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5050057 No.5050057 [Reply] [Original]

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<div class="math"> \sum\limits_{i=1}^n sin((2i - 1)x) = cos(x) + cos(3x) + cos(5x) + ... +cos((2n-1)x) = \frac {sin(nx)cos(nx)}{sin(x)}</div>

>> No.5050078

No.

>> No.5050093

the middle=RHS can be done easily by induction, I think

Can't be fucked to go through it

>> No.5050098

I'm guessing you meant to write sin instead of cos for the very left hand side

>> No.5050102

>>5050098
*cos instead of sin

>> No.5050105

I think this should help:

2cosAsinB=sin(A+B)-sin(A-B)

>> No.5050171

>>5050098
Yeah, its cos((2i-1)x) on the left side.

>>5050093
Induction is the first thing I tried. Great minds think alike!

>>5050105
ITS NOT HELPING

>> No.5050197

Hi OP,
I'll have a go at the middle to RHS.

Notice cos(nx) = Re( exp(inx)), rewrite the middle equation as Re( exp(ix) + exp(3ix) + ...).

You will get a geometric series with some work, look at its convergence, and take Re part after, see if you get RHS. Good luck, hope it works out.

>> No.5050226

>>5050197
I've been trying this for pages, I cant get it to work out. Can you maybe point me in the right direction?

>> No.5050265

>>5050226
Ok.

Seeing now that >>5050171 says cos((2j-1)x) on LHS, lets start from here.

\sum cos((2j-1)x)
= Re[ \sum exp( i(2j-1)x )]
= Re[ \sum exp(-ix)*(exp( 2ix )^j) ]
Now we are summing from 1 to n, find the geometric series formula for this. (Assume convergence)
> http://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series

Use the one from k=m to n, where you put m=1, and n=n. Here r=exp(2ix), and a=exp(-ix), where you are summing over j.

After you've done that, take the real part of your expression. You should get RHS. Good luck

>> No.5050275

>>5050265
whoops, meant this article
> http://en.wikipedia.org/wiki/Geometric_progression

under geometric series.

>> No.5050316

>>5050265
Thanks.

Almost got it. Reduced it to cos(x)sin^2(nx)/sin^2(x). Must have made an algebra mistake somewhere. Oh well, its 2:30 here, I;m going to bed.