# Tag Info

2

You can increase penalty for Gamma and Pochhammer headers: simplify[expr_, n_] := FullSimplify[expr, n ∈ Integers && n > 0, ComplexityFunction -> ((LeafCount@# + 10 Count[#, _Gamma | _Pochhammer, {0, ∞}]) &)]; simplify[RSolveValue[{a[n] == n*a[n - 1] - n!, a[0] == 2}, a[n], n], n] (* -(-2 + n) n (-1 + n)! *)

2

Don't know if this suits your needs or not, but if you are certain that the argument to Gamma is a non-negative integer, then just make the replacement manually RSolveValue[{a[n] == n*a[n - 1] - n!, a[0] == 2}, a[n], n] (* 2 Pochhammer[1, n] - n Pochhammer[1, n] *) Simplify[%, Element[n, Integers]] (* -(-2 + n) Pochhammer[1, n] *) FunctionExpand[%] (* ...

6

Abstract (Please see section 3.1 for a short answer to the original question of the OP.) We present here the complete solution of the problem of the OP, which consists of analyzing and solving the ODE numerically and symbolically. It turns out that the essential parameter for the classification of the solutions is the energy w of the equivalent mechanical ...

1

Using the expressions derived in this paper, we have the following: SetAttributes[aiPrimeZero, Listable]; aiPrimeZero[s_Integer, prec_: MachinePrecision] := With[{t = N[3 π (4 s - 3)/8, prec]}, FixedPoint[# - AiryAiPrime[#]/(# AiryAi[#]) &, -t^(2/3) Fold[#1/t^2 + #2 &, {18683371/1244160, -181223/207360, ...

0

Looking at ContourPlot, we should be able to get an idea of where the roots are: ContourPlot[ f[r, \[Phi]] == g[r, \[Phi]], {r, 1, 10}, {\[Phi], 0, 2 Pi}] We can then use With and Table to get the roots. In[16]:= Table[ With[{r = r1}, FindRoot[f[r, \[Phi]] == g[r, \[Phi]], {\[Phi], 0.5}]], {r1, 1, 4, 0.5}] Out[16]= {{\[Phi] -> 0.785398}, ...

1

Just changing range of $\phi$ from $-\pi$ to $\pi$: cp = ContourPlot[f[x, y] - g[x, y], {x, 0, 10}, {y, -Pi, Pi}, Contours -> {0}, ContourShading -> None]; fun = Cases[cp, Line[x__] :> x, -1]; pts = cp[[1, 1]]; t = pts[[fun[[1]]]]; {xd, yd} = Transpose[t]; xf = ListInterpolation[xd, {0, 1}] yf = ListInterpolation[yd, {0, 1}] You can recover ...

1

maybe this give you some Idea : Plot[Evaluate@ Table[Evaluate[f[x, y] /. {a -> 2, k -> 3}] - Evaluate[g[x, y] /. {a -> 4}], {y, 1, 10}], {x, 0.01, 10}] at least we can guess that for 0<x<10 the y answer should be between 0 and 4 for these parameter's value. I know this is not an answer but I don't know how can I post my comments ...

6

I discussed the prime zeta function at some length in this math.SE answer. In particular, the infinite Möbius inversion $$P(s)=\sum_{k=1}^\infty \frac{\mu(k)}{k}\log\zeta(ks)$$ is the actual computational formula used, as recommended in Fröberg's paper. (It is also noted there that numerical evaluation becomes more difficult at values near the imaginary ...

1

It seems the current version (10.3) is now aware of the Meijer $G$ expressions for the order derivatives (see this math.SE answer as well): Derivative[1, 0][StruveL][0, z] // FunctionExpand BesselK[0, z] - MeijerG[{{1/2, 1/2}, {}}, {{0, 0, 1/2, 1/2}, {}}, z/2, 1/2]/(2 π^2) (The last version I used, version 8, was unable to do this, if memory ...

2

(This is more of a long comment than an answer.) I distinctly recall having the feeling of disappointment in trying to use the derivative of $\eta(\tau)$ as an intermediate in computing $E_2(\tau)$. Among other things, I tried this alternative formula: $$\frac1{\eta(\tau)}\frac{\mathrm d}{\mathrm d\tau}\eta(\tau)=-\zeta(\pi i\mid-\pi i,-\pi i \tau)$$ ...

