Taylor's Theorem with Lagrange Remainder

Taylor's Theorem generalizes the Mean Value Theorem to higher-order approximations. The Lagrange form of the remainder is proved using Rolle's theorem applied to an auxiliary function. This result is essential for error analysis in numerical methods and understanding convergence of Taylor series.

Statement

Theorem5.1Taylor's Theorem

Let $f : [a, b] \to \mathbb{R}$ be $(n+1)$ times differentiable. For $x_0, x \in [a, b]$ , there exists $c$ between $x_0$ and $x$ such that

$f(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k + \frac{f^{(n+1)}(c)}{(n+1)!}(x - x_0)^{n+1}.$

Proof Sketch

Proof

Define $T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k$ and $R_n(x) = f(x) - T_n(x)$ . We want to show $R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - x_0)^{n+1}$ for some $c$ between $x_0$ and $x$ .

Define the auxiliary function

$g(t) = f(x) - \sum_{k=0}^n \frac{f^{(k)}(t)}{k!}(x - t)^k - K(x - t)^{n+1},$

where $K$ is chosen so that $g(x_0) = 0$ . Then $g(x) = 0$ as well (since the sum telescopes to $f(x)$ when $t = x$ ).

By Rolle's theorem applied repeatedly, there exists $c$ between $x_0$ and $x$ such that $g^{(n+1)}(c) = 0$ . Computing $g^{(n+1)}(c)$ yields

$g^{(n+1)}(c) = -f^{(n+1)}(c) + K(n+1)! = 0,$

so $K = \frac{f^{(n+1)}(c)}{(n+1)!}$ . Substituting back into $g(x_0) = 0$ gives the desired formula.

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Summary

Taylor's Theorem extends the MVT to polynomial approximations of arbitrary order. The Lagrange remainder provides error estimates essential for numerical analysis. See Taylor's Theorem concept for applications.