Help with Proof: Sum of Cosines Up to n Terms

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Discussion Overview

The discussion revolves around a mathematical proof concerning the sum of cosines up to n terms, specifically exploring the expression $$\sum_{k=0}^{k=n}(nCk * cos(kx))$$ and its relation to other mathematical identities involving complex exponentials.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the initial expression for the sum of cosines and proposes a formula involving $$cos(nx/2)$$ and $$2cos(x/2)$$ raised to the power of n.
  • Another participant introduces the substitution $$z=exp(ix)$$ and relates it to the expression $$(1+z)^n$$, suggesting it leads to a cosine identity.
  • Two participants claim to have arrived at the answer but do not provide details on their methods.
  • A later reply indicates that the answer can be derived by taking the real part of a complex expression involving $$e^{ix}$$.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method or the final expression, as multiple approaches are presented without resolution of which is correct.

Contextual Notes

The discussion includes various mathematical transformations and substitutions, but the assumptions behind these steps and their implications are not fully explored or agreed upon.

shen07
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$$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$
 
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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 .
 

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