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.

Beckett Henry 2022-09-11 Answered
Nested fractional denominator(up to infinity)calculation
Question:
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 assumptions
The pattern repeats 2, 1, 3, 1, 2, 8 infinitely, but 4 comes only once, i.e., at the beginning.
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Answers (1)

Willie Smith
Answered 2022-09-12 Author has 18 answers
Let'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 "infinitary" equation (or rather, it's a normal equation with an "infinitary" term in it). We want to somehow "tame" 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'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'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 "naive" argument above simply doesn't work.
Did you like this example?
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