3 The white points are scaled and rotated *boundaries* of the Mandelbrot set.

Consider the function family given by $f_\lambda(z) = z - p_\lambda(z)/p_\lambda'(z)$ where $p_\lambda(z) = (z^2 - 1)(z - \lambda)$. Every attracting cycle and every rational neutral cycle of $f_\lambda$ attracts the one critical point of $f_\lambda$, which is $\lambda/3$ (see this other MathOverflow question and Alexandre Eremenko's answer for context).

If the sequence of iterates, $f^n_\lambda(\lambda/3)$, converges to a fixed point of $f_\lambda$ then the fixed point is one of the roots of $p_\lambda$, that is $1$, $-1$, or $\lambda$. If we mark the points $\lambda$ in $\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to each of these fixed points then the result is an image that is reminiscent of a Newton basin fractal. Let $K$ be the set of values $\lambda\in\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to a fixed point of $f_\lambda$. My question is:

Is the border, $\partial K$, a Julia set, and if so then for what function family $g$ does $J(g) = \partial K$?

I suspect that the answer is "yes" and that $g$ is a family of iterates of a rational function. In the attached images, the very bright green points are the ones nearest to the border $\partial K$.

A note about the little Mandelbrot sets visible in two of the attached images: In these images the white points indicate parameters $\lambda$ such that $f^n_\lambda(\lambda/3)$ converges to a rational neutral cycle of period greater than $1$. The white points indicate the bifurcation locus of $f_\lambda$ ($J(f_\lambda)$ is not continuously determined, in the sense of the Hausdorff metric, by the parameter at these points). The white points appear to be composed of many scaled and rotated copies of the boundary of the Mandelbrot set and I believe they are contained by $\partial K$. Perhaps that is a useful thing to know when searching for $g$.

2 Fix link to related question and make images link to full size versions

Consider the function family given by $f_\lambda(z) = z - p_\lambda(z)/p_\lambda'(z)$ where $p_\lambda(z) = (z^2 - 1)(z - \lambda)$. Every attracting cycle and every rational neutral cycle of $f_\lambda$ attracts the one critical point of $f_\lambda$, which is $\lambda/3$ (see this other MathOverflow question and Alexandre Eremenko's answer for context).

If the sequence of iterates, $f^n_\lambda(\lambda/3)$, converges to a fixed point of $f_\lambda$ then the fixed point is one of the roots of $p_\lambda$, that is $1$, $-1$, or $\lambda$. If we mark the points $\lambda$ in $\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to each of these fixed points then the result is an image that is reminiscent of a Newton basin fractal. Let $K$ be the set of values $\lambda\in\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to a fixed point of $f_\lambda$. My question is:

Is the border, $\partial K$, a Julia set, and if so then for what function family $g$ does $J(g) = \partial K$?

I suspect that the answer is "yes" and that $g$ is a family of iterates of a rational function. In the attached images, the very bright green points are the ones nearest to the border $\partial K$.

A note about the little Mandelbrot sets visible in two of the attached images: In these images the white points indicate parameters $\lambda$ such that $f^n_\lambda(\lambda/3)$ converges to a rational neutral cycle of period greater than $1$. The white points indicate the bifurcation locus of $f_\lambda$ ($J(f_\lambda)$ is not continuously determined, in the sense of the Hausdorff metric, by the parameter at these points). The white points appear to be composed of many scaled and rotated copies of the Mandelbrot set and I believe they are contained by $\partial K$. Perhaps that is a useful thing to know when searching for $g$.

1

# Is this a Julia set (and if so, for which function family is it the Julia set)?

Consider the function family given by $f_\lambda(z) = z - p_\lambda(z)/p_\lambda'(z)$ where $p_\lambda(z) = (z^2 - 1)(z - \lambda)$. Every attracting cycle and every rational neutral cycle of $f_\lambda$ attracts the one critical point of $f_\lambda$, which is $\lambda/3$ (see this other MathOverflow question and Alexandre Eremenko's answer for context).

If the sequence of iterates, $f^n_\lambda(\lambda/3)$, converges to a fixed point of $f_\lambda$ then the fixed point is one of the roots of $p_\lambda$, that is $1$, $-1$, or $\lambda$. If we mark the points $\lambda$ in $\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to each of these fixed points then the result is an image that is reminiscent of a Newton basin fractal. Let $K$ be the set of values $\lambda\in\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to a fixed point of $f_\lambda$. My question is:

Is the border, $\partial K$, a Julia set, and if so then for what function family $g$ does $J(g) = \partial K$?

I suspect that the answer is "yes" and that $g$ is a family of iterates of a rational function. In the attached images, the very bright green points are the ones nearest to the border $\partial K$.

A note about the little Mandelbrot sets visible in two of the attached images: In these images the white points indicate parameters $\lambda$ such that $f^n_\lambda(\lambda/3)$ converges to a rational neutral cycle of period greater than $1$. The white points indicate the bifurcation locus of $f_\lambda$ ($J(f_\lambda)$ is not continuously determined, in the sense of the Hausdorff metric, by the parameter at these points). The white points appear to be composed of many scaled and rotated copies of the Mandelbrot set and I believe they are contained by $\partial K$. Perhaps that is a useful thing to know when searching for $g$.