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I have the expression

Sum[(2 j \[Pi] Sin[(2 j \[Pi] x)/(r + 1)])/(r + 1)^2, {j, 0, r}] == 0

I want to find the roots in terms of x and r.

After using FullSimplify, and removing expressions that make no contribution to the zeroes (such as integer multiples and the denominator of the fraction that results from FullSimplify), I end up with

Sin[(\[Pi] x)/(r + 1)] Cos[(\[Pi] (x + 2 r x))/(r + 1)] + Sin[2 π x]==0

However, I'm not sure how to proceed from here. @KraZug, who was very kindly helping me here, says that the roots appear to have a closed form - and it is true that plotting the results for increasing r seem to imply some form of convergence... But I have absolutely no idea how to find an algebraic expression for the roots.

I have tried FindRoot, ToRadicals, Solve... But they all produce error messages and no result. I suspect part of this might be to do with a failure to take limits - but that actually should only affect roots at x=n r + 1 (where n is an integer).

Can anyone suggest how I should proceed?

Richard Burke-Ward
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1 Answers1

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Is this closed form by your defintion?

Reduce[Sin[(\[Pi] x)/(r + 1)] Cos[(\[Pi] (x + 2 r x))/(r + 1)] + 
   Sin[2 \[Pi] x] == 0]

$\left(\sin \left(\frac{\pi x}{r+1}\right)=0\land \sin (2 \pi x)=0\right)\lor \\ \left(\sin \left(\frac{\pi x}{r+1}\right)\neq 0\land \cos \left(\frac{\pi (2 r x+x)}{r+1}\right)=\sin (2 \pi x) \left(-\csc \left(\frac{\pi x}{r+1}\right)\right)\right)$

Johu
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  • Hi. Not sure what the V and inverted V mean... But when I try Reduce I get Reduce::naqs: 0 is not a quantified system of equations and inequalities. which then gives Reduce[0]. – Richard Burke-Ward Sep 08 '18 at 15:09
  • ∧ and ∨ are logical $\texttt{and}$ and $\texttt{or}$. Please add the exact code of your attempt to use Reduce to your question such that we could help. If you copy & paste mine, you should get the same output in the standard Mathematica output form. – Johu Sep 08 '18 at 15:25
  • My apologies. I must have made a typo somewhere. Yes, it works. Many thanks. – Richard Burke-Ward Sep 08 '18 at 15:46