\section{The weak rank condition}\label{sec:rank} The size estimates are now available, but Proposition~\ref{prop:criterion} also requires the weak rank condition \eqref{eq:weak-rank}. We prove it first for the derivative rows $Q_k^{[p]}(-1)$, before removing the odd columns. Section~\ref{sec:conclusion} then passes it to the integer forms in even beta values. The rows have zero entries for $i>b+h$: this holds for $C^{[p]}$ and is preserved by \eqref{eq:Q}. We can therefore work with the system truncated at component $b+h$. The proof adapts \cite[Section~4.5]{F26}. A forbidden kernel vector would give solutions whose polynomial combinations have many vanishing derivatives at $-1$. We convert these into remainders for a single polynomial row, prove that the remainders are independent, and apply Proposition~\ref{prop:multiplicity} to contradict their total vanishing. The inequality below is the one obtained from that final count. \begin{proposition}\label{prop:rank} Assume \begin{equation}\label{eq:rank-condition} (h+1)(\kappa-2r)+\omega>a. \end{equation} For all sufficiently large admissible $n$, if $x_0,\ldots,x_{b+h}\in\Q$ satisfy \[ \sum_{i=0}^{b+h}Q_{k,i}^{[p]}(-1)x_i=0 \quad(0\leq p\leq h,\quad 2M+2\leq k\leq\kappa n), \] then $x_0=0$. \end{proposition} The factor $h+1$ records the gain from using all the derivative families: each will supply an independent remainder at each point above $-1$. For fixed $r,\kappa,\omega$, condition \eqref{eq:rank-condition} forces $h$ to grow proportionally to $a$. This is why the extra families, and their larger beta cutoff $a+h$, were introduced. \subsection{The multiplicity estimate and the six points}\label{subsec:multiplicity} The multiplicity estimate bounds the total vanishing of polynomial combinations of solutions in terms of the polynomial degrees. We first state it, then describe the contributions we will need from the six points in our application. Let $Y'=\matA(w)Y$ be a fixed rational differential system of size $q$. For polynomials $S_0,\ldots,S_{q-1}$ of degree at most $m$, write \[ \mathcal R(Y)=\sum_{j=0}^{q-1}S_jY_j. \] We call this scalar function the \emph{remainder} associated with the solution $Y$ and the fixed polynomial row $(S_0,\ldots,S_{q-1})$. The selected remainders must satisfy a common scalar linear differential equation of some order $\mu$, with rational coefficients. In our application such an equation will follow from the finite dimension of the system. At a point $\sigma$, the Nilsson class consists of finite sums of holomorphic functions multiplied by complex powers and nonnegative integer powers of a logarithm of the local coordinate. At infinity the local coordinate is $1/w$. The order is the least real part of a power that occurs after expanding the holomorphic factors; logarithmic factors do not change it. For example, $\ord_0(w^e(\log w)^j)=\operatorname{Re}e$. In particular, $w^N$ has order $-N$ at infinity. Negative orders will therefore occur when we apply the theorem. \begin{proposition}[{\cite[Theorem~3]{F26}; see also \cite[Theorem~3.1]{F18}}]\label{prop:multiplicity} Let $\Sigma$ be a fixed finite set of points of the Riemann sphere containing infinity. At each $\sigma\in\Sigma$, choose a family of local solutions $Y_{\sigma,j}$, indexed by $j\in J_\sigma$, whose remainders are in the Nilsson class and are linearly independent over $\C$. If a nonzero rational differential operator of order $\mu$ annihilates all these remainders, then \[ \sum_{\sigma\in\Sigma}\sum_{j\in J_\sigma} \ord_\sigma\mathcal R(Y_{\sigma,j}) \leq(m+1)(\mu-|J_\infty|)+C. \] The constant $C$ depends only on the differential system and on $\Sigma$. \end{proposition} The solution families at different points need not agree by analytic continuation, and no