# To sum n terms

• Dec 18th 2008, 06:18 PM
Almo
To 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?

Cheers
• Dec 18th 2008, 06:36 PM
Skalkaz
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, 07:07 PM
Prove It
Quote:

Originally Posted by Almo
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?

Cheers

I take it that $i$ in this case is your counter, not the imaginary number $i = \sqrt{-1}$?

$\sum_{i = 1}^n{ix\cos{ix}}$ equal right?
• Dec 18th 2008, 08:41 PM
Almo
G'day

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, 10:13 PM
Mathstud28
$\sum_{n=1}^{m}n\cdot x^n=\frac{x^{m+1}(mx-m-1)+x}{(x-1)^2}$
So your sum of $\sum_{n=1}^{m}xn\cos(nx)=x\Re\left\{\sum_{n=1}^{m} ne^{inx}\right\}$ and apply the above lemma with $x=e^{ix}$