1
$\begingroup$

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 $-\frac{0.04406}{r}$?

Improve this question
$\endgroup$
6
  • 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. $\endgroup$
    – Sjoerd C. de Vries
    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]. $\endgroup$
    – Daniel Lichtblau
    Jan 27, 2013 at 20:50

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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