Iteration of a Matrix
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2
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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
asked 1 hour ago
ithilquessirr
594
add a comment |
up vote
2
down vote
favorite
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
asked 1 hour ago
ithilquessirr
594
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
asked 1 hour ago
ithilquessirr
594
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
asked 1 hour ago
ithilquessirr
594
asked 1 hour ago
ithilquessirr
594
asked 1 hour ago
ithilquessirr
594
asked 1 hour ago
ithilquessirr
594
asked 1 hour ago
ithilquessirr
594
594
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered 1 hour ago
Henrik Schumacher
47.6k466134
add a comment |
up vote
1
down vote
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, {2}]
matrix1
and
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered 1 hour ago
Titus
520317
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered 1 hour ago
Henrik Schumacher
47.6k466134
add a comment |
up vote
5
down vote
accepted
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered 1 hour ago
Henrik Schumacher
47.6k466134
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered 1 hour ago
Henrik Schumacher
47.6k466134
NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]
answered 1 hour ago
Henrik Schumacher
47.6k466134
answered 1 hour ago
Henrik Schumacher
47.6k466134
answered 1 hour ago
Henrik Schumacher
47.6k466134
answered 1 hour ago
Henrik Schumacher
47.6k466134
47.6k466134
add a comment |
add a comment |
up vote
1
down vote
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, {2}]
matrix1
and
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered 1 hour ago
Titus
520317
add a comment |
up vote
1
down vote
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, {2}]
matrix1
and
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered 1 hour ago
Titus
520317
add a comment |
up vote
1
down vote
up vote
1
down vote
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, {2}]
matrix1
and
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered 1 hour ago
Titus
520317
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, {2}]
matrix1
and
Fold[f[#1] &, matrix1, Range[100]]
print the last result.
answered 1 hour ago
Titus
520317
answered 1 hour ago
Titus
520317
answered 1 hour ago
Titus
520317
answered 1 hour ago
Titus
520317
520317
add a comment |
add a comment |
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