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[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)
II. Make a table of Zeta-functions with even arguments
t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==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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