How to formulate arbitrary complex trigonometric polynomial? I know that in real form it is $\displaystyle\sum_{n=1}^k a_n\cos(nx)+b_n\sin(nx)$
How to formulate arbitrary complex trigonometric polynomial?
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fourier-analysis
1 Answers
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$\cos(nx)=\frac{1}{2}(e^{inx}+e^{-inx})$, and $\sin(nx)=\frac{1}{2i}(e^{inx}-e^{-inx})$, so you have a sum of $A_{n}e^{inx}+B_{n}e^{-inx}$, where $A_{n},B_{n}$ are new constants. Then you can just write it as $\sum_{n=-\infty}^{\infty}{c_{n}e^{inx}},$ for some $c_{n}$.
Clearly, this sum includes the possibility of a constant term, which you don't have in your original sum, but you can just set $c_{n}$=0. It should be easy enough to find a formula for $c_{n}$ in terms of $a_{n}$ and $b_{n}$ if you want to.