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Performing a differentiation on a Fourier series

+ 2 like - 0 dislike
54 views

I'm tutoring a set of problem sheet to do with Fourier series and one problem is as follows:

The Fourier series for a sawtooth wave is,

$f(x)=x=-\sum^{\infty}_{n=1}\frac{2(-1)^n\sin(nx)}{n}$ for $-\pi < x<\pi$.

If you differentiate this you get

$1=-2\sum^{\infty}_{n=1}(-1)^n\cos(nx)$ again for $-\pi < x<\pi$

What is wrong with this?

I have the solutions sheet and it says that it does not converge to 1 (fair enough, I plotted it to large $n$ and it sort of converges but oscillates between 0 and 2 in the interval) and then states ...

An assumption has been made that you can interchange the order of summation and differentiation in the result stated.

It then goes to note that you can interchange the order of summation and integration

I don't understand the argument, can anyone shed some light on this?

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user Brendan
asked Jan 27, 2012 in Mathematics by Brendan (20 points) [ no revision ]
This may help: math24.net/…

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user David Mitra
Also, see page 27 onward here: google.com/…

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user David Mitra
If you differentiate once more, you will obtain a linear combination of linearly independent functions equal everywhere to zero. It contradicts to the linear independence of functions.

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user Vladimir Kalitvianski
Looking at the links David left, pg. 670 of the Pete Olver textbook chapter pdf has a good answer. It seems that it is because, even though the limits are stated, the series is still converging to the sawtooth and not x, so the differentiation is converging to the differential of the sawtooth, i.e. $1-2\pi\delta(x-\pi)$. I had considered this but assumed that because in the question was explicit about the limits, this was not an issue and would converge to x ...

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user Brendan

3 Answers

+ 3 like - 0 dislike

As a complement: suitably interpreted, your computation is perfectly correct. That is, while the resulting series will not converge pointwise, there are many other possible (useful!) ways a Fourier series may converge. And, suitably interpreted, term-wise differentiation is always correct. [Edit: typo'd "pointwise" earlier, when I meant "termwise". Sorry!]

Recall that, for two Fourier expansions of "nice" functions $f(x)=\sum_n a_n e^{inx}$ and $g(x)=\sum_n b_ne^{inx}$, we have the Parseval-Plancherel theorem that ${1\over 2\pi}\int_0^{2\pi} f(x)\,\overline{g(x)}\,dx=\sum_n a_n\,\overline{b_n}$. And, yes, Fourier series do converge pointwise for infinitely-differentiable functions, and can be differentiated termwise...

Slightly changing your example, observe that the Fourier series $\delta(x)=\sum_n 1\cdot e^{inx}$ (the Dirac comb) certainly does not converge point-wise, because the terms do not go to $0$. Nevertheless, thinking in terms of Parseval-Plancherel, for nice function $f$ $$ f(0) \;=\; \sum_n a_n\,e^{in\cdot 0} \;=\; \sum_n a_n\cdot 1 $$ which has the same form as ${1\over 2\pi}\int_0^{2\pi} f(x)\cdot \overline{\delta(x)}\,dx$ if the latter were to make sense. That is, the evaluation-at-$0$ functional can be represented as a sort of inner product, under some constraints.

More precisely, for $s\in\mathbb R$, the $s$-th Sobolev space $H^s$ here consists of Fourier series $\sum_n b_ne^{inx}$ with $\sum_n |b_n|^2/(1+|n|)^s < \infty$. The Fourier series for the Dirac comb is in $H^{-{1\over 2}-\epsilon}$ for every $\epsilon>0$. For an "ordinary" function $f$ whose Fourier coefficients $a_n$ have sufficient decay to put $f$ in $H^{{1\over 2}+\epsilon}$, e.g., some smoothness of $f$ itself, $$ |f(0)| \;=\; \Big|\sum_n a_n\cdot 1\Big| \;=\; \Big|\sum_n a_n\cdot (1+|n|)^{{1\over 2}+\epsilon} \cdot {1\over (1+|n|)^{{1\over 2}+\epsilon}}\Big| $$ and by Cauchy-Schwartz the square of this is dominated by $$ \sum_n |a_n|^2(1+|n|)^{1+2\epsilon} \cdot \sum_n {1\over (1+|n|)^{1+2\epsilon}} $$ That is, the evaluation functional $f\rightarrow f(0)$ is continuous in the $H^{{1\over 2}+\epsilon}$ metric topology, and the Dirac comb is a continuous linear functional on it. (This argument actually shows an instance of Sobolev imbedding, namely, that $H^{{1\over 2}+\epsilon}$ consists of continuous functions.)

