Conference rusure::math

>    Can someone give me a strategy for proving this identity?

Try substituting the integral definition of X() and interchanging order of
summation and integration.

Dick

    Also, don't forget the geometric significance of the relation.

    It states that the "length" of the function, viewed as an
    infinite dimensional vector, is unchanged by Fourier transformation.

    So, in a sense the Fourier transform is a kind of rigid rotation of
    coordinates much like a 3 dimensional rotation in space.  It's just
    another angle to look at the function from.

    - Jim

    Consider the right hand side (RHS) as a function which can be
    represented as a Fourier Series:
    
       RHS  =  SUM { a(n) exp(-jwn) }
    
    To find a(n), substitute RHS into the inverse Fourier integral
    
       a(n)  =  1/2p  int { RHS exp(jwn) dw } where the integral
                                              is over one period (-p,p)
    
    After performing this integration, you'll find that a(n) is the
    sampled time series x(nT).
    
    Although it might appear circular, what you've done is express the RHS
    as a Fourier Series and the integration determines the coefficients of
    the series.
    
    Note:  My background is in applied mathematics and digital signal
    processing.  If you have any further questions, just give me a call:
    dtn  327 3208.
    David Gertler

	The formula follows from 1st verfying the simpler case
of T=1 and f=0. This result is verified by noting that the 
sum, S[x](t), of the translates, x(t+n), of x(t) over all integral n, is
periodic, **IF** such a sum converges. Assuming that this is the
case, **AND** that the Fourier series associated to S[x](t) converges
to S[x](t), the formula follows expanding S[x](t) in its Fourier
series, then specializing at t=o and finally cranking
out the Fourier coefficients of S[x](t).
(This latter fact can be stated by saying that the value at 0 
of any "sufficently regular" periodic function can be recovered
by summing its Fourier coefficients.)

One important
class of functions for which these conditions hold are the
(Laurant) Schwartz functions (also known as the "test" functions). They
are defined as those functions, x(t), defined on the real line which
are infinitely differentiable and such that x(t) and any of its
derivatives rapidly decrease, i.e., 

			    n  (m)
	       lim         t  x  (t)  = 0  for any n,m >= 0
	t -> +inf or -inf

A non-trivial example of such functions are the Gaussian
/Normal probability density functions , e.g., 

	pdf (N(u,s)) (t) = (1/sqrt(2*pi*s))exp(-(t-u)^2/s))

Furthermore the linear space generated by Gaussians is invariant
under taking Fourier transforms. These functions and the sum of
their translates, and the Poisson Summation Formula play an important
role in the theory of the heat equation via theta functions as
well as in "trace formula's" in the arithmetic theory of unitary 
representations of Lie Groups, in particular to applications to 
calculating the dimension of the space of automorphic functions. 
(I can provide references if anyone is interested this kind of
stuff.)

Getting back to the result for T=1, f=0,
the Poisson-Summation Formula says that
the distribution, d(Z), corresponding to summing a function 
at its values on the integer lattice, Z, is invariant under 
the action of the Fourier transform. Specifically for an
arbitrary distribution, d, its Fourier transform d-"hat"
is defined by:

	^             ^            ^
	d (x(t)) = d (x(t)), where x(t) is the Fourier transform

of x(t), i.e.,

                 +inf
        ^        __      -i*2*pi*ts
	x(t) =  /   x(s)e           d(s)
              --
              -inf

Symbolically, Poisson Summation Formula says:

	 ^
	d(Z) = d(Z)
							  
More generally for any lattice L, one can define its dual
lattice, L-"star" , with respect to the pairing, Q, on the 
reals defined by:

	Q(x,y) = exp(2*pi*i*xy)

as
	 *
	L  = { w real : Q(w,l) = 1 for all l in L}.

Letting d(L) denote the distribution obtained by summing a function
over its values on L, it follows easily that the Fourier transform of the
this distribution associated to L is just the distribution associated
to its dual:

         ^        *
	d(L) = d(L ).

                                                *
Specifically, for T != 0, the dual lattice, (TZ)  of 
the T-scaled integer lattice, TZ, is just (1/T)Z. 

Finally for arbitrary frequency, f, the formula in the base note
is not correct and must be modified by replacing f on the RHS
with -f. The result now follows by noting that the Fourier transform
changes multiplication of a function by the 2*pi*-f harmonic into 
translation of its spectrum by f.


-Franklin
    

    re -1
    
    Usually "test" functions refer to infinitely differentiable functions
    with compact support rather than rapidly decreasing functions.  See,
    for example, Hoermander (it has an umlaut but I can't make it from
    home).