Re: Re: Can anyone help about this inverse laplace transform problem?
- To: mathgroup at smc.vnet.net
- Subject: [mg33305] Re: [mg33279] Re: [mg33260] Can anyone help about this inverse laplace transform problem?
- From: "david Xia" <xiaxd at hotmail.com>
- Date: Thu, 14 Mar 2002 02:22:10 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Thanks for your suggestion, since the expression of fi is a rational expression, how to use mathematica to seperate fi into different parts according to its poles? Is there a function to do this automatically? To find the poles, for my question, the parameters are all unknown variables, will the find root routine find roots based on the symbolic function? As I understand, the root function works only for numerical solution by newton-raphson method given a initial value, am I right? If so, then how to find root then for a symbolic function? Thanks for any input! David >From: Sseziwa Mukasa <mukasa at jeol.com> To: mathgroup at smc.vnet.net >Subject: [mg33305] [mg33279] Re: [mg33260] Can anyone help about this inverse laplace >transform problem? >Date: Wed, 13 Mar 2002 03:14:49 -0500 (EST) > >david Xia wrote: > > > Hello, > > > > I am a new user of Mathematica, and want to get the following inverse > > laplace transform problem: > > > > I have 8 parameters, a,b,c,d,w1,w2,w3,w4. (Where there is a constrain > > that a+b+c+d=1) > > > > The following is the script I tried. > > ------------------------------------------------------------- > > f4s=a w1/(s+a)+b w2/(s+b)+c w3/(s+c)+d w4/(s+d) > > fi=f4s/(1-f4s) > > InverseLaplaceTransform[fi,s,t] > > ---------------------------------------------------------------- > > I tried to run it, but mathematica runs forever and can not give me >an > > answer. Can it be solved by Mathematica? > > > > Any ideas? > > > >I don't know why Mathematica doesn't return a result, but I played around a >little with your problem and have some suggestions. Since your function is >rational and the denominator is of fourth degree you can calculate the >poles >of the expression explicitly using Roots. You can then perform the inverse >transform explicitly by taking the sum of the residues evaluated at the >poles. This is a complicated expression. Your constraint may allow you to >simplify a few of the terms. > >Regards, > >Sseziwa > > _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp.