Qualitative relationship between a function f(x) and its fourier components In finding the reason for a definition of group velocity like this, $v_p=\frac {d\omega}{dk}$, I found a lecture note here http://www.hep.man.ac.uk/u/roger/PHYS10302/lecture15.pdf
In the last page of this note, the authors give a more general interpretation on the definition of group velocity using Fourier transformation. One thing I can't understand is the qualitative relationship between a function f(x) and its fourier components F(k). He says that 
I wonder why F(k) is broad and has a short wavepacket, and why the contributions all canceled away from the peak. 
My second question is that he says "An infinite sine wave has a well-defined wavelength and thus a well-defined k. F(k) is a delta function." In my opinion, for a sine wave, F(k) should be one delta funtion plus another delta function if we use $exp(e^{-ikx})$ as our base sets. 
 A: The inverse relationship between $\langle (x - \langle x\rangle)^2\rangle$ and $\langle (k - \langle k\rangle)^2\rangle$ is related to the fact that sine and cosine waves have infinite extent. So, the closer to zero the function is far from its middle, the more Fourier components you need for the approximation to average out to zero in those areas where sine and cosine of the low frequency components still have the same amplitude they had near the mean.
And, you are right, that a standing sine or cosine wave has two traveling wave frequency components; $\cos(kx) = (\operatorname{e}^{ikx} + \operatorname{e}^{-ikx})/2$, for example. You can go the other direction, though, and say that a traveling wave is constructed with two standing wave components: $\operatorname{e}^{ikx} = \cos(kx) + i \sin(kx).$
On a related note, you can work out from the facts that cosine is even and sine is odd the following statements:


*

*the real part of the Fourier transform is determined by the real-even and imaginary-odd parts of the function,

*the imaginary part of the Fourier transform is determined by the real-odd and imaginary-even parts of the function, and

*if a function is real, then its Fourier transform will have an even real part and an odd imaginary part.


More fun is to note the relationship between a function's moments, $\langle x^n \rangle$, and the coefficients in the Taylor expansion of the Fourier transform. When there aren't any problems with divergence or analyticity, and a symmetric Fourier transform is used:
$$\langle x^n \rangle = \left. \frac{\partial^n \tilde{f}(k)}{\partial k^n}\right|_{k=0},$$
and vice versa for $\langle k^n\rangle$ and the Taylor series of the function.
