2 Padé-type approximants of Laurent series

Throughout this section, we fix a field K of characteristic $0$ . We denote the formal power series ring of variable $1/z$ with coefficients K by $K[[1/z]]$ and the field of fractions by $K((1/z))$ . We say an element of $K((1/z))$ is a formal Laurent series. We define the order function at $z=\infty $ by

$$ \begin{align*}\mathrm{{ord}}_{\infty}:K((1/z)) \longrightarrow \mathbb{Z}\cup \{\infty\}; \ \sum_{k} \dfrac{a_k}{z^k} \mapsto \min\{k\in\mathbb{Z}\cup \{\infty\} \mid a_k\neq 0\}.\end{align*} $$

Note that for $f\in K((1/z))$ , $\mathrm {{ord}}_{\infty } \, f=\infty $ if and only if $f=0$ . We recall without proof the following elementary lemma.

Lemma 2.1. Let m be a nonnegative integer, $f_1(z),\ldots ,f_m(z)\in (1/z)\cdot K[[1/z]]$ and $\boldsymbol {n}=(n_1,\ldots ,n_m)\in \mathbb {N}^{m}$ . Put $N=\sum _{j=1}^mn_j$ . For a nonnegative integer M with $M\ge N$ , there exist polynomials $(P,Q_{1},\ldots ,Q_m)\in K[z]^{m+1}\setminus \{\boldsymbol {0}\}$ satisfying the following conditions:

$$ \begin{align*} &(i) \ \mathrm{{deg}}P\le M;\\ &(ii) \ \mathrm{{ord}}_{\infty} (P(z)f_j(z)-Q_j(z))\ge n_j+1 \quad \text{for } 1\le j \le m. \end{align*} $$

Let $f(z)=\sum _{k=0}^{\infty } f_k/z^{k+1}\in (1/z)\cdot K[[1/z]]$ . We define a K-linear map $\varphi _f\in \mathrm {{Hom}}_K(K[t],K)$ by

(2-1) $$ \begin{align} \varphi_f:K[t]\longrightarrow K; \quad t^k\mapsto f_k \quad (k\ge0). \end{align} $$

The above linear map extends naturally to a $K[z]$ -linear map $\varphi _f: K[z,t]\rightarrow K[z]$ , and then to a $K[z][[1/z]]$ -linear map $\varphi _f: K[z,t][[1/z]]\rightarrow K[z][[1/z]]$ . With this notation, the formal Laurent series $f(z)$ satisfies the following crucial identities (see [Reference Nikišin and Sorokin23, Equation $(6.2)$ page 60 and $\mathrm{Equation}\;(5.7)$ page 52]):

$$ \begin{align*} &f(z)=\varphi_f \bigg(\dfrac{1}{z-t}\bigg),\quad\! P(z)f(z)-\varphi_f\bigg(\dfrac{P(z)-P(t)}{z-t}\bigg)\in (1/z)\cdot K[[1/z]]\! \quad \text{for any } P(z)\in K[z]. \end{align*} $$

Proof. By the definition of $Q_j(z)$ ,

$$ \begin{align*} P(z)f_j(z)-Q_j(z)=\varphi_{f_j}\bigg(\dfrac{P(t)}{z-t}\bigg)\in (1/z)\cdot K[[1/z]]. \end{align*} $$

The above equality yields that the vector of polynomials $(P,Q_1,\ldots ,Q_m)$ being a weight $\boldsymbol {n}$ Padé-type approximant of $(f_1,\ldots ,f_m)$ is equivalent to the order of the Laurent series

$$ \begin{align*}\varphi_{f_j}\bigg(\dfrac{P(t)}{z-t}\bigg)={\sum_{k=0}^{\infty}}\dfrac{\varphi_{f_j}(t^kP(t))}{z^{k+1}}\end{align*} $$

being greater than or equal to $n_j+1$ for $1\le j \le m$ . This shows the equivalence of items (i) and (ii).

Lemma 2.3 indicates that it is useful to study $\mathrm {{ker}}\, \varphi _f$ for the explicit construction of Padé-type approximants of Laurent series. We are now going to investigate $\mathrm {{ker}}\, \varphi _{f}$ for a holonomic Laurent series $f\in (1/z)\cdot K[[1/z]]$ . We denote the differential operator ${d}/{dz}$ (respectively ${d}/{dt}$ ) by $\partial _z$ (respectively $\partial _t$ ). We describe the action of a differential operator D on a function f by $D\cdot f$ and denote $\partial _z \cdot f$ by $f^{\prime }$ .

To begin with, let us introduce a map

$$ \begin{align*} \iota :K(z)[\partial_z]\longrightarrow K(t)[\partial_t]; \quad \sum_j P_j(z)\partial^j_z \mapsto \sum_j (-1)^j \partial^j_t P_j(t). \end{align*} $$

Note, for $D\in K(z)[\partial _z]$ , $\iota (D)$ is called the adjoint of D and relates to the dual of differential module $K(z)[\partial _z]/K(z)[\partial _z]D$ (see [Reference André2, Exercise III(3)]). For $D\in K(z)[\partial _z]$ , we denote $\iota (D)$ by $D^{*}$ . Notice that we have $(DE)^{*}=E^*D^{*}$ for any $D,E\in K(z)[\partial _z]$ .

Lemma 2.4. For $D\in K[z,\partial _z]$ , there exists a polynomial $P(t,z)\in K[t,z]$ satisfying

$$ \begin{align*}D\cdot \dfrac{1}{z-t}=P(t,z)+D^{*}\cdot \dfrac{1}{z-t}.\end{align*} $$

Proof. Let $m,n$ be nonnegative integers. It suffices to prove the case $D=z^m\partial ^n_z$ . Then,

(2-2) $$ \begin{align} D \cdot \dfrac{1}{z-t}&=\dfrac{(-1)^nn!z^m}{(z-t)^{n+1}}=(-1)^n\sum_{k=0}^{\infty}\dfrac{(n+k)!}{k!}\dfrac{t^k}{z^{k+1+n-m}}. \end{align} $$

We define a polynomial $P(t,z)$ by $0$ if $m\le n$ and

$$ \begin{align*}P(t,z)= (-1)^n {\sum_{k=0}^{m-n-1}}\dfrac{(n+k)!}{k!}t^kz^{m-n-k-1} \end{align*} $$

for $m>n$ . Equation (2-2) implies

$$ \begin{align*} D \cdot \dfrac{1}{z-t}-P(t,z)&=(-1)^n\sum_{k=\max\{m-n,0\}}^{\infty}\dfrac{(n+k)!}{k!}\dfrac{t^k}{z^{k+1+n-m}}\\ &=(-1)^n\sum_{k=0}^{\infty}(k+1+m-n)\cdots(m+k)\dfrac{t^{k+m-n}}{z^{k+1}}. \end{align*} $$

However,

$$ \begin{align*} D^*\cdot \dfrac{1}{z-t}&=(-1)^n\partial^n_t \cdot\dfrac{t^m}{z-t}=(-1)^n\sum_{k=0}^{\infty}\partial^n_t\cdot \dfrac{t^{m+k}}{z^{k+1}}\\ &=(-1)^n\sum_{k=0}^{\infty}(k+1+m-n)\cdots(m+k)\dfrac{t^{k+m-n}}{z^{k+1}}. \end{align*} $$

The above equalities yield

$$ \begin{align*}D\cdot \dfrac{1}{z-t}-P(t,z)=D^*\cdot \dfrac{1}{z-t}.\end{align*} $$

This completes the proof of Lemma 2.4.

We introduce the projection morphism $\pi $ by

$$ \begin{align*} \pi :K[z][[1/z]]\longrightarrow K[z][[1/z]]/K[z]\cong (1/z)\cdot K[[1/z]]; \quad f(z)=P(z)+\tilde{f}(z)\mapsto \tilde{f}(z), \end{align*} $$

where $P(z)\in K[z]$ and $\tilde {f}(z)\in (1/z)\cdot K[[1/z]]$ . Lemma 2.4 allows us to show the following key proposition.

Proof. First, since $\varphi _f$ acts only on the parameter t,

$$ \begin{align*}D\cdot f=D\circ \varphi_f\bigg(\dfrac{1}{z-t}\bigg)=\varphi_f \bigg(D\cdot \dfrac{1}{z-t}\bigg).\end{align*} $$

Lemma 2.4 implies that there exists a polynomial $P(z)$ with

$$ \begin{align*}D\cdot f=P(z)+\varphi_f\bigg(D^{*}\cdot \dfrac{1}{z-t} \bigg)=P(z)+\sum_{k=0}^{\infty}\dfrac{\varphi_f(D^{*}\cdot t^k)}{z^{k+1}}.\end{align*} $$

Note that $P(z)=\varphi _f(P(t,z))$ , where $P(t,z)\in K[t,z]$ is defined in Lemma 2.4. This shows that $\pi (D\cdot f)=\sum _{k=0}^{\infty }\varphi _f(D^{*}\cdot t^k)/{z^{k+1}}$ and therefore

$$ \begin{align*}\varphi_{\pi(D\cdot f)}(t^k)=\varphi_f\circ D^{*}(t^k) \quad \text{for all } k\ge 0.\end{align*} $$

This concludes the proof of Proposition 2.5.

As a corollary of Proposition 2.5, the following crucial equivalence relations hold.

3 Weighted Rodrigues operators

Let K be a field of characteristic $0$ . Let us introduce the weighted Rodrigues operator, which is first defined by A. I. Aptekarev, A. Branquinho and W. Van Assche in [Reference Aptekarev, Branquinho and Van Assche4].

Definition 3.1 See [Reference Aptekarev, Branquinho and Van Assche4, $\mathrm{Equation}\;(2.5)$ ]

Let $l\in \mathbb {N}$ , $a_1(z),\ldots ,a_l(z)\in K[z]\setminus \{0\}$ , $b(z)\in K[z]$ . Put $a(z)=a_1(z)\cdots a_l(z)$ , $D=-a(z)\partial _z+b(z)$ . For $n\in \mathbb {N}$ and a weight $\vec {r}=(r_1,\ldots ,r_l)\in \mathbb {Z}^l$ with $r_i\ge 0$ , we define the weighted Rodrigues operator associated with D by

$$ \begin{align*}R_{D,n,\vec{r}}=\dfrac{1}{n!}\bigg(\partial_z+\dfrac{b(z)}{a(z)}\bigg)^na(z)^n\prod_{v=1}^la_v(z)^{-r_v}\in K(z)[\partial_z].\end{align*} $$

In the case of $\vec {r}=(0,\ldots ,0)$ , we denote $R_{D,n,\vec {r}}=R_{D,n}$ and call this operator the n th Rodrigues operator associated with D.

We denote the generalized Rodrigues operator associated with D with respect to the parameter t by

$$ \begin{align*}\mathcal{R}_{D,n,\vec{r}}=\dfrac{1}{n!}\bigg(\partial_t+\dfrac{b(t)}{a(t)}\bigg)^na(t)^n\prod_{v=1}^la_v(t)^{-r_v} \in K(t)[\partial_t],\end{align*} $$

and $\mathcal {R}_{D,n,\vec {r}}=\mathcal {R}_{D,n}$ in the case of $\vec {r}=(0,\ldots ,0)$ .

Let us show some basic properties of the weighted Rodrigues operator in order to obtain a generalization of Rodrigues formula of Padé approximants of holonomic Laurent series. In the following, for $a(z)\in K[z]$ (respectively $a(t)\in K[t]$ ), we denote the ideal of $K[z]$ (respectively $K[t]$ ), generated by $a(z)$ (respectively $a(t)$ ) by $(a(z))$ (respectively $(a(t))$ ).

Proposition 3.2. Let $a(t),b(t)\in K[t]$ with $a(t)\neq 0$ . Put $\mathcal {E}_{a,b}=\partial _t+b(t)/a(t)\in K(t)[\partial _t]$ .

(i) Let $n,k$ be nonnegative integers. Then there exist integers $(c_{n,k,l})_{0\le l \le \min \{n,k\}}$ with

$$ \begin{align*}c_{n,k,\min\{n,k\}}=(-1)^nk(k-1)\cdots (k-n+1),\end{align*} $$

$$ \begin{align*} t^k\mathcal{E}^n_{a,b}=\sum_{l=0}^{\min\{n,k\}}c_{n,k,l}\mathcal{E}^{n-l}_{a,b}t^{k-l}\in K(t)[\partial_t]. \end{align*} $$

(ii) Assume there exist polynomials $a_1(t),\ldots ,a_l(t)\in K[t]$ with $a(t)=a_1(t)\cdots a_l(t)$ . For an l-tuple of nonnegative integers $\boldsymbol {s}:=(s_1,\ldots ,s_l)$ , we denote by $\mathrm {I}(\boldsymbol {s})$ the ideal of $K[t]$ generated by ${\prod _{v=1}^l}a_v(t)^{s_v}$ . Then for $n\ge 1$ and $F(t)\in \mathrm {I}(\boldsymbol {s})$ ,

(3-1) $$ \begin{align} \mathcal{E}^n_{a,b}a(t)^n\cdot F(t)\in \mathrm{I}(\boldsymbol{s}). \end{align} $$

Proof. (i) We prove the assertion by induction on $(n,k)\in \mathbb {Z}^2$ with $n,k\ge 0$ . In the case of $n=0$ and any $k\ge 0$ , the statement is trivial. Let $n,k$ be nonnegative integers with $n\ge 1$ or $k\ge 1$ . We assume that the assertion holds for any elements of the set $\{(\tilde {n},\tilde {k})\in \mathbb {Z}^2\mid 0\le \tilde {n},\tilde {k} \text { and } \tilde {n}< n \text { and } \tilde {k}\le k\}$ . The equality $t^k\mathcal {E}_{a,b}=\mathcal {E}_{a,b}t^k-kt^{k-1}$ in $K[t,\partial _t]$ implies that we have

(3-2) $$ \begin{align} t^k\mathcal{E}^n_{a,b}&=(\mathcal{E}_{a,b}t^k-kt^{k-1})\mathcal{E}^{n-1}_{a,b} \nonumber \\ &=\mathcal{E}_{a,b}\sum_{l=0}^{\min\{n-1,k\}}c_{n-1,k,l}\mathcal{E}^{n-1-l}_{a,b}t^{k-l}-k\sum_{l=0}^{\min\{n-1,k-1\}}c_{n,k-1,l}\mathcal{E}^{n-1-l}_{a,b}t^{k-1-l} \\ &=\sum_{l=0}^{\min\{n-1,k\}}c_{n-1,k,l}\mathcal{E}^{n-l}_{a,b}t^{k-l}-\sum_{l=0}^{\min\{n-1,k-1\}}kc_{n-1,k-1,l}\mathcal{E}^{n-1-l}_{a,b}t^{k-1-l}. \nonumber \end{align} $$

Note that we use the induction hypothesis in line (3-2). This concludes the assertion for $(n,k)$ .

