Simplifying a complicated continued fraction expression.

Question

This probably relates to continued fractions or numerical methods but I do not see how.

There is probably some kind of induction or telescoping.

$$\sqrt 3 = 1+\dfrac{2+\dfrac{3+\dfrac{4+\cdots}{5+\cdots}}{4+\dfrac{5+\cdots}{6+\cdots}}} {3+\dfrac{4+\dfrac{5+\cdots}{6+\cdots}}{5+\dfrac{6+\cdots}{7+\cdots}}} $$

(The pattern is +1 when going up and +2 when going down).

I tried to transform it into a simple continued fraction but failed.

How to prove this ?

It reminds me of this one

why is $ 2 = \frac{5}{1+\frac{8}{4+\frac{11}{7 + \frac{14}{10 + \dots}}} } $

edit

even more related seems this one :

Is this formula for $\frac{e^2-3}{e^2+1}$ known? How to prove it?

edit 2

Some have claimed the value is in fact larger than $\sqrt 3$ ( I need to check myself )

but it seems that maybe the correct value is this one :

https://oeis.org/A048836

??

edit 3

I added another conjectured value in the comments but neither of them seems to be correct.

The value is a mystery at the moment.

My apologies for the wrongly assumed closed forms.

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Answer 1

Not an answer, more of an extended comment, not really part of the main question, just related observation :

Consider

$$f(x) = x+1+\dfrac{x+2+\dfrac{x+3+\dfrac{x+4+\cdots}{x+5+\cdots}}{x+4+\dfrac{x+5+\cdots}{x+6+\cdots}}} {x+3+\dfrac{x+4+\dfrac{x+5+\cdots}{x+6+\cdots}}{x+5+\dfrac{x+6+\cdots}{x+7+\cdots}}} $$

This function is remarkable.

At first you would think it has poles for some negative integers, then you notice that idea came from truncating and it might not actually have poles there.

Also

$$f(x) = x + 1 + \frac{f(x+1)}{f(x+2)}$$

This implies it has poles when $f(x+2)=0$ and if that is the case , we have $f(x) = x+1 + \infty = \infty$. But that implies that $f(x-2) = (x-2) + 1 = x-1$. And an infinite amount of identities follow. Notice $f(x-2) = (x-2) + 1 = x-1$ implies that $x$ is not real and thus neither is the pole of $f(x+2)$!

I wonder where the fixpoint of $f(z)$ are.

And what the taylor series for $f(z)$ looks like.

But perhaps a more fun idea is what happens when $x$ goes to $\infty$ ?

A first naive method is

$$f(x) - x = 1 + \frac{f(x+1)}{f(x+2)}$$

And then assume

$$\lim (f(x) - x - 1) = L $$

IF correct, this implies in the limit :

$$ 1 + L = 1 + \frac{x + 2 + f(x+2)/f(x+3)}{x+3 + f(x+3)/f(x+4)} = 2$$

by assuming the ratio of $f(x+2)/f(x+3)$ converges at the same rate as $f(x+3)/f(x+4)$.

This implies the ratio converges to $1$.

So we get the stronger

$$\lim (f(x) - x) = 2 $$

It even seems reasonable to assume

$$\lim (f(x + k i) - (x + ki)) = 2 $$

for real $k \ll x$ even if $k$ is also going to infinity.

This implies $f(x)$ is not a polynomial.

Maybe more a shape like $\operatorname{erf}(z)$ for $\operatorname{Re}(z)>1$ based on the above.

Maybe $f(x)$ is close to $2 + a \cdot \operatorname{erf}(z) + b \cdot \operatorname{erf}(z)^2$ for $\operatorname{Re}(z)>1$.

The functional equation for $f(x)$ reminds me of this :

https://mathoverflow.net/questions/348568/about-a-n-fraca-n-1a-n-1-ca-n-2

$ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$?

or maybe fibonacci or somos sequences. It looks a bit similar.

Just some ideas.

Answer 2

I found a slight simplification of the complicated continued fraction involving nesting infinite CFs into the partial numerator arguments of infinite CFs. Although my solution may be inaccurate, this was a fun attempt to simplify the complicated continued fraction.

I thought it would be interesting to attempt to simplify the complicated continued fraction to a general continued fraction form. In this response, I will use the following generalized CF notation (Gauss Continued Fraction Notation):

$$ \underset{n=x}{\overset{y}{ \mathrm K}} \frac{a_n}{b_n} = \frac{a_x}{b_x + \frac{a_{x+1}}{b_{x+1} + \frac{a_{x+2}}{... b_{y-1} + \frac{a_y}{b_y}}}} $$

Where the numerator of the fraction after the K is the partial numerator argument and the denominator is the partial denominator argument. We will also use the following function:

$$T(x)=x+\frac{x+1+\frac{x+2+\cdots}{x+3+\cdots}}{x+2+\frac{x+3+\cdots}{x+4+\cdots}}$$

To be straight to the point, I found a slightly valid solution to “simplify” T(1) (the complicated continued fraction) in a general CF for all of T(x).

$$T(x)=x+\underset{n_1=1}{\overset{\infty}{ \mathrm K}} \frac{(2n_1-1)+x+\underset{n_2=n_1}{\overset{\infty}{ \mathrm K}} \frac{(2n_2)+x+ \underset{n_3=n_2}{\overset{\infty}{ \mathrm K}} \frac{(2n_3+1)+x+\cdots}{(2n_3+2)+x}}{(2n_2+1)+x}}{(2n_1)+x}$$

Where we infinitely nest CFs in their predecessor’s partial numerator argument. This solution above appears to be valid when expanded, acknowledging the nth Gauss Continued Fraction K Notation has an index equal to the previous CF’s index (where I denoted the current index as $I_n$), a partial numerator argument of $((2I_n-2+n)+x+$the next CF$)$ and a partial denominator argument of $(2I_n-1+n)+x$. The value n is the current continued fraction’s index’s subscript. Thus, setting x equal to 1 in the formula above is a slightly valid solution if this can even be considered a simplification. The table below shows outputs for specific upper bounds of all of the continued fractions and the number of Gauss continued fraction notations of T(1). In other words, it is the outputs of: $$1+\underset{n_1=1}{\overset{N}{ \mathrm K}} \frac{(2n_1-1)+1+\underset{n_2=n_1}{\overset{N}{ \mathrm K}} \frac{(2n_2)+1+\cdots \underset{n_y=n_{y-1}}{\overset{N}{ \mathrm K}} +\frac{2n_y-1+y}{2n_y+y} }{(2n_2+1)+1}}{(2n_1)+1} $$ Note that N represent Upper Bounds on the table, y represents the nested CFs, and the approximated outputs are represented by approx on the table. The CF above may be formatted weirdly on some devices as well.

As one can probably see, as N and y increase, their output seems to be around 1.73. It appears that when N is equal to y and both values are large, they seem to converge to about 1.729. Although, I have no proof yet that the CF above even converges when the bounds approach infinity. Overall, this was a fun and interesting problem and I hope the “solution” above can be modified further to produce great solutions.