I need to solve the following problem.

I have the following function: Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2], where x and y are variables and a and b parameters. I need to slightly modify the function (maybe adding an additional parameter, multiplying with the given parameters or modifying denominator, such as, that its major term in the series expansion about infinity for x and y would be 1/x and 1/y, respectively.

This is what I get:

FullSimplify[Series[Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2], {x, Infinity, 2}]]

and the major term in the output would be 1/(Sqrt[a] x)

Also with the y:

FullSimplify[Series[Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2], {y, Infinity,     2}]]1/(Sqrt[a] x)

and the major term in the output would be 1/(Sqrt[a] y)

So I need a function similar to one here (Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2]),

who will give both 1/x and 1/y respectively, for the given mathematica inputs.

Has anybody got an idea how to modify the function? An intuitive way was to multiply it by Square[a b], but I tried and it did not work. I tried the other things, but the problem to me didn't seem trivial. Has anyone got an idea?

Also does anybody know what Mathematica does to make a series of a function of 2 variables around infinity? I know it is called the asymptotic series, but what would be practical way to get it. Having an insight about that some constraint can be found about the function, and therefore the above-mentioned problem can be solved.

  • $\begingroup$ It would be helpful if you stated precisely what kind of result you are expecting. You want a 'simplified' version of $\text{erf}(r)/r$; how should this simplified version relate to the original? $\endgroup$ Jan 17, 2014 at 15:59
  • $\begingroup$ Is there some reason not to change variables e.g. w=a x^2 + b y^2? $\endgroup$ Jan 17, 2014 at 16:42
  • $\begingroup$ @episanty In order to capture the nature of a system I am describing, I needed a function of two variables (lets call it z[x,y] whose 3D plot "would look like a mountain", decaying differently from maximum in two directions: z[x,0] and [y,0]. The function: (Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2]), where x and y are variables, a and b parameters satisfies that that could be seen from this plot: $\endgroup$
    – ogledala
    Jan 20, 2014 at 10:20
  • $\begingroup$ Plot3D[(Erf[Sqrt[2 x^2 + 5 y^2]]/Sqrt[2 x^2 + 5 y^2]), {x, -3, 3}, {y, -3, 3}] But has also to satisfy another condition, to have 1/x and 1/y as a major term in the given two series expansion: FullSimplify[Series[Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2], {y, Infinity,2}]] to get here 1/y as the main term and here FullSimplify[Series[Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2], {x, Infinity,2}]] 1/x is the main term. $\endgroup$
    – ogledala
    Jan 20, 2014 at 10:54

2 Answers 2


Not clear to me what exactly you want, but possibly something along the lines below?

   Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2] /. 
    Thread[{x, y} -> t*{x, y}], {t, Infinity, 2}]] /. t -> 1

(* Out[2281]= E^(-a x^2 - 
  b y^2) (-(1/(Sqrt[\[Pi]] (a x^2 + b y^2))) + E^(a x^2 + b y^2)/Sqrt[
   a x^2 + b y^2]) *)
  • $\begingroup$ Thanks Daniel, was in this output idea to make both x and y to be about infinity? $\endgroup$
    – ogledala
    Jan 20, 2014 at 10:55
  • $\begingroup$ It's a more or less standard tactic for getting a "total degree" series when there are multiple variables. In the case of expansion at infinity I guess it is in a sense treating all variables on an equal footing. $\endgroup$ Jan 20, 2014 at 17:04

I'm not quite sure what you want. You evidently want an expression which obeys some limit properties together with your starting expression, but it's not quite clear how you would phrase those requirements in terms of limits. It's also not clear whether this is the only expression you want to deal with, or whether this is an example of some more general class of functions.

As Daniel Lichtblau points out, there's little point in dealing directly with the combination of $\text{erf}$ and $w=\sqrt{ax^2+by^2}$. Specifically, you should first look for asymptotic expansions of $\text{erf}(w)/w$; this is a standard expression, it can be accessed through Series[Erf[w]/w, {w, Infinity, 3}], and is essentially $$ \frac{\text{erf}(w)}{w} \sim \frac{1}{w} -\frac{e^{-w^2} }{ \sqrt{\pi } }\left(\frac{1}{w^2}-\frac{1}{2w^4}+\cdots\right), $$ where $ f(x)\sim\phi(x)\Leftrightarrow f(x)/\phi(x)\to 1$.

Thus, as far as your particular expression goes, you can simply reduce it to the problem of finding a satisfactory asymptotic expression for $$ \frac1w=\frac1{\sqrt{ax^2+by^2}}. $$ Now, the problem is: how do you want to take the two limits? Can you assume any relation between $x$ and $y$? Or do you simply know that they are both big, but either can be bigger than the other?


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