E0024 proposition ✓ ai notation E0020 → E0001

Taylor formula

(1) Proposition (Taylor approximation). Let \(k \geq 0\) and \(g \in C^k(U, Y)\) on an open \(U \subset X\), \(x \in U\). Write \(T_k = T_*^k(g; x)\) for the reduced Taylor polynomial and \(R_k = \Delta(g; x) - T_k\) for the Taylor residual.

  • Peano residual.

    \[ g(x + v) = g(x) + T_k(v) + R_k(v), \qquad \frac{\|R_k(v)\|}{\|v\|^k} \to 0 \text{ as } v \to 0. \]

  • Lagrange residual (for scalar-valued maps in coordinates). If \(X = \IR^d\), \(Y = \IR\), and \(g \in C^{k+1}\) near \(x\), then for each sufficiently small nonzero \(v\) there exists \(\tau \in (0,1)\) with

    \[ R_k(v) = \sum_{|\alpha| = k+1} \frac{v^{\alpha}}{\alpha!}\, (\partial^{\alpha} g)(x + \tau v). \]

  • Integral residual. If \(g \in C^{k+1}\) near \(x\), then for every sufficiently small \(v\),

    \[ R_k(v) = \frac{1}{k!} \int_0^1 (1-t)^k D^{k+1}(g; x+tv; v, \dots, v)\,dt. \]

    In coordinates, \(X = \IR^d\), this becomes

    \[ R_k(v) = (k+1) \sum_{|\alpha| = k+1} \frac{v^{\alpha}}{\alpha!} \int_0^1 (\partial^{\alpha} g)(x + tv)\, (1-t)^{k}\, dt = \sum_{|\alpha| = k+1} a_\alpha(v) \cdot v^\alpha, \]

    where \(a_\alpha(v) = (k+1)/\alpha!\, \int_0^1 (\partial^\alpha g)(x + tv)\, (1-t)^k\, dt\). The functions \(a_\alpha\) are continuous at \(0\).

  • Multivariate form. For directions \(v_1, \dots, v_s \in X\):

    \[ g(x + {\textstyle\sum_i} t_i v_i) = g(x) + \sum_{1 \leq |\alpha| \leq k} \frac{t^\alpha}{\alpha!}\, D^\alpha(g;\, x;\, v_\bullet) + o(|t|^k). \]

  • Uniqueness. If \(\Delta(g; x) = T + R\) with \(T \in \KP_{\leq k}(X, Y)\), \(T(0) = 0\), and \(\|R(v)\| / \|v\|^k \to 0\), then \(T = T_k\).

Proof. For fixed \(v\), set \(\phi_v(t) := g(x+tv)\). By the chain rule, \(\phi_v^{(j)}(t) = D^j(g; x+tv; v, \dots, v)\). Thus \(\phi_v^{(j)}(0)/j!\) is the value at \(v\) of the degree-\(j\) homogeneous part of \(T_k\). This restriction to affine lines gives the Peano, Lagrange, and integral formulas.

Ad Peano) For \(k = 0\), the claim is continuity of \(g\) at \(x\). Suppose \(k \geq 1\). Iterating the Banach-valued fundamental theorem of calculus along the segment from \(x\) to \(x+v\) gives the integral remainder at order \(k-1\). After subtracting the degree-\(k\) Taylor term, we obtain

\[ R_k(v) = \frac{1}{(k-1)!} \int_0^1 (1-t)^{k-1} \bigl(D^k(g; x+tv)-D^k(g; x)\bigr) [v^{\tensor k}]\,dt. \]

Consequently,

\[ \frac{\|R_k(v)\|}{\|v\|^k} \leq \frac{1}{k!} \sup_{0 \leq t \leq 1} \|D^k(g; x+tv)-D^k(g; x)\|. \]

The supremum tends to zero as \(v \to 0\) by continuity of \(D^k g\) at \(x\). This proves the Peano estimate for arbitrary Banach-valued maps; no mean-value theorem is required.

Ad Lagrange) When \(Y = \IR\), the auxiliary function \(\phi\) is scalar-valued. The one-dimensional Lagrange remainder gives \(\tau \in (0,1)\) such that \(R_k(v) = \phi^{(k+1)}(\tau)/(k+1)!\). For \(X = \IR^d\), expanding \(\phi^{(k+1)}(\tau)\) by the multinomial theorem gives the stated coordinate formula.

Ad Integral) The Banach-valued fundamental theorem of calculus, iterated \(k+1\) times along the segment, gives

\[ R_k(v) = \frac{1}{k!} \int_0^1 (1-t)^k D^{k+1}(g; x+tv; v, \dots, v)\,dt. \]

In coordinates, expand the differential by the multinomial theorem. The coefficient of \(v^\alpha\) is \((k+1)/\alpha!\) times the corresponding integral, giving the factored form. Continuity of \(a_\alpha\) at \(0\) follows from continuity of \(\partial^\alpha g\) at \(x\), uniformly along the shrinking segment.

Ad Multivariate) Set \(v = \sum_i t_i v_i\). The Taylor polynomial \(T_k(v)\) is a polynomial of degree \(\leq k\) in \(v\), hence a polynomial of degree \(\leq k\) in the \(t_i\). Expanding by the multinomial theorem: the monomial \(t^\alpha\) with \(|\alpha| = j\) collects the term \(D^j(g; x; v_\bullet^{\times \alpha}) / j!\). Since \(D^j(g; x; \cdot)\) is symmetric \(j\)-linear, the multinomial coefficient \(j! / \alpha!\) arises from symmetrization, giving coefficient \(D^\alpha(g; x; v_\bullet) / \alpha!\). The Peano remainder satisfies \(\|R_k(v)\| / \|v\|^k \to 0\), and \(\|v\| \leq C |t|\), so \(R_k = o(|t|^k)\).

Ad Uniqueness) Suppose \(\Delta(g; x) = T + R = T_k + R_k\). Then \(p := T-T_k = R_k-R\) is a polynomial of degree at most \(k\) satisfying \(p(v) = o(\|v\|^k)\). For \(k=0\), the condition \(p(0)=0\) immediately gives \(p=0\). If \(k \geq 1\) and \(p \neq 0\), write \(p = \sum_{\ell=j}^k p_\ell\), where \(p_j \neq 0\) is its lowest nonzero homogeneous part, and choose \(v\) with \(p_j(v) \neq 0\). On the one hand, \(t^{-j}p(tv) \to p_j(v)\). On the other hand, \(p(tv) = o(|t|^k)\) gives \(t^{-j}p(tv) = o(|t|^{k-j}) \to 0\), also when \(j=k\). This contradiction proves \(p=0\) and hence \(T=T_k\).

(2) Validation (AI review, 2026-07-19, gpt-5-codex, pass). Peano, Lagrange, integral, and multivariate remainders checked for Banach targets; the case \(k=0\) and the scalar-target restriction were verified. Corrected the vector-valued Lagrange step and the uniqueness argument. No issues remain.

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