LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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A proper rational function with split denominator has a unique repeated-pole partial-fraction expansion

Statement

Let $E$ be a field, let $λ_{1}, \dots, λ_{s} \in E$ be pairwise distinct and nonzero, let $m_{i} \geq 1$ , and put

$Q (x) = \prod_{i = 1}^{s} (1 - λ_{i} x)^{m_{i}} .$

For every $P \in E [x]$ with $P = 0$ or $\deg P < \deg Q$ , there are unique scalars $b_{i j} \in E$ such that

$\frac{P (x)}{Q (x)} = \sum_{i = 1}^{s} \sum_{j = 1}^{m_{i}} \frac{b_{i j}}{(1 - λ_{i} x)^{j}} .$

For $P = 0$ , every $b_{i j}$ is zero.

Facts & Assumptions

Given: A field $E$ , distinct nonzero $λ_{i}$ , positive multiplicities $m_{i}$ , and a proper numerator $P$ for $Q = \prod_{i} (1 - λ_{i} x)^{m_{i}}$ .

[L1]

A split recurrence denominator has factors $(1 - λ_{i} x)^{m_{i}}$ corresponding to the characteristic factors $(t - λ_{i})^{m_{i}}$ (Reciprocal-root convention: $χ (t) = \prod_{i} (t - λ_{i})^{m_{i}}$ corresponds to $Q (x) = \prod_{i} (1 - λ_{i} x)^{m_{i}}$ ).

[L2]

Coprime polynomials $f, g$ over a field admit $A, B$ with $A f + B g = 1$ (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L3]

Splitting permits a factorisation into linear factors with repetitions recording multiplicity (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

Proof

technique · direct

1.1givenL2L3

The powers $q_{i} = (1 - λ_{i} x)^{m_{i}}$ are pairwise coprime: distinct linear factors have no common root, and [L2] then gives a Bezout identity for every pair.

2.1step 1.1L2algebra

Iterating the two-factor Bezout decomposition gives polynomials $U_{i}$ with $\deg U_{i} < m_{i}$ such that $P / Q = \sum_{i} U_{i} / q_{i}$ ; at each stage the remainder modulo $q_{i}$ supplies $U_{i}$ .

3.1step 2.1algebra

Each $U_{i} / q_{i}$ has a unique expansion $\sum_{j = 1}^{m_{i}} b_{i j} (1 - λ_{i} x)^{- j}$ , because the polynomials $1, (1 - λ_{i} x), \dots, (1 - λ_{i} x)^{m_{i} - 1}$ form a basis of the polynomials of degree below $m_{i}$ .

4.1step 1.1step 3.1L2

If the displayed partial-fraction sum is zero, clear denominators and reduce modulo $q_{i}$ ; all terms except the $i$ th vanish, so $U_{i} \prod_{k \neq i} q_{k} \equiv 0 (m o d q_{i})$ . The second factor is invertible modulo $q_{i}$ by [L2], hence $U_{i} = 0$ for every $i$ , and step 3.1 gives every $b_{i j} = 0$ .

5.1step 2.1step 3.1step 4.1L1∎

Steps 2.1 and 3.1 give existence, while step 4.1 gives uniqueness. The zero numerator yields the all-zero coefficients.

Depends on

Used by

Over a named splitting field in characteristic zero, repeated characteristic roots give polynomial-times-exponential closed forms Theorem

Dependency tree · two levels

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Sources

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Theorem 4.1.1 (standard reference, not scraped)
B. E. Sagan, Combinatorics: The Art of Counting, Theorem 3.7.1 (standard reference, not scraped)