RE: Re: fourier transform time
- To: mathgroup at smc.vnet.net
- Subject: [mg36467] RE: [mg36439] Re: fourier transform time
- From: "DrBob" <drbob at bigfoot.com>
- Date: Sun, 8 Sep 2002 03:30:44 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
Even better -- MUCH better -- add Bob Hanlon's Evaluate to the other tricks:
ClearAll[f, b, FS]
L = 2;
f[x_] := UnitStep[x - 1];
b[n_] := b[n] = N[(2*(Cos[(n*Pi)/2] - (-1)^n))/(n*Pi)];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[Evaluate[FS[30, x]], {x, 0, 2}]]
{0.015 Second, â??Graphicsâ??}
ClearAll[f, b, FS]
L = 2;
f[x_] := UnitStep[x - 1];
b[n_] := b[n] = N[(2*(Cos[(n*Pi)/2] - (-1)^n))/(n*Pi)];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[Evaluate[FS[120, x]], {x, 0, 2}]]
{0.063 Second, â??Graphicsâ??}
Bobby Treat
-----Original Message-----
From: DrBob [mailto:drbob at bigfoot.com]
To: mathgroup at smc.vnet.net
Subject: [mg36467] RE: [mg36439] Re: fourier transform time
With the second method below (mine), the times add up to 40% less than with the first method (Jens-Peer's), for what SEEMS to be exactly the same work. Go figure!
Can anybody explain that?
(* Jens-Peer *)
ClearAll[f, b, FS]
L = 2;
f[x_] := UnitStep[x - 1];
b[n_] := b[n] = (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]
{0.547 Second, â??Graphicsâ??}
(* Treat #1 *)
Timing[(2/L)*Integrate[f[x]*Sin[n*Pi*(x/L)], {x, 0, L}]]
{0.016000000000000014*Second, (2*(Cos[(n*Pi)/2] - Cos[n*Pi]))/(n*Pi)}
ClearAll[f, b, FS]
L = 2;
f[x_] := UnitStep[x - 1];
b[n_] := b[n] = (2*(Cos[(n*Pi)/2] - Cos[n*Pi]))/(n*Pi);
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]
{0.297 Second, â??Graphicsâ??}
Even stranger, it SLOWS the Plot if we precompute b before Timing starts:
(* Treat #2 *)
ClearAll[f, b, FS]
L = 2;
f[x_] := UnitStep[x - 1];
b[n_] := b[n] = (2*(Cos[(n*Pi)/2] -
(-1)^n))/(n*Pi);
b /@ Range[30];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)],
{n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]
{0.328 Second, â??Graphicsâ??}
But here's a winner:
(* Treat #3 *)
ClearAll[f, b, FS]
L = 2;
f[x_] := UnitStep[x - 1];
b[n_] := b[n] =
N[(2*(Cos[(n*Pi)/2] - (-1)^n))/
(n*Pi)];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)],
{n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]
{0.204 Second, â??Graphicsâ??}
Apparently, computing b within the Plot causes machine-precision arithmetic to be used, and that saves time. Precomputing b and then converting exact expressions to approximate ones within the Plot seems to take longer. For that to make sense, I think it must be that n is approximate (not Integer) when it is passed to b within Plot.
Bobby Treat
-----Original Message-----
From: Jens-Peer Kuska [mailto:kuska at informatik.uni-leipzig.de]
To: mathgroup at smc.vnet.net
Subject: [mg36467] [mg36439] Re: fourier transform time
Hi,
with your code you compute the expansion coefficents every
time when FS[] is evaluated. Store the values for b[n] with
L = 2;
f[x_] := UnitStep[x - 1];
b[n_] := b[n] = (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]
and you need only a 1-3 seconds (depending on your machine)
Regards
Jens
Steve Story wrote:
>
> For my PDE class we have been calculating Fourier transforms. The instructor
> arrived today with a printout of two plots of a certain Fourier transform,
> done with a different CAS. The first plot was to 30 terms, the second was to
> 120 terms.
> Curious, I translated the functions into Mathematica (4.0 on Windows2000
> on a PIII 700) to see how much time this required to process. I was
> Staggered at how much time it took. Here's the code:
>
> L = 2;
> f[x_] := UnitStep[x - 1];
> b[n_] := (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];
>
> FS[N_, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];
>
> Timing[Plot[FS[30, x], {x, 0, 2}]]
>
> Out[23]=
> {419.713 Second, \[SkeletonIndicator]Graphics\[SkeletonIndicator]}
>
> In this case the number of terms is 30.
>
> The time required per number of terms seems to fit the following polynomial:
>
> y = 0.3926x^2 + 2.2379x
>
> This is a large amount of time. I understand that the code is not optimized,
> and was more or less copied from the code in the other CAS, but is this a
> reasonable amount of time, or is something going wrong? I don't use Mathematica
> because of the speed, but should it be this slow?
>
> Just curious,
>
> Steve Story