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I came accross an interpretation of complex fourier series as vector sum of vectors described by complex numbers $ e^{inx}$, where n ranges over integers.

Although I understand this form but I am not able to account for the following points:

1) While transitioning from cosine or sine wave to complex, it is just the real part that makes sense, right? (This is in context to the interpretation and not the formal proof)

2) If point 1 is true, why don't we take just the real part of the series while writing it? Are we sure that the series always gives out real value?

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Yes, the final result is real.

In fact the big difference with Fourier series expressed with real coefficients $a_n$ and $b_n$ is that indices $n$ of the $c_n$ coefficients, instead of varying between $0$ (or $1$) and $+\infty$ vary between $-\infty$ and $+\infty$. Therefore, opposite index coefficients $c_{-n}$ and $c_n$ are conjugate one to the other, therefore "cooperate" to yield finally real numbers.

Indeed, from the formulas :

$$c_0=constant, \ \ \ c_n=\tfrac12(a_n-ib_n), \ \ \ \ \ c_{-n}=\tfrac12(a_n+ib_n) \ \ \text{for} \ n>0$$

one deduces :

$$c_ne^{inx}+c_{-n}e^{-inx}=\tfrac12(a_n-ib_n)e^{inx}+\tfrac12(a_n+ib_n)e^{-inx}=\tfrac12 \left(a_n\underbrace{(e^{inx}+e^{-inx})}_{2 \cos x}-ib_n\underbrace{(e^{inx}-e^{-inx})}_{2i \sin x} \right)$$

and we find back in this way the usual formulas.

See the nice animations there explaining the interest of considering complex Fourier coefficients

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  • $\begingroup$ Yes, that is the interpretation I made about the series, as is depicted in the animation. Essentially, the animation shows how vectors are added to yield another vector and while plotting the resultant function they just plot the real part of that vector which makes complete sense. But, while writing the series, we don't take the real part of the series but instead use it in its complex form to which you answered that the negative exponents take care of. Can you please help me understand how? $\endgroup$ – SAGALPREET SINGH May 19 '20 at 9:59
  • $\begingroup$ To be precise about the question, I can't understand why should complex part of the resultant series be $0$. Also , in the animation, the resultant part isn't $0$ for all values of $theta$ $\endgroup$ – SAGALPREET SINGH May 19 '20 at 10:03
  • $\begingroup$ I am going to write it in my text. $\endgroup$ – Jean Marie May 19 '20 at 10:07
  • $\begingroup$ Your remark about the animation : I agree that it is a little misleading because we have to take the imaginary part to retrieve a Fourier series, an operation we shouldn't have to do. $\endgroup$ – Jean Marie May 19 '20 at 10:21

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