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