# Tag Info

7

Since your question's been answered, let me tell you about the handy listing of notations used by the Wolfram Functions site. In particular, if I scroll down to the "F" section of this page, you'll see an explanation that you are indeed looking at Kampé de Fériet's function.

11

It is Kampé de Fériet function, introduced in Joseph Kampé de Fériet, "La fonction hypergéométrique.", Mémorial des sciences mathématiques, Paris, Gauthier-Villars. Its definition is given on Notations page: and, in an alternative form, in Wikipedia: $${}^{p+q}f_{r+s}\left( \begin{matrix} a_1,\cdots,a_p\colon b_1,b_1{}';\cdots;b_q,b_q{}'; \\ ... 9 If you want a different 3D visualization, maybe try field lines. Usually, one should be suspicious of field line plots for time-dependent electromagnetic waves (because causality casts doubts on the meaning of plotting a spatially extended field line for a globally fixed time), but there have been recent high-profile papers that show such plots. Purely as ... 5 (This is a math question, not a Mathematica question.) To add to Artes's answer, there is the well-known(!) identity$$\zeta(-n)=\frac{(-1)^n}{n+1}B_{n+1}$$so you might as well ask why$$\begin{align*} -\frac12\times B_2&=-\frac1{14}\times B_{14}\\ -\frac12\times \frac16&=-\frac1{14}\times \frac76 \end{align*} A justification for the ...

5

In order to understand the issue, we should provide the underlying definitions. Mathematica helps in verifying appropriate relations and definitions. The main functional equation relating Riemann's zeta function $\zeta\;$, to Euler's $\Gamma\;$, established in Riemann's famous paper Über die Anzahl der Primzahlen unter einer gegebener Grösse (1859, English ...

1

In fact there are lots of Zetas which produce the same value, e.g. try these: $\{\zeta(-2), \zeta(-4), \zeta(-6),\ldots ,\zeta(-2n)\}$ {Zeta[-2], Zeta[-4], Zeta[-6]} (* {0, 0, 0} *) Using Plot may shed some light on this: Plot[Zeta[x], {x, -15, 0.5}]

8

FullSimplify[(-1)^n*BesselJ[n, z] - BesselJ[-n, z], n ∈ Integers, ComplexityFunction -> (StringLength @ ToString @ # &)] Also: ComplexityFunction -> (Count[#, _BesselJ | _Power, {-2}] &) ComplexityFunction -> (Count[#, _?NumberQ, Infinity] &)

8

A bit of cheating: DifferenceRootReduce[(-1)^n BesselJ[n, z] - BesselJ[-n, z], n] 0 I must admit I'm not sure why FullSimplify[] fails on this, tho.

8

Following this question you can define: invmollweide[{x_, y_}] := With[{theta = ArcSin[y]}, {Pi (x)/(2 Cos[theta]), ArcSin[(2 theta + Sin[2 theta])/Pi]}]; fc[phi_] := Block[{theta}, If[Abs[phi] == Pi/2, phi, theta /. FindRoot[2 theta + Sin[2 theta] == Pi Sin[phi], {theta, phi}]]]; cart[{lambda_, phi_}] := With[{theta = fc[phi]}, {2/Pi*lambda ...

17

Edit: I added more explanations below, because this visualization method is quite different from conventional vector plots For just this purpose I had at some point invented the following visualization technique. I'll reproduce your definition first. It defines a complex vector field on the surface of a unit sphere. Clear[\[Epsilon]];(*Polarization ...

6

As has been noted by ruebenko in the comments, there does seem to be a bug in the handling of infinite-range Bessel function integrals when MinRecursion and MaxRecursion are both set to non-default values. For instance, even the simple NIntegrate[BesselJ[0, x], {x, 0, ∞}, MinRecursion -> 10, MaxRecursion -> 15] chokes with a NIntegrate::minmax ...

8

Try this: In[1]:= Integrate[BesselJ[0, x]/(x + BesselJZero[0, 1]), {x, -Infinity, Infinity}] Out[1]= Pi StruveH[0, BesselJZero[0, 1]]

3

As the other answers have mentioned, you can confirm the solution without specifying values for C[1] and C[2] lhs = q''[x] + 2 x/(x^2 - 1) q'[x] - 4*q[x]/(x^2 - 1); sol = DSolve[lhs == 0, q, x][[1, 1]]; FullSimplify[lhs /. sol] 0 When C[2] is set to zero, FullSimplify is unable to find the right transformations to reduce the expression to zero, and ...

3

"LevinRule" should work splendidly here, I think: NIntegrate[-m Exp[-m] BesselJ[1, m]^2, {m, 0, Infinity}, Method -> "LevinRule", WorkingPrecision -> 20] -0.18196415067209554877 ruebenko's answer has given a closed form for this particular definite integral. Personally, I prefer it when the parameters of the elliptic integrals are ...

4

This is a bug. As a workaround for this specific integral you could use a symbolic solution: Integrate[-m*Exp[-m]*BesselJ[1, m]^2, {m, 0, Infinity}] (* (-3*EllipticE[-4] + 5*EllipticK[-4])/(5*Pi) *)

1

You can do : sol[x_] = q[x] /. First@DSolve[q''[x] + 2 x/(x^2 - 1) q'[x] - 4*q[x]/(x^2 - 1) == 0, q[x], x] FullSimplify[Derivative[2][sol][x] + 2 x/(x^2 - 1) Derivative[1][sol][x] - 4*sol[x]/(x^2 - 1)] (* 0 *)

0

I think you may not chose one parameter to 0. E.g. try c2=2 and c1=1: FullSimplify[ D[LegendreP[1/2 (-1 + Sqrt[17]), x] + 2 LegendreQ[1/2 (-1 + Sqrt[17]), x], {x, 2}] + 2*x/(x^2 - 1)* D[LegendreP[1/2 (-1 + Sqrt[17]), x] + 2 LegendreQ[1/2 (-1 + Sqrt[17]), x], x] - 4*(LegendreP[1/2 (-1 + Sqrt[17]), x] + 2 LegendreQ[1/2 (-1 + Sqrt[17]), x])/(x^2 - 1)] ...

22

(I had been meaning to write a blog entry about this myself, but since this question has come up, I suppose I'll just write about it here instead...) In demonstrating how the quincuncial projection works, consider first the following complex mapping: With[{ω = N[EllipticK[1/2], 20]}, ParametricPlot[{Re[InverseJacobiCN[Tan[φ/2] Exp[I θ], 1/2]], ...

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