I am trying to obtain an approximate expression for the behaviour of the following function for large $x$

$$F(x)=x\frac{ \text{erf}(x)^2}{\text{erf}(2 x)}$$

I know that the $\lim_{x\rightarrow\infty}\frac{1}{x}F(x)=1$ and $\lim_{x\rightarrow\infty}F(x)-x=0$ so that basically one can say that for large $x$, $F(x)\approx x$.

I would like to obtain this result from Series, but invoking Series[F, {x,∞,1}] I obtain

$$\frac{e^{2 x^2} \left(e^{2 x^2} \left(x+O\left(\left(\frac{1}{x}\right)^2\right)\right)+e^{x^2} \left(-\frac{2}{\sqrt{\pi }}+O\left(\left(\frac{1}{x}\right)^1\right)\right)+\left(\frac{1}{\pi x}+O\left(\left(\frac{1}{x}\right)^2\right)\right)\right)}{\left(-\frac{1}{2 \sqrt{\pi } x}+O\left(\left(\frac{1}{x}\right)^2\right)\right)+e^{4 x^2}}$$

Why do I get a fraction? How can I get an expansion at infinity which reflects the linear behaviour of the function plus higher order corrections?

  • $\begingroup$ I suspect that Mathematica is doing what it's supposed to, and that the problem is just that the power series of F doesn't yield the "obvious" answer. The Series expansion is only valid "near" x0 = Infinity, ie, for large x. Your result (and expansions to any order) gives a singularity between x = 0 and x = 1. Doing the expansion around x0 = 0 would give you sensible behaviour for small x, but wouldn't mean much for larger values, and vice-versa for expansion around x0 = Infinity. $\endgroup$ Sep 30, 2017 at 1:06
  • $\begingroup$ aardvark2012- Yes, clearly that's the meaning of a series expansion. But the point is why is it for instance that the terms are not arranged in order of decreasing relevance? Why is there even a fraction? Certainly an expansion is valid only near infinity, but I would have expected something like $F(x)\approx x+\frac{a1}{x}+\frac{a2}{x^2}+O(\frac{1}{x^3})$. Which is indeed the case if you expand the expression in the numerator (I still would have to understand what's the meaning of the denominator though...). But the other question which remains is: why I cannot get the terms ordered? $\endgroup$
    – semola
    Sep 30, 2017 at 1:11

2 Answers 2


You can use the new in M12 function AsymptoticSolve:

AsymptoticSolve[y == x Erf[x]^2/Erf[2x], y, {x, Infinity, 3}]

{{y -> (E^-x^2 (1 - 2 x^2 + E^x^2 Sqrt[π] x^3))/(Sqrt[π] x^2)}}


You can substitute x==1/y and take series at y==0. Do this twice to get a result you expect.

ser10 = Series[F[1/y], {y, 0, 0}, Assumptions -> y > 0] // Normal // 

(*     (2 E^(2/y^2) (-E^((1/y^2)) Sqrt[Pi] + y)^2)/(
        2 E^(4/y^2) Pi y - Sqrt[Pi] y^2)     *)

ser11 = ExpandAll[ser10]

(Series[ser11, {y, 0, 0}] // Normal) /. y -> 1/x // FullSimplify

(*     -((2 E^-x^2)/Sqrt[Pi]) + x     *)

Since Exp[-x^2] vanishes very fast, you have F[x] -> x

  • $\begingroup$ Actually what I get is $\frac{2 e^{2 x^2} \left(\sqrt{\pi } e^{x^2} x-1\right)^2}{2 \pi e^{4 x^2} x-\sqrt{\pi }}$. Anyway, the point is not really obtaining that result but using Series so that it returns the expansion in the right order. I am actually studying a function which is a complicated compositions of terms similar to F and I cannot check by hand how many exponentials are to be ignored... $\endgroup$
    – semola
    Sep 30, 2017 at 14:31

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