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-2

ListPlot[Table[10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]], {n, 0, 99}]] This alters the cognition of what is really going on hopefully. This plot causes some more sensation: ListPlot[Table[10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]], {n, 0, 10}]] How can this happen? It is the very same function on a subinterval. ListPlot[Table[10^-Accuracy[FresnelS[N[8 + 1*^-...


2

Use FunctionExpand $Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global`*"] i1 = Integrate[1/Sqrt[1 + t^4], {t, 0, 1}] (* (2 Gamma[5/4]^2)/Sqrt[π] *) i2 = Integrate[Sec[u]^2/Sqrt[1 + Tan[u]^4], {u, 0, π/4}] (* 1/2 EllipticK[1/2] *) i1 == i2 // FunctionExpand // FullSimplify (* True *)


4

The incorrect result is: FresnelS[N[8+1*^-28, 32]] //InputForm 0.4607524835944079246`3.970167826243401 Note that the precision of the output is 3.97, indicating that the 4th digit may not be accurate, which is exactly what you observe. If you increase the precision: FresnelS[N[8+1*^-28, 43]] //InputForm 0.4607524835944079246`3.970167826243401 You still ...


6

I think this is worth reporting to Support. For instance, using formula 7.5.8 from the DLMF: With[{x = N[8 + 1*^-28, 32]}, With[{ζ = Sqrt[π] (1 - I) x/2}, Im[(1 + I)/2 Erf[ζ]]]] 0.46021421439301448386198863207105 and the result is comparable to evaluating N[FresnelS[8 + 1*^-28], 32]. In theory, one is supposed to use the auxiliary functions $f(z)$ ...


2

It is even worse. While FullSimplify[Product[(1 - q^k z), {k, 0, n - 1}]/Product[(1 - q^k z), {k, 1, n - 1}]] yields 1-z , FullSimplify[QPochhammer[z, q, n]/QPochhammer[q z, q, n - 1]] just returns the input. On the other hand, FullSimplify[(1 - q) QGamma[x + 1, q]/QGamma[x, q]] returns 1 - q^x .


2

Use ReplaceAll to replace expressions that numerically evaluate to zero with zero. Clear["Global`*"] Derivative[1, 0, 0, 0][Hypergeometric2F1][1/2, 1, 3/2, 0] + Derivative[0, 0, 1, 0][Hypergeometric2F1][1/2, 1, 3/2, 0] + Derivative[0, 1, 0, 0][Hypergeometric2F1][1/2, 1, 3/2, 0] + Derivative[1, 0, 0, 0][Hypergeometric2F1][1/2, 1, 5/2, 0] /....


4

I want to add something to the discussion about the ResourceFunction SymbolQ. The OP observed that doing something like: x = 1; symbolQ = ResourceFunction["SymbolQ"]; ResourceFunction["SymbolQ"][x] symbolQ[x] (* True *) (* False *) does not work because the attributes of the resource function are not applied correctly. However, I just ...


2

With my GeometricAlgebra paclet you can work with any geometric number and apply any MatrixFunction supported function on them with MultivectorFunction , that effectively just converts numbers to matrices and back. Quaternions can be simply expressed as GeometricAlgebra[0,2]: PacletInstall["https://wolfr.am/OkONsyY2"] << GeometricAlgebra` ...


3

The root cause of the difficulties encountered by the OP is that MeijerG, as implemented in Mathematica, often is not accurate when evaluated at machine precision. For instance, with the parameters, params = {p -> 1, q1 -> 1, q2 -> 1, q13 -> 1, qi1 -> 1, mu -> 3/2, L -> 3, gammak1 -> 1, BI -> 9, C1 -> 4, Bsd -> 4, M -&...


2

DSolveyields an involved implicit result in terms of elliptic integrals. In order to solve explicitly the problem at hand we should first transform the equation. Let's multiply given differential equation by $\;y'(x)$ $$y'y''+c y' y^2+by y'+a y'=0$$ then integrating it and multiplying by $2$ we get $${y'}^2+\frac{2c}{3}y^3+b y^2+2a y+d=0$$ where $d$ is an ...