5

There is some issue with internal numerics. As the argument gets smaller, the more precision is needed. For t == 0.01, three times $MachinePrecision is sufficient. Table[ N@N[ Derivative[1][DedekindEta][I Round[t, 1/100]], 3$MachinePrecision], {t, .01, .2, .01}] (* {0. - 1.09595*10^-7 I, 0. - 0.00919556 I, 0. - 0.256807 I, 0. - 1.08606 I, 0. - ...

6

Here is a compiled implementation of the Voigt profile function, based on an approximation derived by Chiarella and Reichel and improved by Abrarov, Quine and Jagpal: voigt = With[{n = 24, τ = 12}, With[{d = N[Range[n] π/τ], b = N[Exp[-(Range[n] π/τ)^2]], s = N[PadRight[{}, n, {-1, 1}]], sq = N[Sqrt[2]], sp ...

3

This integral is unknown to Mathematica and indeed it appears that there is no simple solution at all. You can try rooting around the DLMF, and using the relations of $\operatorname{erfi}$ to the incomplete gamma and Kummer $M$ functions to look for similar relations in the standard books of integrals (such as Gradshteyn & Ryzhik, Prudnikov et al., ...

2

Here is another recursive implementation of the Bickley(-Nayler) function $\mathop{\mathrm{Ki}_n}(z)$, using Leonid's method from here: SetAttributes[BickleyKi, Listable]; BickleyKi[n_Integer?NonPositive, z_] := (-1)^n Derivative[0, -n][BesselK][0, z]; BickleyKi[1, z_] := π (1 - z BesselK[{0, 1}, z].StruveL[{-1, 0}, z])/2; BickleyKi[n_Integer, z_] := ...

2

If you plot the left-hand part of your equation for m=2 and play with the kvalues: Manipulate[ Plot[x^2 + BesselJ[m, k*x^2]*x + k*BesselK[m, k], {x, -5, 5}, PlotRange -> {0, pr}], {k, 0.1, 5}, {pr, 0.1, 5}] you will see something like this and playing with the PlotRange fixed by pryou will see that the equation is likely to have no solutions, ...

1

By varying the controls, from the Plot you can see that there are no real roots in the interval. Manipulate[ f[m_, k_, x_] = x^2 + BesselJ[m, k*x^2] x + k BesselK[m, k]; Column[{ NMinimize[{f[m, k, x], 0 <= x <= 5}, x], Plot[f[m, k, x], {x, 0, 5}]}], {{m, 2}, Range[0, 5]}, {{k, 5}, 1, 5, Appearance -> "Labeled"}]

3

Before this gets closed, I want to point out that your code doesn't match your plot, it generates this plot: If you replace tmax with 2.0, then you get the plot you made. Secondly, why do you use such absurd precision in your data files? z[[2 ;; 3]] (* {0.\ 8150081379877160507750606872010435263248846948854663205945457687523565\ ...

12

Since Root objects can be symbolically differentiated, we can find the closed form for its Taylor series (for an explicit n). root = Root[#1^4 - #1 - t &, 1]; coeff = Refine[FunctionExpand[SeriesCoefficient[root, {t, 0, k}]], k >= 0]; Sum[coeff t^k, {k, 0, ∞}] // FullSimplify -t HypergeometricPFQ[{1/4, 1/2, 3/4}, {2/3, 4/3}, -256t^3/27] We can ...

3

I would prefer to have consistency within the framework of Mathematica and hence start off with the original definition of your function as an integral taken here between 1 and z: f = Integrate[1/Sqrt[x (x^2 - 1)], {x, 1, z}, Assumptions -> z > 1] (* Out[35]= -2 EllipticF[ArcSin[1/Sqrt[z]], -1] + 2 EllipticK[-1] *) This is obviously different ...

11

(I just knew someone would ask this someday...) I had talked (ranted?) about this issue at some length here, so I'd like you to read that first. I'll just give an executive summary here: Mathematica uses the parameter convention, while the formula you found on Wikipedia is based on the modulus convention (quickly betrayed by the comma separating the two ...

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