Simultaneous approximation to values of the exponential function over the adeles

Abstract

We show that Hermite’s approximations to values of the exponential function at given algebraic numbers are nearly optimal when considered from an adelic perspective. We achieve this by taking into account the ratio of these values whenever they make sense in the various completions (Archimedean or p-adic) of a number field containing these algebraic numbers.

Appendix A. Recurrence relations

The notation being as in Sect. 2.2 we extend the definition of $f_{\mathbf {n}}(z)$, $P_{\mathbf {n}}(z)$ and $a_{\mathbf {n}}$ to any s-tuple ${\mathbf {n}}\in {\mathbb {Z}}^s$ by setting

$$\begin{aligned} f_{\mathbf {n}}(z)=P_{\mathbf {n}}(z)=0 \quad \text {and}\quad a_{\mathbf {n}}=(0,\dots ,0) \quad \text {if}\quad {\mathbf {n}}\notin {\mathbb {N}}^s. \end{aligned}$$

For each ${\mathbf {n}}\in {\mathbb {N}}_+^s$, we denote by $A_{\mathbf {n}}$ the matrix whose $\ell$th row is ${\mathbf {a}}_{{\mathbf {n}}-{\mathbf {e}}_\ell }$ for $\ell =1,\dots ,s$. In [9, Sect. IX–X], Hermite provides a recurrence formula linking $A_{{\mathbf {n}}+{\mathbf {1}}}$ to $A_{\mathbf {n}}$ where ${\mathbf {1}}=(1,\dots ,1)$. Here we give more general recurrence relations based on the same principle. The formula (A.1) below is due to Hermite [9, Sect. IX, p. 230] when ${\mathbf {n}}\in {\mathbb {N}}_+^s$.

Proposition A.1

Let ${\mathbf {n}}=(n_1,\dots ,n_s)\in {\mathbb {N}}^s$. We have

$$\begin{aligned} a_{\mathbf {n}}= (f_{\mathbf {n}}(\alpha _1),\dots ,f_{\mathbf {n}}(\alpha _s)) + \sum _{j=1}^s n_ja_{{\mathbf {n}}-{\mathbf {e}}_j}\,. \end{aligned}$$

(A.1)

Moreover, if $k,\ell \in \{1,\dots ,s\}$ with $n_k\ge 1$, we also have

$$\begin{aligned} a_{{\mathbf {n}}+{\mathbf {e}}_\ell -{\mathbf {e}}_k} = a_{\mathbf {n}}+ (\alpha _k-\alpha _\ell )a_{{\mathbf {n}}-{\mathbf {e}}_k}. \end{aligned}$$

(A.2)

Proof

Leibniz formula for the derivative of a product gives

$$\begin{aligned} f_{\mathbf {n}}'(z)=\sum _{j=1}^s n_j f_{{\mathbf {n}}-{\mathbf {e}}_j}(z). \end{aligned}$$

Taking the sum of all derivatives on both sides of this equality, we obtain

$$\begin{aligned} P_{\mathbf {n}}(z)=f_{\mathbf {n}}(z)+\sum _{j=1}^s n_j P_{{\mathbf {n}}-{\mathbf {e}}_j}(z) \end{aligned}$$

and (A.1) follows. The formula (A.2) is trivial if $k=\ell$. Suppose that $k\ne \ell$ and $n_k\ge 1$ so that ${\mathbf {n}}-{\mathbf {e}}_k\in {\mathbb {N}}^s$. Then we find

$$\begin{aligned} f_{{\mathbf {n}}+{\mathbf {e}}_\ell -{\mathbf {e}}_k}(z) - f_{\mathbf {n}}(z) = (z-\alpha _\ell )f_{{\mathbf {n}}-{\mathbf {e}}_k}(z)-(z-\alpha _k)f_{{\mathbf {n}}-{\mathbf {e}}_k}(z) = (\alpha _k-\alpha _\ell )f_{{\mathbf {n}}-{\mathbf {e}}_k}(z). \end{aligned}$$

Taking again the sum of the derivatives, this yields

$$\begin{aligned} P_{{\mathbf {n}}+{\mathbf {e}}_\ell -{\mathbf {e}}_k}(z) = P_{\mathbf {n}}(z) + (\alpha _k-\alpha _\ell )P_{{\mathbf {n}}-{\mathbf {e}}_k}(z) \end{aligned}$$

and (A.2) follows. $\square$

Corollary A.2

Let ${\mathbf {n}}=(n_1,\dots ,n_s)\in {\mathbb {N}}_+^s$ and $\ell \in \{1,\dots ,s\}$. Then we have

$$\begin{aligned} A_{{\mathbf {n}}+{\mathbf {e}}_\ell } = M_{{\mathbf {n}},\ell } A_{\mathbf {n}}\end{aligned}$$

where

$$\begin{aligned} M_{{\mathbf {n}},\ell } =\begin{pmatrix} n_1+(\alpha _1-\alpha _\ell ) &{}\quad n_2 &{}\quad \cdots &{}\quad n_s\\ n_1 &{}\quad n_2+(\alpha _2-\alpha _\ell ) &{}\quad \cdots &{}\quad n_s\\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ n_1 &{}\quad n_2 &{}\quad \cdots &{}\quad n_s+(\alpha _s-\alpha _\ell ) \end{pmatrix}. \end{aligned}$$

Proof

As the entries of ${\mathbf {n}}$ are positive, the polynomial $f_{\mathbf {n}}$ vanishes at all points $\alpha _1,\dots ,\alpha _s$ and the formulas of Proposition A.1 yield

$$\begin{aligned} a_{{\mathbf {n}}+{\mathbf {e}}_\ell -{\mathbf {e}}_k} = (\alpha _k-\alpha _\ell )a_{{\mathbf {n}}-{\mathbf {e}}_k} + \sum _{j=1}^s n_j a_{{\mathbf {n}}-{\mathbf {e}}_j} \quad (1\le k\le s). \end{aligned}$$

When $s=2$, this provides a quick way of computing the matrices $A_{n,n}$.

Corollary A.3

Suppose that $s=2$, $\alpha _1=0$ and $\alpha _2=\alpha \in K{\setminus }\{0\}$. Then, for each $n\in {\mathbb {N}}_+$, we have

$$\begin{aligned} A_{n,n} = \begin{pmatrix} P_{n-1,n}(0) &{}\quad P_{n-1,n}(\alpha )\\ P_{n,n-1}(0) &{}\quad P_{n,n-1}(\alpha ) \end{pmatrix} = (n-1)! C_nC_{n-1}\cdots C_1 \end{aligned}$$

(A.3)

where

$$\begin{aligned} C_i =\begin{pmatrix} 2i-1-\alpha &{}\quad 2i-1\\ 2i-1 &{}\quad 2i-1+\alpha \end{pmatrix} \quad (i\in {\mathbb {N}}_+). \end{aligned}$$

Proof

We find that $P_{0,1}(z)=z+1-\alpha$ and $P_{1,0}(z)=z+1$, thus $A_{1,1}=C_1$. In general, for an integer $n\ge 1$, the formulas of the preceding corollary give

$$\begin{aligned} A_{n+1,n+1} = \begin{pmatrix} n &{}\quad n+1\\ n &{}\quad n+1+\alpha \end{pmatrix} \begin{pmatrix} n-\alpha &{}\quad n\\ n &{}\quad n \end{pmatrix} A_{n,n} = n C_{n+1} A_{n,n} \end{aligned}$$

and the conclusion follows by induction on n. $\square$