Infinitely Large Fraction

Algebra Level 3

$\large \dfrac{1}{\dfrac{1}{\dfrac{1}{\ldots}+\dfrac{1}{\ldots}}+\dfrac{1}{\dfrac{1}{\ldots}+\dfrac{1}{\ldots}}}$

Find the closed form of the above expression to 3 decimal places.

The answer is 0.707.

1 solution

Tapas Mazumdar
Jun 10, 2017

Let $\dfrac{1}{\dfrac{1}{\dfrac{1}{\ldots}+\dfrac{1}{\ldots}}+\dfrac{1}{\dfrac{1}{\ldots}+\dfrac{1}{\ldots}}} = x$ and so we have

$\dfrac{1}{x+x} = x \implies 2x^2 = 1 \implies x = \pm \dfrac{1}{\sqrt{2}}$

Since the continued fraction is always positive, so we get

$x = \dfrac{1}{\sqrt{2}} = \boxed{0.707}$

Your solution is currently incorrect, as it doesn't explain why $-\dfrac{1}{\sqrt{2}}$ is incorrect properly (why is a continued fraction always positive?). Furthermore, this expression can also be written as

$a_{n+1} = \dfrac{1}{a_n+\frac{1}{2a_n}} = \dfrac{2a_n}{2a_n^2+1}$

This expression is different to yours.

You should analyse this question properly if you want to ensure your thinking is correct because currently, it is flawed.

Sharky Kesa - 4 years ago

Can you give a more detailed reasoning? I didn't understand how you constructed this sequence

$a_{n+1} = \dfrac{2 a_n}{2 a_n^2 +1}$

My original thinking was this

If

$\dfrac{1}{\dfrac{1}{\dfrac{1}{\ldots}+\dfrac{1}{\ldots}}+\dfrac{1}{\dfrac{1}{\ldots}+\dfrac{1}{\ldots}}} = x$

Then by observation

$\dfrac{1}{ \underbrace{\boxed{\dfrac{1}{\dfrac{1}{\ldots}+\dfrac{1}{\ldots}}}}_{= x} + \underbrace{\boxed{\dfrac{1}{ \dfrac{1}{\ldots}+\dfrac{1}{\ldots}}}}_{=x} }$

This is as per what I've learnt in my academics, for infinite continuations like these, we substitute as above.

Tapas Mazumdar - 4 years ago

ok, consider the sequence, $a_1=1$ , $a_{n+1} = \dfrac{1}{a_n+\frac{1}{a_n+a_n}}$ . Then, we have

$\begin{aligned} a_1 &= 1\\ a_2 &= \dfrac{1}{1+\frac{1}{1+1}}\\ a_3 &= \dfrac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\frac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\dfrac{1}{1+\frac{1}{1+1}}}}\\ a_4 &= \dfrac{1}{\dfrac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\frac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\dfrac{1}{1+\frac{1}{1+1}}}}+\frac{1}{\dfrac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\frac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\dfrac{1}{1+\frac{1}{1+1}}}}+\dfrac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\frac{1}{\dfrac{1}{1+\frac{1}{1+1}}+\dfrac{1}{1+\frac{1}{1+1}}}}}} \ldots \end{aligned}$

This sequence also tends to the above expression.

For infinite continuations, we must do an analysis to answer the question. Otherwise, we can get these sorts of contradictions:

$\begin{aligned} x&=1+2+4+8+\ldots\\ 2x&=2+4+8+\ldots\\ 2x+1&=x\\ x&=-1 \end{aligned}$

which is incorrect.

Sharky Kesa - 4 years ago

@Sharky Kesa – I take issue with the claim that "This sequence also tends to the above expression." The "rate of convergence along each branch" could affect the answer, and thus we have to carefully define how the fraction is obtained.

Since this is not a standard continued fraction, can you define what sequence we're taking the limit of?

Calvin Lin Staff - 3 years, 12 months ago

Infinitely Large Fraction

The answer is 0.707.

1 solution

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