(ii) Let us prove the statement by induction on n. In the case of $n=1$ , since

$$ \begin{align*} \mathcal{E}_{a,b}a(t)\cdot F(t)&=(\partial_t a(t)+b(t))\cdot F(t) =a^{\prime}(t)F(t)+a(t)F^{\prime}(t)+b(t)F(t), \end{align*} $$

using the Leibniz formula, we obtain Equation (3-1). We assume Equation (3-1) holds for $n\ge 1$ . In the case of $n+1$ ,

(3-3) $$ \begin{align} \mathcal{E}^{n+1}_{a,b}a(t)^{n+1}\cdot F(t)&=\mathcal{E}_{a,b}\mathcal{E}^n_{a,b}a(t)^n\cdot a(t)F(t). \end{align} $$

Note that we have $a(t)F(t)\in \mathrm {I}(\boldsymbol {s}+\boldsymbol {1})$ , where $\boldsymbol {s}+\boldsymbol {1}:=(s_1+1,\ldots ,s_d+1)\in \mathbb {N}^d$ . Relying on the induction hypothesis, we deduce $\mathcal {E}^n_{a,b}a(t)^n\cdot a(t)F(t)\in {\mathrm {I}}(\boldsymbol {s}+\boldsymbol {1})$ . Thus, there exists a polynomial $\tilde {F}(t)\in {\mathrm {I}}(\boldsymbol {s})$ with $\mathcal {E}^n_{a,b}a(t)^n\cdot a(t)F(t)=a(t)\tilde {F}(t)$ . Substituting this equality into Equation (3-3), by using a similar argument to the case of $n=1$ , we conclude $\mathcal {E}^{n+1}_{a,b}a(t)^{n+1}\cdot F(t)\in {\mathrm {I}}(\boldsymbol {s})$ .

Corollary 3.3. (i) Let $a(z)\in K[z]\setminus \{0\}$ and $b(z)\in K[z]$ . We put $D=-a(z)\partial _z+b(z)$ . Let $f(z)\in (1/z)\cdot K[[1/z]]\setminus \{0\}$ with $D\cdot f(z)\in K[z]$ . Put $\mathcal {E}_{a,b}=\partial _t+b(t)/a(t)\in K(t)[\partial _t]$ . Then, for $n,k\in \mathbb {Z}$ with $0\le k<n$ ,

$$ \begin{align*}t^k\mathcal{E}^n_{a,b}\cdot (a(t)^n)\subseteq \mathrm{{ker}}\,\varphi_f.\end{align*} $$

(ii) Let $d,l \in \mathbb {N}$ , $(n_1,\ldots ,n_d)\in \mathbb {N}^d$ and $a_1(t),\ldots ,a_l(t)\in K[t]\setminus \{0\}$ . Put $a(t)=a_1(t)\cdots a_l(t)$ . For $b_1(t),\ldots ,b_d(t)\in K[t]$ and l-tuple of nonnegative integers $\vec {r}_j=(r_{j,1},\ldots ,r_{j,l}) (1\le j \le d)$ , we put $D_j=-a(z)\partial _z+b_j(z)$ and

$$ \begin{align*}{{\mathcal{R}_{j,n_j}=\mathcal{R}_{D_j,n_j,\vec{r}_j}}}=\dfrac{1}{n_j!}\mathcal{E}^{n_j}_{a,b_j}a(t)^{n_j}\prod_{v=1}^la_v(t)^{-r_{j,v}}\in K(t)[\partial_t].\end{align*} $$

Let $s_1,\ldots ,s_d$ be nonnegative integers and $F(t)\in ({\prod _{v=1}^l}a_v(t)^{s_v+\sum _{j=1}^dr_{j,v}})$ . Then,

$$ \begin{align*}\prod_{j=1}^d\mathcal{R}_{j,n_j} \cdot F(t)\in \bigg({\prod_{v=1}^l}a_v(t)^{s_v}\bigg) \end{align*} $$

(The statement in the first term holds for any order of product of operators $(\mathcal {R}_{j,n_j})_j$ .)

Proof. (i) By the definition of D, we have $D^{*}=\mathcal {E}_{a,b}a(t)$ . Since we have $\mathcal {E}_{a,b}\cdot (a(t))\subseteq \mathrm {{ker}}\,\varphi _f$ , by Corollary 2.6, it suffices to show $t^k\mathcal {E}^n_{a,b}\cdot (a(t)^n)\subset \mathcal {E}_{a,b}\cdot (a(t))$ . Relying on Proposition 3.2 (i), there are $\{c_{n,k,l}\}_{0\le l \le k}\subset {{\mathbb {Z}}}$ with

(3-4) $$ \begin{align} t^k\mathcal{E}^n_{a,b}=\sum_{l=0}^k c_{n,k,l}\mathcal{E}^{n-l}_{a,b}t^{k-l}. \end{align} $$

For an integer l with $0\le l \le k$ ,

$$ \begin{align*} \mathcal{E}^{n-l}_{a,b}t^{k-l} \cdot (a(t)^n)\subset \mathcal{E}_{a,b}\mathcal{E}^{n-l-1}_{a,b}\cdot (a(t)^n). \end{align*} $$

The Leibniz formula allows us to get $\mathcal {E}^{n-l-1}_{a,b}\cdot (a(t)^n)\subset (a(t))$ . Combining Equation (3-4) and the above relation gives

$$ \begin{align*}t^k\mathcal{E}^n_{a,b}\cdot (a(t)^n)\subset \mathcal{E}_{a,b}\cdot (a(t)).\end{align*} $$

This completes the proof of item $(i)$ .

(ii) It suffices to prove the assertion in the case of $d=1$ . By the definition of $\mathcal {R}_{1,n_1}$ ,

$$ \begin{align*} \mathcal{R}_{1,n_1}\cdot F(t)&=\dfrac{1}{n_1!}\mathcal{E}^{n_1}_{a,b_1}a(t)^{n_1}\prod_{v=1}^{{l}} a_v(t)^{-r_{1,v}}\cdot F(t) \in \mathcal{E}^{n_1}_{a,b_1}a(t)^{n_1}\cdot \bigg(\prod_{v=1}^la_v(t)^{s_v}\bigg). \end{align*} $$

Using Proposition 3.2(ii), we conclude that $\mathcal {E}^{n_1}_{a,b_1}a(t)^{n_1}\cdot (\prod _{v=1}^la_v(t)^{s_v})\subset (\prod _{v=1}^la_v(t)^{s_v})$ . This completes the proof of item (ii).

4 Rodrigues formula of Padé approximants

Lemma 4.1. Let $a(z),b(z)\in K[z]$ with $a(z)\neq 0$ , $\mathrm {{deg}}\, a=u$ and $\mathrm {{deg}}\,b=v$ . Put

$$ \begin{align*}D=-a(z)\partial_z+b(z)\in K[z,\partial_z], \ \ a(z)=\sum_{i=0}^ua_iz^i, \ \ b(z)=\sum_{j=0}^vb_jz^j,\end{align*} $$

and $w=\max \{u-2,v-1\}$ . Assume $w\ge 0$ and

(4-1) $$ \begin{align} a_u(k+u)+b_v\neq 0 \quad \text{for all } k\ge0 \quad \text{if } u-2=v-1. \end{align} $$

Then there exist $f_0(z),\ldots ,f_w(z)\in (1/z)\cdot K[[1/z]]$ that are linearly independent over K and satisfy $D\cdot f_l(z)\in K[z]$ for $0\le l \le w$ .

Proof. Let $f(z)=\sum _{k=0}^{\infty }f_k/z^{k+1}\in (1/z)\cdot K[[1/z]]$ be a Laurent series. There exists a polynomial $A(z)\in K[z]$ that depends on the operator D and f with $\mathrm {{deg}}\,A\le w$ , satisfying

$$ \begin{align*} D\cdot f(z)=A(z) +\sum_{k=0}^{\infty}\dfrac{\sum_{i=0}^ua_i(k+i)f_{k+i-1}+\sum_{j=0}^vb_jf_{k+j}}{z^{k+1}}. \end{align*} $$

Put

$$ \begin{align*}\sum_{i=0}^ua_i(k+i)f_{k+i-1}+\sum_{j=0}^vb_jf_{k+j}=c_{k,0}f_{k-1}+\cdots+c_{k,w}f_{k+w}+c_{k,w+1}f_{k+w+1} \quad \text{for } k\ge 0,\end{align*} $$

with $c_{0,0}=0$ . We remark that $c_{k,l}$ depends only on $a(z),b(z)$ . Notice that $c_{k,w+1}$ is $a_u(k+u)$ if $u-2>v-1$ , $b_v$ if $u-2<v-1$ and $a_u(k+u)+b_v$ if $u-2=v-1$ . Then by Equation (4-1), we have $\min \{k\ge 0\mid c_{k',w+1}\neq 0 \text { for all } k'\ge k\}=0$ and thus the K-linear map:

$$ \begin{align*}K^{w+1}\longrightarrow \{f\in (1/z)\cdot K[[1/z]]\mid D\cdot f\in K[z]\}; \quad (f_0,\ldots,f_w)\mapsto \sum_{k=0}^{\infty} \dfrac{f_k}{z^{k+1}},\end{align*} $$

where, for $k\ge w+1$ , $f_k$ is determined inductively by

$$ \begin{align*} &\sum_{i=0}^ua_i(k+i)f_{l,k+i-1}+\sum_{j=0}^vb_jf_{l,k+j}=0 \quad \text{for } k\ge 0 \end{align*} $$

is an isomorphism. This completes the proof of Lemma 4.1.

Let us state a generalization of the Rodrigues formula for Legendre polynomials to Padé approximants of certain holonomic Laurent series, which gives a generalization of [Reference Aptekarev, Branquinho and Van Assche4, Theorem $1$ ]. In the following theorem, we construct Padé approximants of the family of Laurent series considered in Lemma 4.1.

Theorem 4.2. Let $l,d\in \mathbb {N}$ , $(a_1(z),\ldots ,a_l(z))\in (K[z]\setminus \{0\})^l$ and $(b_{1}(z),\ldots ,b_{d}(z))\in K[z]^d$ . Put $a(z)=a_{1}(z)\cdots a_{l}(z)$ . Put $D_{j}=-a(z)\partial _z+b_{j}(z) \in K[z,\partial _z]$ and $w_{j}=\max \{\mathrm {{deg}}\, a-2, \mathrm {{deg}}\, b_{j}-1\}$ . Assume $w_j\ge 0$ for $1\le j \le d$ and Equation (4-1) for $D_j$ . Let $f_{j,0}(z),\ldots ,f_{j,w_{j}}(z)\in (1/z)\cdot K[[1/z]]$ be formal Laurent series that are linearly independent over K satisfying

$$ \begin{align*}D_{j}\cdot f_{j,u_{j}}(z)\in K[z] \quad \text{for } 0\le u_{j} \le w_{j}.\end{align*} $$

(The existence of such series is ensured by Lemma 4.1.) Let $(n_1,\ldots ,n_d)\in \mathbb {N}^d$ . For an l-tuple of nonnegative integers $\vec {r}_j=(r_{j,1},\ldots ,r_{j,l})\ (1\le j \le d)$ , we denote by $R_{j,n_j}$ the weighted Rodrigues operator $R_{D_{j},n_j,\vec {r}_j}$ associated with $D_j$ . Assume

$$ \begin{align*} &R_{{j_1},n_{j_1}} R_{{j_2},n_{j_2}}=R_{{j_2},n_{j_2}} R_{{j_1},n_{j_1}} \quad \text{for } 1\le j_1,j_2\le d. \end{align*} $$

Take a nonzero polynomial $F(z)$ that is contained in the ideal $(\prod _{v=1}^{l}a_{v}(z)^{\sum _{j=1}^d r_{j,v}})$ and put

$$ \begin{align*} &P(z)=\prod_{j=1}^dR_{j,n_j}\cdot F(z),\\ &Q_{j,u_{j}}(z)=\varphi_{f_{j,u_{j}}}\bigg(\dfrac{P(z)-P(t)}{z-t}\bigg) \quad \text{for } 1\le j \le d, \ 0\le u_{j}\le w_{j}. \end{align*} $$

Assume $P(z)\neq 0$ . (We need to assume $P(z)\neq 0$ . For example, in the case of $d=1$ , $D=-\partial _z z^2=-z^2\partial -2z$ and $n=1$ , we have $P(z)=(\partial _z-2/z)z^2\cdot 1=0$ .) Then the vector of polynomials $(P(z),Q_{j,u_{j}}(z))_{\substack { \\ 1\le j \le d \\ 0\le u_{j}\le w_{j}}}$ is a weight $(\boldsymbol {n}_1,\ldots ,\boldsymbol {n}_m)\in \mathbb {N}^{\sum _{j=1}^d(w_{j}+1)}$ Padé-type approximants of $(f_{j,u_{j}}(z))_{\substack { 1\le j \le d \\ 0\le u_{j}\le w_{j}}}$ , where $\boldsymbol {n}_j=(n_j,\ldots ,n_j)\in \mathbb {N}^{w_j+1}$ for $1\le j \le d$ .

Proof. By Lemma 2.3, it suffices to prove that any triple $(j,u_{j},k)$ with $1\le j \le d$ , $0\le u_{j} \le w_{j}$ , $0\le k \le n_j-1$ satisfies $t^kP(t)\in \mathrm {{ker}}\,\varphi _{f_{j,u_{j}}}$ . Put $\mathcal {R}_{j,n_j}=\mathcal {R}_{D_{j},n_j,\vec {r}_{j}}$ . Then we have $P(t)={\prod _{j=1}^d}\mathcal {R}_{j,n_j}\cdot F(t)$ and thus

(4-2) $$ \begin{align} t^kP(t)=t^k\mathcal{R}_{j,n_j}\prod_{j^{\prime}\neq j}\mathcal{R}_{j^{\prime},n_{j^{\prime}}}\cdot F(t). \end{align} $$

Since $F(t)\in (\prod _{v=1}^{l}a_{v}(z)^{r_{j,v}+\sum _{j'\neq j}^d r_{j',v}})$ , using Corollary 3.3(ii),

$$ \begin{align*}{\prod_{j^{\prime}\neq j}}\mathcal{R}_{j^{\prime},n_{j^{\prime}}}\cdot F(t)\in \bigg(\prod_{v=1}^{l}a_{v}(t)^{r_{j,v}}\bigg).\end{align*} $$

Combining Equation (4-2) and the above relation yields

$$ \begin{align*} t^kP(t)\in t^k\mathcal{R}_{j,n_j}\cdot \bigg(\prod_{v=1}^{l}a_{v}(t)^{r_{j,v}}\bigg)\subseteq t^k\mathcal{E}^{n_j}_{a,b_{j}}\cdot (a(t)^{n_j})\subseteq \mathrm{{ker}}\,\varphi_{f_{j,u_{j}}}. \end{align*} $$

Note that the last inclusion is obtained from Corollary 3.3(i) for $D_{j}\cdot f_{j,u_{j}}(z)\in K[z]$ .

4.1 Commutativity of differential operators

In this subsection, we give a sufficient condition under which weighted Rodrigues operators commute. We denote $\partial _z\cdot c(z)$ by $c^{\prime }(z)$ for any rational function $c(z)\in K(z)$ .

