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
>