Which functions are their own Fourier transform?
David H. Bailey
transform is a scaled Gaussian distribution. From some checking this
appears to be true, although I can't seem to reconstruct a rigorous
proof. Does anyone know how this is proved? To be specific, I am
seeking solutions f(t) to the identity
f(x) = Integral_-Inf^Inf f(t) e^{-2 pi i t x} dt
Thanks,
David H. Bailey
dba...@nas.nasa.gov
Jacob Hirbawi
> I recall once hearing that the only function that is its own Fourier
> transform is a scaled Gaussian distribution. From some checking this
> appears to be true, although I can't seem to reconstruct a rigorous proof.
I don't think that the Gaussian function is the only function that is its
own Fourier Transform. I looked into this some time ago. Unfortunately the
only notes I can recover are these:
(1) Some functions which are invariant under the Fourier Transform:
exp( -pi x^2 ) ( Guassian )
sech( x ) ( sech(x) is the hyperbolic secant = 1/cosh(x) )
cosh(x/2) / cosh(x) ( cosh(x) is the hyperbloic cosine )
( you might want to verify these with a *good* integration package )
(2) The Prolate Spheroidal Functions (PSF) give a whole family of functions
that *almost* satisfy the requirement. The Fourier Transform of
a PSF is a PSF over a finite region. The PSF's are defined as the
eigen-functions of the integral equation:
sin(c t)/t (*) PSF(c,t) = e(c) PSF(c,t)
where f(t) (*) g(t) = integral( f(t-s)g(s) ds ) ( convolution )
I hope this helps.
Jacob Hirbawi
jcb...@CERF.NET
Allan Adler
I don't think that the Gaussian function is the only function that is its
own Fourier Transform. I looked into this some time ago. Unfortunately the
only notes I can recover are these:
(1) Some functions which are invariant under the Fourier Transform:
exp( -pi x^2 ) ( Guassian )
sech( x ) ( sech(x) is the hyperbolic secant = 1/cosh(x) )
cosh(x/2) / cosh(x) ( cosh(x) is the hyperbloic cosine )
( you might want to verify these with a *good* integration package )
(2) The Prolate Spheroidal Functions (PSF) give a whole family of functions
that *almost* satisfy the requirement. The Fourier Transform of
a PSF is a PSF over a finite region. The PSF's are defined as the
eigen-functions of the integral equation:
sin(c t)/t (*) PSF(c,t) = e(c) PSF(c,t)
where f(t) (*) g(t) = integral( f(t-s)g(s) ds ) ( convolution )
On a self-dual locally compact abelian group G with its self-dual measure
(e.g. the additive group of reals), the Fourier transform has order 4
as an operator on L2(G), i.e. F(F(F(F(g)))) = g for all g in L2(G),
where F is Fourier Transform. Therefore, L2(G) is the direct sum of
the spaces H0,H1,H2,H3 where Hn is the i^n-eigenspace of Fourier transform.
As a corollary of all this, if g is any element of L2(G),
g+F(g)+F(F(g))+F(F(F(g))) is its own Fourier transform.
Allan Adler
a...@altdorf.ai.mit.edu
Vincent Broman
eigenvalues of 1, i, -1, -i. After suitably adjusting the scale
factors, one finds that the subspace of eigenvectors associated with
the eigenvalue of 1 has a basis of Hermite polynomials of order 4*n
multiplied by a gaussian.
Vincent Broman, code 572 bayside, = o
Naval Command Control and Ocean Surveillance Center, RDT&E Div. = _ /- _
San Diego, CA 92152-5000, USA = (_)> (_)
Internet Email: bro...@nosc.mil Fax: +1 619 553 1635 Phone: +1 619 553 1641
Fred Schwab
See also E. M. Stein, ``Harmonic Analysis on R^n'', in J. M. Ash, Ed.,
Studies in Harmonic Analysis, MAA Studies in Math Vol. 13, 1976.
In particular, on pp. 104-105 the Fourier self-transform properties of
the spherical harmonic polynomials in R^n are described.
For functions which, like the PSWF, are finite Fourier self-transforms,
see Fedowtowsky & Boivin, ``Finite Fourier Self-Transforms'',
Quart. Appl. Math, 30 (1972) 236. (A finite Fourier self-transform
is a function which, if truncated (i.e., set to zero outside of an
interval), and Fourier transformed, replicates itself (apart from
some scaling factor).
John C. Baez
>In article <920622060...@nic.cerf.net> jcb...@CERF.NET (Jacob Hirbawi) writes:
>
> I don't think that the Gaussian function is the only function that is its
> own Fourier Transform.
Yes; as Allan notes there is an infinite-dimensional space of such
functions.
>On a self-dual locally compact abelian group G with its self-dual measure
>(e.g. the additive group of reals), the Fourier transform has order 4
>as an operator on L2(G), i.e. F(F(F(F(g)))) = g for all g in L2(G),
>where F is Fourier Transform. Therefore, L2(G) is the direct sum of
>the spaces H0,H1,H2,H3 where Hn is the i^n-eigenspace of Fourier transform.
>
>As a corollary of all this, if g is any element of L2(G),
>g+F(g)+F(F(g))+F(F(F(g))) is its own Fourier transform.
In the case when the group is R (or R^n, but lets keep life simple)
there is quite explicit way to construct an orthonormal basis of the
space of L2 functions which are their own Fourier transform. Start with
the (correctly scaled, and normalized) Gaussian, which we'll call |0>,
following the notation of physicists. Then define operators
q = multiplication by x
p = -i d/dx
and then define creation and annihilation operators
a = (q+ip)/sqrt(2)
c = (q-ip)/sqrt(2) .
