# Tag Info

10

The identity does not hold for x < -1: Plot[ProductLog[x*Exp[x]], {x, -5, 5}] FullSimplify[ProductLog[x*Exp[x]], x >= -1] x (* result in 10.1.0 under Windows *)

9

There is also a newer package, HolonomicFunctions, that has an implementation of Chyzak's generalization of Zeilberger's algorithm. To perform the desired task, use the following commands: smnd = Simplify[ G /. HoldPattern[HypergeometricPFQ[pl_List, ql_List, x_]] :> (Times @@ (Pochhammer[#, k] & /@ pl)) / (Times @@ (Pochhammer[#, k] & /@ ql)...

8

Mathematica returns Sum[BesselJ[n, x]^2/(n - k), {n, -Infinity, Infinity}] unevaluated. However, BesselJ[n, x]^2 and BesselJ[-n, x]^2 are equal, so the Sum can be rewritten as Simplify[-BesselJ[0, x]^2/k + 2 k Sum[BesselJ[n, x]^2/(n^2 - k^2), {n, 1, Infinity}]] (* -π BesselJ[-k, x] BesselJ[k, x] Csc[k π] *) which is the desired result. Its plot, here ...

7

Because NIntegrate evaluates the integrands before starting the actual integration, in some cases (like this one) it is better to define the integrand function F with the signature F[x_?NumericQ]. BF[n_?NumericQ, x_?NumericQ] := BesselJ[n, x] NIntegrate[BF[9/2, x], {x, 0, 1}] (* 0.000148473 *) Integrate[BesselJ[9/2, x], {x, 0, 1}] %% // N (* Sqrt[2/\[...

5

You probably entered ClebschGordan[{1/2,1/2},{0,0},{1/2,-1/2}] and got the warning message. This happens when you use angular-momentum quantum numbers that don't satisfy the conservation laws. In this case, the rule $m_1+m_2=m$ is not met, see the documentation for ClebschGordan. However, Mathematica still produces the correct result, i.e., 0. This just ...

5

The equation x Exp[x] == y has multiple solutions for x. For example, evaluating tab = Table[{x -> ProductLog[i, 1]}, {i, 0, 5}] Exp[x] x /. tab N[tab] gives {1, 1, 1, 1, 1, 1} and {{x -> 0.567143}, {x -> -1.53391 + 4.37519 I}, {x -> -2.40159 + 10.7763 I}, {x -> -2.85358 + 17.1135 I}, {x -> -3.16295 + 23.4277 I}, {x -> -3....

5

A bit speculative, but I think we can see why 171 is a magic number, if we factor the Gamma[n] from the integral: Table[ {n, NIntegrate[Gamma[x] /Gamma[x + n], {x, 1, 2}]} , {n, 168, 173}] {{168, 7.20924*10^-304}, {169, 4.26119*10^-306}, {170, 2.41873*10^-308}, {171, 0.}, {172, 0.}, {173, 0.}} 171 is where the integral fails as you can see the ...

5

People dealing with special functions for the first time ask questions like this a lot; I suppose it's time to write something on this topic. Let's go back to a much simpler case of the polylogarithm: $$\mathrm{Li}_1(x)=\sum_{k=1}^\infty\frac{x^k}{k}=-\log(1-x)$$ Following your train of thought, $\mathrm{Li}_1\left(\frac{11}{10}\right)$ should also be ...

5

PowerExpand[ProductLog[x Exp[x]]] x This assumes $x\ge0$

4

Result Let f[x_] = QPochhammer[x, x]; It can be easily shown that for x close to 1 f[x] has the leading behaviour $$\text{fa}(\text{x})=\exp \left(-\frac{\pi ^2}{6 (1-x)}\right)$$ Comparision of function and leading term Plot[{1, Log[f[x]]/Log[fa[x]]}, {x, 0.9, 1}] Very close to x = 1 the numerical precision becomes poor but still the result is ...

4

The operator in the question can be simplified by first applying it to an arbitrary function f: Simplify[(1/Sin[θ]) D[Sin[θ] D[f[θ,ϕ],θ],θ]+(1/Sin[θ]^2) D[f[θ,ϕ],{ϕ,2}]] $$f^{(2,0)}(\theta ,\phi )+\cot (\theta ) f^{(1,0)}(\theta ,\phi )+\csc ^2(\theta ) f^{(0,2)}(\theta ,\phi )$$ This can be recognized as the negative of the squared angular ...

