how to convert continued fractions into normal fractions?












1












$begingroup$


i couldnt find anything on google so i just tried opening it normally and recording each step. so i got:



[d,c,b,a] = ((((a)*b+1)*c+A)*d+B)/C.



[e;d,c,b,a]=(((((a)*b+1)*c+A)*d+B)*e+C)/D.



etc..



(X is the vaulue of the whole shell that x is a part of)



is this the standard way to do it? is there a better way to notate what i did in the example? are there other methods?



edit:



i also tried factorising, so thats another method, there is a pattern here but i cant see it completly yet



[a;b] = (ab+b)/b



[a;b,c] = (abc+a+c)/(bc+1)



[a;b,c,d] = (abcd+ab+ad+cd+1)/(bcd+b+d)



(im not sure what branch of math this is so sorry if i tagged it wrong)






sequences-and-series fractions continued-fractions







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    i couldnt find anything on google so i just tried opening it normally and recording each step. so i got:



    [d,c,b,a] = ((((a)*b+1)*c+A)*d+B)/C.



    [e;d,c,b,a]=(((((a)*b+1)*c+A)*d+B)*e+C)/D.



    etc..



    (X is the vaulue of the whole shell that x is a part of)



    is this the standard way to do it? is there a better way to notate what i did in the example? are there other methods?



    edit:



    i also tried factorising, so thats another method, there is a pattern here but i cant see it completly yet



    [a;b] = (ab+b)/b



    [a;b,c] = (abc+a+c)/(bc+1)



    [a;b,c,d] = (abcd+ab+ad+cd+1)/(bcd+b+d)



    (im not sure what branch of math this is so sorry if i tagged it wrong)






    sequences-and-series fractions continued-fractions







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      i couldnt find anything on google so i just tried opening it normally and recording each step. so i got:



      [d,c,b,a] = ((((a)*b+1)*c+A)*d+B)/C.



      [e;d,c,b,a]=(((((a)*b+1)*c+A)*d+B)*e+C)/D.



      etc..



      (X is the vaulue of the whole shell that x is a part of)



      is this the standard way to do it? is there a better way to notate what i did in the example? are there other methods?



      edit:



      i also tried factorising, so thats another method, there is a pattern here but i cant see it completly yet



      [a;b] = (ab+b)/b



      [a;b,c] = (abc+a+c)/(bc+1)



      [a;b,c,d] = (abcd+ab+ad+cd+1)/(bcd+b+d)



      (im not sure what branch of math this is so sorry if i tagged it wrong)






      sequences-and-series fractions continued-fractions







      share|cite|improve this question











      $endgroup$




      i couldnt find anything on google so i just tried opening it normally and recording each step. so i got:



      [d,c,b,a] = ((((a)*b+1)*c+A)*d+B)/C.



      [e;d,c,b,a]=(((((a)*b+1)*c+A)*d+B)*e+C)/D.



      etc..



      (X is the vaulue of the whole shell that x is a part of)



      is this the standard way to do it? is there a better way to notate what i did in the example? are there other methods?



      edit:



      i also tried factorising, so thats another method, there is a pattern here but i cant see it completly yet



      [a;b] = (ab+b)/b



      [a;b,c] = (abc+a+c)/(bc+1)



      [a;b,c,d] = (abcd+ab+ad+cd+1)/(bcd+b+d)



      (im not sure what branch of math this is so sorry if i tagged it wrong)






      sequences-and-series fractions continued-fractions




      sequences-and-series fractions continued-fractions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 20:33







      Mettek

















      asked Jan 23 at 19:44









      MettekMettek

      537




      537






















          1 Answer
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          1












          $begingroup$

          Set



          $$h_{-2} = 0, quad h_{-1} = 1\
          k_{-2} = 1, quad k_{-1} = 0\$$



          and recursively define



          $$h_n = a_nh_{n-1} + h_{n-2}\
          k_n = a_nk_{n-1} + k_{n-2}$$



          Then your number is



          $$[a_0; a_1, ldots, a_n] = frac{h_n}{k_n}$$



          The Wikipedia article on continued fractions is a good reference.



          Example: let's turn [1; 2, 2, 2] into a normal fraction. Start with a table like this:



          $$
          left[
          begin{array}{c|cc}
          n&-2&-1&0&1&2&3\
          hline
          a_n& & &1&2&2&2\
          h_n&0&1\
          k_n&1&0
          end{array}
          right]$$



          Then fill out $h_n$ and $k_n$ from left to right using the formulas above:



          $$
          left[
          begin{array}{c|cc}
          n&-2&-1&0&1&2&3\
          hline
          a_n& & &1&2&2&2\
          h_n&0&1&1&3&7&17\
          k_n&1&0&1&2&5&12
          end{array}
          right]$$



          So $[1;2,2,2] = frac{17}{12}$, which incidentally is close to $sqrt2$.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Set



            $$h_{-2} = 0, quad h_{-1} = 1\
            k_{-2} = 1, quad k_{-1} = 0\$$



            and recursively define



            $$h_n = a_nh_{n-1} + h_{n-2}\
            k_n = a_nk_{n-1} + k_{n-2}$$



            Then your number is



            $$[a_0; a_1, ldots, a_n] = frac{h_n}{k_n}$$



            The Wikipedia article on continued fractions is a good reference.



