y$, certainly "lifting $f$ to $Z$" must be a map $h: Z \to X$ such that $fh = g$, not $h: X \to Z$. I don't believe this.2012-07-30
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    No, they have it completely right (despite the fact that it's a stub). It's about *lifting* the morphism $f$ *over* $g$ and that's what they say it is. Just as in homological algebra: you lift a morphism $f\colon P \to B$ from a projective $P$ over every epimorphism $g\colon A \twoheadrightarrow B$. We should stop now, we already auto-generated a flag for having posted too many comments (and the last few comments are decidedly off-topic).2012-07-30
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    @MattN: funny you should mention the article about lifts; I [used it in an answer](http://math.stackexchange.com/questions/176442/what-is-a-lift/176447) about this yesterday (including an explanation why we call *this* a lift and not as you say.)2012-07-30

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