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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

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.

Log in to reply

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.

Log in to reply

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).

Log in to reply

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?

Log in to reply