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As the title shows,we know that there is some points the series not approaching to the function.

Now,take the convergence theorem into consideration.As there is some the first break-points,the series is still convergent.And,the Gibbs phenomenon always takes place on the first break-points.

Why does Gibbs phenomenon take place?What does it show the nature of Fourier Series?

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I'm having trouble understanding the 3rd sentence. Could you explain further what you mean? –  Yemon Choi Mar 7 '10 at 19:15
    
@Yemon Choi: 我也没看明白。 –  Sunni Mar 8 '10 at 1:27
    
this means the Gibbs phenomenon always takes place on the first break-points. –  Gu Yejun May 19 '10 at 8:34
    
Do you mean discontinuities? –  Trevor Alexander Mar 28 at 2:04
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up vote 8 down vote accepted

A Fourier series truncated to order $n$ is the best approximation to the given function in the $L^2$ sense using trigonometric polynomials of order $n$. As such, small rapid deviations don't matter much. Since there is a limit to how big the derivatives of a trigonometric polynomial of fixed order can be (without the coefficients being big), in order to fit such a polynomial to a discontinuity it pays to overshoot a bit on each side of the discontinuity in order to “gather speed” so you can get from one value to the other fast. When I say it “pays”, i mean to say that you what you lose by not approximating the function too well at the overshoot, you more than gain back by doing the jump faster.

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There's a nice discussion of the Gibbs phenomenon (among many other things) in Korner's book Fourier Analysis - but I'm sure other textbooks ought to say something about it –  Yemon Choi Mar 7 '10 at 19:16
    
1.6 of Dym and Mckean covers it. –  Steve Huntsman Mar 7 '10 at 19:20
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Hi there,

I think there is a bit more to answering your question than considering just the strict $L^2$ convergence however. The Gibbs phenomenon is important when considering the pointwise convergence of the partial sums of the fourier series. When $f^{\prime}$ is continuous on a compact interval, you will get pointwise convergence of the partial sums $S_N$ so $S_N(x) \to f(x)$ as $N \to \infty$. (You only get uniform convergence if in addition the function $f$ is compatible with the boundary conditions for your expansion but this is not your question anyway).

Now what happens when $f$ is discontinuous? It turns out that $S_N(x) \to \frac{1}{2}[f(x_-) + f(x_+)]$, the average of the left and right limits of the function. However, this is really only true for $N \to \infty$. Otherwise for any finite $N$ there is a small width of order $1/N$ around your discontinuous point $x$ where your partial sums are uniformly bounded away from either value $f(x_+)$, $f(x_-)$ by some fixed percentage (I recall $9$% of the jump size or something but don't quote me on that). Check out the photos at:http://en.wikipedia.org/wiki/Gibbs_phenomenon That little wiggle of the wave where the jump of $f$ occurs stays uniformly bounded away from the value of the function but the width of this region ($1/N$) goes to zero as $N \to \infty$ so that technically you still get the full pointwise convergence.

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