10
Square brackets are only used for functions
x = Range[4]
(* {1, 2, 3, 4} *)
y = NestList[RotateLeft, x, Length[x] - 1]
(* {{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}} *)
If the extra brackets are intended
x = {Range[4]}
(* {{1, 2, 3, 4}} *)
y = {NestList[RotateLeft, x[[1]], Length[x[[1]]] - 1]}
(* {{{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2},...
9
Permute[Range[4], CyclicGroup[4]]
{{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}}
You can also use RotateLeft:
RotateLeft[Range[4], #] & /@ Range[0, 3]
{{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}}
9
Permute exists for reordering lists.
Permute[list, Cycles[{{1, 4}, {2, 5}, {3, 6}}]]
This swaps entries $1 \leftrightarrow 4$, $2 \leftrightarrow 5$ and $3 \leftrightarrow 6$.
The necessary permutation can be found using FindPermutation:
FindPermutation[Range@9, {4, 5, 6, 1, 2, 3, 7, 8, 9}]
Cycles[{{1, 4}, {2, 5}, {3, 6}}]
8
You can check whether the Head of the expression is Plus, which is associated with a non-factored polynomial:
list = { 1 + x^2,
(-1 + x) (1 + x + x^2),
(1 + x)^2,
2 (2 + x + x^2),
2 (2 + x)^2
};
ClearAll[isF]
isF[Times[_Integer, poly_]] := isF[poly]
isF[poly_] := Head[poly] =!= Plus
SetAttributes[isF, Listable]
isF@...
8
x = Range[4];
Partition[x, Length[x], 1, 1]
{{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}}
7
What you want is called "currying".
Not involving new operators you may try:
Clear[f, g]
f[x_, y_] := x^2 + y^3
g[x_] = Evaluate[f[x, #]] &;
f[3, 2] == 17 == g[3][2]
?? g
However, MMA has a special operator CurryApplied for this:
Clear[f, g]
f[x_, y_] := x^2 + y^3
g = CurryApplied[f, 2];
f[3, 2] == 17 == g[3][2]
?? g
You may want to also ...
7
At least shorter:
Flatten@Partition[list, 3][[{2, 1, 3}]]
6
Edit
SetAttributes[CircleDot, {Flat, OneIdentity}];
CircleDot[{x_, y_}, {a_, b_}] := {x ** θ[y] ** a, y ** b};
{g1,h1}⊙{g2, h2}⊙{g3, h3}
{g1 ** θ[h1] ** g2 ** θ[h1 ** h2] ** g3, h1 ** h2 ** h3}
Update
SetAttributes[CircleDot, {Orderless, Flat, OneIdentity}];
CircleDot[{x_, y_}, {a_, b_}] := {x*θ[y]*a, y*b}
{g1, h1}⊙{g2, h2}⊙{g3, h3}
{g1 g2 g3 θ[h2] θ[h1 ...
5
aa = Table[{tab1[[i]][[1]], tab1[[i]][[2]], tab2[[j]][[1]], tab2[[j]][[2]]},
{i, 1, Length[tab1], 1}, {j, 1, Length[tab2], 1}]; // AbsoluteTiming // First
0.146368
You can use Outer:
bb = Outer[Join, tab1, tab2, 1]; // AbsoluteTiming // First
aa == bb
0.040488
True
Or, if you have to execute it more often, you can Compile the code:
cf = Compile[{{...
5
How about this?
DeterminedQ[x_List] := And @@ DeterminedQ /@ x;
DeterminedQ[x_] := NumericQ[x];
5
Id suggest using TakeList
TakeList[ist, {{4, 6}, 3, All}]
TakeList[ist, {{4, 6}, {1, 3}, {1, 3}}]
The indices need to be relative to those elements that are not yet taken.
If you have absolute indices given:
ist = Range[20];
parts = {{5, 6}, {9, 13}, {1, 4}, {19, 20}, {14, 18}};
Catenate[ist[[# ;; #2]] & @@@ parts]
4
Here is an example. The condition is that k be at least equal to a+b. Since we cannot increment k we set the variable t to the value of k. The While increments t until it is equal or greater than a+b. At this point, just using t by itself will return the value of t.
Clear[f];
f[a_, b_, k_ : 0] :=
Block[ {t},
t = k;
While[t < a + b, t++];
t
...
4
ParametricPlot[{((1 - x^2)*Cos[θ] Sin[θ])/(Cos[θ]^2 + Sin[θ]^2*x^2), -(x/(Cos[θ]^2 + Sin[θ]^2*x^2))}, {θ, 0, π/2}, {x, 1, 10}, MeshFunctions -> {#3 &, #4 &},
MeshStyle -> {Red, Cyan}, Mesh -> 20,
PlotStyle -> {Opacity[.1], Yellow}, PlotPoints -> 80]
4
Clear["Global`*"]
Show[
ListLinePlot[{{0, 1}, {1, 1}, {10, 10}},
PlotStyle -> Directive[Red, Dashed, Thickness[0.008]],
Filling -> Top,
FillingStyle -> Lighter[Gray, 0.85]],
Plot[{2 Sin[x], 3 x - 2, x^2}, {x, 0, 3}, GridLines -> Automatic],
PlotRange -> {{0, 3}, Automatic}]
EDIT: To obscure the lines by the Filling(adjust ...
4
Instead of dealing with indices, try
Sum[Piecewise[{{f[m, n], m^2 + n^2 != 0}, {0, True}}], {m, -Infinity, Infinity}, {n, -Infinity, Infinity}]
It should be noticed the notation n,k is preferable over n,m which are similar.
