Why in the Fourier series for $e^x,A_n=C_n+C_{-n}$ and $B_n=i(C_n+C_{-n})$, where $A_n,B_n$ are the coefficients of full Fourier series and $C_n$ is the coefficient of full Fourier series in complex form ?

Does this always holds, i.e. the coefficients of the full Fourier series can be obtained from $C_n$ ? A yes or no it's enough as an answer (I'll do the rest).

I already tried to add both coefficients I get




You have for $n \geq 0$: $$C_n e^{i n x}+C_{-n} e^{-i n x}=C_n \left( \cos(nx)+i\sin(nx) \right)+C_{-n} \left( \cos(nx)-i\sin(nx) \right)=\left( C_n +C_{-n} \right) \cos(nx)+i \left( C_n-C_{-n} \right) \sin(nx)$$ you can then sum over $n$.

  • $\begingroup$ As you give a hint I assume $B_n=i(C_n+C_{-n})$ and $A_n=(C_n+C_{-n})$ always hold? $\endgroup$ – MagicConchShell Apr 29 '18 at 3:13
  • 1
    $\begingroup$ @Isa Yes but it seems that $B_n=i(C_n \color{blue}{-}C_{-n})$ :-) $\endgroup$ – Delta-u Apr 29 '18 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.