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Let $f(x)=x\sin(x)$, $x\in[-\pi,\pi$

Find the Fourier series of $g(x)=f'(x)=\sin(x)+x\cos(x)$, $x\in(-\pi,\pi)$ and show that it converges pointwise to $g$.

Try

I found the FS of the function $f(x)=x\sin(x)$, $x\in[-\pi,\pi)$ (in the just previous problem) to $1-\frac{1}{4}\cos(x)+\sum_{n=2}^{\infty} \left (\frac{2(-1)^{n+1}}{n^2-1} \right) \cos(nx)$ and shown that it converges uniformly to $f$. (or written as $$1-\frac{1}{2}\cos(x)-2\sum_{k=2}^{\infty}\frac{(-1)^k}{(k^2-1)}\cos(kx)$$

Is there a trick to finding the Fourier of the derivative of my function given that I have the Fourier for my function? How do I show pointwise convergence to $g$?

*I know about convolutions as I suspect I need that in this problem. I think pointwise convergence is true if it is continuous and periodic.

  • Do you know how to write the Fourier coefficients of $f'$ in terms of the Fourier coefficients of $f$? – Kavi Rama Murthy Jun 17 '21 at 11:37
  • Think about it this way - the pointwise convergence is a property of the Fourier series, not the function, so continuity and periodicity of the function don't make sense as sufficient conditions for the pointwise convergence of the series. What you need is a condition on the convergence of the series, could you guess what that might be? – Ninad Munshi Jun 17 '21 at 11:38
  • @KaviRamaMurthy I do not know the relation between the two. Is there a theorem for this? I got the one for $f$ in 2 forms. –  Jun 17 '21 at 13:16
  • @NinadMunshi Cauchy? I think there is an easyer method so must not be that. I know the conditions for convergence but for pointwise I am not sure. –  Jun 17 '21 at 13:18
  • @user879295 See: https://math.stackexchange.com/questions/1754033/integration-and-differentiation-of-fourier-series – Jose Avilez Jun 17 '21 at 13:22

2 Answers2

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Let $f(x) = x\sin(x)$ for $ x\in [-\pi,\pi)$, and similarly, denote by $f$ its periodisation. Recall that, since $f$ is differentiable over $(-\pi,\pi)$ and is $2\pi$ periodic function, we can write it as a Fourier series $$ f(x) = \sum_{n=-\infty}^{\infty} c_n(f) e^{-inx} $$ with Fourier coefficients defined as $$ c_n(f) := \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx} dx, \quad n \in \mathbb{Z}. $$ Then, approaching similarly, we can define the Fourier coefficients of the derivative of $f$ as $$ c_n(f') = \frac{1}{2\pi}\int_{-\pi}^{\pi} f'(x) e^{inx} dx \quad n \in \mathbb{Z}. $$ with a corresponding Fourier series.

Relating Fourier Coefficients

Applying integration by parts, and assuming that $f$ is finite at the boundary, we find that $$ \begin{align} c_n(f') &= \frac{1}{2\pi}\int_{-\pi}^{\pi} f'(x) e^{inx} dx \\ &= \frac{1}{2\pi}\left[f(x)e^{inx}\Big\vert_{-\pi}^{\pi} - in\int_{-\pi}^{\pi}f(x)e^{inx} dx\right] \\ \text{$n$ an integer so $e^{in\pi} = (-1)^n$, giving}\\ &= \frac{1}{2\pi}\left[\left(f(\pi^-)e^{in\pi} - f(-\pi^+)e^{-in\pi}\right) - in\int_{-\pi}^{\pi}f(x)e^{inx} dx\right] \\ &= \frac{1}{2\pi}\left[(-1)^n\left(f(\pi^-) - f(-\pi^+)\right) - in\int_{-\pi}^{\pi}f(x)e^{inx} dx\right]\\ \text{if $f$ periodic, then}\\ &=\frac{-in}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx} dx\\ &= -inc_n(f). \end{align} $$ This gives the nice relation between the Fourier coefficients between a function and its derivative, that being $$ c_n(f') = -in c_n(f), \quad $$ Note: This holds with the current definition of the Fourier coefficient, and may different up to a constant using other definitions.

Using the above you can proceed to find a solution to your question. We can start by first computing the Fourier coefficients of the periodisation of your function $f(x) = x\sin(x)$ when $x\in (-\pi,\pi)$. Indeed, we find that $$ c_n(f) = \begin{cases} 1 & \text{if } n = 0\\ -\frac{1}{4} & \text{if } n = \pm 1\\ -\frac{(-1)^{n}}{(n^2 - 1)} & \text{otherwise}. \end{cases} $$ Using the above work, we therefore find that $$ c_n(g) = \begin{cases} 0 & \text{if } n = 0\\ \frac{i}{4} & \text{if } n = 1\\ -\frac{i}{4} & \text{if } n = -1\\ \frac{in(-1)^{n}}{(n^2 - 1)} & \text{otherwise}. \end{cases} $$ Hence the Fourier series for $g(x)$ would be

$$ \begin{align} g(x) &= \sum_{n=-\infty}^{\infty}c_n(g) e^{-inx} \\ &= \frac{i}{4}(e^{-ix} - e^{ix}) + \sum_{n=2}^{\infty}\frac{in(-1)^{n}}{(n^2-1)}e^{-inx} + \sum_{n=2}^{\infty}\frac{i(-n)(-1)^{-n}}{((-n)^2-1)}e^{inx} \\ &= \frac{i}{4}(e^{-ix} - e^{ix}) + \sum_{n=2}^{\infty}\frac{in(-1)^{n}}{(n^2-1)}(e^{-inx} - e^{inx}) \\ \text{using standard identities}\\ &= \frac{\sin(x)}{2} + 2\sum_{n=2}^{\infty}\frac{n(-1)^{n}}{(n^2-1)}\sin(nx), \end{align} $$ as required.

