Re: Help with project needed
- To: mathgroup at smc.vnet.net
- Subject: [mg96649] Re: Help with project needed
- From: dh <dh at metrohm.com>
- Date: Wed, 18 Feb 2009 04:24:22 -0500 (EST)
- References: <gne6uk$phc$1@smc.vnet.net>
Hi Aaron, if I understand correctly, your operator L can be written: L= f1[eps,alpha] d/dr + f2[eps,alpha] d/dalpha now you can write f1 and f2 as series in epsilon. E.g. f[eps_,al_]:=Exp[eps+al]; Series[f[eps,al],{eps,0,5}] This gives an expansion of L in epsilon. hope this helps, Daniel Aaron Fude wrote: > Hi, > > I'm about to attempt a project in which so many Mathematica related > things are unclear to me that I will describe the project, rather than > try to ask individual questions. > > Suppose that C is a function of alpha, but it also depends on a > parameter epsilon, so I think of it as C[epsilon][alpha] -- but I'm OK > with it if the implementation treats C as a function of two variables. > This C is given. It's complicated, but one could easily obtain a > series for it in epsilon. > > Now, I have the following operator that acts on functions of "r", and > "alpha" (d is partial): > > L = C''(1-C')/(1+C^2) * d/dr + C'(1-C'') /(1-C')* d/dalpha > > For a given function f[r, alpha], I need to take successive orders in > epsilon of the operator and apply it to f. > > I realize that for a given function f, I can form L[f] and the do > Series[L[f], {epsilon, 0, 5}]. However, I'm very interested in seeing > the operator decomposed into a series. I want to be able to say, > here's the epsilon term of the operator, here's the epsilon squared > term, and so forth. Can this be done. Can I decompose L into a series: > > L = L0 + L1 + L2 + .... > > and then be able to apply Ln to f? > > Many thanks in advance, > > Aaron >