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Hot answers tagged special-functions

9

_ matches one thing. __ matches one or more things. ___ matches zero or more things.

6

My first guess when I see parenthesized exponents is that this represents a derivative of a multivariate expression. To check this guess I try FullForm[%] and look at what that shows me. That confirms there is a Derivative wrapped around your HypergeometricPFQ If I use NIntegrate instead of Integrate to get an approximate result instead of an exactly ...

6

Compare Plot[BesselI[0,x]BesselK[0,x]-BesselI[1,x]BesselK[1,x],{x,725,742}] versus Plot[BesselI[0,x]BesselK[0,x]-BesselI[1,x]BesselK[1,x],{x,725,742},WorkingPrecision->64] If you are curious about why this might be happening then you could inspect Table[{BesselI[0,x],BesselK[0,x],BesselI[1,x],BesselK[1,x]},{x,725.,742.}] and see the magnitudes of ...

5

One can make use of Simplify with Assumptions I. Compute the sum s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}] (* -(1/63) π^6 Zeta-13/15 π^4 Zeta-55 π^2 Zeta+644 Zeta *) II. Make a table of Zeta-functions with even arguments t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]] (* {ζ==-(1/2),ζ==π^2/6,ζ==π^4/90,ζ==π^6/945,ζ==π^8/9450,ζ[...

4

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, ...

4

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[...

4

I am running into issues where machine precision is lost and the norm for Hankel function becomes too big. Exactly. So you already know the answer: Don't use machine precision but ensure your computations are numerically sound by giving Mathematica enough digits to work with. I believe this answer about controlling precision is a good start. In general, ...

3

search through the Help Menu but cant really get a lead to the plotting It is really no different than other type of Plot? Plot[UnitStep[t - 4]*t, {t, -10, 10}, PlotStyle -> {Thick, Red}] Or, if you want the engineer plot, you can do Plot[UnitStep[t - 4]*t, {t, -10, 10}, PlotStyle -> {Thick, Red}, Exclusions -> None] Similarly, Plot[...

2

Why not just use Solve? Solve[rho[r] == z && r>0, r, Reals] {{r -> ConditionalExpression[Sqrt[1 + z^2], z > 0]}} For values of q other than -1 you will also need to specify the value of z. For example: q = 1/2; Quiet @ Solve[rho[r] == 2 && r>0, r, Reals] //N {{r -> 1.45108}} You can use ConditionalExpression with ...

2

I think you have to be mindful of the radius of convergence of the series. From the Mejer reference you cited, if $f(z)=x+x^p$, then define $$h(z)=\sum_{k=0}^\text{kMax} \frac{(-1)^k}{pk-k+1}\binom{pk}{k}z^k$$ However, convergence of the series is restricted to the region of convergence $$R=\frac{(p-1)^{p-1}}{p^p}$$ so that if \$0\leq y\leq R^{\frac{1}{p-1}...

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