## The Case of Square Roots

We wish to examine the behavior of the iterated radical expression $R_a(n)=\underbrace{\sqrt{a+\sqrt{a+...\sqrt{a+R_a(0)}}}}_{n\textrm{ radicals}}$ Let $A=\lim_{n \rightarrow \infty} R_a(n)$ Then clearly $A^2=a+A$ And so $A=\tfrac{1+\sqrt{1+4a}}{2}$ In order to determine the nature of the convergence to this limit, let us examine a function defined as follows: $f(x/q)=\sqrt{a+f(x)}$ Where q is a value yet to be determined. Clearly $f(0)=A$, and it is not hard to see that $R_a(n)=f\left ( \tfrac{f^{-1}(R_a(0))}{q^n} \right )$ Thus the behavior of f, as well as the value of q, will determine the convergence of $R_a(n)$. We rearrange the above relation to get $f^2(x)=a+f(qx)$ Let us expand f in a Taylor series. $f(x)=A+b_1 x +b_2 x^2 +b_3 x^3+...$ We can substitute this into our functional equation to get $A^2+2Ab_1 x+(2A b_2+b_1^2)x^2+(2Ab_3+2b_1b_2)x^3+...=a+A+qb_1x+q^2b_2x^2+q^3b_3x^3+...$ By equating coefficients, we find that $q=2A$. Note that changing $b_1$ only affects the scaling of the function. Assuming we want the inverse to be positive as we approach from below, $b_1$ must be negative, thus we simply set $b_1=-1$. Now the rest of the coefficients can be found algorithmically in sequence. In general, the coefficient of $x^k$ will be $b_k=\frac{1}{(2A)^k-2A}\sum_{j=1}^{k-1}b_jb_{k-j}$ And thus $f\left ( \tfrac{x}{2A} \right )=\sqrt{a+f(x)}$$f^2(x)=a+f(2Ax) \\ R_a(n)=f\left ( \tfrac{f^{-1}(R_a(0))}{(2A)^n} \right )$ Where f is defined by the polynomial with the given coefficients. It follows that $\lim_{n \rightarrow \infty} (2A)^n(A-R_a(n))=\lim_{n \rightarrow \infty} (2A)^n(f(0)-f(f^{-1}(R_a(0))/(2A)^n))$$\lim_{n \rightarrow \infty} (2A)^n(A-R_a(n))=-f'(0)f^{-1}(R_a(0))=f^{-1}(R_a(0))$$\lim_{n \rightarrow \infty} (2A)^n \left (A-\underbrace{\sqrt{a+\sqrt{a+...\sqrt{a+z}}}}_{n\textrm{ radicals}} \right )=f^{-1}(z)$ Another way to construct $f(x)$ is by the following approach, which converges fairly quickly: Let $f_0(x)=A-x$. We define $f_{k+1}(x)= f_k^2\left (\frac{x}{2A} \right )-a$ Then $\lim_{k \rightarrow \infty}f_k(x)=f(x)$

### A Special Trigonometric Case

For the case of $a=2$, it is easy to show by induction that $b_k=2(-1)^k\frac{1}{(2k)!}$ Which would imply that $f(x)=2\cos(\sqrt{x})$ Therefore $\lim_{n \rightarrow \infty} 4^n \left( 2-\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2}}}}_{n \textrm{ radicals}}\right)=\pi^2/4$

### An Infinite Product

Beginning with $f^2(x)=a+f(2Ax)$ Let us differentiate to obtain $f(x)f'(x)=Af'(2Ax)$ Thus, if we define $g(x)=-xf'(x)$ Then we easily see that $g(2Ax)=2g(x)f(x)$ Clearly $g(0)=0, g'(0)=1$. Then $g(x)=2f\left (\tfrac{x}{2A} \right )g\left (\tfrac{x}{2A} \right )=2^2f\left (\tfrac{x}{2A} \right ) f\left (\tfrac{x}{(2A)^2} \right ) g\left (\tfrac{x}{(2A)^2} \right )$$g(x)=2^N g\left (\tfrac{x}{(2A)^N} \right )\prod_{k=1}^{N} f\left (\tfrac{x}{(2A)^k} \right )$$g(x)=(2A)^N g\left (\tfrac{x}{(2A)^N} \right )\prod_{k=1}^{N} \tfrac{1}{A}f\left (\tfrac{x}{(2A)^k} \right )$ Taking the limit $g(x)=\underset{N \to \infty}{\lim}(2A)^N g\left (\tfrac{x}{(2A)^N} \right )\prod_{k=1}^{N} \tfrac{1}{A}f\left (\tfrac{x}{(2A)^k} \right )=x\prod_{k=1}^{\infty} \tfrac{1}{A}f\left (\tfrac{x}{(2A)^k} \right )$ Thus $-f'(x)=\prod_{k=1}^{\infty} \tfrac{1}{A}f\left (\tfrac{x}{(2A)^k} \right )$ Thus we need only examine the zeros of f to find the zeros of f'. In fact, if f has zeros $\left \{z_1,z_2,z_3,... \right \}$ Then f will have extrema at $\bigcup_{k=1}^{\infty}\left \{(2A)^kz_1,(2A)^kz_2,(2A)^kz_3,... \right \}$

