$$
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Taylor composition
Taylor polynomials compose under truncation: the Taylor polynomial of a composite is the truncated composite of the Taylor polynomials.
(1) Theorem. Let \(g \in C^n(U, Y)\) and \(f \in C^n(V, Z)\) with \(g(x) \in V\). Then \(f \circ g \in C^n(U, Z)\) and
\[
T_*^n(f \circ g;\, x)
= \pi_{\leq n}(T_*^n(f;\, y) \circ T_*^n(g;\, x)),
\]
where \(\pi_{\leq n}\) truncates a polynomial to degree \(\leq n\).
Proof. The composite is \(C^n\) near \(x\) by the chain rule and induction on \(n\). Write \(g(x + v) = y + T_*^n(g; x; v) + R_g(v)\) and \(f(y + w) = z + T_*^n(f; y; w) + R_f(w)\) with Peano remainders E0024. Substituting \(w = T_*^n(g; x; v) + R_g(v)\):
\[
(f \circ g)(x + v)
= z + T_*^n(f; y; T_*^n(g; x; v) + R_g(v))
+ R_f(T_*^n(g; x; v) + R_g(v)).
\]
Since \(T_*^n(g; x; v) = O(\|v\|)\) and \(R_g(v) = o(\|v\|^n)\), expanding \(T_*^n(f; y; \cdot)\) and collecting terms of degree \(\leq n\) gives \(\pi_{\leq n}(T_*^n(f; y) \circ T_*^n(g; x))(v)\), with all remaining terms \(o(\|v\|^n)\). By uniqueness of the Taylor polynomial E0024, the result follows.
(2) Validation (AI review, 2026-07-19, claude-fable-5, pass). Expansion bookkeeping verified: cross terms \(D^\ell(f; y; P^{\ell-i}, R_g^i)\), \(i \geq 1\), are \(o(\|v\|^n)\) by boundedness; degree-\(>n\) polynomial part is \(O(\|v\|^{n+1})\); \(R_f(w) = o(\|v\|^n)\) via \(\|w\| \leq C\|v\|\); truncated composite has zero constant term as uniqueness requires. Added the (standard) chain-rule sentence for \(f \circ g \in C^n\). No issues found.
Used by: