Find closed formulas for the following two sums:
1 + cosX + cos 2X + . . . + cos nX
sinX + sin 2X + . . . + sinnX
Hint: Use the sum of the geometric series with the general term eikX.
Hi Tonio,
Thanks for the help, I did what you said and I got the sum of both series:
by proving its a geometric series ...
$\displaystyle \sum= \frac{cos[(n+1)x] + isin[(n+1)x] -1}{cosx + isinx -1}$
What do you mean by separating real and imaginary parts, How can I do it here? and will it give me the sum of the original series:
sinx+sin2x+...+sinnx , even if i changed it to a kind of a complex seris?
Thanks,
Ohad.
It will give you the sum of BOTH series, cosines and sines:
$\displaystyle \frac{\cos (n+1)x+i\sin (n+1)x-1}{\cos x+i\sin x-1}$ $\displaystyle =\frac{(\cos (n+1)x-1)+i\sin (n+1)x}{(\cos x-1)+i\sin x}\,\frac{(\cos x-1)-i\sin x}{(\cos x-1)-i\sin x}=$
$\displaystyle =\frac{\cos (n+1)x(\cos x-1)+\sin (n+1)x\sin x}{(\cos x-1)^2+\sin^2x}$ $\displaystyle -\frac{\cos (n+1)x\sin x-\sin (n+1)x(\cos x-1)}{(\cos x-1)^2+\sin^2x}\,i$
And there you have the right side divided in real and imaginary part. Now use a little trigonometry to deduce that the real part is $\displaystyle \frac{\cos nx-\cos (n+1)x}{2(1-\cos x)}$ ,
and the imaginary part is $\displaystyle \frac{\sin nx - \sin (n+1)x}{2(1-\cos x)}$.
OTOH, in the left side we have the sum $\displaystyle \sum\limits_{k=1}^ne^{kxi}=\sum\limits_{k=0}^n(\co s kx+i\sin kx)$ , so again separate in real and imaginary parts this sum.
Tonio