Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

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

> 




  • Prev by Date: Help with Mathematica 7.0
  • Next by Date: Re: Extract Integrate values
  • Previous by thread: Help with project needed
  • Next by thread: Re: Help with project needed