Re: Fourier Transforms
- To: mathgroup at smc.vnet.net
- Subject: [mg64805] Re: Fourier Transforms
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Sat, 4 Mar 2006 02:35:16 -0500 (EST)
- Organization: The University of Western Australia
- References: <du6nkt$5mv$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <du6nkt$5mv$1 at smc.vnet.net>,
"Ben C" <benjamin.chamberlain at seh.ox.ac.uk> wrote:
> On the first of March I posted an appeal for help with some Fourier
> transforms. Since then a couple of people have suggested I post the
> actual transforms. I am trying to inverse Fourier transform the
> functions
>
> p / (sqrt(1+p^2 + sqrt(1+p^2 )) and 1/(sqrt(1+p^2 - sqrt(1+p^2 ))
>
> from p to x space.
>
> Any advice would again be extremely gratefully received,
Your expressions do not make sense (unmatched parens). As Sseziwa Mukasa
suggested, I assume that you are trying to compute the inverse Fourier
transform of
p/Sqrt[1+p^2+Sqrt[1+p^2]]
and
1/Sqrt[1+p^2-Sqrt[1+p^2]]
I would be interested to know how these functions arose in the first
place -- and I would be surprised if either had a closed-form solution.
Nevertheless, we can compute a series expansion solution.
First, note that powers of p can be generated by differentiation with
respect to x.
D[InverseFourierTransform[f[p], p, x], x] ==
-I InverseFourierTransform[p f[p], p, x]
Next, note that
InverseFourierTransform[1/Sqrt[p^2 + 1], p, x]
Sqrt[2/Pi] BesselK[0, x Sign[x]]
and that 1/Sqrt[p^2 + 1] is quite a good approximation to
1/Sqrt[1+p^2+Sqrt[1+p^2]]. One can use (asymptotic) series expansion to
improve the approximation:
ser[n_][p_] := Normal[Series[1/Sqrt[1+p^2+Sqrt[1+p^2]] /.
p -> Sqrt[x^2 - 1], {x, Infinity, n}]] /. x -> Sqrt[p^2 + 1]
Here are the first 3 terms of this series:
ser[3][p]
1/Sqrt[p^2 + 1] - 1/(2 (p^2 + 1)) + 3/(8 (p^2 + 1)^(3/2))
Importantly, the inverse Fourier transform of each term in this series
is straightforward:
inv[a_][x_] = InverseFourierTransform[(p^2 + 1)^a, p, x]
(2^(a + 1) Abs[x]^(-a - 1/2) BesselK[a + 1/2, x Sign[x]])/Gamma[-a]
It is instructive to compare plots of ser[n][p] to that of
1/Sqrt[1+p^2+Sqrt[1+p^2]].
<<Graphics`
DisplayTogether[
Plot[Evaluate[Table[ser[n][p], {n, 3, 8}]],
{p, -10, 10}, PlotStyle -> Table[Hue[i], {i, 0, 1, 1/6}]],
Plot[1/Sqrt[1 + p^2 + Sqrt[1 + p^2]], {p, -10, 10}]]
The convergence is good except for -2 < p < p. Empirically, the
approximation is greatly improved if the coefficient of the
highest-order term (the first term in this case) is reduced by 1/2.
DisplayTogether[
Plot[Evaluate[Table[ser[n][p]-First[ser[n][p]]/2, {n, 3, 8}]],
{p, -10, 10}, PlotStyle -> Table[Hue[i], {i, 0, 1, 1/6}]],
Plot[1/Sqrt[1 + p^2 + Sqrt[1 + p^2]], {p, -10, 10}]]
Hence, an accurate series approximation to the inverse Fourier transform
of 1/Sqrt[1+p^2+Sqrt[1+p^2]] is given by
(ser[10][p]-First[ser[10][p]]/2)/. (p^2 + 1)^(a_) :> inv[a][x]
This expansion simplifies for x > 0 or x < 0. For example, collecting
terms for x > 0, one obtains,
Collect[Simplify[%, x > 0], {BesselK[_, _], Exp[_]}, Simplify]
Finally, the inverse Fourier transform of p/Sqrt[1+p^2+Sqrt[1+p^2]] is
obtained by differentiation with respect to x. (It is easiest to work
with the x > 0 and x < 0 cases separately.)
To compute the inverse Fourier transform of 1/Sqrt[1+p^2-Sqrt[1+p^2]]
there is an extra complexity (which makes me wonder if your original
expression is correct. Is p missing from the numerator?); it is singular
at p = 0. The singularity there is
Normal[Series[1/Sqrt[1 + p^2 - Sqrt[1 + p^2]], {p, 0, 0}]]
Sqrt[2]/p
Now, since
InverseFourierTransform[1/p, p, x]
-I Sqrt[Pi/2] Sign[x]
we can handle this problem by computing the series expansion of
1/Sqrt[1+p^2-Sqrt[1+p^2]] - Sqrt[2/p^2]
exactly as we did above, but now the convergence is not as nice.
Cheers,
Paul
_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul