# What is this function related with continued fractions?

Playing with continued fractions, I came with the idea of iterating the limit of the simplest one:

$$1 + \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}}}\ = \Phi$$

And then I thoght about iterating the result: $$\Phi + \cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cdots}}}}}\ = ?$$

And after that keep iterating again and again. Essentially, what I am doing is solving this:

$$x = a + \frac{1}{x}$$

for different values of $a$. Whose solution is:

$$x = \frac{a \pm \sqrt{a^2 + 4}}{2}$$

And I'm just staying with the positive solution. At the end, I'm just exploring this sequence defined recursively:

$$f(1) = \Phi \\ f(n+1) = \frac{f(n) + \sqrt{f(n)^2 + 4}}{2}$$

When you see the plot of the terms of the sequence, you get this picture:

I would like to know which is the function that generates that curve. It really seems to be a continuos function (notice that the picture is a set of points so close, that may seem to be continuous, but it's not).

So... What's the function underlying this curve and how can we find it?

• When $a=1$, then it converges to the golden-ratio, when $a=2$ it converges to the silver-ratio (en.wikipedia.org/wiki/Silver_ratio) And more generaly for $a=n$ it converges to the $n$'th metalic number. I would be very tempted to say that this always converges for $a>0$ to $\frac{a+\sqrt{a^2+4}}{2}$ as in en.wikipedia.org/wiki/Metallic_mean. Sep 12 '18 at 11:30
• The recursion can be written like $$f(n+1) - f(n) = \frac{\sqrt{f(n)^2 + 4} - f(n)}{2} = \frac{2}{\sqrt{f(n)^2 + 4} + f(n)}$$ If one at this stage removes the 4 in the denominator and solves a continuous approximation $f' = \frac{1}{f}$ i.e. $(f^2)' = 2$, we end up with a square root shape, which roughly agrees with the graph wolframalpha.com/input/?i=plot+2sqrt(x)+from+0+to+5000 Sep 12 '18 at 11:47
• I certainly can't (haven't) prove this, but it looks like something of the form $a\sqrt{bx}$ for constants $a$ and $b$. Sep 12 '18 at 11:54
• The next correction after Calvin's formula is $f'=\frac1f-\frac1{f^3}$ which gives $2n=f^2+\log*(f^2-1)$ Sep 12 '18 at 12:19
• ... or $f(n)=\sqrt{2n-\ln(2n)}$ Sep 12 '18 at 12:27

Using what has been given in comments.

If we use $$2n=f^2+\log(f^2-1)\implies 2n-1=(f^2-1)+\log(f^2 -1)$$ let $t=(f^2-1)$ to make $$2n-1=t+\log(t)\implies t\, e^t=e^{2n-1}\implies t=W\left(e^{2 n-1}\right)$$ where appears Lambert function making the final solution to be
$$f=\sqrt{1+W\left(e^{2 n-1}\right)}\tag1$$

If we use $$n=\frac{1}{4} f \left(f+\sqrt{f^2+4}\right)+\sinh ^{-1}\left(\frac{f}{2}\right)$$ and expand the rhs as a series for infinitely large values of $f$, we should get $$n=\frac{f^2}{2}+\frac{1}{2}+\log \left({f}\right)+O\left(\frac{1}{f^2}\right)=\frac{f^2}{2}+\frac{1}{2}+\frac{1}{2}\log \left({f^2}\right)+O\left(\frac{1}{f^2}\right)$$ and, ignoring the higher order terms, the same process would lead to $$f=\sqrt{W\left(e^{2 n-1}\right)}\tag2$$

Now, for large values of $n$, you could use the asymptotic series given in the linked Wikipedia page $$W(x)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$ and get, as a very first approximation $$f\approx \sqrt{(2n-1)-\log(2n-1)}$$

To allow you to compare with your numbers, I generated a table of numerical values $$\left( \begin{array}{cc} n & (1) \\ 500 & 31.5135 \\ 1000 & 44.6363 \\ 1500 & 54.6991 \\ 2000 & 63.1800 \\ 2500 & 70.6504 \\ 3000 & 77.4035 \\ 3500 & 83.6131 \\ 4000 & 89.3925 \\ 4500 & 94.8203 \\ 5000 & 99.9539 \end{array} \right)$$

