Re: Need help to a beginner.

• To: mathgroup at smc.vnet.net
• Subject: [mg9676] Re: [mg9642] Need help to a beginner.
• From: "C. Woll" <carlw at u.washington.edu>
• Date: Fri, 21 Nov 1997 01:31:13 -0500
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Shinichiro,

Here are two possible approaches:

Series[1 / Sqrt[(Abs[a]-x)^2 + y^2 + z^2], {x,0,2},{y,0,2},{z,0,2}]

or

1 / Sqrt[(Abs[a]-e x)^2 + e^2 y^2 + e^2 z^2] + O[e]^3

Using Abs[a] makes sure that things like Sqrt[a^2] get converted to a.
The first approach theoretically allows you to expand to different
orders in x, y and z, but unfortunately there is a bug which can screw
up your results. The latter approach is cleaner, since you don't have
O[x]^n, O[y]^n and O[z]^n all over the place.

Carl Woll
Dept of Physics
U of Washington

On Mon, 17 Nov 1997, Shinichiro Kondo wrote:

> Hi, all. I am quite new to Mathematica, and am using the older version,
> v. 2.2. I hope someone will help me. My problem is nothing to do with
> homework of a math class, or any sort. I am a physics graduate student,
> and for my research project am trying to have an algebraic expression
> of  potential energy of certain ionic crystalline lattice. The
> electrostatic (i.e.  Coulomb) energy has the expression of 1/r, where r
> is the distance between two  charges.
> I am still far from getting the answer I want because of this problem I
> am facing in the very first stage where I am supposed to be used to
> Mathematica.
>
> 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?
>
> I will greatly appreciate someone's help to this problem. Meanwhile, I
> am trying to find a solution in the manual by myself. If you don't