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My question is as follows. Suppose we have a function $f(r)$ and we want to study its asymptotic behavior at infinity ($r\rightarrow \infty$). For example, the function may reduce to $-\frac{a}{r}$ or $b e^{-cr}$ at infinity. How do I identify the constants $a,b$ and $c$ using Mathematica? Or, more generally, how do I identify the asymptote of a function? Can anybody point out useful built-in functions?

I am interested in the function:

$$ f(r)=-\frac{\sqrt[3]{3} e^{-2 r/3}}{\pi ^{2/3}}-\frac{\sqrt[3]{2 \pi } e^{2 r/3}}{5 \left(\frac{3 \sqrt[3]{\pi } e^{2 r/3} \sinh ^{-1}\left(2 \sqrt[3]{2 \pi } e^{2 r/3}\right)}{5\ 2^{2/3}}+1\right)} $$


f[r_]:=-((3^(1/3) E^(-2 r/3))/\[Pi]^(2/3)) - (E^(2 r/3) (2 \[Pi])^(1/3))/(
 5 (1 + (3 E^(2 r/3) \[Pi]^(1/3)
      ArcSinh[2 E^(2 r/3) (2 \[Pi])^(1/3)])/(5 2^(2/3))))

I expect this function to have -$\frac{1}{r}$-behavior. How do I check it? I am not interested in a numerical value of the limit (which is 0), but rather in a function the original function reduces to at infinity.

P.S. Using Mathematica for a week

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Google for Asymptotic analysis ... there is a whole world in your hands – Dr. belisarius Jan 17 '13 at 17:58
The second part of this question is broad and perhaps--if suitably narrowed--would be more appropriate on the math site. Is there a specific function you would like to analyze? – whuber Jan 17 '13 at 17:59
Closely related question: How to expand a function into a power series with negative powers? – Jens Jan 17 '13 at 18:06
For what it's worth, the expansion I did (with ArcSinh or ArcSin) does show explicitly that the asymptotic behavior is O(1/r). – Daniel Lichtblau Jan 18 '13 at 14:37
I tried N[Normal[Series[f[r], {r, Infinity, 5}]]] /. r -> 100 and N[Normal[Series[f[r], {r, 0, 5}]]] /. r -> r^-1 /. r -> 100 (to get the expansion in negative powers, as suggested above). But two commands return different results. What am I doing wrong? – molkee Jan 18 '13 at 16:10
f[r_] := 3^(1/3)*Exp[-2*r/3]/Pi^(2/3) - (2*Pi)^(1/3)*
         ArcSin[2*(2*Pi)^(1/3)*Exp[2*r/3]]/(5*2^(2/3)) + 1))

In this case you might just observe that there is a commonly appearing expression Exp[2*r/3]. Substitute in a enw variable and expand a series at infinity in both variables. Then replace the substituted variable.

g[r_] := f[r] /. {Exp[-2*r/3] -> 1/y, Exp[2*r/3] -> y}

hh = PowerExpand[
  Normal[Series[g[r], {y, Infinity, 1}, {x, Infinity, 1}]] /. 
   y -> Exp[2*r/3]]

(* -(4/(3*(Pi - (4*I*r)/3 - I*(4*Log[2] + 
                   (2/3)*(Log[2] + Log[Pi]))))) + 
   (3^(1/3)/Pi^(2/3) + (40*2^(2/3))/(9*Pi^(1/3)*
             (Pi - (4*I*r)/3 - 
          I*(4*Log[2] + (2/3)*(Log[2] + Log[Pi])))^
               2))/E^((2*r)/3) *)

This should give a good idea of the behavior at infinity.

There are ways of making this algorithmic. I won't pursue that though.

This result captures the asymptotic behavior quite well, I will say. Here's a quick numeric check.


(* Out[252]= -0.000181229 - 0.00876831 I *)

hh /. r -> 111.

(* Out[263]= -0.000181229 - 0.00876831 I *)

By the way, it would be most helpful if in future you use actual Mathematica notation rather than provide only a LaTeX form that needs to be translated.

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Yes, providing a formula in $\TeX$ is a problem. In this case, for instance, "ArcSinh" inadvertently became "ArcSin" in your answer--that's why you're getting complex values for real values of the argument. – whuber Jan 17 '13 at 21:19
@whuber I don't know if it was inadvertant...that 'h' might have been suicidal. – Daniel Lichtblau Jan 17 '13 at 21:30
yep, you should have ArcSinh in place of ArcSin. Added Mathematica code – molkee Jan 17 '13 at 22:45

You can use the Limit function:

f[r_] := -((3^(1/3) E^(-2 r/3))/\[Pi]^(2/3)) - (E^(2 r/3) (2 \[Pi])^(1/
         3))/(5 (1 + (3 E^(2 r/3) \[Pi]^(1/3) ArcSinh[
         2 E^(2 r/3) (2 \[Pi])^(1/3)])/(5 2^(2/3))))
Limit[f[r], r -> \[Infinity]]

...which returns 0. Plotting shows that this is correct:

Plot[f[r], {r, 1, 10000}, PlotRange -> All]

enter image description here

share|improve this answer
But how do I show that the function behaves like $-\frac{1}{r}$ at infinity? – molkee Jan 18 '13 at 2:19
@molkee You can show that the function behaves like -1/r at infinity via Limit[r*f[r], r->∞]. Note the factor of r multiplying f[r]. – user9892 Oct 8 '13 at 2:14

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