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Is it possible to find an explicit expression for a function, which at integer points takes values:

$$y_1=\sqrt{a}$$

$$y_2=\sqrt{a+\sqrt{a}}$$

$$y_3=\sqrt{a+\sqrt{a+\sqrt{a}}}$$

etc.

We can take $y_0=0$. Then, defining $y$ for negative integers is not a problem:

$$y_n=\sqrt{a+y_{n-1}}$$

$$y_{n-1}=y_n^2-a$$

Thus, for $n=-1,-2,-3,\dots$ we have a quadractic map:

$$y_{-1}=-a$$

$$y_{-2}=a^2-a$$

$$y_{-3}=(a^2-a)^2-a$$


Now I have no idea how to define $y_x$ for $x \notin \mathbb{Z}$. We have an obvious functional equation:

$$y_{x+1}=\sqrt{a+y_x}$$

For $x \to 0$ we have $ y_x \to 0$. But how to find an explicit expression? Maybe in terms of an integral or some special functions? Or how to even approximate this function?

At $x \to 0$ we should have $ y_x \to cx$ for some constant $c$.

This task seems hopeless to me even though such things as Gamma function or general Harmonic numbers exist, despite the fact that interpolation there is not obvious at all.


For the special case $a=2$ @NgChungTak gave an interpolation:

$$y_{x}=2\cos \dfrac{\pi}{2^{x+1}}$$

The plots of the sequence and the function are shown below:

enter image description here

However this kind of function doesn't work with the other values of $a$, even for $x>0$. For $a>0$ and $x>0$ the rate of convergence increases with $a$.

We can plot normalized sequences for different $a>0$.

$$f(n)=\frac{2 y(n)}{1+\sqrt{1+4a}}$$

enter image description here


I think it makes sense to consider a Quadratic map first, since a lot of exact solutions for special cases were found for it, and then extrapolate the solutions to negative indices.

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  • $\begingroup$ I think you'll have to settle for recursion. But you can find a closed form for the limit. $\endgroup$ – quasi Jan 18 '17 at 11:04
  • $\begingroup$ I had no doubt that you knew. Was just pointing that out for the general audience. But for the expression itself, I would forget about closed form. $\endgroup$ – quasi Jan 18 '17 at 11:07
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    $\begingroup$ In particular, $a=2$ and $x\in \mathbb{R}^{+}$, $y_{x}=2\cos \dfrac{\pi}{2^{x+1}}$ $\endgroup$ – Ng Chung Tak Jan 18 '17 at 11:30
  • $\begingroup$ well, in formal power series, if you move the attracting fixpoint to $0$, your transformation is conjugate to a linear map (in a unique way), and you can then interpolate in an obvious way. This also gives you power series that perform "half a step" (or other fractional steps). I have no idea how to prove anything about their radius of convergence though $\endgroup$ – mercio Jan 18 '17 at 13:45
  • $\begingroup$ if $g(x) = x^2-a$, and $z$ is the fixpoint, then you can probably define $y_k = \lim_{n \to \infty} g^{\circ n}(z - \exp((1-k)\log (z-y_n) + k\log (z-y_{n+1})))$ $\endgroup$ – mercio Jan 18 '17 at 13:49
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Let $f(z) = z^2-a$ with $a \neq - \frac 14$.

If $a>0$, $f$ has a repulsive fixpoint at $\tau = (1+\sqrt{1+4a})/2$ (and most of the time, the other fixpoint is repulsive too, but let's take this one). I will note $\lambda$ the derivative of $f$ at $\tau$, which is $\lambda = 2\tau$ (larger than $1$), so near this fixpoint, $f$ acts almost like the dilation $g(z) = \lambda \tau$.

On a small disk $D$ around $\tau$ we can define $f^{-1}$ such that it sends $D$ to itself, and then according to Koenigs, there is a (unique) holomorphic injective function $h$ on $D$ into a small neighbourhood of $0$ such that $h(\tau) = 0, h'(\tau) = 1$, and forall $z \in D$, $h ( f^{-1} (z)) = g^{-1}(h(z))$.

Since $h$ is injective we can define $t = h^{-1}$ on $h(D)$, and then the functional equation can be put in the form $t(g(z)) = f(t(z))$ for $z$ close enough to $\tau$.

Now, since $g$ is a simple dilation and $f$ is entire, this means $t$ has an analytic continuation to the entire complex plane. $t$ has an essential singularity at infinity, so it takes every value infinitely often (with no exception because of the functional equation).

Let $w \in \Bbb C$ be a zero of $t$. It has a corresponding sequence $(y_n)_{n \in \Bbb Z}$ defined by $y_n = t(\lambda^{-n}w)$.
Then $y_0 = 0, y_1 = \pm \sqrt a, y_2 = \pm \sqrt {a \pm \sqrt a} \ldots$ and so on.
If you choose a value for $\log \lambda$, this allows you to interpolate the sequence to all of $\Bbb C$, defined by $y_n = t(w\exp(-n\log(\lambda)))$.

Suppose two zeroes $w_1,w_2$ give rise to the same sequence. Then, by the identity theorem we have $t(w_1z) = t(w_2z)$ forall $z \in \Bbb C$, and differentiating this relation at $z=0$ gives $w_1 = w_2$.
This work for any nonzero value too, so the points $(w,y_0)$ are in a bijective correspondance with the sequences $y_n = f(y_{n+1})$ that eventually converge to $\tau$.

To compute the interpolation approximately, you can plunge into the sequence by applying $f^{-1}$, then once you are close enough to the fixpoint, approximate $f$ with $g$ to interpolate however you need, then go back by iterating $f$ : $y_n = \lim_{m \to \infty} f^{\circ m}(\tau+\lambda^{-n}(y_m-\tau))$.
This works in theory, but you may run into precision issues for both iterations. You may want to instead use a power series to approximate $t$ and $h$ once you're close enough to $\tau$.

If you wonder what the fractional iterates of $f^{-1}$ would look like geometrically, it seems they aren't necessarily algebraic : they should have branch points at all the critical values of $t$.

Differentiating the functional equation of $t$ gets you $g'(z).t'(g(z)) = f'(t(z)).t'(z)$, which means $\lambda t'(g(z)) = 2t(z).t'(z)$, so the critical points of $t$ are the the zeroes of $t$ (critical points of $f$) along with their images by any number of dilations.
Furthermore, $t'$ only has simple zeroes because iterating $f$ on $0$ will never loop back to $0$.

So the critical values are $0,f(0),f((0)),\ldots$, and there $t^{-1}$ (and thus all the fractional iterates, and well everything really) have a square root-like singularity.

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