Lemma 4.3. Let $a(z),b(z)\in K[z]$ and $c(z)\in K(z)$ with $a(z)c(z)\neq 0$ . Let $w(z)$ be a nonzero solution of $-a(z)\partial _z+b(z)$ in some differential extension $\mathcal {K}$ of $K(z)$ and n a nonnegative integer. Put

$$ \begin{align*}R_n=\dfrac{1}{n!}\bigg(\partial_z+\dfrac{b(z)}{a(z)}\bigg)^nc(z)^n\in K(z)[\partial_z].\end{align*} $$

Then, in the ring $\mathcal {K}[\partial _z]$ , we have the following equality:

$$ \begin{align*}R_n=\dfrac{1}{n!}w(z)^{-1}\partial^n_z w(z)c(z)^n=\dfrac{1}{n!}R_{1}(R_{1}+c^{\prime}(z))\cdots (R_{1}+(n-1)c^{\prime}(z)).\end{align*} $$

Proof. The first equality is readily obtained using the identity

$$ \begin{align*}\partial_zw(z)=w(z)\bigg(\partial_z+\dfrac{b(z)}{a(z)}\bigg).\end{align*} $$

The second equality is proved using the identity

$$ \begin{align*} \bigg(\partial_z+\dfrac{b(z)}{a(z)}\bigg){{c(z)^n}}&=\bigg[c(z)^{n-1}\bigg(\partial_z+\dfrac{b(z)}{a(z)}\bigg)+(n-1)c^{\prime}(z)c(z)^{n-2}\bigg]c(z)\\ &=c(z)^{n-1}(R_1+(n-1)c^{\prime}(z)). \end{align*} $$

This completes the proof of Lemma 4.3.

Lemma 4.4. Let $a(z),b_1(z),b_2(z),c(z)\in K[z]$ with $a(z)c(z)\neq 0$ . For a nonnegative integer n and $j=1,2$ ,

$$ \begin{align*}R_{j,n}=\dfrac{1}{n!}\bigg(\partial_z+\dfrac{b_j(z)}{a(z)}\bigg)^nc(z)^n.\end{align*} $$

Assume $\mathrm {{deg}}\,c\le 1$ . Then the following are equivalent.

1. (i) For any $n_1,n_2\in \mathbb {N}$ , we have $R_{1,n_1}R_{2,n_2}=R_{2,n_2}R_{1,n_1}$ .

2. (ii) We have $({b_2(z)-b_1(z)})/{a(z)}c(z)\in K$ .

Proof. Since $\mathrm {{deg}}\,c \le 1$ and therefore $c^{\prime }(z)\in K$ , using Lemma 4.3, we see that item (i) is equivalent to $R_{1,1}R_{2,1}=R_{2,1}R_{1,1}$ . Let us show that the commutativity of $R_{j,1} (j=1,2)$ is equivalent to item (ii). According to the identity,

$$ \begin{align*} R_{1,1}R_{2,1}=R_{2,1}R_{1,1}+(R_{2,1}-R_{1,1})c^{\prime}(z)+\bigg(\dfrac{b_2(z)-b_1(z)}{a(z)}\bigg)'c(z)^{{2}}, \end{align*} $$

the identity $R_{1,1}R_{2,1}=R_{2,1}R_{1,1}$ is equivalent to

$$ \begin{align*} &(R_{2,1}-R_{1,1})c^{\prime}(z)+\bigg(\dfrac{b_2(z)-b_1(z)}{a(z)}\bigg)'c(z)^{{2}}\\&\quad=\bigg(\dfrac{b_2(z)-b_1(z)}{a(z)}c^{\prime}(z)+\bigg(\dfrac{b_2(z)-b_1(z)}{a(z)}\bigg)^{\prime}c(z)\bigg){{c(z)}}\\ &\quad=\bigg(\dfrac{b_2(z)-b_1(z)}{a(z)}c(z)\bigg)^{\prime}{{c(z)}}=0, \end{align*} $$

which means item (ii) holds. This completes the proof of Lemma 4.4.

5 Determinants associated with Padé approximants

Let $f_{j,u_j}(z)$ be the Laurent series in Theorem 4.2. To consider the linear independence results on the values of $f_{j,u_j}(z)$ á la the method of Siegel (see [Reference Siegel30]), we need to study the nonvanishing of determinants of certain matrices. In this section, we compute the determinants of specific matrices whose entries are given by the Padé approximants of $f_{j,u_j}(z)$ obtained in Theorem 4.2.

First, let d be a nonnegative integer and $a_1(z),a_2(z),b_1(z),\ldots ,b_d(z)\in K[z]$ . Put $a(z)=a_1(z)a_2(z)$ , $w_j=\max \{\mathrm {{deg}}\ a-2, \mathrm {{deg}}\,b_j-1\}$ and $W=w_1+\cdots +w_d+d$ .

Assume $w_j\ge 0$ , $\mathrm {{deg}}\,a_1\le 1$ , $a_1$ is a monic polynomial and

$$ \begin{align*}\gamma_{j_1,j_2}=\dfrac{b_{j_1}(z)-b_{j_2}(z)}{a_2(z)}\in K\setminus\{0\} \quad \text{for } 1\le j_1<j_2\le d.\end{align*} $$

Denote $D_j=-a(z)\partial _z+b_j(z)\in K[z,\partial _z]$ and assume Equation (4-1) for $D_j$ . Lemma 4.1 implies that there exist Laurent series $f_{j,0}(z),\ldots ,f_{j,w_j}(z)\in (1/z)\cdot K[[1/z]]$ that are linearly independent over K and satisfy

$$ \begin{align*}D_j\cdot f_{j,u_j}(z)\in K[z] \quad \text{for } 1\le j \le d, \ \ 0\le u_j \le w_j.\end{align*} $$

We now fix these series. For $n\in \mathbb {N}$ , we denote the weighted Rodrigues operator associated with $D_j$ by

$$ \begin{align*}R_{j,n}=\dfrac{1}{n!}\bigg(\partial_z+\dfrac{b_j(z)}{a(z)}\bigg)^na_1(z)^n \quad \text{for } 1\le j \le d.\end{align*} $$

Lemma 4.4 to the case of $a(z)=a_1(z)a_2(z)$ and $c(z)=a_1(z)$ asserts the commutativity of the differential operators $R_{j,n}$ , namely

$$ \begin{align*}R_{j_1,n}R_{j_2,n}=R_{j_2,n}R_{j_1,n} \quad \text{for } 1\le j_1,j_2 \le d.\end{align*} $$

Put $\varphi _{j,u_j}=\varphi _{f_{j,u_j}}$ . For $0\le h \le W$ , we define

$$ \begin{align*} &P_{n,h}(z)=P_h(z)=\prod_{j=1}^dR_{j,n}\cdot [z^ha_2(z)^{dn}],\\ &Q_{n,j,u_j,h}(z)=Q_{j,u_j,h}(z)=\varphi_{j,u_j}\bigg(\dfrac{P_{h}(z)-P_{h}(t)}{z-t} \bigg) \quad \text{for } 1\le j \le d, \ 0\le u_j \le w_j,\\ &\mathfrak{R}_{n,j,u_j,h}(z)=\mathfrak{R}_{j,u_j,h}(z)=P_h(z)f_{j,u_j}(z)-Q_{j,u_j,h}(z) \quad \text{for } 1\le j \le d, \ 0\le u_j \le w_j. \end{align*} $$

Assume $P_h(z)\neq 0$ . Theorem 4.2 yields that the vector of polynomials $(P_h,Q_{j,u_j,h})_{\substack {1\le j \le d \\ {{0}} \le u_j \le w_j}}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^{W}$ Padé-type approximant of $(f_{j,u_j})_{\substack {1\le j \le d \\ 0\le u_j \le w_j}}$ .

First we compute the coefficients of $1/z^{n+1}$ of $\mathfrak {R}_{j,u_j,h}(z)$ .

Lemma 5.1. Let notation be as above. For $1\le j \le d$ , $0\le u_j \le w_j$ and $0\le h \le W$ ,

$$ \begin{align*} \mathfrak{R}_{j,u_j,h}(z)=\sum_{k=n}^{\infty}\dfrac{\varphi_{j,u_j}(t^kP_h(t))}{z^{k+1}} \end{align*} $$

and

$$ \begin{align*}\varphi_{j,u_j}(t^nP_h(t))=\dfrac{(-1)^n}{(n!)^{d-1}}\prod_{\substack{1\le j'\le d \\ j'\neq j}}\bigg[\prod_{k=1}^n(\gamma_{j',j}-k\varepsilon_{a_1})\bigg]\varphi_{j,u_j}(t^ha_1(t)^n\cdot {a}_2(t)^{dn}),\end{align*} $$

where $\varepsilon _{a_1}=1$ if $\mathrm {{deg}}\,a_1=1$ and $\varepsilon _{a_1}=0$ if $\mathrm {{deg}}\,a_1=0$ .

Proof. Since $(\mathfrak {R}_{j,u_j,h})_{j,u_j}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^{W}$ Padé-type approximation of $(f_{j,u_j})_{j,u_j}$ , we have $\mathrm {{ord}}_{\infty }\,\mathfrak {R}_{j,u_j,h}\ge n+1$ and the first equality is obtained by

$$ \begin{align*} \mathfrak{R}_{j,u_j,h}(z)&=\varphi_{j,u_j}\bigg(\dfrac{P_h(t)}{z-t}\bigg) =\sum_{k=n}^{\infty}\dfrac{\varphi_{j,u_j}(t^kP_h(t))}{z^{k+1}}. \end{align*} $$

We prove the second equality. Fix j and put $\mathcal {E}_{a,b_{j'}}=\partial _t+b_{j'}(t)/a(t)$ for $1\le j'\le d$ . Then,

(5-1) $$ \begin{align} \mathcal{E}_{a,b_{j'}}=\mathcal{E}_{a,b_j}+\dfrac{\gamma_{j',j}}{a_1(t)} \end{align} $$

and $\mathcal {R}_{j',n}=({1}/{n!})\mathcal {E}^n_{a,b_{j'}}a_1(t)^n$ . By Proposition 3.2(i), there is a set $\{c_{j,l}\mid l=0,1,\ldots ,n\}$ of integers with $c_{j,n}=(-1)^nn!$ and

$$ \begin{align*}t^n\mathcal{R}_{j,n}=\sum_{l=0}^n\dfrac{c_{j,l}}{n!}\mathcal{E}^{n-l}_{a,b_j}t^{n-l}a_1(t)^n.\end{align*} $$

Note, by the Leibniz formula, the polynomial $\prod _{j'\neq j}\mathcal {R}_{j',n}\cdot [t^h{a}_2(t)^{dn}]$ is contained in the ideal $({a}_2(t)^n)$ . By Corollary 3.3(i),

$$ \begin{align*}\mathcal{E}^{n-l}_{a,b_j} a_1(t)^n \cdot ({a}_2(t)^n)\subseteq \mathrm{{ker}}\,\varphi_{j,u_j} \quad \text{for } 0\le l\le n-1\end{align*} $$

and thus,

(5-2) $$ \begin{align} t^nP_h(t)&=t^n \mathcal{R}_{j,n}\prod_{j'\neq j}\mathcal{R}_{j',n}\cdot [t^h{a}_2(t)^{dn}] \nonumber \\ &=\sum_{l=0}^n\dfrac{c_{j,l}}{n!}\mathcal{E}^{n-l}_{a,b_j}t^{n-l}a_1(t)^n\prod_{j'\neq j}\mathcal{R}_{j',n}\cdot [t^ha_2(t)^{dn}]\nonumber \\ &\equiv (-1)^n a_1(t)^n\prod_{j'\neq j}\mathcal{R}_{j',n}\cdot [t^ha_2(t)^{dn}] \text{ mod }\mathrm{{ker}}\,\varphi_{j,u_j}. \end{align} $$

Equation (5-1) yields

(5-3) $$ \begin{align} a_1(t)^n\mathcal{R}_{j',n}&=\dfrac{a_1(t)^n}{n!}\bigg(\mathcal{E}_{a,b_j}+\dfrac{\gamma_{j',j}}{a_1(t)}\bigg)^na_1(t)^n \nonumber \\ &=\dfrac{1}{n!}\bigg(\mathcal{E}_{a,b_j}a_1(t)^n+(\gamma_{j',j}-n\varepsilon_{a_1})a_1(t)^{n-1}\bigg)\bigg(\mathcal{E}_{a,b_j}+\dfrac{\gamma_{j',j}}{a_1(t)}\bigg)^{n-1}a_1(t)^n \nonumber\\ &\equiv\dfrac{1}{n!}(\gamma_{j',j}-n\varepsilon_{a_1})a_1(t)^{n-1}\bigg(\mathcal{E}_{a,b_j}+\dfrac{\gamma_{j',j}}{a_1(t)}\bigg)^{n-1}a_1(t)^n \text{ mod }\{\mathcal{E}_{a,b_{j}}a_1(t)\cdot K[t,\partial_t]\} \nonumber\\ &\equiv \dfrac{1}{n!}\prod_{k=1}^n(\gamma_{j',j}-k\varepsilon_{a_1})a_1(t)^n \text{ mod }\{\mathcal{E}_{a,b_{j}}a_1(t)\cdot K[t,\partial_t]\}. \end{align} $$

Note that we use the assumption $\mathrm {{deg}}\, a_1\le 1$ and the equality $\partial _t\cdot a(t)=\varepsilon _{a_1}$ in Equation (5-3). Combining the above equality and Equation (5-2) yields

$$ \begin{align*}\varphi_{j,u_j}(t^nP_h(t))=\dfrac{(-1)^n}{(n!)^{d-1}}\prod_{\substack{1\le j'\le d \\ j'\neq j}}\bigg[\prod_{k=1}^n(\gamma_{j',j}-k\varepsilon_{a_1})\bigg]\varphi_{j,u_j}(t^ha_1(t)^n\cdot a_2(t)^{dn}).\end{align*} $$

This completes the proof of Lemma 5.1.

For a nonnegative integer n, we now consider the determinant of the following $(W+1)\times (W+1)$ matrix:

$$ \begin{align*} &\Delta_n(z)=\Delta(z)=\mathrm{{det}} \begin{pmatrix} P_{0}(z) & P_1(z) & \ldots & P_{W}(z)\\ Q_{1,0,0}(z) & Q_{1,0,1}(z) & \ldots & Q_{1,0,W}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{1,w_1,0}(z) & Q_{1,w_1,1}(z) & \ldots & Q_{1,w_1,W}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{d,0,0}(z) & Q_{d,0,1}(z) & \ldots & Q_{d,0,W}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{d,w_d,0}(z) & Q_{d,w_d,1}(z) & \ldots & Q_{d,w_d,W}(z) \end{pmatrix}. \end{align*} $$

Notice that the determinant $\Delta (z)$ is a polynomial.

