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246 results

Questions on the use of numerical functions NIntegrate and NDSolve.

Several minor changes to the last five lines of code resolve the issues in the question. Alpha[t_] := 2 EtaSigmaS[phi1[t], phi2[t]] /. First[numsol]; BBeta[t_] := -2 Epsilon[phi1[t], phi2[t]] + EtaSi …
answered Jun 16 '17 by bbgodfrey
As suggested by Patrick Stevens, it is highly desirable to use exact numbers. B = 25*10^(-7); l = 2/100; V = 1/100; Further, X[t]is entirely negligible for t > 1/10. LogPlot[Evaluate[ReIm[Exp[X[t] …
answered Sep 30 '15 by bbgodfrey
Integrate can be used to handle the DiracDelta, Integrate[Integrand[p, Q2, ν, θ], {p, 0, Infinity}, Assumptions -> p3zero2[Q2, ν, θ] ∈ Reals && p3zero1[Q2, ν, θ] ∈ Reals] (* ((HeavisideTheta[p3z …
answered Aug 26 '15 by bbgodfrey
The code as displayed in the Question runs fine for me using Mathematica 10.0.2.0 under Windows 8.1 (64 bit). However, as I noted in a Comment, z0 = 0 causes z[t] to remain zero. Arbitrarily, I set …
answered Jan 20 '15 by bbgodfrey
This is an extended comment plus a work-around (at least, for the specific functions in the question). According to its documentation, the value of \$MaxPiecewiseCases can impact such functions as Pie …
answered Sep 4 '16 by bbgodfrey
As noted in the Comments, this integration is plagued by precision problems. To proceed, factor the huge constant (h/(((1.989)*10^(30)))*(r^3)) u (* 3.44312*10^51 *) from the integrand and then …
answered Jun 12 '15 by bbgodfrey
For every point {q, b} in the integrand, there is another point with the values of q and b interchanged, and the value of the integrand is the negative of the first point. So, the integral must be ze …
answered Aug 26 '16 by bbgodfrey
A bit of experimentation indicates that "LevinRule" does not work well, and that "MaxErrorIncreases" and MaxRecursion must be increased. Most importantly, WorkingPrecision must be increased substant …
answered Dec 21 '16 by bbgodfrey
The error messages are caused by arguments not being passed to function, by a conditional answer being returned by Integrate, by an infinite recurrence in the definition of zielfun, and by discontinui …
answered Feb 14 '15 by bbgodfrey
Holding NIntegrate Unevaluted also works. Unevaluated[NIntegrate[f, {x, a, b}]] /. {a -> 1, b -> 4, f -> 2/x} (* 2.77259 *)
answered Mar 24 '16 by bbgodfrey
In an effort to answer the question, I Rationalized all constants. Used WhenEvent to stop DSolvebefore z'[ρ] becomes singular. Solved for both z and z' in the second instance of DSolve to improve th …
answered Jul 29 '15 by bbgodfrey
To understand what is happening, consider ComplexExpand[z2, z2] (* I Im[z2] + Re[z2] *) % /. (z2 -> x + I*y) (* -Im[y] + Re[x] + I (Im[x] + Re[y]) *) The code of Nint2, although more complicated, …
answered Feb 18 '17 by bbgodfrey
This ODE system can be solved as outlined in my earlier comment above. For convenience, define eq1 = f''[x] - 4 g'[x] f'[x]; eq2 = g''[x] - λ h'[x]^2; eq3 = g'[x]^2 f[x] - 1/4 f'[x] g'[x] - 1 - λ h' …
answered Sep 11 '18 by bbgodfrey