independence between different families is required. These features are explicit in \cite[Section~3.1]{F18}. We use the formulation in \cite{F26}, which also allows negative orders at infinity, as in its proof of Proposition~3. The constant is uniform in the polynomial row and in the selected solutions. This uniformity is essential when that row and the solutions depend on $n$. The six points will appear after the substitution $z=w^2$. The assumed kernel equations produce high-order zeros at the two points $w=\ii$ and $w=-\ii$ above $z=-1$. The decay of $F_n$ at infinity will supply additional zeros at $w=1$ and $w=-1$. We will also construct $b$ logarithmic remainders at $0$ and infinity, where their orders can be read from the polynomial degrees. The remainders at infinity have a second purpose. In our application the common differential operator has order $\mu\leq b+h+1$. Choosing $|J_\infty|=b$ reduces the factor $\mu-|J_\infty|$ on the right-hand side of the theorem to at most $h+1$. This is essential: the degree cost then involves the number of derivative families, rather than the full size of the system. The negative orders at infinity must still be included on the left. Paired with the orders at $0$, their contribution will be $-2bn$. We will obtain polynomial degrees at most $4rn$ and the inequality \[ 2(h+1)\kappa n+2\omega n-2bn+O(1) \leq 4rn(h+1)+O(1). \] The first two terms count the zeros at $\pm\ii$ and $\pm1$; the third is the combined contribution at $0$ and infinity. After division by $2n$, this contradicts \eqref{eq:rank-condition}. The rest of the section constructs the required remainders, verifies their independence, and justifies this count. \subsection{Taylor corrections and a common polynomial row} Suppose $x_0\neq0$, and scale so that $x_0=1$. Let $g_0,\ldots,g_{b+h}$ solve \eqref{eq:base-system} near $-1$ with $g_i(-1)=x_i$. In particular $g_0=1$. The kernel equations say that certain high derivatives of $\sum_i C_i^{[p]}g_i$ vanish at $-1$. To turn this into a zero of high order, we must also remove the lower Taylor coefficients. Choose $T_p\in\Q[z]$ of degree at most $2M$ to cancel the Taylor polynomial through degree $2M$ of $\sum_i C_i^{[p]}g_i$ at $-1$. The universal identity \eqref{eq:universal-row} and the assumed vanishing of the higher derivatives give \begin{equation}\label{eq:f-zero} f_p(z):=T_p(z)+\sum_{i=1}^{b+h}C_i^{[p]}(z)g_i(z) =O((z+1)^{\kappa n}). \end{equation} The derivatives of orders $0,\ldots,2M$ vanish by the choice of $T_p$; those of orders $2M+1,\ldots,\kappa n-1$ vanish by hypothesis. The functions $f_p$ still have different coefficient rows. The multiplicity theorem instead needs one fixed polynomial row applied to several solutions of the same system. Following Fischler, we arrange this by combining the $f_p$ with powers of a local branch $L(z)=-\log z$ near $-1$. The choice is suggested by $L'=-1/z$, the same coefficient that links $g_j$ to $g_{j-1}$ for $j\geq2$ in \eqref{eq:base-system}. Powers of $L$ can therefore provide the additional chain of components needed to hold the corrections $T_p$. First take $h=1$ and set \[ \rho_0=f_0,\qquad \rho_1=Lf_0+f_1. \] Substituting \eqref{eq:C} gives \begin{align*} \rho_0&=T_0+z^M\sum_{j=1}^b P_jg_j,\\ \rho_1&=T_1+LT_0+z^M\sum_{j=1}^b P_j(Lg_j-jg_{j+1}). \end{align*} Both expressions use the same polynomial row \[ S=(T_1,T_0,z^MP_1,\ldots,z^MP_b). \] Indeed, $\rho_0=SY_0$ and $\rho_1=SY_1$ for the columns \[ \begin{aligned} Y_0&=(0,1,g_1,\ldots,g_b)^T,\\ Y_1&=(1,L,Lg_1-g_2,\ldots,Lg_b-bg_{b+1})^T. \end{aligned} \] The dependence on the family index is now entirely in $Y_0,Y_1$: the multipliers of $T_1,T_0$ and each $P_j$ change, while $S$ stays fixed. In particular, the logarithm belongs to the solutions, so it