(If you have chosen a perverse definition of the Fourier transform,
so that a perversely scaled Gaussian is its own Fourier transform, you
will have to rescale p and q to make things work out.)
Now define elements of L^2 recursively by setting
|n+1> = c|n>/sqrt(n+1).
Up to all sorts of normalizations and differences in convention, the
function |n> is just your Gaussian times the nth Hermite polynomial.
The functions |n> form an orthonormal basis of L2, and if F denotes the
Fourier transform one has
F|n> = (i)^n |n>
There's a slight chance that should be a -i, but who can tell the
difference between i and -i, anyway? :-)
In any event, it follows that the functions |n> where n is a multiple of
4 form an orthonormal basis of the space of functions who are their own
Fourier transform.
The theory behind all this is, in my opinion, far more interesting than
the question of which functions are their own Fourier transform. This
theory is variously called quantum mechanics, quantum field theory, the
free boson field, the canonical commutation relations, the Weyl algebra,
the Heisenberg group, and so on. I recommend taking a look at Folland's
Harmonic Analysis on Phase Space.
Much of this can be generalized to any locally compact abelian group.
Victor Miller
David> I recall once hearing that the only function that is its own Fourier
David> transform is a scaled Gaussian distribution. From some checking this
David> appears to be true, although I can't seem to reconstruct a rigorous
David> proof. Does anyone know how this is proved? To be specific, I am
David> seeking solutions f(t) to the identity
David> f(x) = Integral_-Inf^Inf f(t) e^{-2 pi i t x} dt
David> Thanks,
David> David H. Bailey
David> dba...@nas.nasa.gov
David, As pointed out by others it certainly isn't unique.
Nevertheless, it probably could be made unique by specifying other
properties (such as fast decay). I would suspect that this is related
to the "Central limit Theorem":
I $f$ is a non-negative summable function with $\integral f(x)dx = 1$,
$\integral xf(x)dx = 0$ and $\integral x^2 f(x)dx = 1$, and $f^n$ is
the n-fold convolution of $f$, then
$\lim_{n \rightar \infty} \integral_{a\sqrt{n}}^{b\sqrt{n}} f^n(x)dx =
\integral_a^b \frac{\exp(-x^2/2)){\sqrt{2\pi}}$
for and $-\intfy<a<b<\infty$. For the usual probabilistic
interpretation, one takes for the $f$ the density function of a random
variable $X$ with mean 0, and variance 1. Then $f^n$ is the density
function of the sum of $n$ copies of $X$ independently distributed.
One needs to scale this sum by $\sqrt{n}$ to make sure that things
have a chance of converging. If this approaches a limit at all, it
seems "clear" that the distibution should be closely related to one
with property that you seek.
--
Victor S. Miller
Internet: vic...@watson.ibm.com
IBM, TJ Watson Research Center
Moti M
chap. 5,
The specific answer is that the gauss hermite polynoms are invariant
to within a scaling factor to the fourier transform
the proff is easily derived by inserting the diffrential deffinition
of the polynoms into the integral equation, since the polynoms
constitute an orthonormal basis of L^2 and have four eigenvalues
then any linear combination of them belonging to the same eigen group is
also invariant
hope this helped
moti
Noam Elkies
>On a self-dual locally compact abelian group G with its self-dual measure
>(e.g. the additive group of reals), the Fourier transform has order 4
>as an operator on L2(G), i.e. F(F(F(F(g)))) = g for all g in L2(G),
>where F is Fourier Transform. Therefore, L2(G) is the direct sum of
>the spaces H0,H1,H2,H3 where Hn is the i^n-eigenspace of Fourier transform.
>
>As a corollary of all this, if g is any element of L2(G),
>g+F(g)+F(F(g))+F(F(F(g))) is its own Fourier transform.
All true, though I suppose it leaves open the possibility that some of the
Hi are finite-dimensional. In fact they are all infinite dimensional:
the n-th Hermite function is in the i^n eigenspace (or i^(-n) depending
on how you normalize the Fourier transform). The n-th Hermite function
is essentially a Gaussian multiplied by the n-th Hermite polynomial
(I write "essentially" because the exact formula depends on where you put
the 2*pi factors in your definition of the Fourier Transform); in particular
the zeroth Hermite function is just the Gaussian itself, and for n=0,1,2,3
the i^n-eigenspace of the Fourier transform has an orthogonal basis
consisting of the (n+4k)-th Hermite functions (k=0,1,2,...). People
familiar with the (1-dimensional, nonrelativistic) quantum harmonic
oscillator already know this stuff, at least implicitly: the Fourier
transform of the wave function at time t is the wave function at time
(t+tau/4) where tau is the period (either classical or quantum!) of
the harmonic oscillator.
--Noam D. Elkies (elk...@zariski.harvard.edu)
Dept. of Mathematics, Harvard University
Herman Rubin
> David H. Bailey writes
> > I recall once hearing that the only function that is its own Fourier
> > transform is a scaled Gaussian distribution. From some checking this
> > appears to be true, although I can't seem to reconstruct a rigorous proof.
> I don't think that the Gaussian function is the only function that is its
> own Fourier Transform. I looked into this some time ago. Unfortunately the
> only notes I can recover are these:
[Several examples given.]
There are lots of examples. Here is a method for generating all symmetric
ones. For simplicity, define f*(t) = int (exp(2*pi*i*t*x) * f(x)) dx.
Then if f is symmetric and slightly well behaved, f** = f. So f + f*
is its own Fourier transform.
It is necessary to make this modification, since if f is square integrable,
the usual Fourier transform multiplies the norm by sqrt(2*pi).
--
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hru...@pop.stat.purdue.edu (Internet, bitnet)
{purdue,pur-ee}!pop.stat!hrubin(UUCP)