            Example: let's turn [1; 2, 2, 2] into a normal fraction. Start with a table like this:



            $$
            left[
            begin{array}{c|cc}
            n&-2&-1&0&1&2&3\
            hline
            a_n& & &1&2&2&2\
            h_n&0&1\
            k_n&1&0
            end{array}
            right]$$



            Then fill out $h_n$ and $k_n$ from left to right using the formulas above:



            $$
            left[
            begin{array}{c|cc}
            n&-2&-1&0&1&2&3\
            hline
            a_n& & &1&2&2&2\
            h_n&0&1&1&3&7&17\
            k_n&1&0&1&2&5&12
            end{array}
            right]$$



            So $[1;2,2,2] = frac{17}{12}$, which incidentally is close to $sqrt2$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Set



              $$h_{-2} = 0, quad h_{-1} = 1\
              k_{-2} = 1, quad k_{-1} = 0\$$



              and recursively define



              $$h_n = a_nh_{n-1} + h_{n-2}\
              k_n = a_nk_{n-1} + k_{n-2}$$



              Then your number is



              $$[a_0; a_1, ldots, a_n] = frac{h_n}{k_n}$$



              The Wikipedia article on continued fractions is a good reference.



              Example: let's turn [1; 2, 2, 2] into a normal fraction. Start with a table like this:



              $$
              left[
              begin{array}{c|cc}
              n&-2&-1&0&1&2&3\
              hline
              a_n& & &1&2&2&2\
              h_n&0&1\
              k_n&1&0
              end{array}
              right]$$



              Then fill out $h_n$ and $k_n$ from left to right using the formulas above:



              $$
              left[
              begin{array}{c|cc}
              n&-2&-1&0&1&2&3\
              hline
              a_n& & &1&2&2&2\
              h_n&0&1&1&3&7&17\
              k_n&1&0&1&2&5&12
              end{array}
              right]$$



              So $[1;2,2,2] = frac{17}{12}$, which incidentally is close to $sqrt2$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Set



                $$h_{-2} = 0, quad h_{-1} = 1\
                k_{-2} = 1, quad k_{-1} = 0\$$



                and recursively define



                $$h_n = a_nh_{n-1} + h_{n-2}\
                k_n = a_nk_{n-1} + k_{n-2}$$



                Then your number is



                $$[a_0; a_1, ldots, a_n] = frac{h_n}{k_n}$$



                The Wikipedia article on continued fractions is a good reference.



                Example: let's turn [1; 2, 2, 2] into a normal fraction. Start with a table like this:



                $$
                left[
                begin{array}{c|cc}
                n&-2&-1&0&1&2&3\
                hline
                a_n& & &1&2&2&2\
                h_n&0&1\
                k_n&1&0
                end{array}
                right]$$



                Then fill out $h_n$ and $k_n$ from left to right using the formulas above:



                $$
                left[
                begin{array}{c|cc}
                n&-2&-1&0&1&2&3\
                hline
                a_n& & &1&2&2&2\
                h_n&0&1&1&3&7&17\
                k_n&1&0&1&2&5&12
                end{array}
                right]$$



                So $[1;2,2,2] = frac{17}{12}$, which incidentally is close to $sqrt2$.






                share|cite|improve this answer











                $endgroup$



                Set



                $$h_{-2} = 0, quad h_{-1} = 1\
                k_{-2} = 1, quad k_{-1} = 0\$$



                and recursively define



                $$h_n = a_nh_{n-1} + h_{n-2}\
                k_n = a_nk_{n-1} + k_{n-2}$$



                Then your number is



                $$[a_0; a_1, ldots, a_n] = frac{h_n}{k_n}$$



                The Wikipedia article on continued fractions is a good reference.



                Example: let's turn [1; 2, 2, 2] into a normal fraction. Start with a table like this:



                $$
                left[
                begin{array}{c|cc}
                n&-2&-1&0&1&2&3\
                hline
                a_n& & &1&2&2&2\
                h_n&0&1\
                k_n&1&0
                end{array}
                right]$$



                Then fill out $h_n$ and $k_n$ from left to right using the formulas above:



                $$
                left[
                begin{array}{c|cc}
                n&-2&-1&0&1&2&3\
                hline
                a_n& & &1&2&2&2\
                h_n&0&1&1&3&7&17\
                k_n&1&0&1&2&5&12
                end{array}
                right]$$



                So $[1;2,2,2] = frac{17}{12}$, which incidentally is close to $sqrt2$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 23 at 21:11

























                answered Jan 23 at 21:02









                ThéophileThéophile

                20.2k13047




                20.2k13047






























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