does not spoil the polynomial row. The same combinations also preserve the differential chain. Put $Z_j=Lg_j-jg_{j+1}$ for $1\leq j\leq b$. For $j\geq2$, the product rule and \eqref{eq:base-system} give \begin{align*} Z_j'&=-\frac{g_j}{z} +L\left(-\frac{g_{j-1}}z-\frac{g_j}{2z}\right) -j\left(-\frac{g_j}z-\frac{g_{j+1}}{2z}\right)\\ &=-\frac{Lg_{j-1}-(j-1)g_j}{z} -\frac{Lg_j-jg_{j+1}}{2z}\\ &=-\frac{Z_{j-1}}z-\frac{Z_j}{2z}. \end{align*} The term $L'g_j=-g_j/z$ changes the coefficient $j$ to $j-1$, producing the previous component $Z_{j-1}$. At the first link, the same calculation gives $Z_1'=2L/[z(1-z)]-Z_1/(2z)$. Together with $L'=-1/z$ and the equations for $g$, these identities show that both columns solve the $h=1$ case of \eqref{eq:enlarged-system} below. For general $h$, set \[ \rho_q(z)=\sum_{p=0}^q\binom qp L(z)^{q-p}f_p(z) \qquad(0\leq q\leq h). \] Each $\rho_q$ retains the zero of order at least $\kappa n$ at $-1$. The common polynomial row will have $h+1$ entries for the corrections $T_p$ and $b$ entries for the original polynomials $P_i$: \begin{equation}\label{eq:S-common} S_i(z)= \begin{cases} T_{h-i}(z)/(h-i)!,&0\leq i\leq h,\\ z^M P_{i-h}(z),&h+1\leq i\leq b+h, \end{cases} \end{equation} The dependence on $q$ is placed in the solution vectors. Define \begin{equation}\label{eq:Yq} y_{i,q}(z)= \begin{cases} 0,&ih. \end{cases} \end{equation} Then $\rho_q=\sum_i S_i y_{i,q}$. All columns $Y_q=(y_{0,q},\ldots,y_{b+h,q})^T$ solve the same system \begin{align} y_0'&=0,\notag\\ y_i'&=-y_{i-1}/z &&(1\leq i\leq h),\notag\\ y_{h+1}'&=\frac{2}{z(1-z)}y_h-\frac1{2z}y_{h+1}, \label{eq:enlarged-system}\\ y_i'&=-y_{i-1}/z-y_i/(2z) &&(i\geq h+2).\notag \end{align} The first $h+1$ components are powers of $L$; the components with $i>h$ contain the original solution $g$. On differentiating the latter components, terms from the derivatives of $L^{q-p}$ combine with those from the derivatives of $g_{i-h+p}$. After reindexing, their coefficients simplify by the identity \[ (j)_p-p(j)_{p-1}=(j-1)_p \qquad(j\geq2,\ p\geq1). \] The $p=0$ factor is $1$. This gives the equation at index $i=h+j$. When $j=1$, the corresponding factors vanish for $p\geq1$; the remaining source is $2L^q/[z(1-z)]$, giving the equation for $y_{h+1}$. The diagonal term remains $-y_i/(2z)$ in every summand. The equations for $i\leq h$ follow by differentiating powers of $L$. \subsection{Quadratic change of variable} The integrating factor in Section~\ref{sec:arithmetic} removed the diagonal term locally to control Taylor denominators. Here we need the same cancellation in a form that preserves a polynomial row and leaves a system suitable for the residue argument. Multiplying the components $i>h$ by $\sqrt z$ removes their common diagonal term $-y_i/(2z)$. We introduce $w$ with $z=w^2$ so that this change is rational in the new variable and the row remains polynomial. Set \[ v_i(w)= \begin{cases} y_i(w^2),&i\leq h,\\ w\,y_i(w^2),&i>h. \end{cases} \] The diagonal terms cancel, leaving \begin{align} v_0'&=0,\notag\\ v_i'&=-2v_{i-1}/w &&(i\geq1,\ i\neq h+1),\label{eq:lifted-system}\\ v_{h+1}'&=4v_h/(1-w^2).\notag \end{align} The common coefficient polynomials become \begin{equation}\label{eq:lifted-polynomials} \widetilde S_i(w)= \begin{cases} S_i(w^2),&i\leq h,\\ w^{2M-1}P_{i-h}(w^2),&i>h. \end{cases} \end{equation} Their degrees are at most $4M$: those with $i\leq h$ have degree at most $4M$, and those with $i>h$ have degree at most $2(M+n)-1\leq4M-1$. For the solutions obtained from $Y_q$, the remainder $\sum_i\widetilde S_i v_i$ equals $\rho_q(w^2)$. It vanishes to order at least $\kappa n$ at each of $\ii$ and $-\ii$. The change of variable increases the polynomial degrees, but it also gives two points at which to count the high-order zeros. The