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3

The indeterminate can be overcome using the full identity for $\Gamma(nz)$: $$\Gamma(nz)=(2\pi)^{(1-n)/2}n^{nz-1/2}\prod_{k=0}^{n-1}\Gamma(z+\frac{k}{n})$$ and taking the limit as $z\rightarrow 0$: $Version (* "12.0.0 for Linux x86 (64-bit) (April 15, 2019)" *) Limit[Product[Gamma[z + k/n], {k, 1, n - 1}], z -> 0] (* (2 \[Pi])^(1/2 (-1 + n))/Sqrt[n] *)... 1 At least for this case, one can also consider getting the eigenvalues of the Jacobi matrix associated with the Hermite polynomials. Recall that these matrices are constructed from the coefficients of the three-term recurrences that generate the corresponding orthogonal polynomial. Applied to this case, we have: With[{n = 18}, s1 = Sort[Eigenvalues[N[... 8 Workaround: $$\Gamma \left(\frac{k}{n}\right)=\frac{\Gamma \left(\frac{k}{n}+1\right)}{\frac{k}{n}}$$$Version (* "12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)" *) Product[Gamma[k/n + 1]/(k/n), {k, 1, n - 1}] (* (2 \[Pi])^(1/2 (-1 + n))/Sqrt[n] *)

3

As already noted, it is simply possible that FullSimplify[]'s (and even FunctionExpand[]'s) collection of transformations are not enough to deal with this problem. (See this for another example.) In this case, however, one can enlist DifferentialRootReduce[] to prove what is needed: DifferentialRootReduce[2/33 (-11 HypergeometricPFQ[{5/6}, {3/2, 11/6}, -(x^...

1

Here is a slightly indirect method using MeijerGReduce[]: MeijerGReduce[DSolveValue[x y''[x] + (b - x) y'[x] - a y[x] == 0, y[x], x], x] // Activate (C Gamma[-a + b] Hypergeometric1F1Regularized[a, b, x])/Gamma[1 - a] + C HypergeometricU[a, b, x] FunctionExpand[%] (C Gamma[-a + b] Hypergeometric1F1[a, b, x])/(Gamma[1 - a] Gamma[b]) + C[...

1

The classical method for converting a polynomial to an orthogonal basis is Salzer's algorithm. Adapted to the Legendre case, here is how to use it for conversion: bb = CoefficientList[6 x^3 - x^2 + x, x] {0, 1, -1, 6} Clear[a]; n = Length[bb] - 1; a[0, 0] = a[1, 1] = bb[[n + 1]]; a[0, 1] = bb[[n]]; Do[a[0, k + 1] = bb[[n - k]] + a[1, k]/3; Do[a[m, k +...

1

Assuming your ode is well defined(YY still isn't defined ) you can use NDSolveValue(I changedDtto D in your original ode andYY=1) : F = NDSolveValue[{D[f[z], {z, 2}] == 1, f == 1, f' == 0}, f , {z, 0, 6*10^(-4)}] Now you can use F[z] in your calculations! F[.0002] (*1.*) Plot[ F[z] , {z, 0, 6*10^(-4)}, PlotRange -> All]

0

The code in the answer b3m2a1 linked to was intended for 3D curves, which is why it had a lot of machinery embedded. For plane curves, things are vastly simpler: osculatingCircle[fun_?VectorQ, {t_, tvalue_}] := Module[{ka, nv, tv}, {{ka}, {tv, nv}} = FrenetSerretSystem[fun, t] /. t -> tvalue; Circle[(fun /. t -> tvalue) + nv/ka, 1/Abs[ka]]] ...

0

One method is to set the function as J[n_, x_] := (1/(2*Pi))*Integrate[Exp[I*(x*Sin[u] - n*u)], {u, -Pi, Pi}]; As a simple use case the script J[n_, x_] := (1/(2*Pi))*Integrate[Exp[I*(x*Sin[u] - n*u)], {u, -Pi, Pi}]; Sum[J[n, 0], {n, 0, Infinity}] provides the result of 1, the desired result. Since the Bessel function of the first kind is known to be ...

4

The discrepancy you observe is mostly due to the fact that Mathematica's choice of branch cuts for Log[] (and thus Power[] as well) results in the observation that in general, $$\exp(i m \pi)^\nu\ne\exp(i m \nu \pi)$$ For instance, using formula 10.11.1 as an example (as suggested by Carl in a comment): With[{m = 3, ν = 53/10, z = 38/10}, N[{BesselJ[...

3

Many excellent answers has already been given. They do generate the hypergeometric function. I guess the motivation to know it is computational convenience. But let me represent the final result in even simpler form. Maybe it will be useful for you $$\int_z^0\frac{1}{\sqrt{1+x^4}}dx=-\frac{1}{2}F\!\left(2 \arctan(z)\,\big|\frac{1}{2}\right).$$ The integral ...

1

A quick-and dirty method is to use complex-step differentiation: With[{a = 1/10, b = 1/5, t = Exp[I 2 π/3], z = 1/10, h = 10^-9, prec = 20}, N[(SiegelTheta[{{a}, {b}}, {{t}}, z + I h] - SiegelTheta[{{a}, {b}}, {{t}}, z - I h])/(2 I h), prec]] -0.2567239264794337275 + 1.4709617732598025465 I where even a modest-sized step size can yield a ...

4

This is a slightly simpler version of what Michael and Roman did, through leveraging some knowledge about elliptic integrals. First, some observations (using Roman's simplification): D[{(-1)^(1/4) EllipticF[I ArcSinh[(-1)^(1/4) z], -1], -z Hypergeometric2F1[1/4, 1/2, 5/4, -z^4]}, z] // Simplify {-(1/Sqrt[1 + z^4]), -(1/Sqrt[1 + z^4])} {(-1)^(1/4) ...

1

Here's a way to see the solution in Bessel function format: Simplify[FunctionExpand[Activate[MeijerGReduce[FunctionExpand[ DSolveValue[y''[x] + x^2 y[x] == 0, y[x], x], x]]]]] ((-1 - I) Sqrt[2 π] x BesselJ[1/4,x^2/2] (C + I C) + 2 Sqrt[π] Sqrt[x^2] BesselJ[-(1/4), x^2/2] (C + C))/(2 2^(3/4) (x^2)^(1/4)) Collect[%, _C, ...

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