A Sequence of Cauchy Sequences Which Is Conjectured to Converge to the Imaginary Parts of the Zeros of the Riemann Zeta Function

TL;DR

This paper discusses a conjectured sequence of Cauchy sequences whose convergence could potentially prove the Riemann hypothesis through a specific transcendental equation criterion.

Contribution

It introduces a new sequence of Cauchy sequences conjectured to converge to the imaginary parts of the zeros of the Riemann zeta function, linking to the Riemann hypothesis.

Findings

Conjecture of convergence of the sequence.

Potential proof of the Riemann hypothesis if convergence is established.

Connection to LeClair and França's transcendental equation criteria.

Abstract

The convergence of a sequence of Cauchy sequences is conjectured; which if shown to be true, would prove the Riemann hypothesis by way of LeClair and Fran\c{c}a's transcendental equation criteria.

Equations

$$\vartheta (y_n)+S(y_n)=\left(n-\frac{3}{2}\right)\pi$$

$$\mathrm{Z}(y_n)=0\mathrm{and}\zeta \left(\frac{1}{2}+i{y}_n\right)=0\forall n\in \text{mathbbm}{Z}^{+}$$

$$\tilde{\vartheta }(t)=\frac{t}{2}\ln \left(\frac{t}{2\pi e}\right)-\frac{\pi }{8}+O(t^{-1})$$

$$\frac{t_n}{2\pi }\ln \left(\frac{t_n}{2\pi t}\right)+S(t_n)=n-\frac{11}{8}$$

$$n=\tilde{\vartheta }(t)+\frac{\pi }{8}-\frac{5}{8}+S(t)$$

$$n-2=\tilde{\vartheta }(t)+\frac{\pi }{8}-\frac{5}{8}+S(t)$$

$$\underset{\delta \to 0^{+}}{\lim }\arg \left(\zeta \left(\frac{1}{2}+\delta +it\right)\right)$$

$$N_0(t)=\frac{t}{2\pi }\ln \left(\frac{t}{2\pi e}\right)+\frac{7}{8}+S(t)+O(t^{-1})$$

$$\zeta (t)=0$$

$$N(t)=\frac{t}{2\pi }\ln \left(\frac{t}{2\pi e}\right)+\frac{7}{8}+S(t)+O(t^{-1})$$

$$f(\alpha )=\alpha$$

$${\lambda }_f(\alpha )=\dot{f}(\alpha )$$

$$f(x)\in [a,b]\forall x\in [a,b]$$

$$\frac{|f(x)-f(y)|}{x-y}⩽c$$

$$Y_{n,m}(t)=\left\{\begin{array}[]{ll}t&m=0\\ t+h_{n,m}\cos(\pi n)\tanh\left(\frac{Z(Y_{n,m-1}(t))}{|\Omega(t)|\prod_{k=1}^{n-1}\tanh(Y_{n,m-1}(t)-y_{k})}\right)&m\geqslant 1\end{array}\right.$$

$$\Omega(t)=\left\{\begin{array}[]{ll}1&t=e\\ e^{\frac{3}{4}\sqrt{\frac{\log(t)}{\log(\log(t))}}}&t\neq e\end{array}\right.$$

$$\underset{0⩽s⩽t}{\max }|Z(s)|>\Omega (t)\forall t⩾45.590\dots$$

$$\frac{|Z(t)|}{\Omega (t)}>0\forall t⩾45.590\dots$$

$$h_{n,m}=\left\{\begin{array}[]{ll}1&m\leqslant 2\\ h_{n,m-1}&\operatorname{sign}(\Delta Y_{n,m-2}(t))=\operatorname{sign}(\Delta Y_{n,m-1}(t))\\ \frac{h_{n,m-1}}{2}&\operatorname{sign}(\Delta Y_{n,m-2}(t))\neq\operatorname{sign}(\Delta Y_{n,m-1}(t))\end{array}\right.$$

