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Consider the sum

\begin{equation} S = \sum_{i=1}^N\cos(\kappa i) \end{equation} where $\kappa$ is an irrational number and $N$ is very large ($\mathcal{O}(10^{23})$ as this is a physics problem). In some sense, I think of the sum as "zero" because with $\phi=ki\mod 2\pi$ you sample $\phi\in[0,2\pi]$ densely. But this is "Physics maths". I can see that the sum doesn't actually converge, but was wondering whether anything can be proved about its lack of convergence, e.g. bounds on the partial sum (I don't hope for an analytic expression for the partial sum).

dsfkgjn
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  • This is by no means an answer but maybe heuristics as to why you could rather attach the number $-1/2$: Consider $f(x) = \frac{1}{2}(\pi-x)$ if $x\in[0,2\pi)$ and consider the $2\pi$-periodic extension of $f$. Then the Fourier-Series of $f$ is given by $$\sum_{k=1}^\infty \frac{\sin(kx)}{k}.$$ Differentiation $f$ gives $f'(x)=-1/2$ almost everywhere and the termwise differentiated Fourier series, (which does not converge) is formally given by $$x\mapsto \sum_{k=1}^\infty \cos(kx).$$ – frog Apr 13 '21 at 09:31
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    Actually, you can rewrite this sum as $$ \sum_{i=1}^N\cos(\kappa i) = {\sin{N\kappa\over2} \over \sin{\kappa \over 2}} \cos\left({(N+1)\kappa\over2}\right) $$

    See https://math.stackexchange.com/a/152025/521468

     – Yalikesi Apr 13 '21 at 09:37
  • That's completely correct, not sure how I missed that. – dsfkgjn Apr 13 '21 at 12:50

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