To compute $\Delta (z)$ , we define the determinant of following $W\times W$ matrix:

$$ \begin{align*} &\Theta_n=\Theta=\mathrm{{det}}\\ &\begin{pmatrix} \varphi_{1,0}(a_1(t)^na_2(t)^{dn}) & \varphi_{1,0}(ta_1(t)^na_2(t)^{dn}) & \ldots & \varphi_{1,0}(t^{W-1}a_1(t)^na_2(t)^{dn})\\ \vdots & \vdots & \ddots & \vdots\\ \varphi_{1,w_1}(a_1(t)^na_2(t)^{dn}) & \varphi_{1,w_1}(ta_1(t)^na_2(t)^{dn}) & \ldots & \varphi_{1,w_1}(t^{W-1}a_1(t)^na_2(t)^{dn})\\ \vdots & \vdots & \ddots & \vdots\\ \varphi_{d,0}(a_1(t)^na_2(t)^{dn}) & \varphi_{d,0}(ta_1(t)^na_2(t)^{dn}) & \ldots & \varphi_{d,0}(t^{W-1}a_1(t)^na_2(t)^{dn})\\ \vdots & \vdots & \ddots & \vdots\\ \varphi_{d,w_d}(a_1(t)^na_2(t)^{dn}) & \varphi_{d,w_d}(ta_1(t)^na_2(t)^{dn}) & \ldots & \varphi_{d,w_d}(t^{W-1}a_1(t)^na_2(t)^{dn}) \end{pmatrix}. \end{align*} $$

Notice that $\Theta \in K$ . Replace the coefficient of $z^{(n+1)W}$ of the polynomial $P_W$ by $p_{W}$ , that is,

$$ \begin{align*}p_W=\dfrac{1}{[(n+1)W]!}\partial^{(n+1)W}_z\cdot P_{W}(z).\end{align*} $$

Then we have the following proposition.

Proposition 5.2. $\Delta (z)\in K$ . More precisely,

$$ \begin{align*}\Delta(z)=\bigg(\dfrac{{{-1}}}{(n!)^{d-1}}\bigg)^W p_W\cdot \prod_{j=1}^d\bigg[\prod_{\substack{1\le j' \le d \\ j'\neq j}}\prod_{k=1}^n(\gamma_{j',j}-k\varepsilon_{a_1})\bigg]^{{{w_j+1}}}\cdot \Theta,\end{align*} $$

where $\varepsilon _{a_1}$ is the real number defined in Lemma 5.1.

Proof. First, by the definition of $P_{l}(z)$ ,

(5-4) $$ \begin{align} \mathrm{{deg}}\,P_{l}\le nW+l. \end{align} $$

For the matrix in the definition of $\Delta (z)$ , for $1\le j \le d, 0\le u_j\le w_j$ , adding $-f_{j,u_j}(z)$ times the first row to the $(w_1+\cdots +w_{j-1})+u_j+1$  th row,

$$ \begin{align*} \Delta(z)=(-1)^{W}\mathrm{{det}} {\begin{pmatrix} P_{0}(z) & \dots &P_{W}(z)\\ \mathfrak{R}_{1,0,0}(z) & \dots & \mathfrak{R}_{1,0,W}(z)\\ \vdots & \ddots & \vdots \\ \mathfrak{R}_{1,w_1,0}(z) & \dots & \mathfrak{R}_{1,w_1,W}(z)\\ \vdots & \ddots & \vdots \\ \mathfrak{R}_{d,0,0}(z) & \dots & \mathfrak{R}_{d,0,W}(z)\\ \vdots & \ddots & \vdots \\ \mathfrak{R}_{d,w_d,0}(z) & \dots & \mathfrak{R}_{d,w_d,W}(z)\\ \end{pmatrix}}. \end{align*} $$

We denote the $(s,t)$  th cofactor of the matrix in the right-hand side of the above equality by $\Delta _{s,t}(z)$ . Then we have, developing along the first row,

(5-5) $$ \begin{align} \Delta(z)=(-1)^{W}\bigg(\sum_{l=0}^{{{W}}}P_{l}(z)\Delta_{1,l+1}(z)\bigg). \end{align} $$

Since

$$ \begin{align*}\mathrm{{ord}}_{\infty}\, {{\mathfrak{R}}}_{l,h}(z)\ge n+1\quad \text{for } 1\le j \le d, 0\le u_j \le w_j, 0\le h \le W,\end{align*} $$

$$ \begin{align*} \mathrm{{ord}}_{\infty}\,\Delta_{1,l+1}(z)\ge (n+1)W \quad \text{for } 0\le l \le W. \end{align*} $$

Combining Equation (5-4) and the above inequality yields

$$ \begin{align*}P_{l}(z)\Delta_{1,l+1}(z)\in (1/z)\cdot K[[1/z]] \quad \text{for } 0\le l \le W-1,\end{align*} $$

and

$$ \begin{align*}P_{W}(z)\Delta_{1,{{W+1}}}(z)\in K[[1/z]].\end{align*} $$

Note that in the above relation, the constant term of $P_{W}(z)\Delta _{1,{{W+1}}}(z)$ is

(5-6) $$ \begin{align} {{p_W}} \cdot \text{`Coefficient of} \ 1/z^{(n+1)W} \ \text{of} \ \Delta_{1,{{W+1}}}(z)\text{'}. \end{align} $$

Equation (5-5) implies $\Delta (z)$ is a polynomial in z with nonpositive valuation with respect to $\mathrm {{ord}}_{\infty }$ . Thus, it has to be a constant. At last, by Lemma 5.1, the coefficient of $1/z^{(n+1)W}$ of $\Delta _{1,{{W+1}}}(z)$ is

$$ \begin{align*} &\mathrm{{det}} \begin{pmatrix} (-1)^n\varphi_{1,0}(t^nP_0(t)) & \ldots & (-1)^n\varphi_{1,0}(t^{n}P_{W-1}(t)) \\ \vdots & \ddots & \vdots\\ (-1)^n\varphi_{1,w_1}(t^nP_0(t)) & \ldots & (-1)^n\varphi_{1,w_1}(t^{n}P_{W-1}(t)) \\ \vdots & \ddots & \vdots\\ (-1)^n\varphi_{d,0}(t^nP_0(t)) & \ldots & (-1)^n\varphi_{d,0}(t^{n}P_{W-1}(t)) \\ \vdots & \ddots & \vdots\\ (-1)^n\varphi_{d,w_d}(t^nP_0(t)) & \ldots & (-1)^n\varphi_{d,w_d}(t^{n}P_{W-1}(t)) \end{pmatrix}\\ &\quad=\bigg(\dfrac{{{1}}}{(n!)^{d-1}}\bigg)^W\prod_{j=1}^d\bigg[\prod_{\substack{1\le j' \le d \\ j'\neq j}}\prod_{k=1}^n(\gamma_{j',j}-k\varepsilon_{a_1})\bigg]^{{w_j+1}}\cdot \Theta. \end{align*} $$

Combining Equations (5-5), (5-6) and the above equality yields the assertion. This completes the proof of Proposition 5.2.

6 Examples

In this section, let us describe some examples of the application of Theorem 4.2 and Proposition 5.2.

Example 6.1. Let us give a generalization of the Chevyshev polynomials (see [Reference Andrews, Askey and Roy3, $\mathrm{Section}\;5.1$ ]). Let $u\ge 2$ be an integer. Put $D=-(z^u-1)\partial _z-z^{u-1}\in K[z,\partial _z]$ . The Laurent series

$$ \begin{align*}f_l(z)=\sum_{k=0}^{\infty}\dfrac{(\frac{1+l}{u})_k}{(\frac{u+l}{u})_k}\dfrac{1}{z^{uk+l+1}}=\dfrac{1}{z^{l+1}}\cdot {}_2F_1\bigg(\frac{1+l}{u},1,\frac{u+l}{u}\bigg|\frac{1}{z^u}\bigg) \quad \text{for } 0\le l \le u-2\end{align*} $$

are linearly independent over K and satisfy $D\cdot f_l(z)\in K[z]$ . Note that $f_0(z)= (z^u-1)^{-1/u}$ . We denote $\varphi _{f_l}=\varphi _l$ . For $h,n\in \mathbb {N}$ with $0\le h \le u-1$ , we define

$$ \begin{align*} &P_{n,h}(z)=P_h(z)=\dfrac{1}{n!} \bigg(\partial_z-\dfrac{z^{u-1}}{z^u-1}\bigg)^n(z^u-1)^n \cdot z^h,\\ &Q_{n,l,h}(z)=Q_{l,h}(z)=\varphi_{l}\bigg(\dfrac{P_{h}(z)-P_h(t)}{z-t}\bigg) \quad \text{for } 0\le l \le u-2. \end{align*} $$

Theorem 4.2 yields that the vector of polynomials $(P_h,Q_{j,h})_{0\le j \le u-2}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^{u-1}$ Padé-type approximant of $(f_0,\ldots ,f_{u-2})$ . Define

$$ \begin{align*}\Delta_n(z)= \mathrm{{det}}\begin{pmatrix} P_0(z) & \cdots & P_{u-1}(z)\\ Q_{0,0}(z) & \cdots & Q_{0,u-1}(z)\\ \vdots & \ddots & \vdots \\ Q_{u-2,0}(z) & \cdots & Q_{u-2,u-1}(z) \end{pmatrix}. \end{align*} $$

The determinant $\Delta _n(z)$ is computed in Lemma 7.2.

Example 6.2. In this example, we give a generalization of the Bessel polynomials (see [Reference Grosswald16]). Let $d,n$ be nonnegative integers and $\gamma _1,\ldots ,\gamma _d\in K$ that are not integers less than $-1$ with

$$ \begin{align*}\gamma_{j_2}-\gamma_{j_1}\notin \mathbb{Z} \quad \text{for } 1\le j_1<j_2\le d.\end{align*} $$

Put $D_j=-z^2\partial _z+\gamma _j z-1$ ,

$$ \begin{align*}f_j(z)=\sum_{k=0}^{\infty}\dfrac{1}{(2+\gamma_j)_k}\dfrac{1}{z^{k+1}}\end{align*} $$

and $\varphi _{f_j}=\varphi _j$ . A straightforward computation yields $D_j\cdot f_j(z)\in K$ . Put

$$ \begin{align*}R_{j,n}=\dfrac{1}{n!}\bigg(\partial_z+\dfrac{\gamma_jz-1}{z^2}\bigg)^nz^n.\end{align*} $$

Lemma 4.4 yields

$$ \begin{align*}R_{j_1,n_1}R_{j_2,n_2}=R_{j_2,n_2}R_{j_1,n_1} \quad \text{for } 1\le j_1,j_2 \le d \text{ and } n_{j_1},n_{j_2}\in \mathbb{N}.\end{align*} $$

For $h\in \mathbb {Z}$ with $0\le h \le d$ , we define

$$ \begin{align*} &P_{n,h}(z)=P_h(z)=\prod_{j=1}^dR_{j,n}\cdot z^{dn+h},\\ &Q_{n,j}(z)=Q_{j}(z)=\varphi_{j}\bigg(\dfrac{P_h(z)-P_h(t)}{z-t}\bigg) \quad \text{for } 1\le j \le d. \end{align*} $$

Then Theorem 4.2 yields that the vector of polynomials $(P_h,Q_{j,h})_{1\le j \le d}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^d$ Padé approximant of $(f_1,\ldots ,f_d)$ . By the definition of $P_d(z)$ ,

(6-1) $$ \begin{align} P_d(z)=\dfrac{\prod_{j=1}^d {{(d(n+1)+\gamma_j+1)_n}}}{(n!)^d}z^{d(n+1)}+{(\mathrm{lower \ degree \ terms})}. \end{align} $$

Define

$$ \begin{align*} \Delta_n(z)&=\mathrm{{det}} \begin{pmatrix} P_{0}(z) & P_1(z) & \ldots & P_{d}(z)\\ Q_{1,0}(z) & Q_{1,1}(z) & \ldots & Q_{1,d}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{d,0}(z) & Q_{d,1}(z) & \ldots & Q_{d,d}(z)\\ \end{pmatrix}, \ \ \\\Theta_n&=\mathrm{{det}} \begin{pmatrix} \varphi_{1}(t^{(d+1)n}) & \ldots & \varphi_{1}(t^{(d+1)n+d-1})\\ \vdots & \ddots & \vdots\\ \varphi_{d}(t^{(d+1)n}) & \ldots & \varphi_{d}(t^{(d+1)n+d-1}) \end{pmatrix}. \end{align*} $$

Let us compute $\Theta _n$ . By the definition of $\varphi _j$ and the properties of determinants,

$$ \begin{align*} \Theta_n&= \mathrm{{det}} \begin{pmatrix} \dfrac{1}{(2+\gamma_1)_{(d+1)n}} & \ldots & \dfrac{1}{(2+\gamma_1)_{(d+1)n+d-1}}\\ \vdots & \ddots & \vdots\\ \dfrac{1}{(2+\gamma_d)_{(d+1)n}} & \ldots & \dfrac{1}{(2+\gamma_d)_{(d+1)n+d-1}} \end{pmatrix}\\ &=\prod_{j=1}^d\dfrac{1}{(2+\gamma_j)_{(d+1)n+d-1}}\cdot \mathrm{{det}} \begin{pmatrix} (2+\gamma_1+(d+1)n)_{d-1}& \ldots & (2+\gamma_1+(d+1)n)_{0}\\ \vdots & \ddots & \vdots\\ (2+\gamma_d+(d+1)n)_{d-1} & \ldots & (2+\gamma_d+(d+1)n)_{0} \end{pmatrix}. \end{align*} $$

Here, by using the properties of the determinant again,

$$ \begin{align*} &\mathrm{{det}} \begin{pmatrix} (2+\gamma_1+(d+1)n)_{d-1}& \ldots & (2+\gamma_1+(d+1)n)_{0}\\ \vdots & \ddots & \vdots\\ (2+\gamma_d+(d+1)n)_{d-1} & \ldots & (2+\gamma_d+(d+1)n)_{0} \end{pmatrix}\\ &\quad= (-1)^{{(d-1)d}/{2}} \mathrm{{det}} \begin{pmatrix} (2+\gamma_1+(d+1)n)_{0} & \ldots & (2+\gamma_1+(d+1)n)_{d-1}\\ \vdots & \ddots & \vdots \\ (2+\gamma_d+(d+1)n)_{0} & \ldots &(2+\gamma_d+(d+1)n)_{d-1} \end{pmatrix}\\ &\quad=(-1)^{{(d-1)d}/{2}} \mathrm{{det}} \begin{pmatrix} 1& \gamma_1 & \ldots &\gamma^{d-1}_1\\ \vdots & \ddots & \vdots\\ 1& \gamma_d & \ldots &\gamma^{d-1}_d \end{pmatrix}. \end{align*} $$

Since the last determinant is a Vandermonde determinant,

$$ \begin{align*}\Theta_n=\prod_{j=1}^d\dfrac{1}{(2+\gamma_j)_{(d+1)n+d-1}}\cdot {{(-1)^{{(d-1)d}/{2}}}} \prod_{1\le j_1<j_2\le d}(\gamma_{j_2}-\gamma_{j_1}).\end{align*} $$

Proposition 5.2 and Equation (6-1) imply that

$$ \begin{align*} \Delta_n(z)&={{(-1)^{{(d-1)d}/{2}}}}\bigg(\dfrac{{{-1}}}{(n!)^{d}}\bigg)^d\cdot \prod_{j=1}^d\bigg[\prod_{\substack{1\le j' \le d \\ j'\neq j}}\prod_{k=1}^n(\gamma_{j'}-\gamma_j-k)\bigg]\\ &\quad \times \prod_{j=1}^d\dfrac{{{(d(n+1)+\gamma_j+1)_n}}}{(2+\gamma_j)_{(d+1)n+d-1}} \cdot \prod_{1\le j_1<j_2\le d}(\gamma_{j_2}-\gamma_{j_1}). \end{align*} $$

Especially, we have $\Delta _n(z)\in K\setminus \{0\}$ .

