8
myTable = Array[Subscript[a, Row[{##}]] &, {5, 10}];
MatrixForm @ myTable
functions = {foo, bar};
columns = {2, 5};
1. Part assignment:
myTable[[All, columns]] = {foo@#, bar@#2} & @@@ myTable[[All, columns]];
myTable // MatrixForm
2. MapAt
myTable = Array[Subscript[a, Row[{##}]] &, {5, 10}];
myTable = Fold[MapAt[First@#2, #, {All, Last@#2}] &...
8
A little too long for a comment, and I have marked this as 'Community Wiki' (not an original contribution). In addtion, I think the Query method should not be 'buried' in a comment.
Query
Sjoerd Smit has posted a neat method for applying a function to a matrix column using Query.
Query[All, {1 -> (#+1/137&),3->(2 Pi# + 42&)}]@myTable2//...
7
a[1] = 2;
a[n_] := a[n] = 4 Sum[a[i], {i, n - 1}]
Table[a[n], {n, 1, 10}]
{2, 8, 40, 200, 1000, 5000, 25000, 125000, 625000, 3125000}
6
kglr's ReplacePart method is completely general, but I think in many situations writing the rules explicitly is easier and gives more readable code. So I would write:
data = Array[f, {4, 4}];
ReplacePart[data, {{r_, 1} :> data[[r, 1]] + 137, {r_, 3} :> 2 π data[[r, 3]]}]
{{137 + f[1, 1], f[1, 2], 2 π f[1, 3], f[1, 4]},
{137 + f[2, 1], f[2, 2], 2 π f[...
6
Table[Plot[{Re[ReleaseHold@f] /. j -> 3, Im[ReleaseHold@f] /. j -> 3}, {n, 0, 15},
GridLines -> Automatic,
PlotRange -> All, ImageSize -> 350,
PlotLegends -> {"Real", "Imaginary"},
PlotLabel -> f],
{f, {HoldForm@Sum[Sinc[Pi*(n - i*j)], {i, 1, Floor[n]}],
HoldForm[(-1 + E^(2*I*Pi*n))/(j*(-1 + E^((2*I*...
5
Table[Plot[f, {x, 1, 10}], {f, {1/x, x^2}}]
Table[Plot[f[x], {x, 1, 10}], {f, {1/# &, #^2 &}}]
Table[Plot[f[x], {x, 1, 10}], {f, {1/x &, x^2 &}}]
all give
4
Sum[f, {t, tlist}]
E^(-20/T) + E^(-18/T) + E^(-16/T) + E^(-14/T) + E^(-12/T) + E^(-10/ T) +
E^(-8/T) + E^(-6/T) + E^(-4/T) + E^(-2/T)
3
You can introduce a memory variable b[n] and solve for both a[n] and b[n].
RecurrenceTable[{a[n + 1] == 4 b[n], b[n] == b[n - 1] + a[n],
a[1] == b[1] == 2}, {a, b}, {n, 1, 10}][[All, 1]]
(* {2, 8, 40, 200, 1000, 5000, 25000, 125000, 625000, 3125000} *)
2
Clear["Global`*"]
In your code, you have the wrong syntax for the Sum. However, even with the correct syntax
rt = RecurrenceTable[{a[n + 1] == 4 Sum[a[i], {i, 1, n}], a[1] == 2},
a, {n, 1, 10}]
As stated in the error message, all instances of a[_] must have arguments of the form n + integer
Amplifying on the answer by Suba Thomas
a[1] = 2;
a[...
2
You can do
FindInstance[n^2 == 9 (x^3 (y - 2)^2 + 3 x^2 (y - 2) - 2 x (y - 45) (y - 2) + 7 (y - 1)^2),
{x, y, n}, PositiveIntegers]
(* {{x -> 3, y -> 5, n -> 102}} *)
to find an exemplary instance.
If you want all solutions up to $x,y\le s$, you can do
With[{s = 20},
Solve[{n^2 == 9 (x^3 (y - 2)^2 + 3 x^2 (y - 2) - 2 x (y - 45)...
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 \[...
1
f[n_]:=2*(3n+6-1)(n+2)+Total[9Table[Binomial[n,j]+n,{j,0,n}] ]
f[0]
f[1]
f[2]
f[3]
29
84
178
320
Also here is the hole Table:
Table[Flatten[Riffle[Table[{3n+6-1,3n+6-1},n+2],9Table[Binomial[n,j]+n,{j,0,n} ]]],{n,0,3}]
{{5,5,9,5,5},{8,8,18,8,8,18,8,8},{11,11,27,11,11,36,11,11,27,11,11},{14,14,36,14,14,54,14,14,54,14,14,36,14,14}}
1
The working examples are all using the pattern iterating over a given list from the section Scope on the documentation page for Table.
Table[Sqrt[x], {x, {1, 4, 9, 16}}]
({1, 2, 3, 4})
Plot prefers the pure function as input.
In the section Generalizations & Extensions
Table[a[x]!, {a[x], 6}]
({1, 2, 6, 24, 120, 720})
The variables need not just be ...
Only top voted, non community-wiki answers of a minimum length are eligible