$$\Delta Y_{n,m}(t)=Y_{n,m}(t)-Y_{n,m-1}(t)$$

$$|Y_{n,m}(t)-Y_{n,m}(s)|⩽M|t-s|$$

$$M=\underset{x}{\sup }\left|\frac{\mathrm{d}}{\mathrm{d}x}f(x)\right|$$

$$y_n=\underset{m\to \infty }{\lim }{Y}_{n,m}(s_n)$$

$$s_{n}=\left\{\begin{array}[]{ll}14&n=1\\ 21&n=2\\ \frac{y_{n-1}+y_{n-2}}{2}&n\geqslant 3\end{array}\right.$$

$$Y^{+}_{n,m}(t)=\left\{\begin{array}[]{ll}t&m=0\\ t+h_{n,m}\cos(\pi n)\tanh\left(\frac{Z(Y_{n,m-1}(t))}{|\Omega(t)|\prod_{k=1}^{n-1}\tanh(Y_{n,m-1}(t)-y_{k})}\right)&m\geqslant 1\end{array}\right.$$

$$Y^{-}_{n,m}(t)=\left\{\begin{array}[]{ll}t&m=0\\ t-h_{n,m}\cos(\pi n)\tanh\left(\frac{Z(Y_{n,m-1}(t))}{|\Omega(t)|\prod_{k=1}^{n-1}\tanh(Y_{n,m-1}(t)-y_{k})}\right)&m\geqslant 1\end{array}\right.$$

$$z_n=\min (\underset{m\to \infty }{\lim }{Y}_{n,m}^{+}(t),\underset{m\to \infty }{\lim }{Y}_{n,m}^{-}(t))$$

$$f(t)=c(t-\alpha {)}^{m_f(t)}+(\mathrm{higher}\mathrm{order}\mathrm{terms}\dots )$$

$$m_f(t)=\frac{1}{1-{\lambda }_{N_f}(t)}$$

$${\lambda }_{N_f}(t)=\frac{f(t)\ddot{f}(t)}{\dot{f}(t)}$$

Full text

1. 1 Introduction

2. 1.1 Transcendental Equations Satisifed By The Nontrivial Riemann Zeros

3. 2 Iterated Function Systems

4. 2.1 Fixed-Points of Functions

5. 2.1.1 An Iteration Function Which Successively Removes Roots

6. 2.1.2 Indifferent Fixed-Points

7. 2.1.3 Alternating Attractive and Repulsive Fixed-Points

8. 2.2 Contraction Mappings

9. 2.2.1 Sequential Convergence to the Nearest Fixed-Points

10. 3 Appendix

11. 3.1 Definitions

1 Introduction

LeClair and França established criteria for the Riemann hypothesis in [1]. Here, a sequence of Cauchy sequences based on complex dynamical systems involving the Hardy Z function is constructed which explicitly shows that a solution to $\vartheta(y_{n})+S(y_{n})=\left(n-\frac{3}{2}\right)\pi$ should exist for all values of $n$ if it is always possible to choose a small enough Lipschitz constant.

1.1 Transcendental Equations Satisifed By The Nontrivial Riemann

Zeros

Definition 1

The exact equation for the $n$-th zero of the Hardy $Z$ function $y_{n}$ is given by [1, Equation 20]

[TABLE]

where $y_{n}$ enumerate the zeros of $Z$ on the real line and the zeros of $\zeta$ on the critical line

[TABLE]

where $\mathbbm{Z}^{+}$ denotes the positive integers. [1, Equation 14]

By replacing the $\ln\Gamma$ function in (34) with Stirling’s asymptotic expansion as in [1, Equation 13] we get

[TABLE]

and substitute $\vartheta(t)$ with $\tilde{\vartheta}(t)$ in Equation 1 which leads to

Definition 2

The asymptotic equation for the $n$-th zero of the Hardy $Z$ function

[TABLE]

[1, Equation 20]**

Remark 1

The fact that the exact and asymptotic equations have two solutions when $n=1$ can be understood by noting that Equations (1) and (4) are derived from the equation

[TABLE]

which has a minimum in the interval $(-2,-1)$ and thus $n\geqslant-1$ so that, in order to follow the convention that the zeros are enumerated by the positive integers, the substituion $n\rightarrow n-2$ is made in Equation (5) so that

[TABLE]

[1, Equation 12]

Theorem 1

If the limit

[TABLE]

is exists and is well-defined $\forall t$ then the left-hand side of Equation (4) is well-defined $\forall t$, and due to monotonicity, there must be a unique solution for every $n\in\mathbbm{Z}^{+}$. [1, II.A]

Corollary 1

The number of solutions of Equation (4) over the interval $[0,t]$ is given by

[TABLE]

which counts the number of zeros on the critical line.

