       Re: Help with project needed

• To: mathgroup at smc.vnet.net
• Subject: [mg96700] Re: Help with project needed
• From: dbreiss at gmail.com
• Date: Sat, 21 Feb 2009 19:39:12 -0500 (EST)
• References: <gne6uk\$phc\$1@smc.vnet.net> <gngk3s\$jlj\$1@smc.vnet.net>

```A small bit of advice:  C is a reserved name in Mathematica (as is
K).  So, to avoid conflicts and perhaps puzzling results, use a

--David

On Feb 18, 4:24 am, dh <d... at metrohm.com> wrote:
> 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

```

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