Note: You should keep in mind that different Fourier coefficient definitions also correspond to different Fourier series definitions.

spaceman
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  • Is the sign on the first term in the last series not incorrect? ie $\sin(x)/2$ and arent we forgetting the $n$ in the second term in the last series? i.e $\sin(nx)$ –  Jun 17 '21 at 14:20
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    @user879295 Correct, have made the fixes now. Forgot to write $e^{-inx}$ instead of $e^{inx}$ in the Fourier series – spaceman Jun 17 '21 at 14:34
  • Thanks a ton. This is what I was looking for. For pointwise convergence I am still a bit unsure. What would the argument be? I know that $f$ is piecewise $C^1$ but is $C^0$. How would I finish with that –  Jun 17 '21 at 14:37
  • @user879295 Regarding pointwise convergence, see Theorem 5.5 of https://web.math.ucsb.edu/~grigoryan/124B/lecs/lec5.pdf, and use the argument used there to prove pointwise convergence within the open interval $(-\pi,\pi)$. Note that we do not have pointwise convergence of $g$ on the boundary, i.e. $x = \pm \pi$ due to jump discontinuities, although we will have convergence to the average of the boundary values. – spaceman Jun 17 '21 at 14:52
  • So since $g(x)$ is continuous and $g'(x)$ is piecewise $C^1$ on $[-\pi,\pi]$ (since $f(x)$ is piecewise $C^1$) then the Fourier series of $g(x)$ converges uniformly to $g(x)$, except for at the boundary values where this is the average –  Jun 17 '21 at 15:02
  • @user879295 Precisely, all we need for the pointwise convergence for the derivative is the fact that $g$ is piecewise continuous. Then, this means that the Fourier series of the periodisation of $g$ converges pointwise to $\frac{g(x^-) + g(x^+)}{2}$ for every point $x \in \mathbb{R}$. But since $g$ is continuous within the open interval $(-\pi,\pi)$, it follows that we have pointwise convergence of the Fourier series of $g$ to $g(x)$ for every point $x \in (-\pi,\pi)$. Another reference, with proof can be found at http://www.math.chalmers.se/Math/Grundutb/CTH/mve030/1617/Proof-TheoryList.pdf. – spaceman Jun 17 '21 at 15:16
  • @mjw, that is true for $f$, but not for $g$. $g(x) = \sin(x) + x\cos(x)$, so $g(-\pi) = \pi \neq -\pi = g(\pi)$, giving us the discontinuity. – spaceman Jun 17 '21 at 16:07
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    Oh yes! you are right! Thanks – mjw Jun 17 '21 at 16:08
  • @Spaceman your $c_n(f)$is missing a multiple of 2 right? –  Jun 25 '21 at 14:17
  • @user879295 I don't believe so. Could you specify where you think this is? – spaceman Jun 25 '21 at 15:41
  • That is my bad. If you were to write the series in terms of $c_n(f)$ how would this look? –  Jun 25 '21 at 16:08
  • And the last series; Why is it $n(-1)^{n+1}$ and not $n(-1)^{n}$ as the other answer states? –  Jun 25 '21 at 16:25
  • @user879295 Fixed now, added slightly more detail also. I was missing a -1 in Fourier coefficients $c_n(f)$ for $|n| \geq 2$. – spaceman Jun 28 '21 at 08:30
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$$f(x) \sim 1-\frac{1}{2}\cos(x)-2\sum_{k=2}^{\infty}\frac{(-1)^k}{(k^2-1)}\cos(kx)$$

Differentiating term-by-term:

$$f^\prime(x) \sim \frac{1}{2}\sin x + 2 \sum_{k=2}^\infty (-1)^k\frac{ k}{k^2-1} \sin k x $$

The series converges in $(-\pi,\pi)$ to $f^\prime(x)$, and to $\frac{1}{2} (f^\prime(\pi)+ f^\prime(-\pi))$ at $\{-\pi,\pi\}$ and converges to the periodic extension outside this interval.

You should see that this agrees with the answer by @Spaceman, namely $c_n(f^\prime) = - i n c_n(f),\, $ (with his definition of Fourier series which is $f(x) \sim \sum c_n e^{-inx}$).

mjw
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  • Got it. For pointwise convergence of the Fourier series of $g$ to $g$; I need toshow that $g$ is of "bounded variation" - what does that mean? –  Jun 17 '21 at 14:10
  • https://mathworld.wolfram.com/BoundedVariation.html – mjw Jun 17 '21 at 14:11
  • Ah, so when you say converges it is actually pointwise just because $f'(x)=0$ whenever $x=m\pi$? –  Jun 17 '21 at 14:28
  • https://mathworld.wolfram.com/DirichletFourierSeriesConditions.html – mjw Jun 17 '21 at 15:53
  • And the last series; Why is it $n(-1)^{n}$ and not $n(-1)^{n+1}$ as the other answer states? –  Jun 25 '21 at 16:27
  • @user879295, there is a sign error at the other answer. That's probably why. And only saying that very gently after checking with Mathematica. Of course, I may have made a mistake ... – mjw Jun 25 '21 at 17:37
  • Yes I am seeing that too. –  Jun 25 '21 at 17:43