Conjecture 1

(The Riemann hypothesis) All solutions $t$ of the equation

[TABLE]

besides the trivial solutions $t=-2n$ with $n\in\mathbbm{Z}^{+}$ have real-part $\frac{1}{2}$, that is, $\operatorname{Re}(t)=\frac{1}{2}$ when $\zeta(t)=0$ and $t\neq-2n$.

Definition 3

The Riemann-von-Mangoldt formula makes use of Cauchy’s argument principle to count the number of zeros inside the critical strip $0\char 60\relax\operatorname{Im}(\rho_{n})\char 60\relax t$ where $\zeta(\sigma+i\rho_{n})$ with $0\char 60\relax\sigma\char 60\relax 1$

[TABLE]

and this definition does not depend on the Riemann hypothesis(Conjecture 1). This equation has exactly the same form as the asymptotic Equation 4. [1, Equation 15]

Lemma 1

If the exact Equation (1) has a unique solution for each $n\in\mathbbm{Z}^{+}$ then Conjecture 1, the Riemann hypothesis, follows.

Proof.

If the exact equation has a unique solution for each $n$, then the zeros obtained from its solutions on the critical line can be counted since they are enumerated by the integer $n$, leading to the counting function $N_{0}(t)$ in Equation (8). The number of solutions obtained on the critical line would saturate counting function of the number of solutions on the critical strip so that $N(t)=N_{0}(t)$ and thus all of the non-trivial zeros of $\zeta$ would be enumerated in this manner. If there are zeros off of the critical line, or zeros with multiplicity $m\geqslant 2$, then the exact Equation (1) would fail to capture all the zeros on the critical strip which would mean $N_{0}(t)\char 60\relax N(t)$.  [1, IX] ∎

Corollary 2

The Riemann hypothesis(RH) is not necesarily false if the exact Equation (1) does not have a unique solution for every $n$, since the solutions could still be on the critical line but not necessarily simple, that is, a root on the critical line could have multiplicity $m\geqslant 2$ and the RH would still be true.

Corollary 3

The Riemann hypothesis is true and all of the zeros on the critical line are simple if the exact Equation (1) has a unique solution for each $n\in\mathbbm{Z}^{+}$. [1, IX]

2 Iterated Function Systems

2.1 Fixed-Points of Functions

Definition 4

A fixed-point $\alpha$ of a function $f(x)$ is a value $\alpha$ such that

[TABLE]

[6, 3.]**

Definition 5

The multiplier $\lambda_{f}(\alpha)$ of a fixed point $\alpha$ of a map $f(x)$ is equal to the derivative $\dot{f}(\alpha)$ of the map evaluated at the point $\alpha$ which is the first term in the Taylor expansion at that point

[TABLE]

If $|\lambda_{f}(\alpha)|\char 60\relax 1$ then $\alpha$ is a said to be an attractive fixed-point of $f(x)$. If $|\lambda_{f}(\alpha)|=1$ then $\alpha$ is an indifferent fixed-point of $f(t)$ also known as as neutral fixed-point, and if $|\lambda_{f}(\alpha)\char 62\relax 1|$ then $\alpha$ is a repelling fixed-pint of $f(t)$. When $|\lambda_{f}(\alpha)|=0$ the fixed-point $\alpha$ is said to be superattractive fixed-point of $f(t)$[6, 3.]

Lemma 2

The Banach Fixed-Point Theorem

If $f(x)$ is a continuous function defined on $[a,b]$ and

[TABLE]

and there exists some constant $0\char 60\relax c\char 60\relax 1$ such that

[TABLE]

then $f(x)$ has a unique fixed-point $x\in[a,b]$ and the sequence $f(x_{0}),f(f(x_{0})),f(f(f(x_{0}))),\ldots$ converges to the unique fixed-point of $f(x)$ in the interval $[a,b]$.

