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Multivariable Taylor formula with a Lagrange remainder along a line segment

Statement

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

f(a+h)=Tkf(a;h)+α=k+1Dαf(a+θh)ι(α!)hα.f(a+h)=T_kf(a;h)+\sum_{|\alpha|=k+1}\frac{D^\alpha f(a+\theta h)}{\iota(\alpha!)}h^\alpha.

Facts & Assumptions

Given: The hypotheses of the statement.

[L1]

Convexity keeps the segment a+tha+th in UU for 0t10\le t\le1 (A convex subset of Rm\mathbb{R}^m contains every line segment between two of its points).

[L2]

On an open interval containing [0,1][0,1], the derivatives of tf(a+th)t\mapsto f(a+th) have the multi-index expansion through order k+1k+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+1k+1 on [0,1][0,1] with the required endpoint continuity, then some θ(0,1)\theta\in(0,1) satisfies

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

by the Lagrange remainder formula.

Proof

technique · direct
1.1

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

L1L2
2.1

Apply [L3] to gg between 00 and 11.

step 1.1L3choose
3.1

Substitute the formula of [L2] for g(j)(0)g^{(j)}(0) and g(k+1)(θ)g^{(k+1)}(\theta); the degree-kk part is the definition of Tkf(a;h)T_kf(a;h).

step 2.1L2algebra

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