What does this converge to and why?

What does the below expression converge to and why? $\cfrac{2}{3 -\cfrac{2}{3-\cfrac{2}{3-\cfrac2\ddots}}}$

Setting it equal to $x$ , you can rewrite the above as $x = \dfrac{2}{3-x}$ , which gives the quadratic equation $x^2 - 3x + 2 = 0$ , and the roots are $1$ and $2$ , both positive. How do we know which to reject?

#Algebra

Easy Math Editor

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Math	Appears as
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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

As for me, I guess we can make a good observation by iterating it repeatedly for a number of times and observe where the numbers lead to.

For instance, if we evaluate $\frac{2}{3-2}$ , that will give us $2$ .

Then, if we evaluate $\frac{2}{3 - \frac{2}{3-2}}$ , it will also give us $2$ .

Then again, if we evaluate $\frac{2}{3 - \frac{2}{3 - \frac{2}{3-2}}}$ , it will also give us $2$ .

No matter how many times this will be done, the answer remains to be $2$ .

I don't know if this is a valid approach in finding the value of such infinite nested fraction, but I guess it's a good way to determine the behavior of the function.

Efren Medallo - 4 years, 1 month ago

Okay, so if you watch the video that Anirudh posted, it turns out that the way you evaluated the continued fraction is "wrong". The way you're "supposed" to do it is to look at the sequence $\dfrac{2}{3}, \dfrac{2}{3 - \dfrac{2}{3}}, \dfrac{2}{3- \dfrac{2}{3 - \dfrac{2}{3}}}$ and so on; that way, we see that the sequence seems to converge to $1$ . Why we do it that way, I'm still not clear.

Hobart Pao - 4 years, 1 month ago

It's just by definition that it's evaluated that way.

E.g We define $\sum_{n=1}^\infty$ as the limit of the partial sums $\sum_{n=1}^k$ . This means that the value of $1 - 1 + 1 - 1 + 1 -1 \ldots$ does not exist under our definition.

Sometimes, we may choose to extend the definition to cases where it makes sense (or even when it doesn't).

Calvin Lin Staff - 4 years, 1 month ago

Hmm I don't think that can tell you what my expression converges to, but it does give a hint of behaviour. Unfortunately with the method you used, by replacing the 2 in the denominator with 1, you get the answer of 1, which is why I'm not sure which one is the right root because obviously this expression can't have two different values can it?

Hobart Pao - 4 years, 1 month ago

@Hobart Pao You can get some insight from this video, https://www.youtube.com/watch?v=leFep9yt3JY&feature=youtu.be&t=2m59s

Anirudh Sreekumar - 4 years, 1 month ago

Thank you! This really helped. The answer is indeed $\boxed{1}$ .

Hobart Pao - 4 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$