4
I believe the relation you wrote is only valid for $n>2$. For $n=0,1,2$ you have to use the identities of eq.(2.31) in [1705.02909],
\begin{equation}
\begin{aligned}
t_{0,n_1,n_2\dots} &= Nt_{n_1,n_2},\\
t_{1,n_1,n_2\dots} &= 0,\\
t_{2,n_1,n_2\dots} &= \frac{N^2-1+n_1+n_2+\dots}{2}t_{n_1,n_2,\dots}.
\end{aligned}
\end{equation}
The code below ...
4
I think one way to do this is:
Select[mytup,
!(#[[8]] == -1
&& #[[9]] == 1
&& MemberQ[mytup, # * {1, -1, 1, 1, -1, 1, 1, -1, -1}])
&]
This keeps exactly the ones that fail the criterion, i.e. deletes the ones that do satisfy having their 8th part be -1, their 9th part be 1, and such that the list contains some member equal ...
3
Why not use InterpolatingPolynomial? For your example:
Expand @ InterpolatingPolynomial[
{{0, 1}, {1, 2}, {2, 4}, {3, 8}, {4, 16}},
x
]
1 + (7 x)/12 + (11 x^2)/24 - x^3/12 + x^4/24
3
Legended[Show[
Plot[{2 Sin[x], 3 x - 2, x^2}, {x, 0, 3}, GridLines -> Automatic],
ListLinePlot[{{0, 1}, {1, 1}, {10, 10}},
PlotStyle -> Directive[Red, Dashed, Thickness[0.008]],
Filling -> Top, FillingStyle -> Opacity[0.75, LightGray]]],
Placed[Column@{SwatchLegend[{Opacity[.5, LightGray], White},
{"Above", "...
3
This is what immediate assignments are for:
f[x_] = D[x^100*E^(2*x^5)*Cos[x^2], {x, 137}]
(* long output (can be suppressed with semicolon) *)
f[3.]
(* -8.90666869434476*10^664 *)
See here for a tutorial on the distinction between = and :=.
3
You get no result because Vi is a function.
If you replace Vi, say Vi0 by a constant you will get a solution that is not unique. But if you restrict the solutions to Reals you will get a unique results.
Here is the procedure:
Clear["Global`*"];
q = (((1602176634)/10^9))*10^(-19);
k = (((1380649)/10^6))*10^(-23);
\[Eta] = 3/2;
Td = 20 + (5463/20);
...
3
As it was suggested by @CarlWoll and @thorimur in the comments, I evaluated:
(9.2 - 8)/1.2 // FullForm
which returns:
0.9999999999999994`
I'm not really sure why, but I suspect that this is related to the order in which Mathematica evaluates each operation. Since 9.2/1.2 and 8/1.2 have no finite decimal expansion, apparently Mathematica truncates this ...
3
If you include a list of time series as first argument of DateListPlot you will get a plot with multiple curves.
Here is an example where I plot "TTM" and "Quarterly" of the given company in the same plot. For clarity, I first create the time series separately and subsequently feed them to DateListPlot:
ds1 = Entity["Company", &...
3
The U.S. National Imagery and Mapping Agency (NIMA) (formerly the Defense Mapping Agency) adopted a special grid for military use throughout the world called the Universal Transverse Mercator (UTM) grid. In this grid, the world is divided into 60 north-south zones, each covering a strip 6° wide in longitude. These zones are numbered consecutively beginning ...
2
Include Print["opt=", OptionValue@symbols]; in your function and you will see, that the options are already evaluated. Therefore, you need a function with one of the Hold Attributs: HoldFirst, HoldRest, HoldAll,... to get the symbole name, not its value.
Here is one way to do it:
ClearAll[freshStart];
SetAttributes[freshStart, HoldFirst]
freshStart[...
2
Conversion of comments to a community wiki answer:
Michael E2:
Try = instead of := or use Evaluate in pad[t_] := Evaluate[PadeApproximant[..]].
thorimur:
If you're wondering why that suggestion of @MichaelE2 works, it's because := does not evaluate its right hand side before substituting. So pad[1] substitutes 1 verbatim into PadeApproximant[t Log[t], {t, 1, ...
2
tup4a = Join @@@ Tuples[{tup1, tup3}];
tup4a // Short
{{0,-1,-1,0,-1,-1,d,0,0},{0,-1,-1,0,-1,-1,d,0,1},<<2912>>,
{1,1,1,1,1,1,d,-1,1},{1,1,1,1,1,1,d,-1,-1}}
tup4b = Distribute[{tup1, tup3}, List, List, List, Join];
tup4a == tup4b
True
"to get that directly from tup1 and tup2":
tup4c = Join @@@ Tuples[{tup1, {{d}}, tup2}];
tup4a ...
2
FilledCurve work for Line.
line = Line[{{0, 1}, {1, 1}, {10, 10}, {0, 10}}];
Plot[{2 Sin[x], 3 x - 2, x^2}, {x, 0, 3}, GridLines -> Automatic,
Epilog -> {{Gray, Opacity[.5], FilledCurve[line]},
Directive[Red, Dashed, Thickness[0.008]],
Line[{{0, 1}, {1, 1}, {10, 10}}]}]
2
Construct a polygon using the coordinates of the epilog line:
Plot[{2 Sin[x], 3 x - 2, x^2}, {x, 0, 3}, GridLines -> Automatic,
Epilog -> {Directive[Red, Dashed, Thickness[0.008]],
Line[{{0, 1}, {1, 1}, {10, 10}}],
LightGray, Opacity[.5],
Polygon[{{0, 1}, {1, 1}, {10, 10},
Scaled[{0, 1}, {10, 10}],
Scaled[{0, 1}, {0, 1}]}]}]
2
You can do
Sum[If[n == 0 && m == 0,0, f[m,n]],{m,-Infinity,Infinity},{n,-Infinity,Infinity}]
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