---
tag: 75XCD68557
kind: definition
name: Padé approximant
msc: ["41A21", "30E10", "30B70"]
defines: 1F16G4V64X
uses: [072W67KFYK, 0WCZH4AFMT, 0DPF6WRZW6, 0HFSFVMWQB]
sources:
  - url: https://en.wikipedia.org/wiki/Pad%C3%A9_approximant
---

## Definition

Let $f(z) = \sum_{k \ge 0} c_k z^k$ be a [[formal power series|072W67KFYK]] with complex coefficients, and let $m, n \ge 0$ be integers.
Consider pairs $(P, Q)$ of polynomials with complex coefficients such that:

1. $P$ has [[degree|0DPF6WRZW6]] at most $m$;
2. $Q$ is not the zero polynomial and has degree at most $n$;
3. the series $Q(z) f(z) - P(z)$ has no terms of degree at most $m + n$, that is, $Q f - P = O(z^{m+n+1})$.

Such pairs exist, and the [[rational function|0WCZH4AFMT]] $P/Q$ is the same for all of them.
This rational function is the *$[m/n]$ Padé approximant* of $f$, written $[m/n]_f$.

## Remarks

Existence: condition 3 for the coefficients of $z^{m+1}, \dots, z^{m+n}$ is a system of $n$ homogeneous linear equations in the $n + 1$ coefficients of $Q$.
So a nonzero $Q$ exists, and then $P$ is the truncation of $Q f$ at degree $m$.
Uniqueness: for two pairs, $P Q' - P' Q = Q'(Q f - P) - Q(Q' f - P')$ is a polynomial of degree at most $m + n$ that is $O(z^{m+n+1})$, so it is $0$.
Example: for $f = e^z$, $m = n = 1$, take $Q = 1 - z/2$ and $P = 1 + z/2$.
Then $Q f = 1 + z/2 + 0 \cdot z^2 - z^3/12 + \cdots$, so $Q f - P = O(z^3)$ and $[1/1]_{e^z} = (2 + z)/(2 - z)$.
Non-example: $P = 1 + z$ and $Q = 1$ satisfy clauses 1 and 2 for $m = n = 1$, but $Q f - P = z^2/2 + \cdots$ fails clause 3.
For $n = 0$ the approximant $[m/0]_f$ is the [[Taylor polynomial|0HFSFVMWQB]] of degree $m$.
When $Q(0) \ne 0$ one usually normalizes $Q(0) = 1$, and then $f - P/Q = O(z^{m+n+1})$; when $Q(0) = 0$ the approximant may match fewer coefficients.