Hi,

iS THERE Any one out there who can provide an expression for the sum of n terms:

i*X* COS i*X

where X is in radians, and i varies from 1 to n?

Thanks in advance for your help.

Cheers

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- Dec 18th 2008, 05:18 PMAlmoTo sum n terms
Hi,

iS THERE Any one out there who can provide an expression for the sum of n terms:

i*X* COS i*X

where X is in radians, and i varies from 1 to n?

Thanks in advance for your help.

Cheers - Dec 18th 2008, 05:36 PMSkalkaz
Dear Almo,

you can put out the x front of a bracket, I mean:

sum i*x*cos(i*x) = x*sum i*cos(i*x)

It can be interesting eventuell the derivation of sin(i*x) is. - Dec 18th 2008, 06:07 PMProve It
- Dec 18th 2008, 07:41 PMAlmotHANKS FOR YOUR HELP
G'day

Many thanks for your interest.

Yes you are correct that i is NOT an imaginary number i.e. sqrt(-1).

It is as you say the sum of the terms whereby i is initialy 1 then increases by 1 to the limit n. - Dec 18th 2008, 09:13 PMMathstud28
How about this use the following lemma

$\displaystyle \sum_{n=1}^{m}n\cdot x^n=\frac{x^{m+1}(mx-m-1)+x}{(x-1)^2}$

So your sum of $\displaystyle \sum_{n=1}^{m}xn\cos(nx)=x\Re\left\{\sum_{n=1}^{m} ne^{inx}\right\}$ and apply the above lemma with $\displaystyle x=e^{ix}$ - Dec 19th 2008, 08:48 PMAlmoThanks
Thanks for your suggestion. I'm afraid I don't understand if your last line is my desired answer. Would you please rephrase your solution in terms of my i and X?

Cheers