LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-01

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Higher-order Rolle theorem

Statement

Let $n \in N$ with $n \geq 1$ , let $x_{0} < \dots < x_{n}$ , and let $f$ be continuous on $[x_{0}, x_{n}]$ and $n$ -times differentiable on $(x_{0}, x_{n})$ . If $f (x_{j}) = 0$ for every $j \leq n$ , then some $c \in (x_{0}, x_{n})$ satisfies $f^{(n)} (c) = 0$ .

Facts & Assumptions

Given: The ordered zeros and the stated regularity.

[L1]

Rolle's theorem produces a zero of $h^{'}$ between two zeros of a continuous, interior-differentiable function $h$ (Rolle's theorem: if $a < b$ , $f$ is continuous on $[a, b]$ , differentiable at every point of $(a, b)$ , and $f (a) = f (b)$ , then $f^{'} (c) = 0$ for some $c \in (a, b)$ ).

[L2]

Differentiability at a point implies continuity there (A function differentiable at $c$ is continuous at $c$ ), and induction applies to natural numbers (The principle of mathematical induction).

Proof

technique · induction

1.1

For $n = 1$ , Rolle's theorem on $[x_{0}, x_{1}]$ gives $c \in (x_{0}, x_{1})$ with $f^{'} (c) = 0$ .

baseL1

1.2

For $n \geq 2$ , apply Rolle on each $[x_{j - 1}, x_{j}]$ to obtain $y_{j} \in (x_{j - 1}, x_{j})$ with $f^{'} (y_{j}) = 0$ , so $y_{1} < \dots < y_{n}$ .

givenL1choose

2.1

The function $f^{'}$ is continuous on $[y_{1}, y_{n}]$ , because those points lie in $(x_{0}, x_{n})$ and the existence of $f^{''}$ gives continuity there; it is $(n - 1)$ -times differentiable on $(y_{1}, y_{n})$ . Apply the induction hypothesis of order $n - 1$ to $f^{'}$ and the $n$ ordered zeros $y_{1}, \dots, y_{n}$ . This gives $c \in (y_{1}, y_{n}) \subset (x_{0}, x_{n})$ with $(f^{'})^{(n - 1)} (c) = f^{(n)} (c) = 0$ .

step 1.2L2ih

3.1

The claim follows for every $n \geq 1$ .

step 1.1step 2.1L2discharge-induction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

J. Lebl, Basic Analysis I, Taylor's theorem and related calculus (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 lecture notes (standard reference, not scraped)
University of Florida note, Generalized Rolle's theorem (standard reference, not scraped)