# Evaluate sum(requires use of complex numbers)?

• Apr 30th 2009, 06:15 AM
fardeen_gen
Evaluate sum(requires use of complex numbers)?
Evaluate:
$\displaystyle \sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\}$

Spoiler:
$\displaystyle \frac{2n\sin(2nx) - (2n - 1)\sin\{(2n + 1)x\} - 1}{4\sin^2 \frac{x}{2}}$

I do not know how the problem is done. Any ideas?
• Apr 30th 2009, 07:56 AM
running-gag
Hi

I do not find the same and your answer seems not to work with values like $\displaystyle \frac{\pi}{4}$ for instance ...
• Apr 30th 2009, 07:44 PM
fardeen_gen
Quote:

Originally Posted by running-gag
Hi

I do not find the same and your answer seems not to work with values like $\displaystyle \frac{\pi}{4}$ for instance ...

Hello running-gag,
I checked the answer in the text again and it is the apparent answer. But it doesn't work with $\displaystyle \frac{\pi}{4}$ as you said. Can you post the correct solution?
• Apr 30th 2009, 11:57 PM
running-gag
Let $\displaystyle f(x) = \sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\} = \sum_{k=1}^{n} (2k-1) \sin{(2k-1)x}$

Let F(x) one antiderivative of f

Then $\displaystyle F(x) = -\sum_{k=1}^{n} \cos{(2k-1)x}$

$\displaystyle F(x) = -Re\left(\sum_{k=1}^{n} e^{i(2k-1)x}\right)$

After some calculations I can find

$\displaystyle f(x) = \frac{\sin(2nx) \cos x - 2n \cos(2nx) \sin x}{2 \sin^2x}$

But there are many other possible expressions for f(x)
• Apr 30th 2009, 11:59 PM
fardeen_gen
We cannot get a single unique answer?
• May 1st 2009, 12:11 AM
running-gag
If you use $\displaystyle \sin{(2n+1)x} = \sin(2nx) \cos x + \cos(2nx) \sin x$

you can substitute $\displaystyle \sin(2nx) \cos x = \sin{(2n+1)x} - \cos(2nx) \sin x$

or even $\displaystyle \cos(2nx) \sin x = \sin{(2n+1)x} - \sin(2nx) \cos x$

in the below expression of f

$\displaystyle f(x) = \frac{\sin(2nx) \cos x - 2n \cos(2nx) \sin x}{2 \sin^2x}$

At first I thought that this was the reason why I did not find the same result as yours but then I realized that your solution is not correct