Nested fractional denominator(up to infinity)calculationQuestion: 4 + 1 2 + 1 1 + 1 3 + 1 1 + 1 2 + 1 8 + 1 ⋱ = A Find the positive integer A in the equation above.Details and assumptionsThe pattern repeats 2, 1, 3, 1, 2, 8 infinitely, but 4 comes only once, i.e., at the beginning.

Question

Nested fractional denominator(up to infinity)calculationQuestion:  4  +            1              2        +                              1                          1              +                                                1                                      3                    +                                                                  1                                                  1                          +                                                                                    1                                                              2                                +                                                                                                      1                                                                          8                                      +                                                                                                                        1                                          ⋱                                                                                                                                                                                                                                                                                                                                                                                                                                      =      A  Find the positive integer A in the equation above.Details and assumptionsThe pattern repeats 2, 1, 3, 1, 2, 8 infinitely, but 4 comes only once, i.e., at the beginning.

Willie Smith · Accepted Answer

Let&#039;s take a simpler example first: say,      A    =  1  +            2              3        +                              2                          3              +              .              .              .                                            .Here the repetition   2  ,  3  ,  2  ,  3  ,  .  .  . continues forever after the   1.The problem is that this looks on the face of it like some sort of &quot;infinitary&quot; equation (or rather, it&#039;s a normal equation with an &quot;infinitary&quot; term in it). We want to somehow &quot;tame&quot; the right-hand side.This is where the repetition, or self-similarity, comes in. Let   x  =            2              3        +                              2                          3              +              .              .              .                                          . Then if we take the reciprocal of both sides, we see something interesting:            1      x        =                    3        +        x            2        .Solving for x yields      x    2    +  3  x  −  2  =  0  ,which has as its solutions            −      3      ±              9        +        8                    2      ,      that is,  x  =                              17                −        3            2              or      x  =                    −                  17                −        3            2        .Since x is clearly positive∗, it has to be the former.Plugging back into the original expression, we have      A    =  1  +  x  =                              17                −        1            2        .Squaring both sides finishes the problem.Can you see how to follow a similar argument here? (It&#039;s definitely going to be messier since your patter takes longer to repeat, but the basic idea still applies.)An interesting feature of this argument is: it totally breaks down if we don&#039;t have repetition! To see what I mean, try to calculate      1          2      +                        3                      4            +                                          5                                  6                  +                  .                  .                  .                                                                        .      In general, things are much more complicated and the &quot;naive&quot; argument above simply doesn&#039;t work.

Nested fractional denominator(up to infinity)calculation Question: 4 + (1)/(2 + (1)/(1 + (1)/(3 + (1)/(1+(1)/(2 + (1)/(8 + (1)(...))))))) = sqrt(A) Find the positive integer A in the equation above.

Answered question

Answer & Explanation

New Questions in Pre-Algebra