• A little problem with your equation (1). It does not satisfy the defining recursion $f(n+1) =(f(n) + \sqrt{f(n)^2 + 4})/2$ and thus it is not clear to me how this relates to $f(n)$ in the question.. Sep 13 '18 at 17:02
• @Somos. As I wrote at the start of my answer, I just took what had already be written in comments. No more ! For sure, neither $(1)$ or $(2)$ satisfy the defining recursion but they seem to be quite close to. Sep 14 '18 at 2:11

This question is similar to MSE question 1072256 "$a_{n+1} = \log(1+a_n), a_1>0.$ ..." and to MSE question 3215 "Convergence of $\sqrt{n}x_n$ where $x_{n+1}=\sin(x_n)$". Because of the similarities, it is natural to assume that there is a power series $\, f(n) \approx f^*(n) := \sqrt{2n} \, (1 + c_1 x - c_2 x^2 + c_3 x^3 - c_4 x^4 + O(x^5)) \,$ where $\, x := 1/n \,$ and $\, c_k \,$ is a polynomial of degree $k$ in $\, y := \log(x). \,$ Using the defining recursion for $\, f(n) \,$ we can solve for the coefficients $\, c_k \,$ to get $$c_1 = c + y/8, \quad c_2 = (1+4c)^2/32 + (1+4c)y/32 + y^2/128,$$ $$c_3 \!=\! \frac{11 \!+\! 120 c \!+\! 384 c^2 \!+\! 384 c^3}{768} \!+\! \frac{5 \!+\! 32 c \!+\! 48 c^2}{256} y \!+\! \frac{1 \!+\! 3 c}{128} y^2 \!+\! \frac{y^3}{1024},$$ where $\, c\,$ is a constant depending on $\, f(0). \,$ If we use the initial values $\, f(0) = 0, f(1) = 1, \, f(2) = \phi, \dots, \,$ then $\, c \approx -0.291131527. \,$ Note that since $\, \sqrt{1/2}-1 \approx -0.292 \,$ the formula is already close for $\, n=1.$

To judge the accuracy, we get $\, f(1) = 1, \,$ $f^*(1) \approx 1.00269, \,$ $f(2) \approx 1.61803, \,$ $f^*(2) \approx 1.61826, \,$ $f(3) \approx 2.09529, \,$ and $f^*(3) \approx 2.09535. \,$

I think you can set up a differential equation and solve it.

\begin{eqnarray} 2f'(n) &= -f(n)+\sqrt{f(n)^2+4}\\ \text{d}n&=\frac{2\, \text{d}\! f}{-f+\sqrt{f^2+4}} \\ n &= \frac{1}{4} (f (f + \sqrt{4 + f^2} + 4 \sinh^{-1}\left(\frac{f}{2}\right) \end{eqnarray} Solving for $f$ should result in the function you wanted. Wolfram plot

• Wouldn't this just be an approximation, because $f'(n) \sim f(n+1) - f(n)$ but it is not equal unless $f$ is linear. Sep 12 '18 at 15:31
• I think you can define a discrete derivative this way, no need for $f$ to have the property of linearity. Sep 12 '18 at 15:35
• right but then you would have to do a discrete integral (i.e. a summation) instead of a continuous integral. Sep 12 '18 at 15:37
• I thought this would suffice as an approximation, but now I think you are correct. If I do not think of a good reason, I will delete the answer after some time Sep 12 '18 at 15:47

