In fourier series...?

Question

why is the complex fourier series "parts" e^p added from N= -infinity to infinity, whereas the real fourier series parts are added only from N=0 to N=infinity?

damn, it's so obvious now. Thanks guys, wish I could give you both best answers

Answer 1

The discrete real fourier series is of the form

∑(A Sin(nx) + B Cos(nx))

If n were to vary from -≈ to ≈, then we'd end up with:

∑(A' Sin(nx) + B' Cos(nx))

where n varies from 0 to ≈. In other words, the terms where n is negative aren't indepedent from the terms where n is positive.

On the other hand, the discrete complex fourier series is of the form:

∑(C e^(inx))

so that e^(inx) and e^(-inx) do not add or cancel out. Thus, all values of n from -∞ to ∞ are independent "orthogonal" components.

Answer 2

What R U doin buddy? This isn't rocket science, the science of shooting rockets into black holes in an attempt to compress the universe into a 1cm by 1 mole cube of butane. No, this is far more similar to laserology and laseronomy. the medical practice of curing the common cold by firing a series of laser bursts at the patient's skull. Assuming amputation of the butt was done beforehand.

Answer 3

They are the exact same thing. Suppose you are using a_ne^{-inx) from -infinity to + infinity

Then the "n" term is
a_n cos nx + i a_n sin nx

you also have the "-n" term which is
a_{-n} cos (-nx) + i a_{-n} sin (-nx)
= a_{-n} cos (nx) - i a_{-n} sin (-nx)

Combining these terms you have
a_n e^{-nx} + a_{-n} e^{nx}
= a_n cos nx + i a_n sin nx
+ a_{-n} cos (nx) - i a_{-n} sin (nx)
= (a_n + {-n}) cos(nx) + i (a_n - a_{-n}) sin (nx)

so you see, you can take the series from -infinity to + infinity of e^inx and turn it into a series from 0 to infinity of sines and cosines.

Answer 4

mhhm mhhm mhm mhmm
yes yes very intresting the only logical explanation is as you see below
f=dt=h-d*kfkh jg;ih a;ofughiah;oahio'=N
for shorter version n=pudding