Crazy Continued Fractions

Calculus Level 3

After reading the following, determine who is correct, Alice or Bob:

Alice and Bob want to find the value of x = 2 3 − 2 3 − 2 3 − 2 3 − 2 ⋱ . x = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}}. Alice thinks the answer is 2. Here is her reasoning: 2 = 2 3 − 2 = 2 3 − 2 3 − 2 = 2 3 − 2 3 − 2 3 − 2 = 2 3 − 2 3 − 2 3 − 2 3 − 2 = 2 3 − 2 3 − 2 3 − 2 3 − 2 ⋱ . \begin{aligned}2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}}. \end{aligned} Bob thinks it's 1. Here is his reasoning: 1 = 2 3 − 1 = 2 3 − 2 3 − 1 = 2 3 − 2 3 − 2 3 − 1 = 2 3 − 2 3 − 2 3 − 2 3 − 1 = 2 3 − 2 3 − 2 3 − 2 3 − 2 ⋱ . \begin{aligned} \small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}}. \end{aligned}

Alice Bob Both are correct, 1 = 2 1 = 2 Neither reasoning is correct

Agnishom Chattopadhyay by Agnishom Chattopadhyay , 22, India

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1 solution

Claudio Brot
Jan 24, 2018

With looking at the equation for x we can spot x on the RHS as well and we write:

x = 2 3 − x x=\frac{2}{3-x}

This is, of course, assuming there is a unique solution for x - we will run with this assumption for now and see, where it leads us. Let us solve for x:

x ( 3 − x ) = 2 ⇔ x 2 − 3 x + 2 = 0 x(3-x)=2 \Leftrightarrow x^2-3x+2=0

Solve this with completing the square:

⇒ ( x − 3 2 ) 2 − 9 4 + 2 = 0 ⇔ ( x − 3 2 ) 2 = 1 4 ⇔ x − 3 2 = ± 1 2 \Rightarrow (x-\frac{3}{2})^2-\frac{9}{4}+2=0 \Leftrightarrow (x-\frac{3}{2})^2=\frac{1}{4} \Leftrightarrow x-\frac{3}{2}=\pm\frac{1}{2}

Therefore we get two solutions: x 1 = 2 x_1=2 and x 2 = 1 x_2=1

But this is a contradiction to our assumption of a unique solution above, therefore the expression is meaningless.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

I have updated the answer to "Neither argument is correct".

The value of this (negative) continued fraction is 1. The way we evaluate continued fractions is by taking the limit of the partially truncated fractions, IE 2 3 , 2 3 − 2 3 , … \frac{2}{3}, \frac{2}{ 3 - \frac{2}{3} } , \ldots . This sequence converges to 1, which is the value of the continued fraction.

Remember that we are not allowed to do the "x" substitution until we have proved the limit exists. Even then, all that gives us are possibilities of what the answer can be. Of course, if there is only one possibility, then that has to be the answer (conditional on the limit existing).

Thus, what you have shown so far is that we are in one of 3 possibilities:
1. The limit does not exist
2. The limit is equal to 1
3. The limit is equal to 2

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