Why are these $\sum \cos$ and $\csc$ equivalent?

Question

Mathematica 'simplifies' this formula

$$\sum_{k=1}^R \cos \frac{2k \pi x}{R}$$

to this

$$\frac{1}{2} \biggl(\csc \frac{\pi x}{R} \sin \frac{(2R+1) \pi x}{R}-1\biggr)$$

A graphical plot of the two formulae generates two identical continuous curves - but why?

Surely $\csc \frac{\pi x}{R}$ is discontinuous, with poles at $x={0,R,2R,3R...}$? So, how can $\frac{1}{2} \bigl(\csc \frac{\pi x}{R} \sin \frac{(2R+1) \pi x}{R}-1\bigr)$ produce a continuous curve?

I'd be grateful for:

A proof of this equivalence
An explanation for why the resulting curve is continuous despite $\csc \frac{\pi x}{R}$ being discontinuous.

I get the poles match the zeroes of the sine, because $\csc x=\frac{1}{\sin x}$... But how does multiplying this by a cosine function create a continuous curve? — Richard Burke-Ward, Apr 18 '18 at 14:26
I take your point, but surely $x/x$ is discontinuous if $x$ is undefined or infinite? — Richard Burke-Ward, Apr 18 '18 at 14:50

cansomeonehelpmeout · Accepted Answer · 2018-04-18T14:51:50.840

1

Assuming you're a bit familiar with complex numbers: $$e^{ix}=\cos(x)+i\sin(x)\tag{1}$$ Let $a=\frac{2\pi}{R}$, then you want to find the sum $$\sum_{k=1}^{R}\cos(akx)\tag{2}$$ Notice that it is way easier to first compute (which we later can relate to $\cos(akx)$) $$\sum_{k=1}^{R}e^{akix}=\sum_{k=1}^{R}\left (e^{aix}\right )^k\tag{3}$$ which is a geometric series. This evaluates to $$\sum_{k=0}^R \left (e^{aix}\right )^k=\frac{e^{aix(R+1)}-1}{e^{aix}-1}\tag{4}$$

But since your series starts at $k=1$ we have to subtract the first term, so that $$\sum_{k=1}^R\left (e^{aix}\right )^k=\frac{e^{aix(R+1)}-1}{e^{aix}-1}-1\tag{5}$$

Now substitute eq. $(1)$ back into $(5)$ to get $$\sum_{k=1}^R\cos(akx)+i\sum_{k=1}^R\sin(akx)=\dots$$ Can you take it from here?

edited Apr 18 '18 at 14:51

answered Apr 18 '18 at 14:25

cansomeonehelpmeout

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I think it might be safer to treat me as a total noob! I know (1), but can't follow (2). (Also, see note about the sum being from $k=1$.) – Richard Burke-Ward Apr 18 '18 at 14:36
@RichardBurke-Ward The second one is a finite geometric series. – John Apr 18 '18 at 14:43
OK, I am sure that I can put this lot together and get there! I appreciate your patience. – Richard Burke-Ward Apr 18 '18 at 14:52
@Jam Sorry, I edited my answer, so that the tags changed. – cansomeonehelpmeout Apr 18 '18 at 14:52
@cansomeonehelpmeout No worries! – Jam Apr 18 '18 at 14:53
@RichardBurke-Ward Do that! Feel free to comment if you get stuck, and I'll try to help you more! – cansomeonehelpmeout Apr 18 '18 at 14:55

score 1 · Answer 2 · answered Apr 18 '18 at 14:25

1

Hint:

Evaluate $$\Re\left(\sum_{k=0}^Re^{i2\pi kx/R}\right)=\Re\left(\sum_{k=0}^R\left(e^{i2\pi x/R}\right)^k\right),$$ which involves a geometric sum.

answered Apr 18 '18 at 14:25

Could you tell me what that gothic-looking R stands for? – Richard Burke-Ward Apr 18 '18 at 14:30
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@RichardBurke-Ward: real part. – Apr 18 '18 at 14:32
Humble apologies, but I got the OP wrong. It's the sum from $k=1$... Have amended the question. Does it affect your hints? – Richard Burke-Ward Apr 18 '18 at 14:35
@RichardBurke-Ward: obviously not. – Apr 18 '18 at 14:36

score 1 · Answer 3 · answered Apr 18 '18 at 14:37

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This is because of the well-known formula for the sum of cosines of angles in arithmetic progression: $$1+\cos \theta+\cos 2\theta+\dots\cos R\theta=\frac{\sin\dfrac{R+1}2\,\theta}{\sin\cfrac\theta2}\,\cos\frac{R\theta}2,$$ where we set $\;\theta=\dfrac{2\pi x}R$, and using a linearisation trigonometry formula.

answered Apr 18 '18 at 14:37

Bernard

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OK. How come there are no discontinuities from $\sin \frac{\theta}{2}=0$? – Richard Burke-Ward Apr 18 '18 at 14:43
It's purely formal. Of course the formula is valid only if $\theta\not\equiv 0 \mod 2\pi$. – Bernard Apr 18 '18 at 14:47
So, if I am trying to find a way to simplify the original formula that is analytically continuous - i.e., has a meaningful value for all real $x$, then the $\csc$ simplification is a bad idea...? – Richard Burke-Ward Apr 18 '18 at 14:49
You have to determine whether there's a limit at the poles At $0$, one obtains readily a limit equal to $\frac{R+1}{2R}$. Personally I would stick to a formula with sines and cosines, as we have more tools. – Bernard Apr 18 '18 at 14:58

Why are these $\sum \cos$ and $\csc$ equivalent?

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