New answers tagged version-8
3
"LevinRule" should work splendidly here, I think:
NIntegrate[-m Exp[-m] BesselJ[1, m]^2, {m, 0, Infinity},
Method -> "LevinRule", WorkingPrecision -> 20]
-0.18196415067209554877
ruebenko's answer has given a closed form for this particular definite integral. Personally, I prefer it when the parameters of the elliptic integrals are ...
4
This is a bug. As a workaround for this specific integral you could use a symbolic solution:
Integrate[-m*Exp[-m]*BesselJ[1, m]^2, {m, 0, Infinity}]
(* (-3*EllipticE[-4] + 5*EllipticK[-4])/(5*Pi) *)
1
To get an actual answer in this case, you could simply retreat to the numerical evaluation of this sum. It works by simply changing Sum to NSum in your code:
NSum[((-1)^(n) Log[1 + 2 n])/(1 + 2 n), {n, 0, Infinity}]
(* ==> -0.192901 *)
1
If you just want to do a simple cartoon (as opposed to evaluating the solution components of the Lorenz equations at particular values), you can just directly extract the points generated by NDSolve[]. Here's one way to go about it:
sol[σ_] :=
Transpose[Through[{x, y, z}["ValuesOnGrid"] /.
First @ NDSolve[{x'[t] == σ (y[t] - x[t]),
...
3
You can define a solution function depending on your parameter and then use it for the animation :
sol[sigma_] := {x, y, z} /.
NDSolve[{x'[t] == sigma (y[t] - x[t]), y'[t] == 28 x[t] - y[t] - x[t] z[t],
z'[t] == x[t]*y[t] - 8/3 z[t], x[0] == z[0] == 0, y[0] == 2},
{x, y, z}, {t, 0, 25}][[1]]
Animate[
With[{f = sol[sigma]},
...
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