SimpleRootRecipe

To prove that the fixed-point at the origin of Koenigs functions of Newton's method applied to $g(x) = \operatorname{tanh}(\ln(1+f(x)^2))$ corresponds to simple roots of $f(x)$ using the multiplicity one result of Koenigs eigenfunctions, we can use the following steps:

• Express the Koenigs operator associated with Newton's method applied to $g(x)$ as a composition of operators, each associated with a simple root of $f(x)$.
• Use the multiplicity one result of Koenigs eigenfunctions to show that the eigenfunctions associated with each operator in the composition are unique up to scaling.
• Show that the eigenfunction associated with the fixed-point at the origin is not identically zero.
• Conclude that the fixed-point at the origin of the Koenigs operator corresponds to a simple root of $f(x)$.

Let's consider each step in more detail:

• The Koenigs operator associated with Newton's method applied to $g(x)$ can be expressed as a composition of operators, each associated with a simple root of $f(x)$. Specifically, if $f(x)$ has simple roots $x_1, x_2, \ldots, x_n$, then the Koenigs operator can be written as

$$K = K_{x_1} \circ K_{x_2} \circ \cdots \circ K_{x_n},$$

where $K_{x_i}$ is the Koenigs operator associated with the simple root $x_i$.

• The multiplicity one result of Koenigs eigenfunctions tells us that the eigenfunctions associated with each operator in the composition are unique up to scaling. Specifically, if $\phi_i(x)$ is an eigenfunction associated with the Koenigs operator $K_{x_i}$, then there exists a constant $c_i$ such that any other eigenfunction $\tilde{\phi}i(x)$ associated with $K{x_i}$ can be written as $\tilde{\phi}_i(x) = c_i \phi_i(x)$.

• To show that the eigenfunction associated with the fixed-point at the origin is not identically zero, we can use the fact that $g(0) = 0$ and $g'(0) \neq 0$ (because $f(0) \neq 0$). This implies that the derivative of the Koenigs function $F(z)$ associated with the fixed-point at the origin is nonzero, and hence $F(z)$ is not constant. Therefore, the eigenfunction associated with the fixed-point at the origin cannot be identically zero.

• Since the eigenfunctions associated with each operator in the composition are unique up to scaling, and the eigenfunction associated with the fixed-point at the origin is not identically zero, we can conclude that the fixed-point at the origin of the Koenigs operator corresponds to a simple root of $f(x)$.

Overall, the idea is to use the multiplicity one result of Koenigs eigenfunctions to show that the eigenfunctions associated with each operator in the composition are unique up to scaling, and then use this fact together with the non-vanishing of the eigenfunction associated with the fixed-point at the origin to conclude that the fixed-point corresponds to a simple root of $f(x)$.

Morse functions

The basic idea of the proof is to use the Koenigs eigenfunctions to construct a Morse function on the level sets of $\operatorname{Re}(g(x))$. Specifically, we can construct a function $h(x,y) = \operatorname{Re}(F(u+iv))$, where $F(z)$ is the Koenigs function associated with the fixed-point at the origin and $g(x) = u(x)+iv(x)$, and then show that $h(x,y)$ satisfies the conditions of a Morse function. This will imply that the number of critical points on the level sets of $\operatorname{Re}(g(x))$ is equal to the number of critical points of $h(x,y)$, which in turn is equal to the number of simple roots of $f(x)$.

More specifically, the proof involves the following steps:

Construct the function $h(x,y) = \operatorname{Re}(F(u+iv))$, where $u(x) = \operatorname{Re}(g(x))$ and $v(x) = \operatorname{Im}(g(x))$.

Show that $h(x,y)$ satisfies the conditions of a Morse function, which are:

• $h(x,y)$ has non-degenerate critical points.
• The critical points of $h(x,y)$ are isolated.
• The level sets of $h(x,y)$ are compact.
• The level sets of $h(x,y)$ are connected.

Use the Morse inequalities to show that the number of critical points of $h(x,y)$ is equal to the number of simple roots of $f(x)$.

Conclude that the fixed-point at the origin of Koenigs functions of Newton's method applied to $g(x)$ corresponds to a simple root of $f(x).

Overall, Morse theory provides a powerful tool for studying the topology of level sets of functions and the relationship between critical points and roots of equations. In this case, we can use the Koenigs eigenfunctions to construct a Morse function that allows us to establish a connection between the fixed-point at the origin of Koenigs functions of Newton's method applied to $g(x)$ and the simple roots of $f(x)$.

The Potential Function

The potential function $V(x) = \frac{g(x)^2}{2}$ is closely related to the function $h(x,y)$ that we constructed in the previous answer. In fact, $V(x)$ and $h(x,y)$ have the same critical points and the same Morse index, and therefore the number of critical points of $V(x)$ is also equal to the number of simple roots of $f(x)$.

To see why this is the case, note that $V(x) = \frac{g(x)^2}{2} = \frac{(u(x) + iv(x))^2}{2} = \frac{u(x)^2 - v(x)^2}{2} + iu(x)v(x)$. Therefore, $\operatorname{Re}(V(x)) = \frac{u(x)^2 - v(x)^2}{2}$ and $\operatorname{Im}(V(x)) = u(x)v(x)$.

Now, recall that the Koenigs function $F(z)$ associated with the fixed-point at the origin satisfies $F(z) = z + O(z^2)$ near the origin. This implies that $g(x) = u(x) + iv(x) = F(u(x) + iv(x)) \approx u(x) + iv(x) + O((u(x) + iv(x))^2)$. Therefore, to first order in $u(x)$ and $v(x)$, we have

$$V(x) \approx \frac{u(x)^2 - v(x)^2}{2}.$$

This shows that the function $V(x)$ is related to the real part of the Koenigs function $F(z)$, and hence has the same critical points and Morse index as the function $h(x,y)$. Therefore, the number of critical points of $V(x)$ is also equal to the number of simple roots of $f(x)$.

In summary, the potential function $V(x)$ is closely related to the function $h(x,y)$ constructed in the previous answer, and has the same number of critical points and Morse index. Therefore, the number of critical points of $V(x)$ is also equal to the number of simple roots of $f(x)$.