2

Not an answer, but long comment and some analysis. The problem can be more clearly seen as follows. The solution to the ODE contains elliptic integrals, whose solution contain elliptic special functions. The problem is evaluating this at $\infty$ and its derivative also, to solve for the constant of integrations. And this it can not do. The solutions to your ...


6

We can integrate in 3 steps: Integrate[(yp/(b + yp^2)^(3/2)), {yp, 0, Infinity}, Assumptions -> b > 0]*(x - xp) /. {b -> g (x - xp)^2} //Simplify Out[]: (x - xp)/Sqrt[g (x - xp)^2] So we have intyp=1/Sqrt[g] as results and it means that Q[g,x] not depends on x. Next step: Integrate[(Exp[-ypp^2/(8 T)])*(-1 + ypp^2/(8 T)) (ypp/(g*(xp - ...


1

To reduce the number of integration in NIntegrate seems reasonable. The effects are somehow dependent on the choices of options for NIntegrate. Choices are Values for lower integral bounds larger than zero. Values replacing the infinite upper bound of the integral to a meaningful numerical integration value. The default method is GlobalAdaptive. This can be ...


2

Workaround: $$\Gamma \left(\frac{k}{n}+\frac{z}{n}\right)=\frac{\Gamma \left(\frac{k}{n}+\frac{z}{n}+1\right)}{\frac{k}{n}+\frac{z}{n}}$$ $Version (* 12.1.1 for Microsoft Windows (64-bit) (June 9, 2020) *) Product[Gamma[k/n + z/n + 1]/(k/n + z/n), {k, 0, n - 1}] // FunctionExpand (* n^(1/2 - z) (2 \[Pi])^(-(1/2) + n/2) Gamma[z] *)


3

With version 12.1.0, for any positive integer value of $\,n\,$ you can use FunctionExpand to verify with ex[n_] := Product[Gamma[(k - 1 + z)/n], {k, n}] == Gamma[z](2Pi)^(n/2 - 1/2)n^(1/2 - z) // FunctionExpand For example, ex[3] returns True. However, using ex[n] does not evaluate to a truth value.


10

I'd probably use x_Symbol in a function argument to control evaluation. Otherwise, one might do the following (thanks to @Leonid for pointing out an oversight). If the argument is to be evaluated before testing: SymbolQ = MatchQ[#, t_Symbol /; AtomQ[t]] & If the argument is not to be evaluated: SymbolQ = Function[s, MatchQ[Unevaluated@s, t_Symbol /; ...


7

How about: SymbolQ[_Symbol] = True SymbolQ[_] = False ?


7

I found my answer thanks to Sjoerd Smit who referenced me to the Mathematica Function Repository. And yes it is appropriately called SymbolQ which is used like the following... xxx = 123 ResourceFunction["SymbolQ"][xxx] (* returns True *) A little bit ugly and long but it works. But why not fix if it isn't broken? And that is what I tried to do... ...


3

The following problem is hidden by NIntegrate, but which is one of the common sources of the NIntegrate::izero message you get from the OP's code. Exp[-I*k*s]*s^(1/2 + 2)*Exp[-s]*Cos[f/2]* Exp[-Sqrt[8*s*u]/Cos[f/2]] /. {k -> 0, u -> 0.1} /. {f -> 1., s -> 1000.} General::munfl: Exp[-1032.23] is too small to represent as a normalized machine ...


1

Your 'dumping' constant is a shift of $a$ parameter as it can be seen from your pics. To some get actual dumping your DE should be of the form: $$y'' + g y' + (a-2 q \cos(2 t)) y = 0$$ This equation is linear with periodic coefficients, stability info can be obtained from eigenvalues of the monodromy matrix. (* t-map *) map = DSolveValue[{y''[t] +g y'[t] + (...


5

This is not a bug. It's an improvement. The integral is divergent. The V5 Oscillatory method is defunct. NIntegrate chooses the "ExtrapolatingOscillatory" method (which is the method it chooses for this integral if Method -> Automatic). This method checks convergence, and the amplitude of the oscillations goes to infinity. Therefore you ...


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