2.1.1 An Iteration Function Which Successively Removes Roots

Definition 6

Let

[TABLE]

denote the $m$-th iterate of the $n$-th iteration function corresponding to the $n$-th zero of the Hardy $Z$ function where

[TABLE]

is a lower bound for the running maximum of $|Z(s)|$

[TABLE]

ensuring that

[TABLE]

which normalizes the range of $Z(t)$ which is known to grow in both maximum and average value as $t\rightarrow\infty$ and $h_{n,m}$ is factor which influences the rate of convergence

[TABLE]

where

[TABLE]

is the $1$-st difference of the $m$-th iterate for the $n$-th zero. [7, Theorem 3.2.3]

Lemma 3

The roots of $Z(t)$ are fixed-points of $Y_{n,m}(t)\forall n,m\in\mathbbm{Z}^{+}$.

Proof.

If $Z(t)=0$ then $\tanh\left(\frac{Z(t)}{|\Omega(t)|\prod_{k=1}^{n-1}\tanh(t-y_{k})}\right)=\tanh\left(\frac{0}{|\Omega(t)|\prod_{k=1}^{n-1}\tanh(t-y_{k})}\right)=\tanh(0)=0$ so that $Y_{n}(t)=t+\cos(\pi n)0=t+0=t$ when $Z(t)=0$. ∎

2.1.2 Indifferent Fixed-Points

Theorem 2

$Y_{n,m}(t)$* has indifferent fixed-points at each point $y_{k}$ where $k=1\ldots n-1$*

Proof.

The product in the denominator $\prod_{k=1}^{n-1}\tanh(t-y_{k})\rightarrow 0$ smoothly as $t$ approaches any $y_{k}\in\bigcup_{k=1}^{n-1}y_{k}$ since $\tanh(0)=0$ and $\tanh$ is a smooth function. When any element of the product is zero the value of the product is zero regardless of the values of any other elements of the product. Since $\frac{1}{s}\rightarrow\infty$ as $s\rightarrow 0$ and $\tanh(|x|)\rightarrow 1$ as $|x|\rightarrow\infty$ we have $\tanh(\infty)=1$ and $\tanh(-\infty)=-1$ so that $Y_{n}(t)=t+\cos(\pi n)\forall t\in\bigcup_{k=1}^{n-1}y_{k}$. Since $Y_{n}(t)=t\pm 1\forall t\in\bigcup_{k=1}^{n-1}y_{k}$ when $n$ is an integer, we see that $\frac{\mathrm{d}}{\mathrm{d}t}Y_{n}(t)=\frac{\mathrm{d}}{\mathrm{d}t}(t\pm 1)=1$ so that the multiplier $\lambda_{Y_{n}(t)}=\left|\frac{\mathrm{d}}{\mathrm{d}t}Y_{n}(t)\right|=1\forall t\in\bigcup_{k=1}^{n-1}y_{k}$. ∎

Theorem 3

$Y_{n,m}(t)$* has indifferent fixed points at each trivial zero $-\frac{i}{2}(-4n-1)$ where $Z\left(-\frac{i}{2}(-4n-1)\right)=0\forall n\in\mathbbm{Z}^{+}$.*

Proof.

Since $\frac{\mathrm{d}}{\mathrm{d}t}(f(t)+g(t))=\dot{f}(t)g(t)+f(t)\dot{g}(t)$ and $Z(t)=e^{i\vartheta(t)}\zeta\left(\frac{1}{2}+it\right)$ it suffices to show that $\lim_{t\rightarrow-\frac{i}{2}(-4n-1)}\left|\frac{\mathrm{d}}{\mathrm{d}t}e^{i\vartheta(t)}\right|=\infty$. Since $\frac{\mathrm{d}}{\mathrm{d}t}e^{i\vartheta(t)}=i\left(\frac{\frac{i\Psi\left(\frac{1}{4}-\frac{it}{2}\right)}{2}-\frac{i\Psi\left(\frac{1}{4}+\frac{it}{2}\right)}{2}}{2}-\frac{\ln(\pi)}{2}\right)e^{i\vartheta(t)}$ we only have to check that $\lim_{t\rightarrow\frac{i}{2}(-4n-1)}\left|\Psi\left(\frac{1}{4}+\frac{it}{2}\right)\right|=\lim_{t\rightarrow-\frac{i}{2}(-4n-1)}\left|\Psi\left(\frac{1}{4}+\frac{it}{2}\right)\right|=\infty$ which is true since $\Psi(t)$ has poles at $t=1-n\forall n\in\mathbbm{Z}^{+}$ where $\Psi(t)=\frac{\dot{\Gamma}(t)}{\Gamma(t)}$ and $\Gamma(t)$ has poles at $t=1-n\forall n\in\mathbbm{Z}^{+}$. Since $|\tanh(\infty)|=1$ the multiplier is equal to 1 at each $-\frac{i}{2}(-4n-1)$. ∎

