How to prove $t+\sum\limits_{n=1}^{t-1} 2(t-n)\cos(nx) = \frac{1-\cos(tx)}{1-\cos(x)}$

Question

How to prove $t+\sum\limits_{n=1}^{t-1} 2(t-n)\cos(nx) = \dfrac{1-\cos(tx)}{1-\cos(x)}$

I am not so sure how to prove the about equation to be equal, should I use $\ 2\cos(x) = \exp(ix)+\exp(-ix)$ or identities for trigonometry, I tried both but not sure how to proceed.

One possibility could be to note that the $\sum$ part is just twice the real part of $\sum\limits_{n=1}^{t-1}(t-n)e^{inx}$. If you know the geometric series formula and how to find sums like $\sum\limits_{n=1}^{t-1} nz^n$, you could split up this sum and proceed from here. — Minus One-Twelfth, Mar 05 '19 at 12:38

Bernard · Answer 1 · 2019-03-05T21:28:22.023

1

Hint:

$$\sum_{n=1}^{t-1} \cos nx =\operatorname{Re}\Bigl(\sum_{n=1}^{t-1} \mathrm e^{inx}\Bigr), \qquad \sum_{n=1}^{t-1} n\cos nx =\sum_{n=1}^{t-1}(\sin nx)'=\operatorname{Im}\Bigl(\sum_{n=1}^{t-1} \mathrm e^{inx}\Bigr)'$$ so it comes down to computing the sum of consecutive terms of a geometric series.

edited Mar 05 '19 at 21:28

answered Mar 05 '19 at 12:38

Bernard

175,478

sorry, but what is the meaning of that ' at the end? – Davidfufu Mar 05 '19 at 12:44
It's the derivative with respect to x. And there is probably an imaginary part instead of a real part, since you get an $i$ by derivating. – Statistic Dean Mar 05 '19 at 12:47
Oops, I was out of my mind… Thanks for pointing it! – Bernard Mar 05 '19 at 13:01

How to prove $t+\sum\limits_{n=1}^{t-1} 2(t-n)\cos(nx) = \frac{1-\cos(tx)}{1-\cos(x)}$

1 Answers1