Summations using complex numbers

[RK1992]

Homework Statement

express in terms of n and r:

\sum { ^{n} C _{r} } \times cos(rx)<br />
from r=0 up to n

Homework Equations

well i know de moivre's theorem
also, in class we have been finding ways to represents a series of cosines added up as the sum of a geometric series so i tried it on this but got stuck

The Attempt at a Solution

i tried summing two sequences:
C = 1 + n.cosx + nC2.cos(2x) + nC3.cos(3x) ...
and
S = n.i.sin(x) + n.i.C2.sin(2x) + n.i.C3.sin(3x) ...

so then computed C + iS and got:
c + iS = 1 + n[cos(x) + i.sin(x)] + nC2[cos(2x) + sin(sx)] + ...

then using eulers formula, you can make that into:
c + iS = 1 + n e^ix + nC2 e^2ix ...

but i don't see a common ratio between the terms...

am i going about this the wrong way? i don't think so seeing as the problem sheet is entitled "further pure 2 review sheet - complex numbers"... have i missed something?

 

[dextercioby]

Could you, please, fix your LaTeX code, cause it's very difficult to understand what you're asking help for ?

 

[RK1992]

do you mean there is a problem with my latex, or you want the whole thing in latex?

 

[dextercioby]

No, just the equation you need to prove.

 

[RK1992]

it looks fine to me now.. in case my latex won't work (it often doesn't refresh on my laptop), i need:

the sum of "nCr multiplied by cos(rx)" from r=0 up to r=n

 

[dextercioby]

Well, so you want

\sum_{r=0}^{n} nC_{r}\cos rx ? (please note my code and compare it to yours).

Well, who are the C_r-s ? They seem to mess things up.

 

[RK1992]

Well, so you want

\sum_{r=0}^{n} nC_{r}\cos rx ? (please note my code and compare it to yours).

Well, who are the C_r-s ? They seem to mess things up.

what you have written is fine except that nCr is take n choose r... defined:
nCr = n! / (r!(n-r)!

maybe you write it like a matrix (n r)?

is this:
\sum_{r=0}^{n} ( ^{n}_{r} )<br /> <br /> cos(rx)
how you would write what I am describing?

 

[dextercioby]

AAh, that's something else ! You got a good idea by wanting to calculate actually

S= \sum_{r=0}^{n} ( ^{n}_{r} )cos(rx) = \mathfrak{Re} \sum_{r=0}^{n} ( ^{n}_{r} )e^{irx} (1)

I don't see a solution to express the sum of products as a closed solution, even though you know the sum of individual terms in the products.

I would say that (1) is already a solution to your problem.

 

[RK1992]

AAh, that's something else ! You got a good idea by wanting to calculate actually

S= \sum_{r=0}^{n} ( ^{n}_{r} )cos(rx) = \mathfrak{Re} \sum_{r=0}^{n} ( ^{n}_{r} )e^{irx} (1)

I don't see a solution to express the sum of products as a closed solution, even though you know the sum of individual terms in the products.

I would say that (1) is already a solution to your problem.

i guess it is a solution. it asked for it in terms of n and r, and what youve written is in terms of functions, n and r... can it definitely not be simplified any more? i will be annoyed if it cant, I've been trying for about 2 hours :P

 

[dextercioby]

The r is summed over, S it will not depend on it. It will only depend on n and x, as required ("express in terms of n and x").

Still, did you miss out a \pi in the argument of \cos ?

 

[RK1992]

The r is summed over, S it will not depend on it. It will only depend on n and x, as required ("express in terms of n and x").

oops.. there's my error.. its supposed to be in terms of n and r, not x..

 

[RK1992]

okay, i did it today in my free :)

i had to do the C + jS series and realize that the result was the expansion of (1 + e^(jx) )^n , then use de moivre's theorem to express it in the form a+bi and then take the real part of that.

there was also a previous result in the question which id forgotten about which was: e^ix = 2cos(x/2).e^(ix/2) so yeah i got the final answer in the end :)