3

This is not a solution but may provide a good first step. The inner integral (a Green's function, I presume) can be performed symbolically. g = Simplify[Integrate[q^2*((BesselJ[0, t*q] - (2*BesselJ[1, t*q])/(t*q))* BesselJ[0, k*q] + ((6*BesselJ[1, t*q])/(t*q) - 2*BesselJ[0, t*q])* BesselJ[1, k*q]/(k*q)), {q, 0, Infinity}, Assumptions -> k > 0 ...

3

As noted by b.gatessucks, a preliminary application of FunctionExpand[] allows the confirmation of this identity: FullSimplify[FunctionExpand[Derivative[1, 0][BesselK][1/2, z]] == Sqrt[π/2] ExpIntegralE[1, 2 z] Exp[z]/Sqrt[z]] True FullSimplify[FunctionExpand[Derivative[1, 0][BesselK][-1/2, z]] == -Sqrt[π/2] ExpIntegralE[1, 2 z]...

3

Expressing the entire business in terms of SphericalBesselJ[] cures the problem: NIntegrate[r With[{x = r BesselJZero[15/2, 1]}, Sqrt[2 x/π] SphericalBesselJ[7, x]]^2, {r, 0, 1/50}] 1.1879560281974252*^-27 The nice thing about SphericalBesselJ[] is that it does not auto-evaluate to a potentially numerically unstable combination of ...

3

Fixing it ... Is there any settings that I can turn on to fix this? Does using larger working precision produce results you expect? F[n_] := NIntegrate[(Gamma[x] Gamma[n])/Gamma[x + n], {x, 1, 2}, WorkingPrecision -> 60, MinRecursion -> 6, MaxRecursion -> 20] In[10]:= F[170] Out[10]= 0....

3

You are doing quite a few things wrong. It is really quite simple, though. ParametricPlot[{FresnelC[t], FresnelS[t]}, {t, -3, 3}, AxesLabel -> {x, y}] Let's enumerate your errors. Trying to promote an inherently 2D plot to a 3D plot. Trying to define functions with SetDelayed inside a plot's first argument. Using Set would have worked, but making ...

2

Let us have a single integrand function first: F[x_?NumericQ, xp_?NumericQ] := ( EllipticK[(4 x xp)/(x + xp)^2] Sinh[x] Sinh[ xp])/((x + xp) (Cosh[202] + Cosh[2 x]) (Cosh[202] + Cosh[2 xp])); obtained by expanding f[x, 101.] f[xp, 101.] g[x, xp]. For me (with Mathematica v. 10.3), the integral fails with NIntegrate::inumri: "The integrand (....

2

I tried to make it a comment and then can't control the words. I think the problem is arising from the fact that the value of the function and its derivative is too small to near x=0. Plot[BesselJ[9/2, x], {x, 0, .01}] Plot[Evaluate[D[BesselJ[9/2, x], x]], {x, 0, 0.01}] As you can see the derivative is highly oscillatory at this small range. When ...

1

Just trying to evaluate the function fun a few times on some different y1 and x1 values like fun/. y1 -> 3 /. x1 -> 2 // Simplify ((-4 + x2) (1 + 2 u + x2))/((-3 + x2) (2 u + x2)) One can infer that the whole hypergeometric function actually in general reduces to just ((-1 + x2 - y1) (-2 + 2 u + x2 + y1))/((-1 + x2 - x1) (-2 + 2 u + x1 + x2)) ...

1

Defining f = 1/r - (StruveH[0, r/2] - BesselY[0, r/2]) Pi/4 the maximum value that f r^2 assumes is NMaximize[{f r^2, r > 0}, r] (* {0.498508, {r -> 2.58341}} *) Thus, the critical value of γ is 0.498508. The following illustrates this maximum. Plot[r^2 f, {r, 0, 10}, AxesLabel -> {r, "f"}]

1

Let me point out my paper http://arxiv.org/pdf/1301.6617.pdf (also http://iopscience.iop.org/article/10.1088/1751-8113/46/44/445302/pdf). Figure 3--containing SIX hypergeometric functions--in it is the result of an application of FindSequenceFunction, as explained in the paper. The displayed output was obtained using various transformation rules from the ...

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