Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is from Woll's "Functions of Several Variables," but there's no proof.

If $g$ is of class $C^k$ ($k \ge 2$) on a convex open set $U$ about $p$ in $\mathbb{R}^d$, then for each $q \in U$,

$ g(q) = g(p) + \sum_{i=1}^d \frac{\partial g}{\partial r_i} \bigg|_p (r_i(q) - r_i(p)) + \sum_{i,j} (r_i(q) - r_i(p)) (r_j(q) - r_j(p)) \int_0^1 (1-t) \frac{\partial^2g}{\partial r_i \partial r_j} \bigg|_{p + t(q - p)} dt. $

It looks like Taylor's or mean value theorem. I especially don't understand the integral part.

share|cite|improve this question
up vote 2 down vote accepted

Consider function $$ h(t)=g(p+t(q-p)) $$ and its Taylor series with the remainder in the integral form $$ h(1)=h(0)+h'(0)+\int\limits_{0}^{1}(1-t)h''(t)dt $$ Now note that $$ h(0)=g(p) $$ $$ h'(0)=\sum\limits_{i=1}^d\frac{\partial g}{\partial r_i}\biggl|_p(r_i(p)-r_i(q)) $$ $$ h''(t)=\sum\limits_{i=1}^d\sum\limits_{j=1}^d\frac{\partial^2 g}{\partial r_i\partial r_j}\biggl|_{p+t(q-p)}(r_i(p)-r_i(q))(r_j(p)-r_j(q)) $$

share|cite|improve this answer
Thanks! $ $ $ $ – Steven Li Jun 17 '12 at 4:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.