Re: fourier transform time

*To*: mathgroup at smc.vnet.net*Subject*: [mg36451] Re: [mg36425] fourier transform time*From*: BobHanlon at aol.com*Date*: Sat, 7 Sep 2002 02:54:11 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

In a message dated 9/6/02 5:14:56 AM, sbstory at unity.ncsu.edu writes: >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, Your definition of b recalculates the integral for every call. To evaluate the integral once include Evaluate. Clear[f, b, FS]; L = 2; f[x_] := UnitStep[x - 1]; b[n_] := Evaluate[ (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}]]; In the case of FS it is best to wait for an integer value of N prior to performing the Sum. FS[N_Integer, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}]; Now Evaluate the argument of the Plot to cause the Sum to be done once. Timing[Plot[Evaluate[FS[120, x]], {x, 0, 2}]][[1]] 0.366667 Second While my computer may be faster than yours, this result for N=120 is 1000 times faster than your result for N=30. Bob Hanlon