Complex Fourier Series: n from -∞ to +∞?

[Tosh5457]

Hi, I don't understand why does n goes from -∞ to +∞ in the complex Fourier series, but it goes from n=1 to n=+∞ in the real Fourier series?

 

[EWH]

I'm not exactly sure either, (or even if that is strictly correct) but I think
http://en.wikipedia.org/wiki/Negative_frequency should give some insight.

http://en.wikipedia.org/wiki/Single-sideband_modulation is also relevant, but best to read the first link first

 

[HallsofIvy]

"Real Fourier Series" are in the form \sum a_ncos(nx)+ b_nsin(nx)<br /> cosine is an even function and sine is an odd function so that if we <b>did</b> use negative values for n, it wouldn't give us anything new: a_{-n}cos(-nx)+ b_n sin(-nx)= a_{-n}cos(nx)- b_{-n} sin(nx) and would can then combine that with the corresponding "n" term: ( a_n+ a_{-n})cos(nx)+ (b_n- b_{-n})sin(nx)<br /> <br /> Another, but equivalent, way of looking at it is that cos(nx)= (e^{inx}+ e^{-inx})/2 and sin(nx)= (e^{inx}+ e^{-inx})/2i so that sin(nx) and cosine(nx) with only positive n includes exponentials with both positive and negative n.

 

[Tosh5457]

"Real Fourier Series" are in the form \sum a_ncos(nx)+ b_nsin(nx)<br /> cosine is an even function and sine is an odd function so that if we <b>did</b> use negative values for n, it wouldn't give us anything new: a_{-n}cos(-nx)+ b_n sin(-nx)= a_{-n}cos(nx)- b_{-n} sin(nx) and would can then combine that with the corresponding "n" term: ( a_n+ a_{-n})cos(nx)+ (b_n- b_{-n})sin(nx)<br /> <br /> Another, but equivalent, way of looking at it is that cos(nx)= (e^{inx}+ e^{-inx})/2 and sin(nx)= (e^{inx}+ e^{-inx})/2i so that sin(nx) and cosine(nx) with only positive n includes exponentials with both positive and negative n.

<br /> <br /> Thanks <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />