Trigonometric Polynomial Form: How Does the +1 Factor Fit in T_n(x)?

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SUMMARY

The discussion focuses on the trigonometric polynomial form T_n(x) and its representation as T_n(x)=\sum_{k=-n}^n c_k e^{ikx}. The specific equation T_n(x)=\left(\frac{1+\cos(t-a)}{b}\right)^n is analyzed, particularly the inclusion of the +1 factor. The contributor successfully derives the term \left(\frac{\cos(t-a)}{b}\right)^n by setting c_{\pm n}=\left(\frac{e^{\mp ia}}{2b}\right)^n and identifies that the +1 factor can be understood through the binomial theorem, leading to a clearer exponential representation of each term.

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quasar987
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Homework Statement


Maybe I'm just blind, but how is

T_n(x)=\left(\frac{1+\cos(t-a)}{b}\right)^n

of the form

T_n(x)=\sum_{k=-n}^nc_ke^{ikx}

?

I can get

\left(\frac{\cos(t-a)}{b}\right)^n

by setting

c_{\pm n}=\left(\frac{e^{\mp ia}}{2b}\right)^n

and the rest of the c_k= 0, but how does the +1 comes in?
 
Last edited:
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](1+ a)^n= 1+ a+ _nC_2a^2+ ...+ a^{n-1}+ a^n[/itex] by the binomial theorem. Now use what you give to write each term as an exponential.
 
Ah!

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