Hot answers tagged

10

For your first question, if we gather the factors into a single variable z, there's a simple hypergeometric function: f[z_] = (-1)^(1/4) EllipticF[I ArcSinh[(-1)^(1/4) z], -1]; g[z_] = -z Hypergeometric2F1[1/4, 1/2, 5/4, -z^4]; These two functions are the same, even though FullSimplify cannot prove it: Series[f[z] - g[z], {z, 0, 100}] (* O[z]^101 *) Plot[...


8

Match up power series and solve for parameters for Hypergeometric2F1[a, b, c, d x]: ClearAll[reduce2F1, iReduce2F1]; Options[reduce2F1] = {"ExtraTerms" -> 2}; reduce2F1[expr_, x_, opts : OptionsPattern[]] := reduce2F1[expr, x, True, OptionsPattern[]]; reduce2F1[expr_, x_, assum_: True, OptionsPattern[]] := With[{res = iReduce2F1[expr, x, assum, ...


3

Note that Log[2 - Sqrt[3]] == -Log[2 + Sqrt[3]].


3

An alternative to @Jens' solution is to add the phase to the argument of the Log instead, and then remove the associated rotation in the output of the Log: logbranch[σ_] := With[{phase = I (σ + Pi)}, Function[Log[# Exp[-phase]] + phase] ] I used a slightly different definition so that you just replace Log with logbranch[ϕ] instead of replacing Log[z] ...


3

Just because the origin is an essential singularity doesn't mean that the residue does not exist. The sum of the residues of all of the singularities is 0. Three of the singularities have residues that are easy to compute: f[x_] := (x^3 Exp[1/x])/(1-x^2) res1minus = Residue[f[x], {x, -1}] res1 = Residue[f[x], {x, 1}] res∞ = Residue[f[x], {x, ∞}] -(1/(2 ...


3

Not what you are hoping for but is there a pole at x = 0? Limit[(x^3 Exp[1/x])/(1 - x^2), x -> 0] (* Indeterminate *) I think you have an essential singularity which takes all values as x -> 0. Hope that helps


2

I am not sure I understand what is the aim. I post this as motivation. Apologies for errors and misunderstanding. Manipulate andDynamicModule are useful for dynamic/interactive exploration. Locator can allow you to explore different z. Slider2D can also be useful. Here is an example: Manipulate[ Module[{r = Norm[pt]^(1/n), pts}, pts = Table[ r {...


1

There is a bug in the imaginary part in the equation: Tan@z-(value[[1]]+I start[[2]]) Change this and it will work, except when FindRoot does not converge.


1

You are getting an empty contour plot because the surface z[x_, y_] = ComplexExpand[Abs[x + I y - 6] + Abs[x + I y + 6]] doesn't have an intersection with the plane z = 1. In fact, it does not go below z = 12. You can verify this by observing the behavior z[x, y] with planes of constant z with the following Manipulate expression. Manipulate[ Plot3D[{z[...


Only top voted, non community-wiki answers of a minimum length are eligible