Let [ 1 ; 2 , 1 ¯ be a continued fraction. Find the corresponding number.

Question

Let   [  1  ;            2      ,      1        ¯   be a continued fraction. Find the corresponding number.

antantil0 · Accepted Answer

For something like this, it becomes  x  −  1  =                                                  1                                                                  2          +                                                                                                                  1                                                                                                                                                  x                                                                            ⟹    x  −  1  =                                                  1                                                                                                                                                                          2                  x                  +                  1                                                                                                                                                  x                                                                          =      x          2      x      +      1        ,Then, you can use the regular technique. If this is   x, then  x  −  1  =                                                  1                                                                  2          +                                                                                                                  1                                                                                                                                                  x                                                                            ⟹    x  −  1  =                                                  1                                                                                                                                                                          2                  x                  +                  1                                                                                                                                                  x                                                                          =      x          2      x      +      1        ,then you can solve.In general, for                     b        1            ,              b        2            ,      ⋯      ,              b        n              ¯  , instead of the denominators outside of the infinite fractions being continues   b as suggested by       b    ¯  , it keeps cycling between       b    1    ,      b    2    ,      b    3    ,  ⋯  ,      b    n  .

Brooke Kramer · Accepted Answer

Let  A  =  1  +                                                  1                                                                  2          +                                                                                                                  1                                                                                                                                                  1                  +                                                                                                                                                                                  1                                                                                                                                                                                                                                  2                          +                                                                                                                                                                                                                                                  1                                                                                                                                                                                                                                                                                                                  ⋱                                                                                                                                                                                                                                                                                Then notice that  A  =  1  +                                                  1                                                                  2          +                                                                                                                  1                                                                                                                                                  A                                                                        This gives  (  A  −  1  )  (  2  A  +  1  )  =  AYou can solve this quadratic.

Let [ 1 ; <mover> 2 , 1 </mrow> <mo accent="false"

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