Example 6.3. In this example, we give a generalization of the Laguerre polynomials (see [Reference Andrews, Askey and Roy3, $\mathrm{Section}\;6.2$ ]). Let $d,n\in \mathbb {N}$ , $\gamma _1,\ldots ,\gamma _d\in K\setminus \{0\}$ be pairwise distinct and $\delta \in K$ be a nonnegative integer. Put $D_{j}=-z\partial _z-\gamma _jz+\delta $ ,

$$ \begin{align*}f_{j}(z)={\sum_{k=0}^{\infty}}(1+\delta)_k\bigg(\dfrac{1}{\gamma_jz}\bigg)^{k+1}\end{align*} $$

and $\varphi _{f_{j}}=\varphi _{j}$ . A straightforward computation shows $D_{j}\cdot f_{j}(z)\in K$ . Put

$$ \begin{align*}R_{j,n}=\dfrac{1}{n!}\bigg(\partial_z-\dfrac{\gamma_j z-\delta}{z}\bigg)^{n}.\end{align*} $$

By Lemma 4.4,

$$ \begin{align*}R_{j_1,n_{j_1}}R_{j_2,n_{j_2}}=R_{j_2,n_{j_2}}R_{j_1,n_{j_1}} \quad \text{for } 1\le j_1,j_2\le d, \ \ n_{j_1},n_{j_2}\in \mathbb{N}.\end{align*} $$

For $h\in \mathbb {Z}$ with $0\le h \le d$ , we define

$$ \begin{align*} &P_{n,h}(z)=P_h(z)=\prod_{j=1}^dR_{j,n}\cdot z^{dn+h},\\ &Q_{n,j}(z)=Q_{j}(z)=\varphi_{j}\bigg(\dfrac{P_h(z)-P_h(t)}{z-t}\bigg) \quad \text{for } 1\le j \le d. \end{align*} $$

Then Theorem 4.2 yields that the vector of polynomials $(P_h,Q_{j,h})_{1\le j \le d}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^d$ Padé-type approximant of $(f_{j})_{\substack {1\le j \le d}}$ . By the definition of $P_d(z)$ ,

(6-2) $$ \begin{align} P_d(z)=\dfrac{\prod_{j=1}^d\gamma^n_j}{(n!)^d}z^{d(n+1)}+{(\mathrm{lower \ degree \ terms})}. \end{align} $$

Define

$$ \begin{align*} &\Delta_n(z)=\mathrm{{det}} \begin{pmatrix} P_{0}(z) & P_1(z) & \ldots & P_{d}(z)\\ Q_{1,0}(z) & Q_{1,1}(z) & \ldots & Q_{1,d}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{d,0}(z) & Q_{d,1}(z) & \ldots & Q_{d,d}(z)\\ \end{pmatrix}, \ \ \Theta_n=\mathrm{{det}} \begin{pmatrix} \varphi_{1}(t^{dn}) & \ldots & \varphi_{1}(t^{d(n+1)-1})\\ \vdots & \ddots & \vdots\\ \varphi_{d}(t^{dn}) & \ldots & \varphi_{d}(t^{d(n+1)-1}) \end{pmatrix}. \end{align*} $$

We now compute $\Theta _n$ . By the definition of $\varphi _j$ and the properties of the determinant,

$$ \begin{align*} \Theta_n&= \mathrm{{det}} \begin{pmatrix} \dfrac{(1+\delta)_{dn}}{\gamma^{dn+1}_1} & \ldots & \dfrac{(1+\delta)_{d(n+1)-1}}{\gamma^{d(n+1)}_1}\\ \vdots & \ddots & \vdots\\ \dfrac{(1+\delta)_{dn}}{\gamma^{dn+1}_d} & \ldots & \dfrac{(1+\delta)_{d(n+1)-1}}{\gamma^{d(n+1)}_d} \end{pmatrix}\\ &=\prod_{j=1}^d\dfrac{(1+\delta)_{dn+j-1}}{\gamma^{d(n+1)}_j}\cdot \mathrm{{det}} \begin{pmatrix} 1 & \gamma_1 & \ldots & \gamma^{d-1}_1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & \gamma_d & \ldots & \gamma^{d-1}_d\\ \end{pmatrix}. \end{align*} $$

Since the last determinant is nothing but a Vandermonde determinant,

$$ \begin{align*}\Theta_n=\prod_{j=1}^d\dfrac{(1+\delta)_{dn+j-1}}{\gamma^{d(n+1)}_j}\cdot \prod_{1\le j_1<j_2\le d}(\gamma_{j_2}-\gamma_{j_1}).\end{align*} $$

Proposition 5.2 and Equation (6-2) imply that

$$ \begin{align*} \Delta_n(z)=\bigg(\dfrac{{{-1}}}{(n!)^{d}}\bigg)^d\cdot \prod_{j=1}^d\bigg[\prod_{\substack{1\le j' \le d \\ j'\neq j}}(\gamma_{j'}-\gamma_j)^n\bigg]\cdot \prod_{j=1}^d\dfrac{(1+\delta)_{dn+j-1}}{\gamma^{(d-1)n+d}_j}\cdot\! \prod_{1\le j_1<j_2\le d}\!(\gamma_{j_2}-\gamma_{j_1})\in K\setminus \{0\}. \end{align*} $$

Example 6.4. Let us give an alternative generalization of the Laguerre polynomials. Let $d,n\in \mathbb {N}$ , $\gamma \in K\setminus \{0\}$ , and $\delta _1,\ldots ,\delta _d\in K$ be nonnegative integers with

$$ \begin{align*}\delta_{j_1}-\delta_{j_2}\notin \mathbb{Z} \quad \text{for } 1\le j_1<j_2\le d.\end{align*} $$

Put $D_{j}=-z\partial _z-\gamma z+\delta _j$ ,

$$ \begin{align*}f_{j}(z)={\sum_{k=0}^{\infty}}(1+\delta_j)_k\bigg(\dfrac{1}{\gamma z}\bigg)^{k+1},\end{align*} $$

and $\varphi _{f_{j}}=\varphi _{j}$ . Then we have $D_{j}\cdot f_{j}(z)\in K$ . Put

$$ \begin{align*}R_{j,n}=\dfrac{1}{n!}\bigg(\partial_z-\dfrac{\gamma z-\delta_j}{z}\bigg)^{n}z^n.\end{align*} $$

By Lemma 4.4,

$$ \begin{align*}R_{j_1,n_{j_1}}R_{j_2,n_{j_2}}=R_{j_2,n_{j_2}}R_{j_1,n_{j_1}} \quad \text{for } 1\le j_1,j_2\le d, \ \ n_{j_1},n_{j_2}\in \mathbb{N}.\end{align*} $$

For $h\in \mathbb {Z}$ with $0\le h \le d$ , we define

$$ \begin{align*} &P_{n,h}(z)=P_h(z)=\prod_{j=1}^dR_{j,n}\cdot z^h,\\ &Q_{n,j}(z)=Q_{j}(z)=\varphi_{j}\bigg(\dfrac{P_h(z)-P_h(t)}{z-t}\bigg) \quad \text{for } 1\le j \le d. \end{align*} $$

Then Theorem 4.2 yields that the vector of polynomials $(P_h,Q_{j,h})_{1\le j \le d}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^d$ Padé-type approximant of $(f_{j})_{\substack {1\le j \le d}}$ . By the definition of $P_d(z)$ ,

(6-3) $$ \begin{align} P_d(z)=\dfrac{\gamma^{dn}}{(n!)^d}z^{d(n+1)}+{(\mathrm{lower \ degree \ terms})}. \end{align} $$

Define

$$ \begin{align*} &\Delta_n(z)=\mathrm{{det}} \begin{pmatrix} P_{0}(z) & P_1(z) & \ldots & P_{d}(z)\\ Q_{1,0}(z) & Q_{1,1}(z) & \ldots & Q_{1,d}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{d,0}(z) & Q_{d,1}(z) & \ldots & Q_{d,d}(z)\\ \end{pmatrix}, \ \ \Theta_n=\mathrm{{det}} \begin{pmatrix} \varphi_{1}(t^{n}) & \ldots & \varphi_{1}(t^{d+n-1})\\ \vdots & \ddots & \vdots\\ \varphi_{d}(t^{n}) & \ldots & \varphi_{d}(t^{d+n-1}) \end{pmatrix}. \end{align*} $$

Let us compute $\Theta _n$ . By the definition of $\varphi _j$ and the properties of the determinant,

$$ \begin{align*} \Theta_n&= \mathrm{{det}} \begin{pmatrix} \dfrac{(1+\delta_1)_{n}}{\gamma^{n+1}} & \ldots & \dfrac{(1+\delta_1)_{d+n-1}}{\gamma^{d+n}}\\ \vdots & \ddots & \vdots\\ \dfrac{(1+\delta_d)_{n}}{\gamma^{n+1}} & \ldots & \dfrac{(1+\delta_d)_{d+n-1}}{\gamma^{d+n}} \end{pmatrix}\\ &=\prod_{j=1}^d\dfrac{(1+\delta_j)_{n}}{\gamma^{n+j}} \cdot \begin{pmatrix} (n+\delta_1)_0 & \ldots & (n+\delta_1)_{d-1}\\ \vdots & \ddots & \vdots\\ (n+\delta_1)_0 & \ldots & (n+\delta_d)_{d-1} \end{pmatrix}. \end{align*} $$

A similar computation as in Example 6.2 leads us to get

$$ \begin{align*}\Theta_n=\prod_{j=1}^d\dfrac{(1+\delta_j)_{n}}{\gamma^{n+j}}\cdot \prod_{1\le j_1<j_2\le d}(\delta_{j_2}-\delta_{j_1}).\end{align*} $$

Proposition 5.2 and Equation (6-3) imply that

$$ \begin{align*} \Delta_n(z)&=\bigg(\dfrac{{{-1}}}{(n!)^{d}}\bigg)^d\cdot \prod_{j=1}^d\bigg[\prod_{\substack{1\le j' \le d \\ j'\neq j}}\prod_{k=1}^n(\delta_{j'}-\delta_j-k)\bigg]\\ &\quad\times\prod_{j=1}^d\dfrac{(1+\delta_j)_{n}}{\gamma^{j}}\cdot \prod_{1\le j_1<j_2\le d}(\delta_{j_2}-\delta_{j_1}) \in K\setminus \{0\}. \end{align*} $$

Example 6.5. In this example, we give a generalization of the Hermite polynomials (see [Reference Andrews, Askey and Roy3, $\mathrm{Section}\;6.1$ ]). Let $d,n\in \mathbb {N}$ , $\gamma \in K\setminus \{0\}$ and $\delta _1,\ldots ,\delta _d\in K$ be pairwise distinct. Put $D_j=-\partial _z+\gamma z+\delta _j$ ,

$$ \begin{align*}f_j(z)={\sum_{k=0}^{\infty}}\dfrac{f_{j,k}}{z^{k+1}},\end{align*} $$

where $f_{j,0}=1$ , $f_{j,1}=-\delta _j/\gamma $ and

(6-4) $$ \begin{align} f_{j,k+2}=-\dfrac{\delta_jf_{j,k+1}+(k+1)f_{j,k}}{\gamma} \quad \text{for } k\ge 0, \end{align} $$

and $\varphi _{f_j}=\varphi _j$ . Then we have $D_j\cdot f_{j}(z)\in K$ . Put

$$ \begin{align*}R_{j,n}=\dfrac{1}{n!}(\partial_z+\gamma z+\delta_j)^{n}.\end{align*} $$

By Lemma 4.4,

$$ \begin{align*}R_{j_1,n_{j_1}}R_{j_2,n_{j_2}}=R_{j_2,n_{j_2}}R_{j_1,n_{j_1}} \quad \text{for } 1\le j_1,j_2\le d, n_{j_1},n_{j_2}\in \mathbb{N}.\end{align*} $$

For $h\in \mathbb {Z}$ with $0\le h \le d$ , we define

$$ \begin{align*} &P_{n,h}(z)=P_h(z)=\prod_{j=1}^dR_{j,n}\cdot z^h,\\ &Q_{n,j,h}(z)=Q_{j,h}(z)=\varphi_{j}\bigg(\dfrac{P_h(z)-P_h(t)}{z-t}\bigg) \quad \text{for } 1\le j \le d. \end{align*} $$

Then Theorem 4.2 yields that the vector of polynomials $(P_h,Q_{j,h})_{\substack {1\le j \le d}}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^d$ Padé-type approximant of $(f_{j})_{1\le j \le d}$ . By the definition of $P_d(z)$ ,

(6-5) $$ \begin{align} P_d(z)=\dfrac{\gamma^{dn}}{(n!)^d}z^{d(n+1)}+{(\mathrm{lower \ degree \ terms})}. \end{align} $$

Define

$$ \begin{align*} \Delta_n(z)=\mathrm{{det}} \begin{pmatrix} P_{0}(z) & P_1(z) & \ldots & P_{d}(z)\\ Q_{1,0}(z) & Q_{1,1}(z) & \ldots & Q_{1,d}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{d,0}(z) & Q_{d,1}(z) & \ldots & Q_{d,d}(z)\\ \end{pmatrix}, \ \ \Theta_n=\mathrm{{det}} \begin{pmatrix} \varphi_{1}(1) & \ldots & \varphi_{1}(t^{d-1})\\ \vdots & \ddots & \vdots\\ \varphi_{d}(1) & \ldots & \varphi_{d}(t^{d-1}) \end{pmatrix}. \end{align*} $$

Let us compute $\Theta _n$ . By the definition of $\varphi _j$ ,

(6-6) $$ \begin{align} \Theta_n= \mathrm{{det}} \begin{pmatrix} f_{1,0} & f_{1,1}& \ldots & f_{1,d-1}\\ \vdots &\vdots & \ddots & \vdots\\ f_{d,0} & f_{d,1} & \ldots & f_{d,d-1} \end{pmatrix}. \end{align} $$

Here, using Equation (6-4) and the properties of the determinant repeatedly,

(6-7) $$ \begin{align} \mathrm{{det}} \begin{pmatrix} f_{1,0} & f_{1,1}& \ldots & f_{1,d-1}\\ \vdots &\vdots & \ddots & \vdots\\ f_{d,0} & f_{d,1} & \ldots & f_{d,d-1} \end{pmatrix} = \mathrm{{det}} \begin{pmatrix} 1 & \frac{-\delta_1}{\gamma}& \ldots & \bigg(\frac{-\delta_1}{\gamma}\bigg)^{d-1}\\ \vdots &\vdots & \ddots & \vdots\\ 1 & \frac{-\delta_d}{\gamma}& \ldots & \bigg(\frac{-\delta_d}{\gamma}\bigg)^{d-1} \end{pmatrix}. \end{align} $$

