Evaluate sum(requires use of complex numbers)?

**fardeen_gen** · April 30th 2009, 06:15 AM

Evaluate:
$\sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\}$

Answer:

Spoiler:

I do not know how the problem is done. Any ideas?

**running-gag** · April 30th 2009, 07:56 AM

Hi

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

**fardeen_gen** · April 30th 2009, 07:44 PM

Originally Posted by running-gag

Hi

Are you sure about the answer ?
I do not find the same and your answer seems not to work with values like $\frac{\pi}{4}$ 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 $\frac{\pi}{4}$ as you said. Can you post the correct solution?

**running-gag** · April 30th 2009, 11:57 PM

Let $f(x) = \sin x + 3\sin 3x + 5\sin 5x + ... + (2n - 1)\sin \{(2n - 1)x\} = \sum_{k=1}^{n} (2k-1) \sin{(2k-1)x}$

Let F(x) one antiderivative of f

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

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

After some calculations I can find

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

But there are many other possible expressions for f(x)

**fardeen_gen** · April 30th 2009, 11:59 PM

We cannot get a single unique answer?

**running-gag** · May 1st 2009, 12:11 AM

If you use $\sin{(2n+1)x} = \sin(2nx) \cos x + \cos(2nx) \sin x$

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

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

in the below expression of f

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

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

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