Iteration on a matrix
$begingroup$
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 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
$endgroup$
add a comment |
$begingroup$
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 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
$endgroup$
add a comment |
$begingroup$
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 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
$endgroup$
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 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
edited Dec 17 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
edited Dec 17 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
edited Dec 17 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
edited Dec 17 '18 at 14:16
Αλέξανδρος Ζεγγ
4,1541929
4,1541929
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
asked Dec 16 '18 at 20:01
ithilquessirrithilquessirr
644
644
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
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add a comment |
$begingroup$
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 '18 at 21:51
bill sbill s
52.8k375150
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:44
TitusTitus
645417
$endgroup$
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
$begingroup$
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
$endgroup$
add a comment |
$begingroup$
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
$endgroup$
add a comment |
$begingroup$
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
$endgroup$
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
answered Dec 16 '18 at 20:11
Henrik SchumacherHenrik Schumacher
50.8k469145
50.8k469145
add a comment |
add a comment |
$begingroup$
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 '18 at 21:51
bill sbill s
52.8k375150
$endgroup$
add a comment |
$begingroup$
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 '18 at 21:51
bill sbill s
52.8k375150
$endgroup$
add a comment |
$begingroup$
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 '18 at 21:51
bill sbill s
52.8k375150
$endgroup$
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 '18 at 21:51
bill sbill s
52.8k375150
answered Dec 16 '18 at 21:51
bill sbill s
52.8k375150
answered Dec 16 '18 at 21:51
bill sbill s
52.8k375150
answered Dec 16 '18 at 21:51
bill sbill s
52.8k375150
52.8k375150
add a comment |
add a comment |
$begingroup$
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 '18 at 20:44
TitusTitus
645417
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:44
TitusTitus
645417
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:44
TitusTitus
645417
$endgroup$
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 '18 at 20:44
TitusTitus
645417
edited Dec 17 '18 at 13:43
answered Dec 16 '18 at 20:44
TitusTitus
645417
answered Dec 16 '18 at 20:44
TitusTitus
645417
answered Dec 16 '18 at 20:44
TitusTitus
645417
645417
add a comment |
add a comment |
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