Combining Equations (6-6) and (6-7) implies

$$ \begin{align*}\Theta_n=\bigg(\dfrac{-1}{\gamma}\bigg)^{1+2+\cdots+(d-1)}\cdot \prod_{1\le j_1<j_2\le d}(\delta_{j_2}-\delta_{j_1}).\end{align*} $$

Proposition 5.2 and Equation (6-5) imply that

$$ \begin{align*} \Delta_n(z)&=\bigg(\dfrac{{{-1}}}{(n!)^{d}}\bigg)^d\cdot \prod_{j=1}^d\bigg[\prod_{\substack{1\le j' \le d \\ j'\neq j}}(\delta_{j'}-\delta_j)^n\bigg]\\ &\quad\times(-1)^{{(d-1)d}/{2}}\gamma^{dn-{(d-1)d}/{2}}\cdot \prod_{1\le j_1<j_2\le d}(\delta_{j_2}-\delta_{j_1}) \in K\setminus \{0\}. \end{align*} $$

Example 6.6. In this example, we consider a generalization of the Legendre polynomials (see [Reference Andrews, Askey and Roy3, Remark $\mathrm{5.3.1}$ ]). Let $d,m,n\in \mathbb {N}$ , $\alpha _{1},\ldots ,\alpha _{m}\in K\setminus \{0\}$ be pairwise distinct and $\gamma _1,\ldots ,\gamma _d\in K$ be nonnegative integers, satisfying $\gamma _{j_1}-\gamma _{j_2}\notin \mathbb {Z}$ for $1\le j_1<j_2\le d$ . Put $a_2(z)=\prod _{i=1}^m(z-\alpha _i)$ , $D_{j}=-za_2(z)\partial _z+\gamma _ja_2(z)$ ,

$$ \begin{align*}f_{i,j}(z)={\sum_{k=0}^{\infty}}\dfrac{1}{k+1+\gamma_j}\bigg(\dfrac{\alpha_{i}}{z}\bigg)^{k+1} \quad \text{for } 1\le i \le m, 1\le j \le d,\end{align*} $$

and $\varphi _{f_{i,j}}=\varphi _{i,j}$ . Then we have $D_{j}\cdot f_{i,j}(z)\in K[z]$ . Put

$$ \begin{align*}R_{j,n}=\dfrac{1}{n!}\bigg(\partial_z+\dfrac{\gamma_j}{z}\bigg)^nz^n.\end{align*} $$

By Lemma 4.4, we have

$$ \begin{align*}R_{j_1,n_{j_1}}R_{j_2,n_{j_2}}=R_{j_2,n_{j_2}}R_{j_1,n_{j_1}} \quad \text{for } 1\le j_1,j_2\le d, n_{j_1},n_{j_2}\in \mathbb{N}.\end{align*} $$

For $h\in \mathbb {Z}$ with $0\le h \le dm$ , we define

$$ \begin{align*} &P_{n,h}(z)=P_h(z)=\prod_{j=1}^dR_{j,n}\cdot [z^ha_2(z)^{dn}],\\ &Q_{n,i,j,h}(z)=Q_{i,j,h}(z)=\varphi_{i,j}\bigg(\dfrac{P_h(z)-P_h(t)}{z-t}\bigg) \quad \text{for } 1\le i \le m, 1\le j \le d. \end{align*} $$

Then Theorem 4.2 yields that the vector of polynomials $(P_h,Q_{i,j,h})_{\substack {1\le i \le m \\ 1\le j \le d }}$ is a weight $(n,\ldots ,n)\in \mathbb {N}^{dm}$ Padé-type approximant of $(f_{i,j})_{\substack {1\le i \le m \\ 1\le j \le d}}$ . Define

$$ \begin{align*} &\Delta_n(z)=\mathrm{{det}} \begin{pmatrix} P_{0}(z) & P_1(z) & \ldots & P_{dm}(z)\\ Q_{1,1,0}(z) & Q_{1,1,1}(z) & \ldots & Q_{1,1,dm}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{m,1,0}(z) & Q_{m,1,1}(z) & \ldots & Q_{m,1,dm}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{1,d,0}(z) & Q_{1,d,1}(z) & \ldots & Q_{1,d,dm}(z)\\ \vdots & \vdots & \ddots & \vdots\\ Q_{m,d,0}(z) & Q_{m,d,1}(z) & \ldots & Q_{m,d,dm}(z)\\ \end{pmatrix}. \end{align*} $$

The nonvanishing of $\Delta _n(z)$ has been proven in [Reference David, Hirata-Kohno and Kawashima12, Proposition $4.1$ ].

7 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. We prove the more precise theorem that we state below. To state the theorem, we prepare the notation.

Let K be an algebraic number field. We denote the set of places of K by ${{\mathfrak {M}}}_K$ . For $v\in {{\mathfrak {M}}}_K$ , we denote the completion of K with respect to v by $K_v$ and define the normalized absolute value $|\cdot |_v$ as follows:

$$ \begin{align*} &|p|_v=p^{-{[K_v:\mathbb{Q}_p]}/{[K:\mathbb{Q}]}} \text{ if } v\mid p, \quad |x|_v=|\iota_v x|^{{[K_v:\mathbb{R}]}/{[K:\mathbb{Q}]}} \text{ if } v\mid \infty, \end{align*} $$

where p is a prime number and $\iota _v$ the embedding $K\hookrightarrow \mathbb {C}$ corresponding to v.

Let $\beta \in K$ . We define the absolute Weil height of $\beta $ as

$$ \begin{align*} &{{\mathrm{H}}({\beta})=\prod_{v\in {{\mathfrak{M}}}_K} \max\{ 1,|\beta|_v\}.} \end{align*} $$

Let m be a positive integer and $\boldsymbol {\beta }=(\beta _0,{\ldots },\beta _m) \in \mathbb {P}_m(K)$ . We define the absolute Weil height of $\boldsymbol {\beta }$ by

$$ \begin{align*} &{\mathrm{H}}(\boldsymbol{\beta})=\prod_{v\in {{\mathfrak{M}}}_K} \max\{ |\beta_0|_v,\ldots,|\beta_m|_v\}, \end{align*} $$

and the logarithmic absolute Weil height by ${\mathrm {h}}(\boldsymbol {\beta })=\mathrm {\log }\, \mathrm {H}(\boldsymbol {\beta })$ . Let $v\in \mathfrak {M}_K$ , then ${\mathrm {h}}_v(\boldsymbol {\beta })=\log \Vert \boldsymbol {\beta }\Vert _v$ where $\Vert \cdot \Vert _v$ is the sup v-adic norm. Then we have ${\mathrm {h}}(\boldsymbol {\beta })={{\sum _{v\in \mathfrak {M}_K}}}{\mathrm {h}}_v(\boldsymbol {\beta })$ and for $\beta \in K$ , ${\mathrm {h}}(\beta )$ is the height of the point $(1,\beta )\in \mathbb {P}_1(K)$ .

Let u be an integer with $u\ge 2$ . We put $\nu (u)=u\prod _{q:\text {prime}, q|u}q^{1/(q-1)}$ . Let $v_0$ be a place of K, $\alpha \in K$ with $|\alpha |_{v_0}>2$ . In the case where $v_0$ is a nonarchimedean place, we denote the prime number under $v_0$ by $p_{v_0}$ and put $\varepsilon _{v_0}(u)=1$ if u is coprime with $p_{v_0}$ and $\varepsilon _{v_0}(u)=0$ if u is divisible by $p_{v_0}$ . We denote Euler’s totient function by $\varphi $ .

We define real numbers

$$ \begin{align*} \mathbb{A}_{v_0}(\alpha)&=\mathrm{{h}}_{v_0}(\alpha)-\begin{cases} \mathrm{{h}}_{v_0}(2) & \text{if } v_0\mid \infty \\ \dfrac{\varepsilon_{v_0}(u) \log\,|p_{v_0}|_{v_0}}{p_{v_0}-1} & \text{if } v_0\nmid \infty, \end{cases}\\ \mathbb{B}_{v_0}(\alpha)&=(u-1)\mathrm{{h}}(\alpha)+(u+1)\mathrm{{h}}(2)+\dfrac{(2u-1)\log\,\nu(u)}{u}\\&\quad+\dfrac{u-1}{\varphi(u)} -(u-1)\mathrm{{h}}_{v_0}(\alpha)-\begin{cases} (u+1)\mathrm{{h}}_{v_0}(2) & \text{if } v_0\mid \infty \\ \log\,|\nu(u)|^{-1}_{v_0} & \text{if } v_0\nmid \infty, \end{cases}\\ U_{v_0}(\alpha)&=(u-1)\mathrm{{h}}_{v_0}(\alpha)+ \begin{cases} (u+1)\mathrm{{h}}_{v_0}(2)& \text{if } v_0\mid \infty \\ \log\,|\nu(u)|^{-1}_{v_0} & \text{if } v_0\nmid \infty, \end{cases}\\ V_{v_0}(\alpha)&=\mathbb{A}_{v_0}(\alpha)-\mathbb{B}_{v_0}(\alpha). \end{align*} $$

We can now state the following theorem.

Theorem 7.1. Assume $V_{v_0}(\alpha )>0$ . Then, for any positive number $\varepsilon $ with $\varepsilon <V_{v_0}(\alpha )$ , there exists an effectively computable positive number $H_0$ depending on $\varepsilon $ and the given data such that the following property holds. For any $\boldsymbol {\lambda }=(\lambda ,\lambda _{l})_{0\le l \le u-2} \in K^{u} \setminus \{ \boldsymbol {0} \}$ satisfying $H_0\le {\mathrm {H}}(\boldsymbol {\lambda })$ , then

$$ \begin{align*} \bigg|\lambda+\sum_{l=0}^{u-2}\lambda_{l}\cdot \dfrac{1}{\alpha^{l+1}} {}_2F_1\bigg(\frac{1+l}{u},1, \frac{u+l}{u}\,\bigg|\frac{1}{\alpha^u}\bigg)\bigg|_{v_0}>C(\alpha,\varepsilon){\mathrm{H}}_{v_0}(\boldsymbol{\lambda}) {\mathrm{H}}(\boldsymbol{\lambda})^{-\mu(\alpha,\varepsilon)}, \end{align*} $$

where

$$ \begin{align*} \mu(\alpha,\varepsilon)&= \dfrac{\mathbb{A}_{v_0}(\alpha)+{{U}}_{v_0}(\alpha)}{V_{v_0}(\alpha)-\varepsilon} \quad\mbox{and}\quad C(\alpha,\varepsilon)\\&= \exp \bigg(-\bigg(\frac{\log(2)}{V_{v_0}(\alpha)-\varepsilon}+1\bigg)(\mathbb{A}_{v_0}(\alpha)+{{U}}_{v_0}(\alpha))\bigg). \end{align*} $$

We derive Theorem 1.1 from Theorem 7.1.

Proof of Theorem 1.1

Let us consider the case of $K=\mathbb {Q}$ , $v_0=\infty $ and $\alpha \in \mathbb {Z}\setminus \{0,\pm 1\}$ . Then we see that $V_{\infty }(\alpha )=V(\alpha )$ where $V(\alpha )$ is defined in Theorem 1.1. Assume $V(\alpha )>0$ . Choose some $\boldsymbol {\lambda }=(\lambda ,\lambda _{0}\ldots ,\lambda _{u-2})\in \mathbb {Q}^{u}\setminus \{\boldsymbol {0}\}$ such that

$$ \begin{align*}\lambda_0+\sum_{l=0}^{u-2}\lambda_{l}\cdot \dfrac{1}{\alpha^{l+1}} {}_2F_1\bigg(\frac{1+l}{u},1, \frac{u+l}{u}\,\bigg|\frac{1}{\alpha^u}\bigg)=0.\end{align*} $$

If $H(\boldsymbol {\lambda })\geq H_0$ (where $H_0$ is as in Theorem 7.1), there is nothing more to prove. Otherwise, let $m>0$ be a rational integer such that $H(m\boldsymbol {\lambda })\geq H_0$ . Then Theorem 7.1 ensures that

$$ \begin{align*}m\bigg(\lambda_0+\sum_{l=0}^{u-2}\lambda_{l}\cdot \dfrac{1}{\alpha^{l+1}} {}_2F_1\bigg(\frac{1+l}{u},1, \frac{u+l}{u}\,\bigg|\frac{1}{\alpha^u}\bigg)\bigg)\neq 0.\end{align*} $$

This is a contradiction and completes the proof of Theorem 1.1.

Now we start the proof of Theorem 7.1. The proof is relying on the Padé approximants obtained in Example 6.1. In the following, we use the same notation as in Example 6.1.

