Approximaths

Month: August 2025

  • Padé approximants: Convergence II

    In the previous post we gave a modern definition of Montessus’s theorem. Here is the original formulation from R. De Montessus (1902):

    “Il ressort de ces considérations qu’étant donnée une série de Taylor représentant une fonction f(x) dont les p pôles les plus rapprochés de l’origine sont intérieurs à un cercle (C) lui-même intérieur aux pôles suivants, chaque pôle multiple étant compté pour autant de pôles simples qu’il existe d’unités dans son degré de multiplicité, la fraction continue déduite de la ligne horizontale de rang p du Tableau de M. Padé, ce tableau étant composé de réduites normales, représente la fonction f(x) dans un cercle de rayon \displaystyle \lvert \alpha_{p+1} \lvert, où \alpha_{p+1} est l’affixe du pôle le plus rapproché de l’origine parmi tous ceux qui sont extérieurs au cercle (C). Si tous les pôles ont des modules différents, les fractions continues correspondant aux lignes horizontales représentent toute la fonction ; s’il existe simplement des discontinuités dans l’ensemble linéaire des modules des pôles, les fractions continues correspondant à des lignes horizontales convenablement choisies représentent encore la fonction. Si tous les pôles sont simples, la représentation a lieu dans des cercles d’autant plus grands que la ligne horizontale choisie est plus éloignée dans le Tableau. S’il y a des pôles multiples, il y a stationnement, en ce sens que plusieurs lignes horizontales consécutives représentant la fonction ont le même rayon de convergence. S’il y a enfin un point singulier essentiel, le stationnement se prolonge indéfiniment, aucune des fractions continues considérées ne représente la fonction en dehors du cercle sur la circonférence duquel se trouve le point singulier essentiel le plus rapproché de l’origine.”

    References:

    • R. De Montessus, “Sur les fractions continues algébriques”, Bulletin de la S. M. F., tome 30 (1902), p. 28-36.
    • E. B. Saff, “An extension of Montessus de Ballore’s theorem on the convergence of interpolating rational functions”, Journal of Approximation Theory, vol 6, No. 1, July 1972.

  • Padé approximants: Convergence I

    For row sequences on the Padé table, Montessus’s theorem (1902) proves convergence for functions meromorphic on a disk. Before giving the statement of the theorem, we would like to remind the reader of a few definitions:

    Holomorphic function: A holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighborhood of each point in a given domain.

    Analytic function: An analytic function is a function that is locally given by a convergent power series.

    Meromorphic function: A meromorphic function on an open subset D of the complex \mathbb{C}-plane is a function that is holomorphic on all of D except for a set of isolated points. These points are called the ‘poles’ of the function.

    Here is the Montessus’s theorem as stated by E. B. Saff in 1972:

    Let f(z) be analytic at z = 0 and meromorphic with precisely \nu poles (multiplicity counted) in the disk |z| < \tau. Let D be the domain obtained from |z| < \tau by deleting the \nu poles of f(z). Then, for all m sufficiently large, there exists a unique rational function R_{m, \nu} of type (m, \nu), which interpolates f(z) in the point z = 0 considered of multiplicity m+\nu+1. Each R_{m, \nu} has precisely \nu finite poles and, as m \to \infty, these poles approach the \nu poles of f(z) in |z| < \tau. The sequence R_{m, \nu} converges in D to f(z), uniformly on any compact subset of D.

    Alternative formulation:

    Let f(z) be meromorphic in |z| < \tau, analytic at z = 0, and with a total of \nu poles \zeta_1, \zeta_2, \dots, \zeta_{\nu} (with multiplicity included) in |z| < \tau. Then, as m \to \infty, the Padé approximants P(m,\nu) of f converge on:

    \displaystyle S_f := \{ z \in \mathbb{C} \mid |z| < \tau \} \setminus \{ \zeta_1, \zeta_2, \dots, \zeta_\nu \}

    to f, uniformly on every compact subset K of S_f. In particular:

    \displaystyle P(m,\nu)(z) \to f(z)

    The Padé table is represented as follows:

    n=0 n=1 n=2 \dots n=\nu \dots
    m=0 P(0,0) P(0,1) P(0,2) \dots \boxed{P(0,\nu)} \dots
    m=1 P(1,0) P(1,1) P(1,2) \dots \boxed{P(1,\nu)} \dots
    m=2 P(2,0) P(2,1) P(2,2) \dots \boxed{P(2,\nu)} \dots
    \dots \dots \dots \dots \dots
    m \to \infty \dots \dots \dots \boxed{P(m,\nu)} \dots

    The Montessus’s theorem is crucial in approximation theory as it ensures the uniform convergence of Padé approximants for meromorphic functions, enhancing the accuracy of rational approximations.

  • Two properties of Padé Approximants

    Property 1

    Let g(x) = \frac{1}{f(x)}, with f(0) \neq 0, and assume f is at least C^{m+n} at x = 0. If P(m,n)_f = \frac{P(x)}{Q(x)}, then the [n/m] Padé approximant of g(x):

    \displaystyle P(n,m)_g = \frac{Q(x)}{P(x)},

    provided P(0) \neq 0.

    Proof: Given P(m,n)_f = \frac{P(x)}{Q(x)}, we have:

    \displaystyle f(x) Q(x) - P(x) = \epsilon(x) x^{m+n+1}.