2.1.3 Alternating Attractive and Repulsive Fixed-Points

Proposition 1

When $n$ is an odd number, $Y_{n}(t)$ has attractive fixed-points at the odd-numbered roots $y_{2k-1}\forall 2k-1\geqslant n$ and repulsive fixed-points at the even-numbered roots $y_{2k}\forall 2k\geqslant n$.

Proposition 2

When $n$ is an even number, $Y_{n}(t)$ has attractive fixed-points at the even-numbered roots $y_{2k}\forall 2k\geqslant n$ and repulsive fixed-points at the odd-numbered roots $y_{2k-1}\forall 2k-1\geqslant n$.

Remark 2

The function $h_{n,m}$ is defined to be $1$ when $1\leqslant m\leqslant 2$. If $\operatorname{sign}(\Delta Y_{n,m-1}(t))\neq\operatorname{sign}(\Delta Y_{n,m}(t))$ then $h_{n,m+1}=\frac{h_{n,m}}{2}$ so that the convergence rate is  halved when the sign of the difference between successive iterates changes, indicating that it jumped across the root. This prevents the sequence generated by the iteration from getting stuck in an artifical $2$-cycle and jumping back and forth across the root with equal magnitude indefinately when implementing this method with finite-precision arithmetic on a digital computer. Without this successive relaxation, the iterates still converge in theory however the number of iterations required could be several million or higher, while still having the difficulty of possibly getting stuck in a $2$-cycle in computer implementations.

2.2 Contraction Mappings

Theorem 4

The Lipschitz constant $M$ of the map $Y_{n,m}(t)\char 60\relax 1\forall t\char 62\relax e$ therefore $Y_{n,m}(t)$ is a contraction mapping

[TABLE]

Proof.

The Lipschitz constant of a continuous differentiable function $f(x)$ is equal to the maximum absolute value of its derivative

[TABLE]

The derivative of $t-\tanh(t)$ is $\tanh(t)^{2}$. Since the maximum absolute value of $\tanh(t)$ is 1 then the maximum value of its square is also $1$. Since $\Omega(t)\char 62\relax 1$ and $h_{n,m}\leqslant 1$ the derivative $\frac{\mathrm{d}}{\mathrm{d}t}Y_{n,m}(t)$ can never have an absolute value $\geqslant 1$ since that would require $\left|\tanh\left(\frac{Z(t)}{|\Omega(t)|\prod_{k=1}^{n-1}\tanh(t-y_{k})}\right)\right|=1$ which is only possible if $Z(t)=\pm\infty$ which is only the case when $t=\pm\frac{i}{2}$ which corresponds to the pole at $\zeta(1)$. Since $Z(t)\in\mathbbm{R}$ when $t\in\mathbbm{R}$ it can never be the case that $Z(t)=\infty$ so that $\left|\frac{\mathrm{d}}{\mathrm{d}t}Y_{n,m}(t)\right|\neq 1\forall t\in\mathbbm{R}$ and the Lipschitz constant $M$ is strictly less than 1. ∎

2.2.1 Sequential Convergence to the Nearest Fixed-Points

Proposition 3

The limit

[TABLE]

where

[TABLE]

exists and is equal to the $n$-th zero of the Hardy Z function for all integer $n\subset\mathbbm{Z}^{+}$. That is, $Y_{n,m}(z_{n})$ forms a Cauchy sequence, due to the contraction mapping property proved in Theorem 4 whose elements are indexed by $m$ converging to the $n$-th root $y_{n}$ where the $n$-th starting point is defined to be half-way between the $(n-2)$-th and the $(n-1)$-th root $y_{n}$ when $n\char 62\relax 2$ and equal to a point close to the first known zero at $14.134\ldots.$ when $n=1$ and a point close to the $2\operatorname{nd}$ zero at $21.022\ldots$ when $n=2$

