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 > mind, please send your solution to my email address: shink at iastate.edu > > Thank you for your attention. > > >