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12 votes

Series vs Asymptotic in 12.1

Extended comment, I won't accept this as an answer. Here are some cases I've found where Series might be a better choice than ...
imas145's user avatar
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10 votes
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How could this asymptotic expansion be obtained?

Exact expression for $\sigma_n$: b[n_] = BesselJ[1, BesselJZero[0, n]]*BesselJZero[0, n]*StruveH[0, BesselJZero[0, n]]; σ[n_] = π/2*(-1)^n*(b[n + 1] - b[n]); ...
Roman's user avatar
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9 votes
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Asymptotic inverse function?

...
Roman's user avatar
  • 47.9k
8 votes
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Asymptotic Solve

As I understand it, Mathematica has a problem with the order of the series expansion at infinity. The following works in 13.0.0 and produces the required result. ...
user64494's user avatar
  • 26.6k
7 votes
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Asymptotic expansion around infinity for inverse cdf of normal distribution

Using assumptions and an immediate assignment: ...
Roman's user avatar
  • 47.9k
7 votes

Mathematica can't simplify asymptotic expressions containing constant symbols

TrigToExp does the job Assuming[a > 0, Asymptotic[TrigToExp[Sech[a *x]], x -> Infinity]] ...
user64494's user avatar
  • 26.6k
7 votes
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AsymptoticSum does not give any output

How about analytically evaluating the product Product[1 + (3/(4 n)), {n, 2, j}] $$\frac{\Gamma \left(j+\frac{7}{4}\right)}{\Gamma \left(\frac{11}{4}\right) \Gamma (...
yarchik's user avatar
  • 18.7k
7 votes

Asymptotic inverse function?

@Roman's answer is exactly what you asked for, but as a check, here is a second answer using Bessel's series expansion for the solution of the Kepler equation: ...
user293787's user avatar
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6 votes
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Not getting the correct asymptotic behaviour when sending a small parameter to zero

You could try using AsymptoticSolve instead: ...
Carl Woll's user avatar
  • 131k
5 votes
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Asymptotic expansion of an Euler-type integral

Try this: ...
Mariusz Iwaniuk's user avatar
5 votes

Not getting the correct asymptotic behaviour when sending a small parameter to zero

Execute ...
Andreas's user avatar
  • 3,342
5 votes
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Determinant of matrix with asymptotic expansion

The O representation of an expansion point of Infinity is obtained with: O[x, Infinity] (...
Carl Woll's user avatar
  • 131k
5 votes
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Asymptotic[] Doesn't Actually Compute

tao2 = (2*(-(1/4)/E^(2*t^2) + Pi/8 - ((1/4)*Sqrt[Pi]*t)/E^t^2 + (1/4)*Pi*Erf[t] - ((1/4)*Sqrt[Pi]*t*Erf[t])/E^t^2 + (1/8)*Pi*Erf[t]^2))/Pi; ...
Bob Hanlon's user avatar
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5 votes
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AsymptoticDSolveValue returing input

The ode ode = eps y''[x] + (x + x^3)*y'[x] - 2*y[x] == 0 ode /. y->(y[-#]&) (* -2 y[-x] - (x + x^3) Derivative[1][y][-x] +eps (y^\[Prime]\[Prime])[-x] == 0*) ...
Ulrich Neumann's user avatar
5 votes
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Asymptotics and limits of second order ODE which depend on (two) parameters

You could find the solution $h(x)$ and then use Asymptotic on its derivative? ...
Nasser's user avatar
  • 145k
5 votes

AsymptoticIntegrate of a difficult integral

Edit: there's a factor of 2 missing from the RHS of the $\sin(\sin x)$ relation, which follows from a mistake/typo in the Maths.S.E answer for which I provided a ...
bmf's user avatar
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5 votes
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Is there any possibility of obtaining an asymptotic approximation (instead of numerical solutions) of such a 2nd-order homogeneous ODE in Mathematica?

Here's a symbolic solution, whose asymptotic expansion may be obtained: ...
Michael E2's user avatar
  • 236k
5 votes

When to use Series vs Asymptotic?

The definition of Asymptotic can be accessed by GeneralUtilities`PrintDefinitionsLocal, and it seems that part of its code ...
Lacia's user avatar
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4 votes
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Why can we only find the asymptotic expression of the solution of the first implicit function?

The AsymptoticSolve command works with the result of the Solve command. Let us consider these results. ...
user64494's user avatar
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4 votes
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Is there a way to check whether $f(x)=o(g(x))$ for given $f$ and $g$?

Another pre V11.3 method is to use the limit definition: littleO[f_, g_, x_, x0_:Infinity] := PossibleZeroQ[Limit[f/g, x -> x0]] littleO[x, x^2, x] ...
Greg Hurst's user avatar
  • 36.2k
4 votes

How to study asymptotic behavior, built-in functions

Another possibility is to use new in M12 function AsymptoticSolve: sol = AsymptoticSolve[y==f[r],y,{r,Infinity,4}] ...
Carl Woll's user avatar
  • 131k
4 votes

Finding Asymptotics for a Series

As @DanielLichtblau says, Mathematica can do the symbolic sum in terms of LerchPhi: sum = Sum[2^i/i, {i, n}] -I ([Pi] - I 2^...
Carl Woll's user avatar
  • 131k
4 votes

AsymptoticDSolveValue multiple solutions

DSolve does not find $y(x)=x$ either, and I think this is why AsymptoticDSolveValue does not. I am sure they share some core ...
Nasser's user avatar
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4 votes
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Asymptotic inversion of ExpIntegralEi function

You got correct hints on Math.SE. Define: E1[z_] := EulerGamma - ExpIntegralEi[-z]; E2[z_] := -ExpIntegralEi[-z] Exp[z]; They can be inverted like this ...
yarchik's user avatar
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4 votes
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Selecting the negative expression

Select negative solutions. Select[sols, Resolve[Exists[ϵ, ϵ > 0, ForAll[s, Pi - ϵ < s < Pi, (t /. # /. Theta -> s) < 0]]] &] <...
cvgmt's user avatar
  • 74.9k
4 votes

AsymptoticDSolveValue returing input

Mathematica can not do it. May be because it is boundary value problem. Compare ...
Nasser's user avatar
  • 145k
4 votes

How can I investigate limit at infinity with asymptotic series?

Here a solution using Asymptotic ...
Ulrich Neumann's user avatar
4 votes

Series solution does not match numerical solution

[Update: I needed to recalculate the initial $a(0)$ for each different initial $x(0)$, so that the solution starts on the algebraic equation.] If I rework the OP's approach with a differentiated ...
Goofy's user avatar
  • 3,012
4 votes

Asymptotic integral expansion at infinity

Solving this problem analytically is difficult. However, it is possible to get exact values of the integral In in symbolic form for each ...
Vaclav Kotesovec's user avatar
4 votes
Accepted

Is it possible to have the asymptotics of this function?

The following proves your expectations. ...
user64494's user avatar
  • 26.6k

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