#### An Associated Infinite Series

Differentiating the log of both sides of the result above, we find the infinite series: $\frac{d}{dx}\ln\left (-f'(x) \right )=\frac{d}{dx}\ln\left (\prod_{k=1}^{\infty} \tfrac{1}{A}f\left (\tfrac{x}{(2A)^k} \right ) \right )$$\frac{f''(x)}{f'(x)}=\sum_{k=1}^{\infty}\frac{1}{(2A)^k}\frac{f'\left (\tfrac{x}{(2A)^k} \right )}{f\left (\tfrac{x}{(2A)^k} \right )}$

### Zeros of $f(x)$

Below is a plot of the zeros of for different values of a on the vertical axis, plotted semi-logarithmically.

Below is a plot the sign of f (Yellow is positive, blue is negative), from which the zero contours can be seen. However, we can also see that some zeroes of f for certain values of a are multiple roots, as f goes to zero without changing sign.

#### Special Cases

Two special cases bear mentioning. In the case $a=1$, the zeros are given by $z_n=2.1973\cdot (1+\sqrt{5})^{2n}$ for $n \geq 0$. In fact, in this case, after the first zero, f is always between -1 and 0. f is -1 at $x_n=2.1973\cdot (1+\sqrt{5})^{2n+1}$ for $n \geq 0$. For $a=2$, the zeros are at $z_n=\left ( (2n+1)\frac{\pi}{2} \right )^2$ And, in fact, $f(x)=2$ at $x_n=\left ( 2n\pi \right )^2$, and $f(x)=-2$ at $x=\left ( (2n+1)\pi \right )^2$, for $n\geq0$.

### Periodic and Possible Fractal Structure

Although f is generally not very interesting close to zero, it exhibits remarkable behavior on larger scales. We find, namely, that if we take $h(x)=\left | f(x) \right |^{x^{-\log_{2A}(2)}}$ Then h is exponentially periodic, asymptotically. We define $J(x)=h((2A)^x)$ This function has period 1, asymptotically. Below we show the behavior of J for some values of A

Note that the number of zeros remains constant. All seem to be single roots. In fact, the location of the dominant maxima seem constant as well However, within the periodicity, J appears to have a fractal structure. Below we show a zoom of $J(x)$ for $a=3$.

## Complex Behavior

We can take the series and functional definitions of the function and use them to extend the function to the entire complex plane. Below we plot the complex sign of $f(Cz|z|)$ for different values of a, and a certain value of C (this rescaling done to make the regularities more evident). The complex sign is given by the color:
• Dark Blue$\Leftrightarrow\textrm{Re}< 0 ,\textrm{Im} < 0$
• Light Blue$\Leftrightarrow\textrm{Re} < 0 ,\textrm{Im} > 0$
• Orange$\Leftrightarrow\textrm{Re} > 0,\textrm{Im} < 0$
• Yellow$\Leftrightarrow\textrm{Re} > 0,\textrm{Im} > 0$
This allows us to find zeros, which correspond to points where all four colors meet.

We note several remarkable features:
• The function is conjugate-symmetric.
• The function displays remarkable regularity away from the real line. Note the persistent ripples which reach total regularity at $a=2$. There is a structure of "fingers" that gradually join, each finger corresponding to one zero. The position of certain features on the real line remains fixed, e.g. the prominent feature at about 0.8.
• The evolution of the function over a can be broken into three eras.
1. Pre-Saturating: For $a< 1$, there is exactly one real zero.
2. Saturating: For $1\leq a < 2$, zeros join to form pairs of real zeros.
3. Saturated: For $2 \leq a$, all zeros are real.
• The number and larger-scale density of zeros remains roughly constant.
• The function displays quasi-fractal properties, as it becomes increasingly self-similar on larger scales. In a sense, a cross between periodic and fractal behavior, as seen in the other figures.
• The process of the fusing of complex zeros into pairs of real zeros can also be seen in the plots of the real zeros above, giving a new view of the branching features.
• The fingers coalesce along elliptical paths. In fact, these ellipses are of the form $x^2+2y^2=C'^2$

## The Case of Arbitrary Roots

More generally, suppose we examine $R_a(n)=\underbrace{\sqrt[p]{a+\sqrt[p]{a+...\sqrt[p]{a+R_a(0)}}}}_{n\textrm{ radicals}}$ Let $A=\lim_{n \rightarrow \infty} R_a(n)$ Then $f(x/q)=\sqrt[p]{a+f(x)}$ Clearly $f(0)=A$, and it is not hard to see that, again $R_a(n)=f(f^{-1}(R_a(0))/q^n)$ If we do the same analysis as before we find that $q=pA^{p-1}=p(1+a/A)$. Let $f_0(x)=A-x$. We define $f_{k+1}(x)= f_k^p\left (\frac{x}{q} \right )-a$ Then $\lim_{k \rightarrow \infty}f_k(x)=f(x)$ Then similarly we have $\lim_{n \rightarrow \infty} q^n(A-R_a(n))=f^{-1}(R_a(0))$ MATLAB code for evaluating the function for a given a and given radical can be found here.