Iteration on a matrix












3














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.










share|improve this question





























    3














    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.










    share|improve this question



























      3












      3








      3







      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.










      share|improve this question















      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 17 at 14:16









      Αλέξανδρος Ζεγγ

      4,0421928




      4,0421928










      asked Dec 16 at 20:01









      ithilquessirr

      644




      644






















          3 Answers
          3






          active

          oldest

          votes


















          9














          NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]





          share|improve this answer





























            6














            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]






            share|improve this answer





























              4














              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.






              share|improve this answer























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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                9














                NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]





                share|improve this answer


























                  9














                  NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]





                  share|improve this answer
























                    9












                    9








                    9






                    NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]





                    share|improve this answer












                    NestList[x [Function] 2 x + 4, {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, 100]






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Dec 16 at 20:11









                    Henrik Schumacher

                    48.1k466135




                    48.1k466135























                        6














                        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]






                        share|improve this answer


























                          6














                          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]






                          share|improve this answer
























                            6












                            6








                            6






                            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]






                            share|improve this answer












                            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]







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Dec 16 at 21:51









                            bill s

                            52.7k375150




                            52.7k375150























                                4














                                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.






                                share|improve this answer




























                                  4














                                  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.






                                  share|improve this answer


























                                    4












                                    4








                                    4






                                    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.






                                    share|improve this answer














                                    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.







                                    share|improve this answer














                                    share|improve this answer



                                    share|improve this answer








                                    edited Dec 17 at 13:43

























                                    answered Dec 16 at 20:44









                                    Titus

                                    550317




                                    550317






























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