same polynomials \eqref{eq:lifted-polynomials} are used at both points; only the local solution families differ. This is why the Taylor corrections were constructed in $z$ before making the change of variable. \subsection{Independence by residues} To count these zeros with Proposition~\ref{prop:multiplicity}, we must know that no nonzero constant combination of the $h+1$ remainders vanishes identically. Independence of the solution vectors alone would not suffice: a fixed row can annihilate a nonzero combination of vectors. We must rule out such annihilation for our row, whose highest coefficient is nonzero because $P_b\not\equiv0$. The triangular form of the new system makes this possible. Each derivative link has a nonzero residue, whereas the derivative of a rational function has zero residue at every finite point. The following lemma uses this to rule out rational-function relations between successive components. It abstracts the argument in \cite[Section~4.5, Step~4]{F26}. Different links of the system may use different finite poles. \begin{lemma}\label{lem:chain} Suppose $v_0$ is a nonzero constant and $v_j'=a_j(w)v_{j-1}$ for $1\leq j\leq t$, where each rational function $a_j$ has a nonzero residue at some finite point. Then $v_0,\ldots,v_t$ are linearly independent over $\C(w)$. \end{lemma} \begin{proof} Clear the denominators of a putative rational relation and write it as $\sum_{i=0}^{\alpha}U_i v_i=0$, with $U_\alpha\neq0$ and $\alpha\geq1$. Let $D=\deg U_\alpha$. Differentiating $D+1$ times eliminates $v_\alpha$. The coefficient of $v_{\alpha-1}$ is \begin{equation}\label{eq:residue-coefficient} U_{\alpha-1}^{(D+1)} +\sum_{j=0}^{D}(a_\alpha U_\alpha^{(j)})^{(D-j)}. \end{equation} This follows by differentiating $U_\alpha v_\alpha$ successively and retaining the two highest components. All summands except $j=D$ are derivatives of rational functions; the first term is a polynomial. They have zero residue at every finite point. The remaining residue is $U_\alpha^{(D)}\Res(a_\alpha)$, nonzero at a suitable point. Thus \eqref{eq:residue-coefficient} is not the zero function. Clearing denominators gives a polynomial relation with nonzero highest coefficient at index $\alpha-1$. Descent eventually gives a nonzero polynomial times the nonzero constant $v_0$ equal to zero, a contradiction. \end{proof} In \eqref{eq:lifted-system}, the link $-2/w$ has residue $-2$ at $0$, and $4/(1-w^2)$ has residue $-2$ at $1$. Consider a nonzero constant combination of the solutions $Y_q$. If $q_0$ is the largest index with nonzero coefficient, its first nonzero component is at index $h-q_0$ and is a nonzero constant, by \eqref{eq:Yq}. This component remains constant after the change of variable and rescaling. Lemma~\ref{lem:chain}, starting at that index, shows that the remainder cannot vanish identically, because its highest coefficient $\widetilde S_{b+h}=w^{2M-1}P_b(w^2)$ is nonzero. Thus the $h+1$ high-order remainders are linearly independent at each of $\ii$ and $-\ii$. \subsection{The other local remainders} We also need remainders at $0$, infinity and $\pm1$. These will come from solutions whose first $h+1$ components vanish. The remaining equations then integrate to powers of a logarithm. Their combined orders at $0$ and infinity will be independent of $M$, while one of them will inherit additional zeros at $\pm1$ from the decay of $F_n$. For $j=1,\ldots,b$, system \eqref{eq:lifted-system} has a solution whose components below $h+j$ vanish and whose remaining components are \[ \widetilde v_{i,j}(w) =\frac{(-\log(w^2))^{i-h-j}}{(i-h-j)!} \qquad(i\geq h+j). \] Its remainder is \begin{equation}\label{eq:homogeneous-remainders} \widetilde\rho_j(w)=w^{2M-1} \sum_{i=j}^bP_i(w^2) \frac{(-\log(w^2))^{i-j}}{(i-j)!