5
$Version
(* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *)
int[c_] = Integrate[Exp[-c*x + x^2], {x, 0, c}]
(* E^(-(c^2/4)) Sqrt[π] Erfi[c/2] *)
Erfi is real for real c
FunctionDomain[int[c], c]
(* True *)
Plot[int[c], {c, -15, 15}]
For an alternate representation using DawsonF (EDIT: Eliminated unnecessary assumption)
int[c] // ...
3
The above expression for EisensteinE[2, q] is a bit slow. Here is a faster version using double precision arithmetic:
EisensteinE[2, tau_Complex /; Im[tau] > 0] := ee2Upper[tau];
ee2Upper = Compile[
{{tauIn, _Complex}},
Block[{tau = tauIn, f, k, q, qp, s},
tau -= Round[Re[tau]];
If[7 Im[tau] < 6,
f = ee2Upper[-1/...
3
Clear["Global`*"]
f[x_] = x*Sin[x];
data = Table[{x, f[x]}, {x, 0, 10, 0.5}];
model[m_, x_] := Sum[a[n] ChebyshevT[n, 2 x - 1], {n, 0, m}];
nlm[m_Integer?Positive, x_] :=
NonlinearModelFit[data, model[m, x], Array[a, m + 1, 0], x]
Manipulate[
Module[{model = nlm[m, x]},
Column[{
Plot[Evaluate@{f[x], model // Normal}, {x, 0, 10},
...
3
modified answer (11.04.2021)
(Thanks @DanielLichtblau for his helpful comment)
Here my ideas to make it work:
Mathematica evaluates the first integral to
Integrate[Exp[a u^2 + b v^2 +c u v], {v, -∞,∞}, {u, -∞, ∞},Assumptions -> {Element[{a, b, c}, Reals] ]
(*ConditionalExpression[(2 \[Pi])/(Sqrt[-b]Sqrt[-4 a + c^2/b]),
4 a b^2 < b c^2] *)
result is ...
3
In this case it really helps to use an explicit formula for $U(2,b,z)$, which can be expressed in terms of the much more stable exponential integral function:
U2[b_, z_] = HypergeometricU[2, b, z] // FullSimplify
(* (1 - E^z (2 - b + z) ExpIntegralE[2 - b, z])/(-2 + b) *)
U2[-97., 177.]
(* 0.0000130837 *)
Such transformations exist very often ...
3
$Version
(* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *)
Clear["Global`*"]
HypergeometricU[2., -97., 177.]
(* -1.18092*10^64 *)
Any calculation done with machine precision is done with the understanding that "you get what you get." Machine precision is fast but neither tracks nor attempts to control precision. ...
2
There's a rounding error at the initial sampling point that causes the x coordinate to become purely imaginary. One fix is to add Method -> "BoundaryOffset" -> True:
ParametricPlot[
{hR[Tan[α], γ], Areac[Tan[α], γ]},
{α, FBc[γ],
FAc[γ] - 0.001 (*to avoid-∞*)},
PlotRange -> {{0, 1.2}, {-10, -1}}, AxesOrigin -> {0.0, -5},
...
2
You can use a different representation of the skew harmonic numbers in terms of harmonic numbers:
$$
\bar{H}_n=\frac{1}{2} (-1)^n \left(H_{\frac{n-1}{2}}-H_{\frac{n}{2}}\right)+\log (2)
$$
This expression can be expanded around infinity
f[n_] := HarmonicNumber[(n - 1)/2] - HarmonicNumber[n/2]
Series[f[n], {n, Infinity, 6}]
$$-\frac{1}{n}+\frac{1}{2 n^2}-\...
2
Clear["Global`*"]
SphericalHarmonicY[1, 0 , θ, Φ] is real for real {θ, Φ}
FunctionDomain[
SphericalHarmonicY[1, 0, θ, Φ], {θ, Φ}]
(* True *)
The min and max values are
{min, max} = #[{Re@SphericalHarmonicY[1, 0, θ, Φ],
0 <= θ <= Pi,
0 <= Φ <= 2 Pi}, {θ, Φ}] & /@ {MinValue,
MaxValue}
(* {-(Sqrt[(3/π)]/2), Sqrt[3/...
1
Shifted Chebyshev polynomials are orthogonal on {0,1} with the weight function 1/Sqrt[x-x^2].
As the data: dat is not known, but we can create an interpolating function:
fun[x_]=Interpolation[dat]
For an example, we choose funx[x_]=Sin[2Pi x]. We can now expand this function in a series of shifted Chebyshev polynomials. The expansion coefficients are given ...
1
Clear["Global`*"]
Functions should use explicit arguments for all variables.
g[bi_, β_] := (β*BesselJ[1, β]) - (bi*BesselJ[0, β]);
A function that uses a numeric technique (e.g., NSolve) should use SetDelayed rather than Set and have its arguments restricted to numeric values.
EDIT: Used Rationalize on argument and used Solve rather than NSolve
...
1
The best way to get an idea of asymptotic power-law behavior is to make a log-log plot.
LogLogPlot[Abs[curvature] /. k -> 1, {r, 10^-6, 10^6}]
Note that we have to use Abs here because curvature, as defined, takes on both positive and negative values.
The slope of a log-log graph is, of course, the exponent of the power-law behavior of the function. ...
Only top voted, non community-wiki answers of a minimum length are eligible
