The question was inspired by this discussion:

How to expand a function into a power series with negative powers?

I am interested in asymptotic behavior of a function at infinity:

f[r_]:=(0.04962 Exp[-2 r] (-1.000+r))/((0.06119+(Exp[-2 r])^(2/3))^2 r)

or (TeX)

$$ f(r)=\frac{0.04962 e^{-2 r} (r-1.000)}{\left(\left(e^{-2 r}\right)^{2/3}+0.06119\right)^2 r} $$

Tried Series[f[r],{r,0,10}] for expansion in negative powers at infinity (as suggested) and got:

$$ -\frac{0.04406}{r}+0.02146+0.02106 r+0.004405 r^2-0.001355 r^3-0.001205 r^4-0.0003607 r^5-\left(8.402\times 10^{-6}\right) r^6+0.00004149 r^7+0.00001982 r^8+\left(3.921\times 10^{-6}\right) r^9-\left(6.018\times 10^{-7}\right) r^{10}+O\left(r^{11}\right) $$

Seems like the function decays faster than $1/r^n$ and the expansion is meaningless. But what does the term $-\frac{0.04406}{r}$ mean then? The function is strictly positive at infinity and I am kinda confused by that. Does this mean that the asymptotic form of the function is something plus the term $-\frac{0.04406}{r}$ which effectively gives the observed behavior? Can someone clarify it? How can one explain the term $-\frac{0.04406}{r}$?

  • 1
    $\begingroup$ 1) you forgot the underscore after r in the definition of f. Probably not essential in this case. 2) you have round brackets around r in the Series call. 3) In the answer you linked to, the series is developed at infinity, you use 0. $\endgroup$ Jan 27, 2013 at 18:52
  • 1
    $\begingroup$ $-0.04406 = 0.04962(-1.000) / (1 + 0.06119)^2$ exactly as it should: this is the limiting behavior of $r f(r)$ as $r$ approaches $0$. $\endgroup$
    – whuber
    Jan 27, 2013 at 19:08
  • $\begingroup$ Series[f[r],{r,0,10}] is equivalent to Series[f[r],{1/r,Infinity,10}], isn't it? we can not use negative powers in Series, though $\endgroup$
    – molkee
    Jan 27, 2013 at 19:20
  • $\begingroup$ I am just wondering whether or not this term has anything to do with the behavior of the function at infinity. $\endgroup$
    – molkee
    Jan 27, 2013 at 19:34
  • $\begingroup$ It goes to zero ~ 13.25*Exp[-2*r]. $\endgroup$ Jan 27, 2013 at 20:50

1 Answer 1


It is not quite clear, why do you expand the expression around 0, if you want to study its behavior in infinity? I trust that it is not its limit at infinity that you are interested in, since this limit is clear without any calculations. You need probably one or few largest terms. I would in this case go to a new variable x=Exp[-2r] that tends to zero, when r goes to infinity, and rewrite the expression in its terms

f[r_] := (0.04962 Exp[-2 r] (-1.000 + 
  r))/((0.06119 + (Exp[-2 r])^(2/3))^2 r);

g[x_] := f[r] /. {E^(-2 r) -> x, r -> -Log[x]/2} // Simplify

ss=Series[g[x], {x, 0, 2}] // Normal

which yields

x^(5/3) (-433.157 - 866.314/Log[x]) + x (13.2524 + 26.5049/Log[x]) + 
 x^(7/3) (10618.3 + 21236.7/Log[x])

And go back to the variable r.

  ss /. {x -> Exp[-2*r], a_/Log[x] -> a/(-2 r)} // PowerExpand

This brings the following:

E^(-14 r/3) (10618.3 - 10618.3/r) + E^(-2 r) (13.2524 - 13.2524/r) + 
 E^(-10 r/3) (-433.157 + 433.157/r)

You are right that there are both the terms with exponents and with 1/r, but the most importantis the term containing the exponent with the smallest decrement.


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