1
$\begingroup$

I have an expression below and want to approximate it with a simpler form by taking $d$ to infinity

$$ f(s)=\frac{s^{1/p} \Gamma \left(-\frac{1}{p},(d+1)^{-p} s\right)}{p} $$

Taking Asymptotic returns unevaluated.

expr = (t^(1/p) Gamma[-(1/p), (1 + d)^-p t])/p ;
Asymptotic[expr, {d, Infinity, 2}]

Are there tricks in Mathematica to get a simpler expression for this in the infinite $d$ limit?

With[{p = 101/100, d = 10^9},
 expr = (t^(1/p) Gamma[-(1/p), (1 + d)^-p t])/p ;
 Plot[expr, {t, 1, d}, AxesLabel -> {"time", "error"}]
 ]

enter image description here

$\endgroup$
4
  • 2
    $\begingroup$ Adding Assumptions -> d > 0 && p > 0 && t > 0 (assuming those are desired restrictions) results in a response from Asymptotic (Windows 10, Mathematica 13.2.0.0). $\endgroup$
    – JimB
    Apr 19, 2023 at 3:41
  • $\begingroup$ The 2nd order approximation appears to be 1 + d + ((1 + d)^(1 - p) t)/(-1 + p) + ((1 + d)^(1 - 2 p) t^2)/(2 - 4 p) + ((1 + d)^(1 - 3 p) t^3)/(-6 + 18 p) - t^(1/p) Gamma[(-1 + p)/p]. And the form for the $k$-th order approximation is $\sum _{i=0}^{k+1} \frac{(-1)^{i+1} t^i (d+1)^{1-i p}}{i! (i p-1)}-t^{1/p} \Gamma \left(1-\frac{1}{p}\right)$. $\endgroup$
    – JimB
    Apr 19, 2023 at 4:50
  • $\begingroup$ Asymptotic does produce some result, but it doesn't seem to fit the graph in $0<t<d$ regime which is what I'm trying to find $\endgroup$ Apr 19, 2023 at 6:34
  • $\begingroup$ Please include your requirements for a good fit. The following produces a ratio of the true to approximate value from 0.999997 to 1: a = With[{k = 6}, -t^(1/p) Gamma[1 - 1/p] + Sum[(-1)^(i + 1) t^i (1 + d)^(1 - i p)/(i! (i p - 1)), {i, 0, k + 1}]]; Plot[expr/a /. {d -> 10^9, p -> 101/100}, {t, 1, 10^9}, PlotRange -> All]. $\endgroup$
    – JimB
    Apr 19, 2023 at 13:29

1 Answer 1

4
$\begingroup$

First substituting u -> (1 + d)^-p t and assuming u->0 for d->Infinity Mathematica (v12.2 Windows11) evaluates

expr = (t^(1/p) Gamma[-(1/p), (1 + d)^-p t])/p
zw = Asymptotic[Gamma[-(1/p), u], {u, 0, 3}]  
asy = t^(1/p)/p zw /. u -> (1 + d)^-p t // Simplify(* third order asymptote*)

$\frac{t^{\frac{1}{p}} \left(\frac{1}{6} p \left(\frac{t^3 (d+1)^{-3 p}}{3 p-1}+\frac{3 t^2 (d+1)^{-2 p}}{1-2 p}+\frac{6 t (d+1)^{-p}}{p-1}+6\right) \left(t (d+1)^{-p}\right)^{-1/p}+\Gamma \left(-\frac{1}{p}\right)\right)}{p}$

Block[{p = 101/100, d = 10^9}, 
Plot[{expr, asy}, {t, 1, d}, AxesLabel -> {"time", "error"},PlotRange -> {{0, Automatic}, All}[[-1]],PlotStyle -> {Automatic, Dashed}]]

enter image description here

But initial assumption u->0 doesn't hold for z=O[d]! That's why approximation applies only to t=o[d]!

Hope it's what you're looking for!

$\endgroup$
2
  • $\begingroup$ ah, thanks, indeed, I was looking for something tight in the range of $t=o(d)$ $\endgroup$ Apr 20, 2023 at 2:56
  • $\begingroup$ background for this problem is given here $\endgroup$ Apr 21, 2023 at 5:52

Your Answer

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

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