MHB Help with Proof: Sum of Cosines Up to n Terms

  • Thread starter Thread starter shen07
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary
The discussion focuses on deriving the formula for the sum of cosines up to n terms, specifically $$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$. The method involves using the substitution $$z=e^{ix}$$ and manipulating the expression $$ (1+z)^n $$ to arrive at the final result. The key step is to extract the real part of the expression after substituting, leading to the conclusion that the answer is indeed the real part of $$2^n e^{\frac{ixn}{2}}\cos^n \left(\frac{x}{2}\right)$$. The thread emphasizes the importance of sharing solutions for the benefit of others in the forum.
shen07
Messages
54
Reaction score
0
$$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$
 
Physics news on Phys.org
I also know that i have to use

$$
z=exp(ix)
$$
$$
(1+z)^n = 2^n cos^n (x/2) cos (nx/2) $$
 
Got the answer..
 
shen07 said:
Got the answer..

Hi shen07 (Wave),

Welcome to MHB! Sorry we couldn't help you quickly enough this time. I'm sure that in the future you'll find guidance with something you are stuck on. Care to share your answer so others may see it?

Jameson
 
The answer could be obtained by choosing the real part after substituting $z=e^{ix}$

Hence we have

$$\Large (1+e^{ix})^n= e^{\frac{ixn}{2}}\left(e^{\frac{-ix}{2}}+e^{\frac{ix}{2}}\right)^n= 2^n e^{\frac{ixn}{2}}\cos^n \left(\frac{x}{2}\right)$$

Clearly the answer is the real part of the above expression .
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
7
Views
2K