New answers tagged evaluation
4
A zero-order interpolation is a good way to in effect invert the function from positive integers to values in sortedlist.
Timing[
table = Table[N[n^2 + m^2], {n, 1, 3000},
{m, 1, 3000}];
sortedlist = Sort[Apply[Join,
table[[1 ;; 50]]]];
lastpairs =
SplitBy[Join[{{0, 0}},
Transpose[{sortedlist,
Range[Length[sortedlist]]}]],
First][...
3
Here are some suggestions for simpler methods to find the counts, and to make a plot.
Instead of Array[table[[#]]&, 50]] use table[[;; 50]]. Check the documentation for Part.
An easy way to count the number of values that are less or equal to $x$, is to count the values with Tally, and then total the tallies with Accumulate.
table = Table[N[n^2 + m^2], {...
16
Workaround:
This issue seems to be related to caching of previous numerical values. If we clear the cache during each iteration we get a result back:
Table[
ClearSystemCache["Numeric"];
N[Log[Im[ZetaZero[n + 1]]/Im[ZetaZero[n]]]],
{n, 1, 128}
]; // AbsoluteTiming
{6.8104, Null}
What to do
Report this to WRI.
It's strange that N[ZetaZero[...
3
If expr is your expression with the Inactivate, and you're looking for a complex (Wirtinger) derivative, then we can use ResourceFunction["ComplexD"] and simplify assuming t is real valued:
FullSimplify[
ResourceFunction["ComplexD"][expr, t]
, t ∈ Reals]
Which gives us:
(-0.92 sqr Cosh[(sqr t)/2] + (0.8448 + 0.25 sqr^2) Sinh[(sqr t)/2])...
2
If the plot is the main goal, then this is simplest:
xsol = NDSolveValue[eqns, x, {t, t0, tfin}] (* N.B. x, not x[s] *)
ListLinePlot@xsol
See also
Easy way to plot ODE solutions from NDSolve?
1
You can use a Table:
With[{x0 = {1, 3, 6}, t0 = 0,tfin = 5},
solx[s_] = Table[NDSolveValue[{x'[t] == -x[t], x[0] == x0[[i]]}, x[s],
{t, t0, tfin}], {i, 1, Length[x0]}];
Plot[Evaluate[solx[t]], {t, 0, 5}]]
3
You can use Indexed:
Plot[Evaluate @ Table[Indexed[xsol[t], i], {i, 3}], {t, t0, tfin}]
2
Perhaps this is what you want:
x0 = {1, 3, 6};
eqns = {x'[t] == -x[t], x[0] == #} &;
t0 = 0;
tfin = 5;
xsol[s_] = NDSolveValue[eqns@#, x[s], {t, t0, tfin}] & /@ x0
Plot[Evaluate@xsol[t], {t, t0, tfin}]
10
Root Cause
The root cause of the error is that we are attempting to join tensors of incompatible shapes. Namely, a 1x2 array...
Dimensions[{{0,0}}]
(* {1, 2} *)
... with a zero-length one-dimensional array:
Dimensions[ConstantArray[{0, 0}, 0]]
(* {0} *)
Dimensions[{}]
(* {0} *)
While this is permitted by the main evaluator, the compiler applies stricter ...
11
I think it's a bug that DeleteCases does not work; more specifically, it is not permitted to return an empty tensor when deleting elements from an tensor of rank > 1. Compile can return an empty tensor, but certain internal functions cannot. In particular Table in this case, but also Part. It seems a complicated situation, would take a long time to ...
2
You can also hold then release:
Hold[rvec[[2]]] /. rvec -> {x, y} // ReleaseHold
y
Pattern replacement works through hold.
3
Using exact math works:
subst = SetPrecision[{μ -> 1.0649627263045793` mχ,
mχ -> m0,
μF -> 0.126881598`,
σ0 -> 5.137783801543852`*^-18,
nn -> 0.0022767242753033053`,
B0 -> 0.5801875039070804`,
mn -> 0.939`},
...
4
It's instructive to look at the output of TracePrint.
Length[Unevaluated[1 + 1 + 1 + 1]] // TracePrint
(* Out: *)
Length[Unevaluated[1+1+1+1]]
Length
Length[1+1+1+1]
4
Whereas
f[Unevaluated[1 + 1 + 1 + 1]] // TracePrint
(* Out: *)
f[Unevaluated[1+1+1+1]]
f
f[1+1+1+1]
Length[1+1+1+1]
Length
1+1+1+1
Plus
1
1
1
1
4
Length[4]
...
0
Try this (which works in 0.1 second):
Clear[numNeeded, dimension, thevecs, thevertices, theWeights,
numConstraints];
numNeeded = 50;
dimension = 15;
numConstraints = 1000;
thevecs =
Normalize /@ RandomReal[{-1, 1}, {numConstraints, dimension}];
thevertices =
Flatten[Table[Drop[thevecs, {i}], {i, Length[thevecs]}], 1];
theWeights =
Normalize /@ ...
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