Summation of trigonometric identities with the argument squared.

Asked Feb 02 '22 at 18:09

Active Feb 02 '22 at 18:09

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Currently trying to solve the following summations:

$$\sum_{i=1}^n \sin(i^2)$$ $$\sum_{i=1}^n \cos(i^2)$$

I know that the following summations have the solutions: $$\sum_{i=1}^n \sin(i) = \frac{1}{2}(\sin(n) - \cot(\frac{1}{2})\cos(n)+\cot(\frac{1}{2}))$$

and,

$$\sum_{i=1}^n \cos(i) = \frac{1}{2}(\cos(n) + \cot(\frac{1}{2})\sin(n)-1)$$

which were answered in this post. No clue where to even begin.

asked Feb 02 '22 at 18:09

K Bazan

3

I doubt there's a nice form for that. If you had $\sum_{k=0}^{N-1} \sin(2 \pi k^2/N)$ instead, you should be able to do it using Euler's formula and Gauss sums. – Hans Lundmark Feb 02 '22 at 18:18
What make you think that a closed form exists? – mathcounterexamples.net Feb 02 '22 at 18:25
@mathcounterexamples.net It is an assumption. I really don't know if a closed form exists. – K Bazan Feb 02 '22 at 18:28
You can get one sum basically in terms of the other, via $\cos{(k^2)} + i \sin{(k^2)} = e^{2 \pi i k^2}$, but that's on the page that you link to. – Matt Groff Feb 03 '22 at 03:37
Are you ok with an approximation? You should be able to get arbitrarily close if that's what you're after. – Matt Groff Feb 04 '22 at 03:19

0 Answers0