7.1 Computation of determinants

Lemma 7.2. Let n be a positive integer. Put $n=uN+s$ for nonnegative integers $N,s$ with $0\le s \le u-1$ . Then,

$$ \begin{align*} \Delta_{n}(z)=(-1)^{(uN+s+1)(u-1)}\dfrac{((uN+s+1)u-1-uN)_{uN+s}}{(uN+s)!} \prod_{l=0}^{u-2}\dfrac{(\tfrac{u-1}{u})_{uN+s}}{(\tfrac{u+l}{u})_{uN+s}}\in K\setminus \{0\}. \end{align*} $$

Proof. Put

$$ \begin{align*} \Theta_n= \mathrm{{det}}\begin{pmatrix} \varphi_{0}((t^u-1)^n) & \ldots & \varphi_0(t^{u-2}(t^u-1)^n)\\ \vdots & \ddots & \vdots\\ \varphi_{u-2}((t^u-1)^n) & \ldots & \varphi_{u-2}(t^{u-2}(t^u-1)^n) \end{pmatrix}. \end{align*} $$

Proposition 5.2 implies that

$$ \begin{align*}\Delta_n(z)=(-1)^{(u-1)}\times \dfrac{1}{[(n+1)(u-1)]!}\partial^{(n+1)(u-1)}_z\cdot P_{u-1}(z)\times \Theta_n.\end{align*} $$

According to the definition of $P_{u-1}(z)$ ,

$$ \begin{align*} \dfrac{1}{[(n+1)(u-1)]!}\partial^{(n+1)(u-1)}_z\cdot P_{u-1}(z)= \dfrac{((n+1)u-1-n)_n} {n!}. \end{align*} $$

By the definition of $f_l$ ,

$$ \begin{align*} \varphi_l(t^k)=\begin{cases} \dfrac{(\tfrac{1+l}{u})_N}{(\tfrac{u+l}{u})_N} & \text{if } k=uN+l \ \text{for some} \ N\in \mathbb{Z},\\ 0 & \text{ otherwise}. \end{cases} \end{align*} $$

The above equality shows

(7-1) $$ \begin{align} \Theta_{n}&= \mathrm{{det}}\begin{pmatrix} \varphi_{0}((t^u-1)^{uN+s}) & 0 & \ldots & 0 \\ 0& \varphi_{1}(t(t^u-1)^{uN+s}) & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0& \ldots & \varphi_{u-2}(t^{u-2}(t^u-1)^{uN+s}) \end{pmatrix} \nonumber\\ &=\prod_{l=0}^{u-2}\varphi_{l}(t^{l}(t^u-1)^{uN+s}). \end{align} $$

We now compute $\varphi _{l}(t^{l}(t^u-1)^{uN+s})$ . Since we have

$$ \begin{align*}t^{l}(t^u-1)^{uN+s}=\sum_{v=0}^{uN+s}\binom{uN+s}{v}(-1)^{uN+l-v}t^{uv+l},\end{align*} $$

we obtain

$$ \begin{align*} \varphi_{l}(t^{l}(t^u-1)^{uN+s})=\sum_{v=0}^{uN+s}\binom{uN+s}{v}(-1)^{uN+l-v}\dfrac{(\tfrac{1+l}{u})_v}{(\tfrac{u+l}{u})_v}. \end{align*} $$

For positive real numbers $\alpha ,\beta $ with $\alpha <\beta $ and a nonnegative integer v,

$$ \begin{align*} \dfrac{(\alpha)_v}{(\beta)_v} &=\dfrac{\Gamma(\beta)}{\Gamma(\alpha)\Gamma(\beta-\alpha)}\int^1_0 \xi^{\alpha+v-1}(1-\xi)^{\beta-\alpha-1}d\xi. \end{align*} $$

Applying the above equality for $\alpha =(1+l)/u,\beta =(u+l)/u$ , we obtain

(7-2) $$ \begin{align} &\varphi_{l}(t^{l}(t^u-1)^{uN+s})\nonumber\\&=\dfrac{\Gamma(\tfrac{u+l}{u})}{\Gamma(\tfrac{1+u}{u})\Gamma(\tfrac{u-1}{u})}\sum_{v=0}^{uN+s}\binom{uN+s}{v}(-1)^{uN+l-v} \int^1_0\xi^{({1+l})/{u}+v-1}(1-\xi)^{({u-1})/{u}-1}d\xi \nonumber\\&=\dfrac{(-1)^{uN+s}\Gamma(\tfrac{u+l}{u})}{\Gamma(({1+l})/{u})\Gamma(\tfrac{u-1}{u})}\int^1_0\xi^{\tfrac{1+l}{u}-1}(1-\xi)^{uN+s+({u-1})/{u}-1}d\xi \nonumber\\&=\dfrac{(-1)^{uN+s}\Gamma(\tfrac{u+l}{u})}{\Gamma(uN+s+\tfrac{u+l}{u})}\dfrac{\Gamma(uN+s+\tfrac{u-1}{u})}{\Gamma(\tfrac{u-1}{u})} \nonumber\\&=\dfrac{(-1)^{uN+s}(\tfrac{u-1}{u})_{uN+s}}{(\tfrac{u+l}{u})_{uN+s}}. \end{align} $$

Substituting the above equality into Equation (7-1), we obtain the assertion.

7.2 Estimates

Unless stated otherwise, the Landau symbols small o and large O refer when N tends to infinity.

For a finite set S of rational numbers and a rational number a, we define

$$ \begin{align*}\mathrm{{den}}\,(S)=\min\{n\in \mathbb{Z}\mid n\ge 1, ns\in \mathbb{Z} \ \text{for all} \ s\in S\} \quad \text{and} \quad \mu(a)=\mathrm{{den}}\,(a)\prod_{\substack{q:\mathrm{{prime}}\\ q|\mathrm{{den}}(a)}} q^{1/(q-1)}.\end{align*} $$

We now quote an estimate of the denominator of $((a)_k/(b)_k)_{0\le k \le n}$ for $n\in \mathbb {N}$ and $a,b\in \mathbb {Q}$ being nonnegative integers.

Lemma 7.3 [Reference Kawashima and Poëls20, Lemma $5.1$ ]

Let $n\in \mathbb {N}$ and $a,b\in \mathbb {Q}$ be nonnegative integers. Put

$$ \begin{align*}D_n=\mathrm{{den}}\bigg(\dfrac{(a)_0}{(b)_0},\ldots,\dfrac{(a)_n}{(b)_n}\bigg).\end{align*} $$

Then,

$$ \begin{align*}\limsup_{n\to \infty}\dfrac{1}{n}\log\,D_n\le \log\,\mu(a)+\dfrac{\mathrm{{den}}(b)}{\varphi(\mathrm{{den}}(b))},\end{align*} $$

where $\varphi $ is Euler’s totient function.

For a rational number a and a nonnegative integer b, we denote $\binom {a}{b}=(-1)^k(-a)_b/b!$ .

Lemma 7.4. Let $N,l,h$ be nonnegative integers with $0\le l \le u-2$ and $0\le h\le u-1$ .

$(i)$ We have

$$ \begin{align*}P_{uN,h}(z)=(-1)^{uN}\sum_{k=0}^{N(u-1)}\bigg[\sum_{s=0}^k\binom{uN-1/u}{s+N}\binom{u(s+N)+h}{uN}\binom{1/u}{k-s}\bigg](-1)^kz^{uk+h}.\end{align*} $$

$(ii)$ Put $\tilde {\varepsilon }_{l,h}= 1$ if $h<l+1$ and $0$ if $l+1\le h$ . We have

$$ \begin{align*} Q_{uN,l,h}(z)&=(-1)^{uN} \sum_{v=\tilde{\varepsilon}_{l,h}}^{N(u-1)}\bigg(\sum_{k=0}^{(u-1)N-v}(-1)^{k+v}\\ &\quad \times\bigg[\sum_{s=0}^{k+v}\binom{uN-1/u}{s+N}\binom{u(s+N)+h}{uN} \binom{1/u}{k+v-s}\bigg] \dfrac{(\tfrac{1+l}{u})_{k}}{(\tfrac{u+l}{u})_{k}}\bigg)z^{uv+h-l-1}. \end{align*} $$

$(iii)$ Put $\varepsilon _{l,h}=1$ if $l< h$ and $\varepsilon _{l,h}=0$ if $h\le l$ . We have

$$ \begin{align*}\mathfrak{R}_{uN,l,h}(z)=\dfrac{(\tfrac{u-1}{u})_{uN}}{(\tfrac{u+l}{u})_{uN}z^{uN+l-h+1}}{{\sum_{k=\varepsilon_{l,h}}^{\infty}}}\binom{u(N+k)+l-h}{uN}\dfrac{(\tfrac{1+l}{u})_{k}}{(\tfrac{u+l}{u}+uN)_{k}}\dfrac{1}{z^{uk}}. \end{align*} $$

Proof. (i) Put

$$ \begin{align*}w(z)=(1-z^u)^{-1/u}=\sum_{k=0}^{\infty}\binom{-1/u}{k}(-z^u)^{k}\in K[[z]].\end{align*} $$

Then $w(z)$ is a solution of $-(z^u-1)\partial _z-z^{u-1}\in K[z,\partial _z]$ . Lemma 4.3 yields

$$ \begin{align*} \dfrac{1}{(uN)!}\bigg(\partial_z-\dfrac{z^{u-1}}{z^u-1}\bigg)^{uN}(z^u-1)^{uN}=\dfrac{1}{(uN)!}w(z)^{-1}\partial^{uN}_zw(z)(z^u-1)^{uN} \end{align*} $$

and therefore

$$ \begin{align*} P_{uN,h}(z)&=\dfrac{1}{(uN)!}w(z)^{-1}\partial^{uN}_z w(z) (z^u-1)^{uN}\cdot z^h\\ &=\dfrac{(-1)^{uN}}{(uN)!}w(z)^{-1}\partial^{uN}_z\cdot \sum_{k=0}^{\infty}\binom{uN-1/u}{k}(-1)^{k}z^{uk+h}\\ &={(-1)^{uN}}\sum_{k=0}^{\infty}\binom{1/u}{k}(-1)^kz^{uk} \cdot \sum_{k=0}^{\infty}\binom{uN-1/u}{k+N}\binom{u(k+N)+h}{uN}(-1)^{k}z^{uk+h}\\ &=(-1)^{uN}\sum_{k=0}^{\infty}\bigg[\sum_{s=0}^k\binom{uN-1/u}{s+N}\binom{u(s+N)+h}{uN} \binom{1/u}{k-s}\bigg](-1)^kz^{uk+h}. \end{align*} $$

Since $\mathrm {{deg}}\, P_{uN,h}=u(u-1)N+h$ , using the above equality, we obtain the assertion.

(ii) Put $P_{uN,h}(z)=\sum _{k=0}^{u(u-1)N+h}p_kz^{k}$ . Notice that, by item (i),

$$ \begin{align*}p_k= \begin{cases} \displaystyle(-1)^{uN+k'}\sum_{s=0}^{k'}\binom{uN-1/u}{s+N}\binom{u(s+N)+h}{uN} \binom{1/u}{k'-s} & \text{ if there exists}\ k'\ge 0\\[-12pt]& \text{ such that } k=uk'+h,\\ 0 & \text{ otherwise}. \end{cases} \end{align*} $$

Then,

$$ \begin{align*} \dfrac{P_{uN,h}(z)-P_{uN,h}(t)}{z-t}&=\sum_{k'=1}^{u(u-1)N+h}p_{k'}\sum_{v'=0}^{k'-1}z^{v'}t^{k'-v'-1}=\sum_{k'=0}^{u(u-1)N+h-1}p_{k'+1}\sum_{v'=0}^{k'}z^{v'}t^{k'-v'}\\ &=\sum_{v'=0}^{u(u-1)N+h-1}\bigg[\sum_{k'=v'}^{u(u-1)N+h-1}p_{k'+1}t^{k'-v'}\bigg]z^{v'}\\ &=\sum_{v'=0}^{u(u-1)N+h-1}\bigg[\sum_{k'=0}^{u(u-1)N+h-v'-1}p_{k'+v'+1}t^{k'}\bigg]z^{v'}. \end{align*} $$

Since $\varphi _l(t^{k'})=0$ if $k' \not \equiv l \ \mathrm {{mod}} \ u$ , putting $k'=uk+l$ , we obtain

$$ \begin{align*} Q_{uN,l,h}(z)&=\varphi_l\bigg(\dfrac{P_{uN,h}(z)-P_{uN,h}(t)}{z-t}\bigg)\\ &=\sum_{v'=0}^{u(u-1)N+h-1}\bigg[\sum_{k'=0}^{u(u-1)N+h-v'-1}p_{k'+v'+1}\varphi_l(t^{k'})\bigg]z^{v'}\\ &=\sum_{v'=0}^{u(u-1)N+h-1}\bigg[\sum_{k=0}^{(u-1)N+\lfloor (h-v'-l-1)/u\rfloor}p_{uk+l+v'+1}\dfrac{(\tfrac{1+l}{u})_k}{(\tfrac{u+l}{u})_k}\bigg]z^{v'}. \end{align*} $$

Since we have $p_{uk+l+v'+1}=0$ for $0\le v'\le u(u-1)N+h-1$ with $v'\notin u\mathbb {Z}+h-l-1$ , putting $v'=uv+h-l-1$ , we conclude

$$ \begin{align*} Q_{uN,l,h}(z)&=\sum_{v=\tilde{\varepsilon}_{l,h}}^{(u-1)N}\bigg[\sum_{k=0}^{(u-1)N-v}p_{u(k+v)+h}\dfrac{(\tfrac{1+l}{u})_k}{(\tfrac{u+l}{u})_k}\bigg] z^{uv+h-l-1}. \end{align*} $$

This completes the proof of item (ii).

(iii) Lemma 5.1 yields

(7-3) $$ \begin{align} \mathfrak{R}_{uN,l,h}(z)=\sum_{k=uN}^{\infty}\dfrac{\varphi_l(t^kP_{uN,h}(t))}{z^{k+1}}. \end{align} $$

We now compute $\varphi _l(t^kP_{uN,h}(t))$ for $k\ge uN$ . Put $\mathcal {E}=\partial _t-t^{u-1}/(t^u-1)$ . Using Proposition 3.2(i) for $k\ge uN$ , there exists a set of integers $\{c_{uN,k,v}\mid v=0,1,\ldots ,uN\}$ with

$$ \begin{align*} &c_{uN,k,uN}=(-1)^{uN}k(k-1)\cdots (k-uN+1) \ \ \text{and} \\ &t^k \mathcal{E}^{uN}(t^u-1)^{uN}=\sum_{v=0}^{uN}c_{uN,k,v}\mathcal{E}^{uN-v}t^{k-v}(t^u-1)^{uN} \ \ \text{in} \ \ \mathbb{Q}(t)[\partial_t]. \end{align*} $$

Since $\mathcal {E}(t^u-1)\subseteq \mathrm {{ker}}\,\varphi _l$ , using the above relation,

(7-4) $$ \begin{align} \varphi_l(t^kP_{uN,h}(t))&=\varphi_l\bigg(\dfrac{t^k}{(uN)!}\mathcal{E}^{uN}(t^u-1)^{uN}\cdot t^h\bigg)=\varphi_l\bigg(\sum_{v=0}^{uN}\dfrac{c_{uN,k,v}}{(uN)!}\mathcal{E}^{uN-v}t^{k-v}(t^u-1)^{uN}\cdot t^h\bigg) \nonumber\\ &=\varphi_l\bigg(\dfrac{c_{uN,k,uN}}{(uN)!}t^{k-uN}(t^u-1)^{uN}\cdot t^h\bigg)=(-1)^{uN}\binom{k}{uN}\varphi_l(t^{k-uN+h}(t^u-1)^{uN}). \end{align} $$

Note we have $\varphi _l(t^{k-uN+h}(t^u-1)^{uN})=0$ if $k-uN+h \not \equiv l \ \mathrm {{mod}} \ u$ . Let $\tilde{k} \geq 0$ and put $k=u(\tilde {k}+N+\varepsilon _{l,h})+l-h$ . A similar computation which we performed in Equation (7-1) implies

$$ \begin{align*} \varphi_l(t^{k-uN+h}(t^u-1)^{uN})&=\varphi_l(t^{u(\tilde{k}+\varepsilon_{l,h})+l}(t^u-1)^{uN})\\ &=\dfrac{(-1)^{uN}(\tfrac{u-1}{u})_{uN} (\tfrac{1+l}{u})_{\tilde{k}+\varepsilon_{l,h}}}{(\tfrac{u+l}{u})_{uN+\tilde{k}+\varepsilon_{l,h}}}= \dfrac{(-1)^{uN}(\tfrac{u-1}{u})_{uN} (\tfrac{1+l}{u})_{\tilde{k}+\varepsilon_{l,h}}}{(\tfrac{u+l}{u})_{uN}(\tfrac{u+l}{u}+uN)_{\tilde{k}+\varepsilon_{l,h}}}. \end{align*} $$

Substituting the above equality into Equations (7-4) and (7-3), we obtain the desired equality.

In the following, for a rational number a and a nonnegative integer n, we put

$$ \begin{align*}\mu_n(a)=\mathrm{{den}}(a)^n\prod_{\substack{q:\mathrm{prime} \\ q|\mathrm{{den}}(a)}}q^{\lfloor {n}/({q-1})\rfloor} .\end{align*} $$

Notice that $\mu _n(a)=\mu _n(a+k)$ for $k\in \mathbb {Z}$ and

(7-5) $$ \begin{align} \mu_{n_2}(a) \ \ \text{is divisible by} \ \ \mu_{n_1}(a) \quad \text{and} \quad \mu_{n_1+n_2}(a) \ \ \text{is divisible by} \ \ \mu_{n_1}(a)\mu_{n_2}(a) \end{align} $$

for $n,n_1,n_2\in \mathbb {N}$ with $n_1\le n_2$ .

Lemma 7.5. Let K be an algebraic number field, v a place of K and $\alpha \in K\setminus \{0\}$ .

(i) We have

$$ \begin{align*}\max_{0\le h \le u-1}\log\,|P_{uN,h}(\alpha)|_{v}\le o(N)+u(u-1)\mathrm{{h}}_{v}(\alpha)N+ \begin{cases} u(u+1)\mathrm{{h}}_v(2)N & \ \text{if} \ v\mid \infty\\ \log\,|\mu_{uN}(1/u)|^{-1}_{v} & \ \text{if} \ v \nmid \infty. \end{cases}\end{align*} $$

(The function $o(N)$ is equal to $0$ for almost all places v. This also holds in statement (ii).)

(ii) For $0\le l \le u-2$ , put

$$ \begin{align*}D_{N}=\mathrm{{den}}\bigg(\dfrac{(\tfrac{1+l}{u})_k}{(\tfrac{u+l}{u})_k}\bigg)_{\substack{0\le l \le u-2 \\ 0 \le k \le (u-1)N}}.\end{align*} $$

Then,

$$ \begin{align*} \max_{\substack{0\le l \le u-2 \\ 0\le h \le u-1}}\log\,|Q_{uN,l,h}(\alpha)|_{v}&\le o(N)+u(u-1)\mathrm{{h}}_v(\alpha)N\\ &+ \begin{cases} u(u+1)\mathrm{{h}}_v(2)N & \ \text{if} \ v \mid \infty\\ \log\,|\mu_{uN}(1/u)|^{-1}_{v}+\log\,|D_{N}|^{-1}_v & \ \text{if} \ v \nmid \infty. \end{cases} \end{align*} $$

Proof. (i) Let v be an archimedean place. Since

$$ \begin{align*} \binom{uN-1/u}{s+N} \le 2^{uN}, \ \ \ \binom{u(s+N)+h}{uN}\le 2^{u(s+N)+h} \quad \text{and} \quad\bigg|\binom{1/u}{k-s}\bigg|\le 1 , \end{align*} $$

for $0\le k \le N(u-1)$ and $0\le s \le k$ , we obtain

(7-6) $$ \begin{align} \bigg|\sum_{s=0}^k\binom{uN-1/u}{s+N}\binom{u(s+N)+h}{uN}\binom{1/u}{k-s}\bigg|\le 2^{2uN+h}\sum_{s=0}^k2^{us}\le 2^{2uN+h+u(k+1)}. \end{align} $$

Thus, by Lemma 7.4(i),

$$ \begin{align*} |P_{uN,h}(\alpha)|_{v}\le |2^{2uN+h}|_{v}\cdot \bigg|\sum_{k=0}^{N(u-1)}2^{u(k+1)}\alpha^{uk+h}\bigg|_{v}\le e^{o(N)}|2|^{u(u+1)N}_v \max\{1,|\alpha|_v\}^{u(u-1)N}. \end{align*} $$

This completes the proof of the archimedean case.

Second, we consider the case of v is a nonarchimedean place. Note that

$$ \begin{align*}\binom{uN-1/u}{s+N}=\dfrac{(-1)^{s+N}(1/u-uN)_{s+N}}{(s+N)!} \quad \text{and} \quad \binom{1/u}{k-s}=\dfrac{(-1)^{k-s}(-1/u)_{k-s}}{(k-s)!}\end{align*} $$

for $0\le k \le N(u-1), \ 0\le s \le k$ . Combining

$$ \begin{align*}\bigg|\dfrac{(a)_k}{k!}\bigg|_v\le |\mu_n(a)|^{-1}_v \quad \text{for } a\in \mathbb{Q} \quad \text{and} \quad k,n\in \mathbb{N} \ \ \text{with} \ \ k\le n,\end{align*} $$

(see [Reference Beukers9, Lemma $2.2$ ]) and Equation (7-5) yields

$$ \begin{align*}\bigg|\binom{uN-1/u}{s+N}\binom{u(s+N)+h}{uN} \binom{1/u}{k-s}\bigg|_v\le |\mu_{k+N}(1/u)|^{-1}_v \quad \text{for } 0\le k \le (u-1)N.\end{align*} $$

Therefore, the strong triangle inequality yields

(7-7) $$ \begin{align} \max_{0\le k \le N(u-1)}\bigg|\sum_{s=0}^k\binom{uN-1/u}{s+N}\binom{u(s+N)+h}{uN} \binom{1/u}{k-s}\bigg|_v\le |\mu_{uN}(1/u)|^{-1}_v. \end{align} $$

Using Lemma 7.4(i) again, we conclude the desired inequality.

(ii) Let v be an archimedean place. We use the same notation as in the proof of Lemma 7.4(ii). Using Equation (7-6) again, we obtain

$$ \begin{align*} \bigg|\sum_{k=0}^{(u-1)N-v}p_{u(k+v)+h}\dfrac{(\tfrac{1+l}{u})_k}{(\tfrac{u+l}{u})_k}\bigg|_v&\le |2|^{2uN+h+u(v+1)}_{v}\sum_{k=0}^{N(u-1)-v}|2|^{uk}_v\\ &\le |2|^{2uN+u(u-1)N+u+h}_{v}. \end{align*} $$

Lemma 7.4(ii) implies that

$$ \begin{align*} |Q_{uN,l,h}(\alpha)|_v&\le \sum_{v=0}^{(u-1)N}\bigg|\bigg[\sum_{k=0}^{(u-1)N-v}p_{u(k+v)+h}\dfrac{(\tfrac{1+l}{u})_k}{(\tfrac{u+l}{u})_k}\bigg]\bigg|_v |\alpha|^{uv+h-l-1}_v\\ &\le e^{o(N)}|2|^{u(u+1)N}_v \max\{1,|\alpha|_v\}^{u(u-1)N}. \end{align*} $$

Let v be a nonarchimedean place. Then by the definition of $D_N$ ,

$$ \begin{align*} \max_{\substack{0 \le l \le u-2 \\ 0 \le k \le (u-1)N}}\bigg(\bigg| \dfrac{(\tfrac{1+l}{u})_k}{(\tfrac{u+l}{u})_k}\bigg|_v\bigg)\le |D_N|^{-1}_v \end{align*} $$

for all $N\in \mathbb {N}$ . Using the above inequality and Equation (7-7) for Lemma 7.4(ii), we obtain the desire inequality by the strong triangle inequality. This completes the proof of Lemma 7.5.

Lemma 7.6. Let K be an algebraic number field, $v_0$ a place of K, $\alpha \in K$ . Let $N,l,h$ be nonnegative integers with $0\le l \le u-2$ and $0\le h \le u-1$ .

(i) Assume $v_0$ is an archimedean place and $|\alpha |_{v_0}>2$ . We have

$$ \begin{align*}\max_{\substack{0\le l \le u-2 \\ 0\le h \le u-1}} \log\,|\mathfrak{R}_{uN,l,h}(\alpha)|_{v_0} \le -u(\mathrm{{h}}_{v_0}(\alpha)-\mathrm{{h}}_{v_0}(2))N+o(N) .\end{align*} $$

(ii) Assume $v_0$ is a nonarchimedean place and $|\alpha |_{v_0}>1$ . Let $p_{v_0}$ be the rational prime under $v_0$ . Put $\varepsilon _{v_0}(u)=1$ if u is coprime with $p_{v_0}$ and $\varepsilon _{v_0}(u)=0$ if u is divisible by $p_{v_0}$ . We have

$$ \begin{align*} \max_{\substack{0\le l \le u-2 \\ 0\le h \le u-1}} \log\,|\mathfrak{R}_{uN,l,h}(\alpha)|_{v_0} \le -u\bigg(\mathrm{{h}}_{v_0}(\alpha)-\dfrac{\varepsilon_{{v_0}}(u) \log\,|p_{v_0}|_{v_0}}{p_{v_0}-1}\bigg)N+o(N). \end{align*} $$

Proof. (i) For a nonnegative integer k, we have $\binom {u(N+k)+l-h}{uN}\le 2^{u(N+k)+l-h}$ . Thus,

$$ \begin{align*} \bigg|{{\sum_{k=\varepsilon_{l,h}}^{\infty}}}\binom{u(N+k)+l-h}{uN}\dfrac{(\tfrac{1+l}{u})_{k}}{(\tfrac{u+l}{u}+uN)_{k}}\dfrac{1}{\alpha^{uk}}\bigg|_{v_0} &\le|2^{uN+l-h}|_{v_0}{{\sum_{k=\varepsilon_{l,h}}^{\infty}}}\bigg|\dfrac{(\tfrac{1+l}{u})_{k}}{(\tfrac{u+l}{u}+uN)_{k}}\bigg|_{v_0}\bigg|\dfrac{2}{\alpha}\bigg|^{uk}_{v_0}\\ &\le |2^{uN+l-h}|_{v_0}{{\sum_{k=0}^{\infty}}}\bigg|\dfrac{2}{\alpha}\bigg|^{uk}_{v_0}= |2^{uN}|_{v_0}e^{o(N)}. \end{align*} $$

Using the above inequality in Lemma 7.4(ii), we obtain the assertion.

(ii) By [Reference Delaygue, Rivoal and Roques14, Proposition $4$ , Lemma $4$ ] (loc. cit. Section $(6.1)$ , $(6.2)$ ),

$$ \begin{align*} &\max_{0\le l \le u-2}\, \bigg(\bigg| \dfrac{(\tfrac{u-1}{u})_{uN}}{(\tfrac{u+l}{u})_{uN}} \bigg|_{v_0} \bigg) \le |p_{v_0}|^{\varepsilon_{{{v_0}}}(u)v_{p_{v_0}}((uN)!)+o(N)}_{v_0},\\ &\quad\bigg|{\sum_{k=\varepsilon_{l,h}}^{\infty}}\binom{u(N+k)+l-h}{uN}\dfrac{(\tfrac{1+l}{u})_{k}}{(\tfrac{u+l}{u}+uN)_{k}}\dfrac{1}{\alpha^{uk}}\bigg|_v=e^{o(1)}. \end{align*} $$

Combining $v_p((uN)!)=uN/(p-1)+o(N)$ and the above inequality in Lemma 7.4(ii), we obtain the assertion. This completes the proof of Lemma 7.6.

7.3 Proof of Theorem 7.1

We use the same notation as in Theorem 7.1. Let $\alpha \in K$ with $|\alpha |_{v_0}>1$ . For a nonnegative integer N, we define a matrix

$$ \begin{align*}\mathrm{{M}}_N= \begin{pmatrix} P_{uN,0}(\alpha) & \cdots & P_{uN,u-1}(\alpha)\\ Q_{uN,0,0} (\alpha) & \cdots & Q_{uN,0,u-1} (\alpha)\\ \vdots & \ddots & \vdots\\ Q_{uN,u-2,0} (\alpha) & \cdots & Q_{uN,u-2,u-1} (\alpha) \end{pmatrix}\in \mathrm{{M}}_u(K) .\end{align*} $$

By Lemma 7.2, the matrices $\mathrm {{M}}_N$ are invertible for every N. We define functions

$$ \begin{align*} &F_v:\mathbb{N}\longrightarrow \mathbb{R}_{\ge0}; \ N\mapsto u(u-1)\mathrm{{h}}_v(\alpha)N\\ &\qquad +o(N)+ \begin{cases} u(u+1)\mathrm{{h}}_v(2)N & \ \text{if} \ v \mid \infty\\ \log\,|\mu_{uN}(1/u)|^{-1}_{v}+\log\,|D_N|^{-1}_v & \ \text{if} \ v \nmid \infty \end{cases} \end{align*} $$

for $v\in \mathfrak {M}_K$ . By Lemma 7.3,

$$ \begin{align*} &\lim_{N\to \infty}\dfrac{1}{N}\log\,D_N\le (u-1)\bigg(\log\,\nu(u)+\dfrac{u}{\varphi(u)}\bigg), \end{align*} $$

where $D_N$ is the integer defined in Lemma 7.5,

$$ \begin{align*} &\lim_{N\to \infty}\dfrac{1}{N}\bigg(\sum_{v\neq v_0}F_v(N)\bigg)\le u\mathbb{B}_{v_0}(\alpha), \end{align*} $$

and, by Lemma 7.5,

$$ \begin{align*} &\max_{\substack{0\le h \le u-1}}\log\, \max\{|P_{uN,h}(\alpha)|_{v_0}\}\le u U_{v_0}(\alpha)N+o(N) ,\\ &\max_{\substack{0\le l \le u-2 \\ 0\le h \le u-1}}\log\, \max\{|P_{uN,h}(\alpha)|_v, |Q_{uN,l,h}(\alpha)|_{v}\}\le F_{v}(N) \quad \text{for } v\in \mathfrak{M}_K. \end{align*} $$

By Lemma 7.6,

$$ \begin{align*} \max_{\substack{0 \le l \le u-2 \\ 0\le h \le u-1}} \log\, |\mathfrak{R}_{uN,l,h}(\alpha)|_{v_0}\le -u \mathbb{A}_{v_0}(\alpha)N+o(N). \end{align*} $$

Using a quantitative linear independence criterion in [Reference David, Hirata-Kohno and Kawashima11, Proposition $5.6$ ] for

$$ \begin{align*} \theta_l=\dfrac{1}{\alpha^{l+1}} {}_2F_1\bigg(\frac{1+l}{u},1, \frac{u+l}{u}\bigg|\frac{1}{\alpha^u}\bigg) \quad \text{for } 0\le l \le u-2, \end{align*} $$

and the invertible matrices $(\mathrm {{M}}_N)_N$ , and applying the above estimates, we obtain Theorem 7.1.

A Appendix

Denote the algebraic closure of $\mathbb {Q}$ by $\overline {\mathbb {Q}}$ . Let $a(z),b(z)\in \overline {\mathbb {Q}}[z]$ with $w:=\max \{\mathrm {{deg}}\,a-2,\mathrm {{deg}}\,b-1\}\ge 0$ and $a(z)\neq 0$ . Put $D=-a(z)\partial _z+b(z)$ . The Laurent series $f_0(z),\ldots ,f_w(z)$ obtained in Lemma 4.1 for D become G-functions in the sense of Siegel when D is a G-operator (see [Reference André2, Section IV]). Here we refer below to a result due to Fischler and Rivoal [Reference Fischler and Rivoal15] in which they gave a condition so that D becomes a G-operator.