Remark 3

The mid-way point between the nearest neighbors to the left of $y_{n}$ is used as the starting point for the iteration since any point less than $y_{n}$ and greater than $e$ is within the immediate basin of attraction of $y_{n}$. The precise location of any roots $y_{p}$ where $p\char 60\relax n$ cannot be used as a starting point since the map $Y_{p,m}(t)$ is a non-expansive mapping with Lipschitz constant precisely equal to 1 when $t\in\bigcup^{p-1}_{k=1}y_{k}$ so that the hyperbolic tangent has an argument of infinity resulting in a value of 1. Trajectories are neither attracted or repelled to any point $\bigcup^{n-1}_{k=1}y_{k}$ under the action of the map $Y_{n,m}(t)$ however, trajectories started precisely on any point $t\in\bigcup^{n-1}_{k=1}y_{k}$ will never attain a value other than $t$ since any $y_{k}$ is a fixed-point of $Y_{n,m}(t)$.

Note 1

The truth of Propositon 3 has been verified computationally up to $n=800,000$ with a computer program which implements the methods described here using the arbitrary precision complex ball arithmiticlibrary arblib[3] and compares the results against the tables published by Andrew Odlyzko[5].

Theorem 5

The Cauchy sequence $\lim_{m\rightarrow\infty}Y_{n,m}(s_{n})$ will never converge to any $y_{k}$ where $k\char 60\relax n$.

Proof.

All $y_{k}$ are indifferent fixed-points of $Y_{n,m}(t)$ and the trajectories generated by $Y_{n,m}(s_{n})$ are never started from a point $y_{k}$ since $s_{n}\not\in\bigcup_{k=1}^{n-1}y_{k}$ and the only way $Y_{n,m}(t)$ would ${}^{\prime\prime}\operatorname{convege}^{\prime\prime}$ to an indifferent fixed-point is if it was started precisely on one, and $s_{n}$ is by definition equal to the mid-point between successive $y_{n}$. ∎

Theorem 6

The Cauchy sequence $Y_{n,m}(s_{n})$ will never converge to any $y_{n+2k-1}\forall k\in\mathbbm{Z}^{+}$ if Proposition 1 is true.

Proof.

If Propositions 1 is true then $y_{n+2k-1}$ are repelling fixed-points for $Y_{n,m}(t)$. ∎

Note 2

If Propositions 1 and 2 are true then $Y_{n,m}(s_{n})$ will never converge to $y_{q}$ with $q$ odd and $n$ even nor to $y_{r}$ with $r$ even and $n$ odd. It suffices to prove that $Y_{n,m}(s_{n})\char 60\relax y_{n+1}\forall n,m\in\mathbbm{Z}^{+}$ which would mean that $Y_{n,m}(s_{n})$ can never jump across the repelling fixed-point at $y_{n+1}$ to land on any of the attractive fixed-points in $\bigcup_{k=1}^{\infty}y_{n+2k}$

Lemma 4

Let

[TABLE]

[TABLE]

and

[TABLE]

which must exist because there is known to be an infinity of zeros on the critical line.

Proof.

The only way $z_{n}$ would not exist is if all the roots $y_{k}$ were indifferent fixed-points$\forall k\char 62\relax n$ but that is impossible since there are no indifferent fixed-points of $Y_{n,m}(t)$ because for a fixed-point $y_{k}$ to be indifferent would require $\tanh\left(\frac{Z(Y_{n,m-1}(t))}{|\Omega(t)|}\right)=1$ which is only possible if $|Z(Y_{n,m-1}(t))|=\infty$ for some $m\in\mathbbm{Z}^{+}$ and the $Z$ function only takes on the value $\infty$ when $t=-\frac{i}{2}$ which corresponds to the pole at $\zeta(1)$ since $\frac{1}{2}+i\left(-\frac{i}{2}\right)=1$. ∎

