Re: Simple Trigonometric Integrals

*To*: mathgroup at smc.vnet.net*Subject*: [mg32354] Re: [mg32338] Simple Trigonometric Integrals*From*: BobHanlon at aol.com*Date*: Wed, 16 Jan 2002 03:30:06 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

In a message dated 1/15/02 4:40:14 AM, jhelfand at wam.umd.edu writes: >I have a thing about Mathematica. Sometimes I have a real long >expression that involves the integral of the sum of lots of cosines and >sines of some variable let's say 't'. But having done some fancy maths >on my own to reduce it and get into a simple integral from 0 to 2 Pi, >and the sines and cosines all involve some integer multiple of t, the >integration takes for ever, it basically hangs. Now, although the >expression is long, and there are a lot of terms in it, it still just >becomes a simple periodic integral from zero to 2 pi, and all the >trigonometric terms involving t should just drop out. Kind of like what >sometimes can happen if you are playing around with a Fourier series >expansion (by the way, does Mathematica have a built in Fourier Series >expansion? I mean something like Series[], but returns fourier >coefficients?). Uptill now, I have been able to get by with something >like using > >periodicIntegral={Cos[t] -> 0, Cos[2 t] -> 0, Cos[3 t] -> 0, Cos[4 t] -> >0, Cos[5t] -> 0, Cos[6 t] -> 0, Cos[7 t] -> 0, Sin[t] -> 0, Sin[2 t] -> >0, Sin[3 t] -> 0, Sin[4 t] -> 0, Sin[5t] -> 0, Sin[6 t] -> 0, Sin[7 t] >-> 0}; > >and then doing a replace on the expresion, multiplying the result by 2 >Pi. But now I am in a bind where no amount of TrigReduce, TrigExpand, >TrigFactor, etc. will get this big ass expression into the desired form >where the above is approriate (because there are other sines and cosines >of other variables that get put into the terms and stand by >themselves). Still, the expression should be easy to do for the >computer, even I can go through and set these terms to zero, but it will >just take me a long time. An example of what I am talking about, just >try the following: > >In[687]:= >Joe = a c Cos[t]/(g s) + b q Cos[2 t]/(c f) + c Cos[3 t]/(d a) + > d f Cos[4 t]/(h a n) + e q Cos[5t]/(g a) + f l Cos[6 t]/(w r m) + > g b Cos[7 t]/(o n x) + h Sin[t]/(b c) + i Sin[2 t]/(h e r) + > j y Sin[3 t]/(l p) + d k Sin[4 t]/(j c) + l m a Sin[5 t]/(f s b h) >+ > m p Sin[6 t]/(k j) + q n Sin[7 t]/(x c); > >In[688]:= >Integrate[Joe, {t, 0, 2 Pi}] > >and you willl see it takes a long time to integrate. (It will >eventually get done.) I know this is just zero, but why does it take so >long for the computer to figure out? It is true that my expression is >even longer than this one, so essentially it hangs, but basically it is >the same problem. I do not what to be hunting through my equation from >hell setting all the relevant trigonometric terms to zero when the >computer should be able to do this. Well, sorry for the harangue but I >greatly appreciate you reading down so far, really. If you have any >suggestions or comments, point out I am an idiot there is some simple >thing in Mathematica, please send it. Needs["Calculus`FourierTransform`"]; FourierSeries[t, t, 3] E^(2*I*Pi*t)*((1 - I*Pi)/(4*Pi^2) - (1 + I*Pi)/(4*Pi^2)) + (-((1 - I*Pi)/(4*Pi^2)) + (1 + I*Pi)/(4*Pi^2))/ E^(2*I*Pi*t) + ((1 - 2*I*Pi)/(16*Pi^2) - (1 + 2*I*Pi)/(16*Pi^2))/E^(4*I*Pi*t) + E^(4*I*Pi*t)*(-((1 - 2*I*Pi)/(16*Pi^2)) + (1 + 2*I*Pi)/(16*Pi^2)) + E^(6*I*Pi*t)* ((1 - 3*I*Pi)/(36*Pi^2) - (1 + 3*I*Pi)/(36*Pi^2)) + (-((1 - 3*I*Pi)/(36*Pi^2)) + (1 + 3*I*Pi)/(36*Pi^2))/ E^(6*I*Pi*t) % // ExpToTrig Sin[2*Pi*t]/Pi - Sin[4*Pi*t]/(2*Pi) + Sin[6*Pi*t]/(3*Pi) FourierTrigSeries[t, t, 3] Sin[2*Pi*t]/Pi - Sin[4*Pi*t]/(2*Pi) + Sin[6*Pi*t]/(3*Pi) Joe=a c Cos[t]/(g s)+b q Cos[2 t]/(c f)+ c Cos[3 t]/(d a)+d f Cos[4 t]/(h a n)+ e q Cos[5t]/(g a)+f l Cos[6 t]/(w r m)+ g b Cos[7 t]/(o n x)+h Sin[t]/(b c)+ i Sin[2 t]/(h e r)+j y Sin[3 t]/(l p)+ d k Sin[4 t]/(j c)+l m a Sin[5 t]/(f s b h)+ m p Sin[6 t]/(k j)+q n Sin[7 t]/(x c); Head[Joe] Plus Integrate term-by-term and sum the results Tr[Integrate[#, {t,0,2Pi}]& /@ (List @@ Joe)]//Timing {0.08999999999999986*Second, 0} Bob Hanlon Chantilly, VA USA