New answers tagged evaluation
2
Answer revised for Abs[R] > 1
The code for sigx in the question contains a typo, [-1 + Sqrt[R^2]] instead of (-1 + Sqrt[R^2]). With it fixed and R > 1 chosen, consistent with the code in the question, the following returns an answer in just a few minutes for Version 12.1.1
Integrate[x*sigx, x, Assumptions -> 0 < x < 1 && R > 1];
...
0
I don't remember any built-in option for this, so we may need to implement it on our own.
SetOptions[
EvaluationNotebook[],
CellProlog :> With[
{
status = (
SelectionMove[EvaluationNotebook[], Previous, CellGroup, AutoScroll -> False];
NotebookRead[EvaluationNotebook[]][[1,2]]
)
...
4
You may use the variable $Pre that is applied to every input expression before evaluation. Default is not set. If you want to stop logging, you simply say $Pre=. Here is what you do:
ClearAll[log, $Pre]
SetAttributes[log, HoldAll]
log[inp_] := (PutAppend[Unevaluated[inp], "NameOfLogFile"]; inp)
$Pre = log
3
You can get the sum for 169 values of $(i,j)$ by taking the limit of the what you get from Sum:
expr = FullSimplify[Sum[F[m, k, i, j], {m, 0, ∞}, {k, 0, ∞}]];
ss2[iv_, jv_] := ss2[iv, jv] = Limit[expr, {i, j} -> {iv, jv}]
mat = Table[ss2[iv, jv], {iv, 0, 12}, {jv, 0, 12}];
From these values FindSequenceFunction can suggest a closed form that works on ...
5
$Version
(* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *)
Clear["Global`*"]
F[m_, k_, i_, j_] := (-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j]
sum[i_, j_] =
Sum[F[m, k, i, j], {m, 0, Infinity}, {k, 0, Infinity}] // FullSimplify
(* -2^i E^(-3 + I i π) (-1 + j) Binomial[0, i] Binomial[1,
j] (Gamma[1 - i] + i ...
2
I tried to Plot3D it.
Plot3D[Evaluate@ Sum[(-1)^(m + k)/(m!*k!)*2^m*Binomial[m, i]*Binomial[k + 1, j], {m, 0, Infinity}, {k, 0, Infinity}], {i, -5, 5}, {j, -5, 5}, PlotPoints -> 20, PlotRange -> All]
Looks like a lot of discontinuity.
But there are some points return values.
/. {i -> 51/10, j -> 45/10}
-(1/(E^3))(-((544 (-2)^(1/10) E Binomial[...
1
I never played roulette, so check my assumptions: There are 37 wholes, 18 red 18 black, 1 green. If I set on red and the ball stop on red, I get twice my money back, otherwise I lose all. With this assumptions:
We mark "win" by 1 and loose by "0". Then playing n times the numbers of "wins" is:
Count[RandomChoice[{18, 19} -> {...
3
To add to what Kuba said: you can avoid the infinite loop by creating a button that starts the evaluation:
Button["Evaluate notebook",
NotebookEvaluate[EvaluationNotebook[], InsertResults -> True],
Method -> "Queued"
]
Note the Method option. Without it, the FE will freeze up completely.
2
ds = Replace[
assoc, {Entity["Individual", n_] :>
EntityValue[Entity["Individual", n], "PropertyAssociation"]},
Infinity]
or create a list of rules (using Normal) from the Association to use ReplaceAll.
ds = Normal[assoc] /.
Entity["Individual", n_] :>
EntityValue[Entity["Individual", n],...
0
The equation is not a differential equation, but a boundary condition. Without a PDE, the solution space is quite large and can be found by "algebraic" means without any integration.
eq = A h[b, t] + B Derivative[1, 0][h][b, t] == g[t];
a0 = Solve[eq, h[b, t]]
(* {{h[b, t] -> (g[t] - B*Derivative[1, 0][h][b, t])/A}} *)
hEXPR = Normal@...
1
Clear["Global`*"]
\[ScriptH][k_] :=
DifferenceRoot[
Function[{\[FormalY], \[FormalN]}, {-(1 + k)^2 + (3 + 4 k) \[FormalN] -
4 \[FormalN]^2 -
2 (-k - 1 + \[FormalN]) (-k/2 + \[FormalN]) (-k - 1 +
2 \[FormalN]) \[FormalY][\[FormalN]] +
2 (-k - 1 + \[FormalN]) (-k/2 + \[FormalN]) (-k - 1 +
2 \[...
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