Natural Science Forum / Physics / Research / May 2008

Hey again J-boy!

Your eq(5) looks wrong. You can't just add the exponents
like that unless they commute. (d/dB doesn't commute with B^2).

You're guaranteed to get more help with this sort of boring
basic stuff over on www.physicsforums.com (try the quantum
physics forum). Also, it's latex-capable, so you don't need
to rely on auxiliary files like you're doing here on spr.

- LOL from Neuropulp!

P.S: I tried to reply to your previous posting, but it
attempt failed. Don't know why.

P.P.S: Try to think of a more original salutation than
"Pulp-boy". You can do better than that! :-) :-)
(Wrong gender, btw.)

It would seem OK up to a point but it is not the way I would tackle
it.

Now e^x = 1+x+x^2/2! + x^3/3! + etc

Hence e^f(x) = 1+f(x)+f(x)^2/2! + f(x)^3/3! + etc

We know that Int (x^n)dx = x^(n+1)/(n+1) Hence if we have a power
series integration is trivial.

Now f(x) = -x^2 + V(x) we have a power series for f(x) and getting a
term by term evaluation is a matter of a term by term multiplication.
This gives your your equations for what you have labelled as C(n).

You can also constuct a series using Taylor's theorem, that is to say
you differentiate e^(f(x)) repeatedly.

You get f'(x)e^(f(x)) differentiating this again (f''(x) +
(f'(x)^2)e^(f(x))

Can you see one thing? The process of Taylor differentiation is
isomorphic with binomial multiplication (we would be surprised if it
wasn't).

You want clarification. I think the isomorphism of binomial
multiplication and Taylor's theorem does clarify things consiserably.

The physical meaning. There isn't a way to do things that isn't a
little bit messy.

 - Ian Parker