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Question: What is the sum of this infinite series?: $\cos(x)-\cos(3x)+\cos(5x)-\cos(7x) +\cdots$

I know that $\sum \cos(nx)$ is something like $\sum \delta(x-k)$ and $\sum \sin(nx)$ is something like $\cot(k \pi x)$. I tried plotting this in wolfram-alpha and it looks like $\sec(x)$ or perhaps $\sec^2(x)$, but not sure. Any help?

Community
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Srini
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  • Wolfram Alpha https://www.wolframalpha.com/input/?i=sum+of+cos(4k-3)+-+cos(4k-1)+for+k+%3D+1+to+k+%3D+infinity shows that the sum does not converge by the root test. – Toby Mak Jun 25 '17 at 00:27
  • You can get to the sum by grouping each pair of + terms and - terms together, and then considering the pattern of the averages of the constant multipliers (the average of $\cos(x)$ and $-\cos(3x)$ would be $\cos(2x)$, and find a linear equation for that. – Toby Mak Jun 25 '17 at 00:28
  • @TobyMak Perhaps you should check your link, as I don't think WA says it diverges by the root test. Here's what I see: https://i.stack.imgur.com/pB04R.png – Simply Beautiful Art Jun 25 '17 at 00:40
  • @Simply Beautiful Art Changed the post – Toby Mak Jun 25 '17 at 07:09

4 Answers4

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Obviously it doesn't actually converge. We can compute the finite sums, $$ S_m(x) = \sum_{k=0}^m (-1)^k \cos{(2k+1)x} = \frac{1}{2}\sec{x}\big( 1+ (-1)^m \cos{2(m+1)x} \big), $$ by using, e.g., the formulae given in this question. So the function oscillates between $\sec{x}$ and $0$, and the oscillations become more rapid as $m$ increases.

We can, however, sum the series to a finite, well-behaved expression using Fejér sums: $$ \frac{1}{n+1} \sum_{m=0}^n S_n(x) = \frac{1}{2}\sec{x} + \frac{\cos{x}+(-1)^n\cos{(2n+3)x}}{4 (n+1)}\sec^2{x} \to \frac{1}{2}\sec{x} $$ as $n \to \infty$.

Chappers
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  • Fejér's sums? I am more familiar with Cesàro sums. – Simply Beautiful Art Jun 25 '17 at 00:50
  • In the context of trigonometric series, Fejér's name is often attached to the summation process, as well as the kernel and the theorem. – Chappers Jun 25 '17 at 00:53
  • Thanks. This does match my suspicion that it is sec x. The WA reference I made in my post is this: [link] (https://www.wolframalpha.com/input/?i=cosx+-+cos3x+%2B+cos5x+-+cos7x+%2B+cos9x+-+cos11x+%2B+cos13x+-+cos15x+%2B+cos17x+-+cos19x+%2B+cos21x+-+cos23x+%2B+cos25x+-+cos27x+%2B+cos29x+-+cos31x+%2B+cos33x+-+cos35x) – Srini Jun 25 '17 at 00:54
  • Hm, I see. Seems to be naturally the case. – Simply Beautiful Art Jun 25 '17 at 00:54
  • @Srini One needs to be careful about what "is" means here. – Chappers Jun 25 '17 at 00:56
  • @Srini Your link only concerns the partial sum of the infinite series. If this is what you meant, you should update your post to be more clear on this. – Simply Beautiful Art Jun 25 '17 at 00:57
  • No, no, it is clearly the inifinite sum I am after (very explicitly mentioned in the title). I just posted that link to say why my hunch was towards sec(x). I know that it is partial sum that I linked to. But not because I wanted partial sum, but because I didn't know how to input inifite sum functions – Srini Jun 25 '17 at 01:00
  • @Chappers You mean that the sum (as computed infinitely) tends towards sec(x)/2 right? Your reference to "is" is to clarify that it is a limit? – Srini Jun 25 '17 at 01:03
  • @Srini See the first line. It clearly says "Obviously it doesn't actually converge." More info may be found in the first link, or in my answer, which likewise shows this series does not converge in the ordinary sense. – Simply Beautiful Art Jun 25 '17 at 01:06
  • @Srini I mean that the ordinary partial sums do not converge to a limit, but the Fejér sums do converge to a limit (except at the singularities of $\sec{x}$, of course). – Chappers Jun 25 '17 at 01:18
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For $x=\frac{(2n+1)\pi}2,n\in\mathbb Z$, it is easy to see that

$$\sum_{n=0}^\infty(-1)^n\cos((2n+1)x)=0$$

Since every term is zero. However, if $x\ne\frac{(2n+1)\pi}2,n\in\mathbb Z$, then it suffices to prove that

$$\lim_{n\to\infty}\cos((2n+1)x)\ne0$$

Or,

$$\limsup_{n\to\infty}\cos((2n+1)x)=1$$

Hence, it diverges.

If one wishes to regularize this series to a finite values, then one may apply an Abel sum:

$$\begin{align}\operatorname{Abel}(S_x)&=\lim_{t\to-1^+}\sum_{n=0}^\infty t^n\cos((2n+1)x)\\&=\Re\left[\lim_{t\to-1^+}\sum_{n=0}^\infty t^ne^{(2n+1)ix}\right]\\&=\Re\left[\lim_{t\to-1^+}\frac{e^{ix}}{1-te^{2ix}}\right]\\&=\Re\left[\frac{e^{ix}}{1+e^{2ix}}\right]\\&=\frac12\sec(x)\end{align}$$

Simply Beautiful Art
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2

just a hint

The general term of your series is

$$u_n=(-1)^n\cos( (2n+1)x)$$

$$\lim_{n\to+\infty} u_n=0\iff x=\frac {\pi}{2}+k\pi $$

hamam_Abdallah
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For integer $n$, if $\cos x=1,\cos(2n+1)x=1$

and if $\cos x=-1,\cos(2n+1)x=-1$

In both cases, the series oscillates finitely between $-1,1$

If $\cos x=0,\cos(2n+1)x=0$ so, each term of the series $=0$

Else as $(-1)^n\cos(2n+1)x=\cos\{n\pi+(2n+1)x\}$

Using $\sum \cos$ when angles are in arithmetic progression,

$$\sum_{n=0}^{m-1}\cos\{n\pi+(2n+1)x\}=\dfrac{\cos\left\{x+(m-1)\left(\dfrac\pi2+x\right)\right\}\sin m\left(\dfrac\pi2+x\right)}{\sin\left(\dfrac\pi2+x\right)}=\cdots=\dfrac{\sin^2m\left(\dfrac\pi2+x\right)}{\cos x}$$

lab bhattacharjee
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