False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16

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FALSE: A split characteristic polynomial always gives a linear combination of pure exponentials

Statement

False claim. If the characteristic polynomial of a linear recurrence splits, then every solution is a linear combination of pure exponentials $λ^{n}$ , with no polynomial factors in $n$ .

Facts & Assumptions

Given: The sequence $a_{n} = n$ over $Q$ .

[L1]

A root of multiplicity $m$ contributes a polynomial in $n$ of degree below $m$ times the corresponding exponential (Over a named splitting field in characteristic zero, repeated characteristic roots give polynomial-times-exponential closed forms).

[L2]

The double pole $(1 - λ x)^{- 2}$ has coefficients $(n + 1) λ^{n}$ (Repeated poles expand formally as $(1 - λ x)^{- j} = \sum_{n \geq 0} (\binom{n + j - 1}{j - 1}) λ^{n} x^{n}$ ).

Refutation

technique · repeated-root counterexample

1.1givenalgebra

Direct calculation gives $a_{n + 2} - 2 a_{n + 1} + a_{n} = 0$ , so the characteristic polynomial is $(t - 1)^{2}$ , which splits over $Q$ .

2.1step 1.1algebra

A linear combination of pure exponentials supplied only by the characteristic root $1$ is constant, whereas $a_{n} = n$ is not. Thus no such pure-exponential expression exists.

3.1step 1.1L1L2∎

The required degree-one factor is exactly the repeated-root term permitted by [L1]; equivalently, [L2] with $λ = 1$ gives coefficients $n + 1$ , whose one-step shift yields $n$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

B. E. Sagan, Combinatorics: The Art of Counting, Section 3.7 (standard reference, not scraped)