# How can I calculate the asymptotic value of this function correctly?

I'm trying to reproduce the results of a paper which in one part of it, I have to calculate the asymptotic value of a function but I can't reproduce that result exactly. I will be so grateful if someone can tell me where is my fault. According to paper, there is a function as follows

nn = Sqrt[\[Pi]^(3/2) (8 + 5*\[Pi]^(1/2))^-1];
f[re_] =
2*nn^2*Exp[-1/2 re^2] ((\[Pi]/2)^(
1/2) (7/4 + 1/4*re^2 + (re + 1/re) Erf[2^(-1/2) re]) +
Exp[-1/2 re^2]);


and the asymptotic values of $$\frac{\nabla^2f}{f}$$ and $$(\frac{\nabla f}{f})^2$$ in $$r_e\rightarrow \infty$$ are $$(r_e^2-7)$$ and $$-(r_e^2-4)$$ respectively. I tried the following code to reproduce these results, but I can't get $$7$$ and $$4$$!

lap[re_] = D[f[re], {re, 2}] + 2/re *D[f[re], re];
f2[re_] = lap[re]/f[re];
Asymptotic[f2[re], re -> \[Infinity]]
Asymptotic[f3[re], re -> \[Infinity]]
(*re^2*)
(*-re^2*)


When I calculate the asymptotic value of f[re] from the beginning

Asymptotic[f[re], re -> \[Infinity]]
(*(\[ExponentialE]^(-(re^2/2)) \[Pi]^2 re^2)/(Sqrt[2] (16+10 Sqrt[\
\[Pi]])) *)


and continue the process with it I reach an answer which is very similar to main answer other than an extra 1/re^2 term:

f5[re_] = (E^(-(re^2/2)) \[Pi]^2 re^2)/(Sqrt[2] (16 + 10 Sqrt[\[Pi]]))
;
lapE5[re_] = (D[f5[re], {re, 2}] + 2/re *D[f5[re], re])/f5[re] //
Simplify
grad2[re_] = (D[f5[re], re]/f5[re])^2 // Simplify
(*-7+6/re^2+re^2 *)
(*-4+4/re^2+re^2 *)

• Since re is arbitrarily large, the constants 7 or 4 are meaningless in comparison with re or re^2. Apr 1, 2022 at 19:02
• @BobHanlon Thanks a lot. I see but take a look at the addendum which I wrote right now. Apr 2, 2022 at 4:21

Clear["Global*"]

nn = Sqrt[π^(3/2) (8 + 5*π^(1/2))^-1];
f[re_] =
2*nn^2*
Exp[-1/
2 re^2] ((π/2)^(1/2) (7/4 +
1/4*re^2 + (re + 1/re) Erf[2^(-1/2) re]) + Exp[-1/2 re^2]);


EDIT: Corrected typo in definition of f5

f5[re_] = Asymptotic[f[re], re -> Infinity] // Simplify;


Laplacian is a built-in function

lap1 = Asymptotic[Laplacian[f[re], {re}]/f[re], re -> Infinity]

(* re^2 *)

lap2 = Asymptotic[Laplacian[f5[re], {re}]/f5[re], re -> Infinity]

(* re^2 *)


Grad is a built-in function

grad1 = Asymptotic[(Grad[f[re], {re}][[1]]/f[re])^2, re -> Infinity] //
Simplify

(* re^8/(4 + 7 re + 4 re^2 + re^3)^2 *)

Series[grad1, {re, Infinity, 0}] // Normal

(* 34 - 8 re + re^2 *)


• Many thanks. however the answer for laplacian term does not match the paper result, why? And a more question: how you decide to continue the process after gard1 and go to grad2 (and use of Series command)? Apr 2, 2022 at 5:35
• I tried your code: 1- you have a mistypo in part1 where you wrote r->infinity instead of re->infinity, 2- I think the built-in function for Laplacian is different a bit, mine has a 2 factor, however it still returns a 1/re^2 Apr 2, 2022 at 5:48
• grad1 is a rational polynomial (numerator of order 8 and denominator of order 6). The Series` is used to convert the rational polynomial to a simple order 2 polynomial. Apr 2, 2022 at 6:06