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Multivariable Taylor formula with a Lagrange remainder along a line segment

Statement

Let $k \in N$ , let $U \subseteq R^{m}$ be open and convex, $a, a + h \in U$ , and $f \in C^{k + 1} (U)$ . Write $ι : N \to R$ for the canonical-natural map of The multivariable Taylor polynomial in multi-index notation. Then some $θ \in (0, 1)$ satisfies

$f (a + h) = T_{k} f (a; h) + \sum_{∣ α ∣ = k + 1} \frac{D^{α} f (a + θ h)}{ι (α!)} h^{α} .$

Statement references

The multivariable Taylor polynomial in multi-index notation

Facts & Assumptions

Given: The hypotheses of the statement.

[L1]

Convexity keeps the segment $a + t h$ in $U$ for $0 \leq t \leq 1$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L2]

On an open interval containing $[0, 1]$ , the derivatives of $t \mapsto f (a + t h)$ have the multi-index expansion through order $k + 1$ (Repeated derivatives along a line expand by the multinomial formula).

[L3]

By The Lagrange and Cauchy forms of Taylor's remainder, if a one-variable function has derivatives through order $k + 1$ on $[0, 1]$ with the required endpoint continuity, then some $θ \in (0, 1)$ satisfies

$g (1) = \sum_{j = 0}^{k} \frac{g^{(j)} (0)}{ι (j!)} + \frac{g^{(k + 1)} (θ)}{ι ((k + 1)!)}$

by the Lagrange remainder formula.

Proof

technique · direct

1.1L1L2

Put $I : = {t \in R : a + t h \in U}$ and $g (t) : = f (a + t h)$ . The set $I$ is open, contains $[0, 1]$ by [L1], and is an interval because $U$ is convex. Hence [L2] shows that $g$ has the derivatives through order $k + 1$ required by [L3].

2.1step 1.1L3choose

Apply [L3] to $g$ between $0$ and $1$ .

3.1step 2.1L2algebra∎

Substitute the formula of [L2] for $g^{(j)} (0)$ and $g^{(k + 1)} (θ)$ ; the degree- $k$ part is the definition of $T_{k} f (a; h)$ .

Depends on

Dependency tree · 2 levels

Sources

MAT237 notes: Taylor's theorem in several variables