}. \end{equation} These $b$ remainders are independent. In a nonzero constant relation, the least index $j$ with nonzero coefficient supplies the unique highest logarithmic degree, with a nonzero coefficient proportional to $P_b$. This is impossible because $\log(w^2)$ is transcendental over $\C(w)$. At $0$, each remainder has order at least $2M-1$; at infinity its order is at least $-2(M+n)+1$. For the zeros at $\pm1$, consider \begin{equation}\label{eq:Rn-zero} R_n(z):=\sum_{i=1}^b P_i(z)\frac{(-\log z)^{i-1}}{(i-1)!} =O((z-1)^{\omega n-1}) \end{equation} near $1$, on the branch $\log1=0$. The asserted order is another form of the exact Laurent vanishing controlled by $\omega$, rather than the additional smallness controlled by $\Omega$. Indeed, the partial fractions give the Laplace-transform identity \[ F_n(t)=\int_0^\infty e^{-tu} \sum_{i=1}^b P_i(e^{-u})\frac{u^{i-1}}{(i-1)!}\dd u \qquad(t>0). \] Write the factor multiplying $e^{-tu}$ as $G(u)$. The elementary identity \[ \int_0^\infty e^{-tu}\frac{u^k}{k!}\dd u=\frac1{t^{k+1}} \qquad(t>0,\ k\geq0) \] explains how Taylor coefficients at $u=0$ become Laurent coefficients at infinity. Here $G$ is a finite sum of polynomials times exponentials $e^{-ju}$ with $j\geq0$, so repeated integration by parts gives, for fixed $n$ and every fixed integer $K\geq0$, \[ F_n(t)=\sum_{k=0}^K\frac{G^{(k)}(0)}{t^{k+1}} +O(t^{-K-2})\qquad(t\longrightarrow+\infty). \] Comparing with the Laurent expansion yields $A_{k+1}=G^{(k)}(0)$. Thus the exact zeros $A_1,\ldots,A_{\omega n-1}$ in Proposition~\ref{prop:rational} force $G$ to vanish to order at least $\omega n-1$. Finally, $G(u)=R_n(e^{-u})$ and $e^{-u}=1-u+O(u^2)$, so this change of variable preserves the order of vanishing and proves \eqref{eq:Rn-zero}. This is the observation used in \cite[Sections~3.5 and~4.5]{F26}. Now $\widetilde\rho_1(w)=w^{2M-1}R_n(w^2)$. At each of $w=1$ and $w=-1$, choose the local branch for which $\log(w^2)=0$ at that point. The map $w\mapsto w^2$ has nonzero derivative there, so \eqref{eq:Rn-zero} gives order at least $\omega n-1$ at each point. These two local branches need not be restrictions of a single global branch; Proposition~\ref{prop:multiplicity} allows the different local solutions. \subsection{Counting the orders} Apply Proposition~\ref{prop:multiplicity} with \[ \Sigma=\{0,\infty,1,-1,\ii,-\ii\},\qquad m=4M \] to \eqref{eq:lifted-system} and \eqref{eq:lifted-polynomials}. The system is regular singular: its matrix has only simple finite poles and is $O(1/w)$ at infinity. Its solutions therefore have the required Nilsson expansions. The constructed families give the following bounds. \begin{center} \begin{tabular}{@{}lcc@{}} \toprule Point & Independent remainders & Order of each, at least\\ \midrule Each of $\ii,-\ii$ & $h+1$ & $\kappa n$\\ $0$ & $b$ & $2M-1$\\ $\infty$ & $b$ & $-2(M+n)+1$\\ Each of $1,-1$ & $1$ & $\omega n-1$\\ \bottomrule \end{tabular} \end{center} There is a common rational differential operator of order $\mu\leq b+h+1$ annihilating all remainders. Differentiating a remainder iterates the operation $S\mapsto S'+S\matA$ on a row of length $b+h+1$; its first $b+h+2$ iterates are rationally dependent. Since the $b$ remainders at infinity are independent, $\mu\geq b$. The multiplicity estimate gives \begin{align*} \bigl(2(h+1)\kappa-2b+2\omega\bigr)n-2 &\leq(4rn+1)(\mu-b)+C\\ &\leq(4rn+1)(h+1)+C. \end{align*} The constant depends only on the system and $\Sigma$. The finitely many possibilities $1\leq b\leq a$ allow a uniform constant, even when $b$ varies with $n$. Dividing the last inequality by $2n$ would give \[ (h+1)(\kappa-2r)+\omega\leq b+o(1)\leq a+o(1). \] This contradicts \eqref{eq:rank-condition} for sufficiently large $n$ and proves Proposition~\ref{prop:rank}.