Iteration on a matrix
I want to update "matrix1" 100 times. "matrix3" will be new "matrix1" and it will iterate 100 times. Should I use a loop or function? First iteration is:
matrix1 = ( {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (First iteration):
{{6, 8, 10}, {12, 14, 16}, {18, 20, 22}}
matrix3 will be new matrix1
matrix1 = ( {
{6, 8, 10},
{12, 14, 16},
{18, 20, 22}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (Second iteration):
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
And, It will repeat 100 times.
matrix education iteration
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 16 at 20:01
ithilquessirr
644
add a comment |
I want to update "matrix1" 100 times. "matrix3" will be new "matrix1" and it will iterate 100 times. Should I use a loop or function? First iteration is:
matrix1 = ( {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (First iteration):
{{6, 8, 10}, {12, 14, 16}, {18, 20, 22}}
matrix3 will be new matrix1
matrix1 = ( {
{6, 8, 10},
{12, 14, 16},
{18, 20, 22}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (Second iteration):
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
And, It will repeat 100 times.
matrix education iteration
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 16 at 20:01
ithilquessirr
644
add a comment |
I want to update "matrix1" 100 times. "matrix3" will be new "matrix1" and it will iterate 100 times. Should I use a loop or function? First iteration is:
matrix1 = ( {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (First iteration):
{{6, 8, 10}, {12, 14, 16}, {18, 20, 22}}
matrix3 will be new matrix1
matrix1 = ( {
{6, 8, 10},
{12, 14, 16},
{18, 20, 22}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (Second iteration):
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
And, It will repeat 100 times.
matrix education iteration
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 16 at 20:01
ithilquessirr
644
I want to update "matrix1" 100 times. "matrix3" will be new "matrix1" and it will iterate 100 times. Should I use a loop or function? First iteration is:
matrix1 = ( {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (First iteration):
{{6, 8, 10}, {12, 14, 16}, {18, 20, 22}}
matrix3 will be new matrix1
matrix1 = ( {
{6, 8, 10},
{12, 14, 16},
{18, 20, 22}
} );
matrix2 = matrix1*2 - 1;
matrix3 = matrix2 + 5;
matrix3
Output (Second iteration):
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
And, It will repeat 100 times.
matrix education iteration
matrix education iteration
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 16 at 20:01
ithilquessirr
644
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 16 at 20:01
ithilquessirr
644
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
edited Dec 17 at 14:16
Αλέξανδρος Ζεγγ
4,0421928
4,0421928
asked Dec 16 at 20:01
ithilquessirr
644
asked Dec 16 at 20:01
ithilquessirr
644
asked Dec 16 at 20:01
ithilquessirr
644
644
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
add a comment |
Another possibility is to write the problem via a recursion:
f[n_] := 2 f[n - 1] + 4;
f[1] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
To get any power, you then ask for
f[3]
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
or
f[100]
answered Dec 16 at 21:51
bill s
52.7k375150
add a comment |
Just for illustration and learning reasons, three examples with Fold, Do and Table. Define
matrix1 := {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Then
Table[matrix1 = 2*matrix1 + 4, {100}]
prints all intermediate matrices
Do[matrix1 = 2*matrix1 + 4, {100}]
matrix1
and
f[x_] := 2 x + 4
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered Dec 16 at 20:44
Titus
550317
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
add a comment |
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
add a comment |
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
answered Dec 16 at 20:11
Henrik Schumacher
48.1k466135
48.1k466135
add a comment |
add a comment |
Another possibility is to write the problem via a recursion:
f[n_] := 2 f[n - 1] + 4;
f[1] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
To get any power, you then ask for
f[3]
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
or
f[100]
answered Dec 16 at 21:51
bill s
52.7k375150
add a comment |
Another possibility is to write the problem via a recursion:
f[n_] := 2 f[n - 1] + 4;
f[1] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
To get any power, you then ask for
f[3]
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
or
f[100]
answered Dec 16 at 21:51
bill s
52.7k375150
add a comment |
Another possibility is to write the problem via a recursion:
f[n_] := 2 f[n - 1] + 4;
f[1] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
To get any power, you then ask for
f[3]
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
or
f[100]
answered Dec 16 at 21:51
bill s
52.7k375150
Another possibility is to write the problem via a recursion:
f[n_] := 2 f[n - 1] + 4;
f[1] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
To get any power, you then ask for
f[3]
{{16, 20, 24}, {28, 32, 36}, {40, 44, 48}}
or
f[100]
answered Dec 16 at 21:51
bill s
52.7k375150
answered Dec 16 at 21:51
bill s
52.7k375150
answered Dec 16 at 21:51
bill s
52.7k375150
answered Dec 16 at 21:51
bill s
52.7k375150
52.7k375150
add a comment |
add a comment |
Just for illustration and learning reasons, three examples with Fold, Do and Table. Define
matrix1 := {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Then
Table[matrix1 = 2*matrix1 + 4, {100}]
prints all intermediate matrices
Do[matrix1 = 2*matrix1 + 4, {100}]
matrix1
and
f[x_] := 2 x + 4
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered Dec 16 at 20:44
Titus
550317
add a comment |
Just for illustration and learning reasons, three examples with Fold, Do and Table. Define
matrix1 := {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Then
Table[matrix1 = 2*matrix1 + 4, {100}]
prints all intermediate matrices
Do[matrix1 = 2*matrix1 + 4, {100}]
matrix1
and
f[x_] := 2 x + 4
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered Dec 16 at 20:44
Titus
550317
add a comment |
Just for illustration and learning reasons, three examples with Fold, Do and Table. Define
matrix1 := {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Then
Table[matrix1 = 2*matrix1 + 4, {100}]
prints all intermediate matrices
Do[matrix1 = 2*matrix1 + 4, {100}]
matrix1
and
f[x_] := 2 x + 4
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered Dec 16 at 20:44
Titus
550317
Just for illustration and learning reasons, three examples with Fold, Do and Table. Define
matrix1 := {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Then
Table[matrix1 = 2*matrix1 + 4, {100}]
prints all intermediate matrices
Do[matrix1 = 2*matrix1 + 4, {100}]
matrix1
and
f[x_] := 2 x + 4
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered Dec 16 at 20:44
Titus
550317
edited Dec 17 at 13:43
answered Dec 16 at 20:44
Titus
550317
answered Dec 16 at 20:44
Titus
550317
answered Dec 16 at 20:44
Titus
550317
550317
add a comment |
add a comment |
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