Definition 7

The multiplicity $m_{f}(t)$ of a root $\alpha$ is a root $f(\alpha)=0$ such that its Taylor expansion about the point $\alpha$ has the form

[TABLE]

where $c\neq 0$ and $m\geqslant 1$. The multiplicity of a root $t$ is related to the multipler $\lambda_{f}(t)$ through the formula

[TABLE]

where

[TABLE]

is the first derivative of the Newton map of $f(t)$

[TABLE]

Lemma 5

(Milnor’s Lemma) Every simple root of $f(t)$ is a super-attractive fixed-point of $N_{f}(t)$ since a superattractive fixed-point is one such that its multiplier $\lambda_{f}(t)=0$ so that its multiplicity is

[TABLE]

See [4, p.52]

Proof.

Let $\alpha$ be a root $Z(\alpha)=0$ then the multiplier of its Newton map is $\lambda_{N_{Z}}(\alpha)=\lambda_{f}(\alpha)=\frac{Z(\alpha)\ddot{Z}(\alpha)}{\dot{Z}(\alpha)}=0$ since $Z(\alpha)=0$ the entire expression $\frac{Z(\alpha)\ddot{Z}(\alpha)}{\dot{Z}(\alpha)}$ is equal to 0 since due to the ordering of operations the value of $\dot{Z}(t)$ or $\ddot{Z}(t)$ is never required to be known in order to know the value of $\lambda_{N_{Z}}(t)$ when $Z(t)=0$. If any term in the product is 0 then the entire product takes the value 0. The multiplicity is related to the multiplier by $m_{Z}(t)=\frac{1}{1-\lambda_{N_{Z}}(t)}=1$ and therefore simple. Since  $m_{Z}(t)=\frac{1}{1-\lambda_{N_{Z}}(t)}\forall\lambda_{N_{Z}}(t)\neq 1$then it is known that $\lambda_{N_{Z}}(t)=0$ when $Z(t)=0$  therefore the point $\alpha$ is a superattractive fixed-point corresponding to a simple zero at $\alpha$. Since we now know that $m_{Z}=(\alpha)$ and therefore the zero at $Z(\alpha)=0$ is simple, we therefore know that the denominator $\dot{\mathrm{Z}}(t)$ of the multiplier $\lambda_{N_{Z}}(t)$ cannot vanish so that   $\dot{\mathrm{Z}}(\alpha)\neq 0$ since that would imply that $\alpha$ is not a simple root, which would be a contradiction to the already established fact that $m_{Z}(\alpha)=1$ when $Z(\alpha)=0$. ∎

Conjecture 2

The roots generated by the sequence $y_{n}=\lim_{m\rightarrow\infty}Y_{n,m}(t)$ are simple

Conjecture 3

Let

[TABLE]

denote the Lipschitz constant in Formula 14 then it is always possible to choose a small enough positive $\varepsilon$ such that $0\char 60\relax c_{n}(\varepsilon)\char 60\relax 1$.

3 Appendix

3.1 Definitions

Let $\zeta(t)$ be the Riemann zeta function

[TABLE]

and $\vartheta(t)$ be Riemann-Siegel vartheta function

[TABLE]

so that the Hardy $Z$ function[2] can be defined by

[TABLE]

which is real-valued when $t$ is real and satisfies the identity

[TABLE]

where $\ln\Gamma(z)$ is the principal branch of the logarithm of the $\Gamma$ function defined by

[TABLE]

which is analytically continued from the positive real axis when $z\in\mathbbm{C}$ is complex. Each of the points $z\in\mathbbm{Z}=\{0,-1,-2,\ldots\}$ is a singularity and a branch point so that the union of the branch cuts is the negative real axis. On the branch cuts, the values of $\ln\Gamma(z)$ are determined by continuity from above. Let $S(t)$ denote the normalized argument of $\zeta(t)$ on the critical line

[TABLE]

Definition 8

The critical line is the line in the complex plane defined by $\operatorname{Re}(t)=\frac{1}{2}$.

Definition 9

The critical strip is the strip in the complex plane defined by $0\char 60\relax\operatorname{Re}(t)\char 60\relax 1$.