Thus, a Fourier series in $H^{-s}$ (with $s> 0$), even though not pointwise convergent at all, directly gives a (continuous) linear functional on $H^s$ by extending Parseval-Plancherel: $$ \Big(\sum_n b_n\,e^{inx}\Big)\bigg(\Big(\sum_n a_n e^{inx}\Big)\bigg) \;=\; \sum_n a_n\cdot b_n $$

That is, distributions have legitimate Fourier series expansions, though typically not converging pointwise at all. Termwise differentiation of not-very-convergent Fourier series is completely justifiable if construed as distributional derivatives, via integration by parts: after all, differentiation in the Fourier series is termwise multiplication by $in$.

Indeed, Sobolev and others looked at this sort of situation in the 1930s. A discussion in this direction, beginning more-or-less from scratch, is at http://www.math.umn.edu/~garrett/m/mfms/notes/09_sobolev.pdf

It bears repeating that there are many other types of convergence than pointwise.

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user paul garrett
answered Jul 28, 2012 by paul garrett (120 points) [ no revision ]
Indeed, a simple way to justify the distributional interpretation of the convergence of the derivative of the sawtooth wave Fourier series is to integrate the original series term by term. This should generate a series which is uniformly convergent and converges to a locally integrable function $F(x)$, which generates a regular distribution. Differentiating generates the "sawtooth wave" distribution $f(x)$, and again gives $f'(x)$, which is the "Dirac comb" singular distribution, showing that the series mentioned by the OP does converge if considered in the distributional sense.

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user Bitrex
Indeed! :) Good reconnect to the literal question! :)

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user paul garrett
+ 2 like - 0 dislike

"What's wrong with this?" What is wrong is the following:

We are talking about the function $f(x):=x$ $\,(-\pi<x<\pi)$ continued periodically on all of ${\mathbb R}$, and we are presented with a series "representing $f$" which barely converges – in fact we need the oscillations of $\sin$ to make it converge. One cannot expect that such a series, resp. its formal (termwise) derivative, is able to represent $f'$ correctly. The relevant theorem here says: If the termwise differentiated series converges uniformly then it actually represents $f'$. In the case at hand the termwise differentiated series doesn't even converge, so it can't represent anything.

It is a fact of life that the Fourier series of a function $f$ with a jump continuity at a single point is badly convergent at all points, and a fortiori its formal derivatives are not able to represent the derivatives of $f$ even in points $x$ where $f$ is smooth.

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user Christian Blatter
answered Jul 28, 2012 by Christian Blatter (30 points) [ no revision ]
+ 1 like - 0 dislike

My answer in 'Conceptual/Graphical understanding of the Fourier Series' could help.

You are computing the derivative of a function given by $y=x$ in $(-\pi,\pi]$. From the distributions' point of view the derivative will be $1$ except at $x=\pi$ where you'll get $-2\pi \delta(x-\pi)$ (from a jump discontinuity of $-2\pi$ because of the transition from $\pi$ to $-\pi$).
Of course we have to repeat this for every period $2\pi$ so that the full result will be : $$1 -2\pi \sum_{n\in \mathbb{Z}} \delta(x-\pi -2n\pi)$$ (it seems that you found a nice explication in Pete Olver's 'Fourier Series' pdf)

This is a 'Dirac comb' (note that the Fourier series in this link corresponds nearly to your result).
More exactly you got : $$\sum_{n=-\infty}^\infty \delta(x-a-2\pi n)=\frac 1{2\pi}+\frac 1{\pi}\sum_{n=1}^\infty \cos\left(n(x-a)\right)$$ in the special case $a=\pi$ producing $(-1)^n\cos(nx)$ at the right

I should add that this is a 'formal' result since $\sum^{\infty}_{n=1}(-1)^n\cos(nx)$ is not convergent!

Hoping all this clarified things,

This post imported from StackExchange Mathematics at 2014-06-01 19:37 (UCT), posted by SE-user Raymond Manzoni
answered Mar 27, 2012 by Raymond Manzoni (10 points) [ no revision ]

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