Re: Need help to a beginner.
- To: mathgroup at smc.vnet.net
- Subject: [mg9680] Re: [mg9642] Need help to a beginner.
- From: Hugh Walker <hwalker at hypercon.com>
- Date: Fri, 21 Nov 1997 01:31:17 -0500
- Sender: owner-wri-mathgroup at wolfram.com
shink at iastate.edu (Shinichiro Kondo) wrote:
> ... First of all, let me explain my problem. It is known, if you expand
>1/Sqrt[1-x], provided that x^2<<1, you have
>1+(1/2)*x+(3/*)*x^2+(5/16)*x^3+.... Now, let's have a similar
>expression to this: 1/Sqrt[(a-x)^2+y^2+z^2].
>Suppose x, y and z are cartesian coordinates, and r^2=x^2+y^2+z^2. And a
>is some positive constant, and it satisfies a>>r. So factoring the
>denominator by a^2 and kicking it out of the Sqrt as a, I can continue
>this algebra by my hand, and I should end up with:
>(1/a)*(1+x/a-r^2/(2*a^2)+(3*x^2)/(2*a^2)+....) in which I only keep up
>to the 2nd order of r/a (and x/a).
>
>I would like to be able to do this by Mathematica. That is, given this
>sort of a reciprocal of a Sqrt of quadratic expression with x, y and z,
>I'd like it to expand "approximately" so that a resultant expression
>only contain the terms up to a specified order of r/a (thus, x/a, y/a,
>and z/a). How can I do this? This kind of expansion goes on forever,
>but I don't need many higher order terms. How can I specify the maximum
>order that I want to have?
>
Hello Shinichiro:
I think the expansion you want is
f = 1/Sqrt[(a-x)^2+y^2+z^2];
Series[f,{a,Infinity,2}]
If 5 terms are needed, just write 5 instead of 2.
Have fun.
Hugh Walker
Gnarly Oaks