    Since g(x) = \frac{1}{f(x)}, consider:

    \displaystyle g(x) P(x) - Q(x) = \frac{1}{f(x)} P(x) - Q(x) = \frac{P(x) - f(x) Q(x)}{f(x)} = -\frac{\epsilon(x) x^{m+n+1}}{f(x)}.

    Since f(0) \neq 0, \frac{1}{f(x)} is bounded near x = 0, and:

    \displaystyle g(x) P(x) - Q(x) = O(x^{m+n+1}),

    indicating that \frac{Q(x)}{P(x)} is the [n/m] Padé approximant of g(x), as it matches the Taylor series of g(x) up to x^{m+n}. For example, for f(x) = \sqrt{1+x}, the [2/2] Padé approximant can be computed, and g(x) = \frac{1}{\sqrt{1+x}} yields a consistent [2/2] approximant by taking the reciprocal.

    Property 2

    If f is even (f(x) = f(-x)) and at least C^{m+n}, and the [m/n] Padé approximant exists and is unique (guaranteed if the Hankel determinant is non-zero), then P(m,n)_f = \frac{P(x)}{Q(x)} is even, i.e., P(x) = P(-x) and Q(x) = Q(-x).

    Proof: Since f(x) = f(-x), the Taylor series of f contains only even powers. For P(m,n)_f = \frac{P(x)}{Q(x)}, we have:

    \displaystyle f(x) Q(x) - P(x) = O(x^{m+n+1}).

    Evaluate at -x:

    \displaystyle f(-x) Q(-x) - P(-x) = f(x) Q(-x) - P(-x) = O(x^{m+n+1}),

    since f(-x) = f(x), and the error term remains of order x^{m+n+1}. Thus, \frac{P(-x)}{Q(-x)} satisfies the same Padé condition as \frac{P(x)}{Q(x)}. By uniqueness of the [m/n] approximant (assuming non-zero Hankel determinant), we conclude:

    \displaystyle P(x) = P(-x), \quad Q(x) = Q(-x).

  • Padé approximant of 1/(1-x)

    In this post we will have a look at the Padé approximants of \frac{1}{1-x}. The Maclaurin series of \frac{1}{1-x} is:

    \displaystyle 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots

    which converges for x \in ]-1,1[. We can calculate the corresponding P(1,1)

    \displaystyle \frac{A_0 + A_1 x}{1 + B_1 x} = 1 + x + x^2 + \cdots
    \displaystyle = (1 + B_1 x)(1 + x + x^2 + \cdots)
    \displaystyle = 1 + x + x^2 + \cdots + B_1 x + B_1 x^2 + B_1 x^3 + \cdots
    \displaystyle = 1 + (1 + B_1) x + (1 + B_1) x^2 + (1 + B_1) x^3 + \cdots

    Keeping only degrees up to 2:

    \displaystyle = 1 + (1 + B_1) x + (1 + B_1) x^2

    This implies:

    \displaystyle A_0 = 1
    \displaystyle A_1 = 1 + B_1
    \displaystyle 1 + B_1 = 0

    Therefore, A_0 = 1, A_1 = 0 and B_1 = -1 and the P(1,1) is:

    \displaystyle \boxed{P(1,1)(x) = \frac{1}{1-x}}

    This is an exceptional result since the Padé approximant P(1,1) is equal to the function it is supposed to approximate. This result is very attractive since it suggests a way to solve very hard problems using series up to some terms. Then using Padé approximation we may hope to recover the exact solution (or at least a sufficient approximation for a specific application).

    Let’s imagine that we would like to solve the following differential equation:

    \displaystyle y' = y^2 \text{ with initial condition } y(0) = 1

    This differential equation can be solved exactly since it is separable. The solution is y(x) = \frac{1}{1-x}. Let’s pretend for a moment that solving this equation is a very difficult problem because we don’t know how to solve separable differential equations.

    A possible approach is to consider a solution of the form:

    \displaystyle y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots

    Then we have:

    \displaystyle y(x)' = a_1 + 2 a_2 x + 3 a_3 x^2 + \cdots
    \displaystyle y(x)^2 = (a_0 + a_1 x + a_2 x^2 + \cdots)^2
    \displaystyle = a_0^2 + a_0 a_1 x + a_0 a_2 x^2 + \cdots
    \displaystyle + a_0 a_1 x + a_1^2 x^2 + \cdots
    \displaystyle + a_2 a_0 x^2 + \cdots
    \displaystyle = a_0^2 + 2 a_0 a_1 x + (a_1^2 + 2 a_0 a_2) x^2 + \cdots

    We can write the differential equation keeping only terms up to two:

    \displaystyle y(x)' = y(x)^2
    \displaystyle a_1 + 2 a_2 x + 3 a_3 x^2 = a_0^2 + 2 a_0 a_1 x + (a_1^2 + 2 a_0 a_2) x^2

    we obtain:

    \displaystyle a_1 = a_0^2
    \displaystyle 2 a_2 = 2 a_0 a_1
    \displaystyle 3 a_3 = a_1^2 + 2 a_0 a_2

    Setting a_0 = 1 (since y(0) = 1) implies a_1 = a_2 = a_3 = 1. An approximation of the solution to the differential equation is therefore:

    \displaystyle 1 + x + x^2 + x^3

    As presented above, the corresponding P(1,1) = \frac{1}{1-x} which is the exact solution to the differential equation. In fact, the diagonal sequence of Padé approximants (P(1,1), P(2,2), … P(n,n)) recovers \frac{1}{1-x}. This is, of course, a special case.

    So, for this differential equation, we have the following pattern: