# Tag Info

### 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 ...
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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]); ...
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...
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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. ...
• 26.6k
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### Asymptotic expansion around infinity for inverse cdf of normal distribution

Using assumptions and an immediate assignment: ...
• 47.9k

### Mathematica can't simplify asymptotic expressions containing constant symbols

TrigToExp does the job Assuming[a > 0, Asymptotic[TrigToExp[Sech[a *x]], x -> Infinity]] ...
• 26.6k
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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 (...
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### 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: ...
• 11.9k
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### Not getting the correct asymptotic behaviour when sending a small parameter to zero

You could try using AsymptoticSolve instead: ...
• 131k
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Try this: ...
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Execute ...
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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] (...
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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; ...
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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*) ...
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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? ...
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### 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 ...
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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: ...
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### When to use Series vs Asymptotic?

The definition of Asymptotic can be accessed by GeneralUtilities`PrintDefinitionsLocal, and it seems that part of its code ...
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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. ...
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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] ...
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### 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}] ...
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### 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^...
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### 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 ...
• 145k
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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 ...
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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]]] &] <...
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### AsymptoticDSolveValue returing input

Mathematica can not do it. May be because it is boundary value problem. Compare ...
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### How can I investigate limit at infinity with asymptotic series?

Here a solution using Asymptotic ...
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### 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 ...
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### 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 ...
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