At first glance that seems odd. If x is not 0 mod pi, for some large enough n then nx will approximate a multiple of 2pi to arbitrary precision. Given that large enough n, for all k <= n I'd suspect sin(kx) to be reasonably well distributed over [-1, 1] because I'd suspect kx mod 2pi to be well distributed too --- i.e., I'd like to believe this is some sort of Riemann partition in disguise. So why isn't the sum you're getting approaching zero? Of course the easiest explanation is that my suspicions about distribution are wrong (i.e. why do you say that clearly the original series converges only for x = n*pi?). However, just out of curiosity, how is that sum to small values of infinity being calculated? Could the values you're seeing be explained by floating point arithmetic behavior? Specifically I am wondering about the underlying implementation of sin(x) and its argument reduction mechanism. On 12/18/17 14:46 , Dan Asimov wrote:

I recently tried to make a plot of

(*) y = Sum_{1 <= n < oo} sin(nx)

for some small values of infinity, and it distinctly hovered around the apparent value

Sum_{1 <= n < oo} Im( exp(inx))

= Im(Sum_{1 <= n < oo} exp(inx))

=(???) Im( exp(ix) / (1 - exp(ix)) ) = sin(x) / (2 - 2*cos(x)).

Clearly the original series converges only for x = n*pi for some integer n.

Can this function

f(x) = sin(x) / (2 - 2*cos(x))

be the *Cesaro sum* of the original series (*) ???

Or at least for certain values of x ???

—Dan

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