Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am interested in asymptotic behavior of a function at infinity: $$ f(r)=\frac{0.04962 e^{-2 r} (r-1.000)}{\left(\left(e^{-2 r}\right)^{2/3}+0.06119\right)^2 r} $$

Tried Series[f(r),{r,0,10}] (Mathematica) for expansion in negative powers at infinity 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?

Got stuck at this point. Need help(

share|cite|improve this question
I'm a little worried that you have $x$ on the left but $r$ on the right. – Gerry Myerson Jan 27 '13 at 9:16
I expect that $.04406$ comes from $.04962/(1+.06119)^2$, but I can't explain the minus sign. – Gerry Myerson Jan 27 '13 at 9:19
edited the post – molkee Jan 27 '13 at 9:21
up vote 1 down vote accepted

The asymptotic behavior at $\infty$ is determined by the behavior of the dominant term for large values of $r$. In this case, this term is

$$\frac{0.04962 \, e^{-2 r}}{(0.06119)^2}$$

The first correction term is simply the next term in the numerator. This more accurate asymptotic behavior is then

$$\frac{0.04962 \, e^{-2 r}}{(0.06119)^2} \left ( 1 - \frac{1}{r} \right )$$

Further correction terms may be found by looking in the denominator. Be aware that these are very small corrections because of the exponential term. You may determine the full asymptotic behavior by using the expansion

$$(1+y)^{-2} = \sum_{k=0}^{\infty} (-1)^k (k+1) \, y^k$$

where $y=e^{-4 r/3}/(0.06119)^2$. That is, the full asymptotic behavior of the above expression at $\infty$ is

$$\frac{0.04962 \, e^{-2 r}}{(0.06119)^2} \left ( 1 - \frac{1}{r} \right ) \sum_{k=0}^{\infty} (-1)^k \frac{(k+1)}{(0.06119)^{k}} \, e^{-\frac{4}{3} k r}$$

NB The Mathematica series you derived is for small $r$, so it will not give the correct behavior at $\infty$.

share|cite|improve this answer
BTW, Series[f(r),{r,0,10}] is equivalent to expansion in negative powers at infinity – molkee Jan 27 '13 at 16:31
@molkee: OK, but at $\infty$, exponentials have essential singularities and are not expandable in powers, which is why the expansion I provided is the one I think you want, unless I am missing something. – Ron Gordon Jan 27 '13 at 16:36
Yes, you are right. But what is the origin of the 1/r term then? As far as I understand, if the expansion were not valid, I would get ONLY positive powers. – molkee Jan 27 '13 at 16:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.