One way to solve this is to use the theory of continuous iterations, which is often best tackled using power series around fixed points. Let $g(z) = \frac{z + \sqrt{z^2 + 4}}2$. Then notice $$g^{-1}(w) = \frac{w^2 - 1}w$$ This has no fixed points except at infinity, but we can actually make use of that if we conjugate by $w\to \frac1w$, since $h(w) = \frac1{g^{-1}(1/w)}$ will have a fixed point at $w=0$. We can see the series for $h(w)$ is given by $$h(w) = \frac{w}{1-w^2} = w + w^3 + w^5 + ...$$ Formally composing this series with itself is fairly straightforward, for example: \begin{eqnarray} h(h(w)) &=& (w + w^3 + w^5 +...) + (w + w^3 + w^5 +...)^3 + (w + w^3 + w^5 +...)^5 + ... \\ &=& w + 2w^3 + 2w^5 + ... \end{eqnarray} In general, we will have $$h^n(x) = w + p_3(n)w^3 + p_5(n)w^5 + ...$$ where $p_i(n)$ satisfy a recurrence given by $$w + p_3(n+1) w^3 + p_5(n+1)w^5 + ... = h^{n+1}(x) = h(h^n(x)) \\ = (w + p_3(n)w^3 + p_5(n)w^5 + ...) + (w + p_3(n)w^3 + p_5(n)w^5 + ...)^3 + (w + p_3(n)w^3 + p_5(n)w^5 + ...)^5 + ... \\ = w + (p_3(n) + 1) w^3 + (p_5(n) + 3p_3(n) + 1) w^5 + ...$$ We get recursions like $p_3(n+1) = p_3(n) + 1$. Clearly, this is solved by $p_3(n) = n$. The $w^5$ coefficient is $p_5(n+1) = p_5(n) + 3n + 1$, which is solved by $p_5(n) = n + \frac32 n(n+1)$. In general, you can actually find a unique polynomial for each $p_i(n)$ by using Faulhaber's formula since each one is defined by a recursion $p_i(n+1) = p_i(n) + q_i(n)$ for some appropriate polynomial $q_i(n)$. Notice this recursion will still be satisfied for noninteger $n$, so we have a function $H(t,w) = w + p_3(t)w^3 + p_5(t)w^5 +...$ satisfying: $$H(t+1,w) = h(H(t,w))$$ OK so what does this have to do with the original question? Well, we have \begin{eqnarray} H(t+1,w) &=& \frac1{g^{-1}(1/H(t,w))} \\ \frac1{H(t+1,w)} &=& g^{-1}\left(\frac1{H(t,w)}\right) \\ g\left(\frac1{H(t+1,w)}\right) &=& \frac1{H(t,w)} \end{eqnarray} substituting $-n = t+1$, and recalling the definition of $g$, we find: \begin{eqnarray} \frac1{H(-(n+1),w)} &=& g\left(\frac1{H(-n,w)}\right) \\ \frac1{H(-(n+1),w)} &=& \frac{\frac1{H(-n,w)} + \sqrt{\frac1{H(-n,w)^2} + 4}}2 \end{eqnarray} Then, as you can see, we will have $f(n) = \frac1{H(-n,w)}$, where $w$ is the starting value. I do not expect this function to have a nice closed form.

This formula for $f$ is only helpful on a small scale, but it does demonstrate we can draw a smooth (hence continuous) function through those points. If you want to know about the asymptotic behavior of $f(n)$ as $n\to\infty$, somos has already covered that in his answer.

• What you describe won't really solve the problem. That is, you won't get the correct behavior $\, f(n) \sim \sqrt{2n}.$ In particular this approach won't give the logarithms in the series expansion. Sep 12 '18 at 16:23
• Yea I guess it doesn't give any information about the asymptotics of $f$. Sep 12 '18 at 16:28
• The main thing this solution gives is a smooth interpolation that still satisfies the recursion Sep 12 '18 at 16:31
• Yes, it does, but only locally. I have used similar methods myself for this kind of problem, but I know it has severe limitations. It is similar to polynomial interpolation. It doesn't work well for extrapolation. Sep 12 '18 at 16:36
• Definitely a valid point! Of course, given $f$ locally you can use the recursion to extend it, but you need an asymptotic expansion to know what that will look like. I don't really have anything to contribute with regards to asymptotics, I think you covered that in your answer perfectly well. Sep 12 '18 at 16:49