Taylor's Theorem
1. Statement and Proof Strategy
Theorem
Suppose that \(f\) is a real-valued function of class \(C^k\) on some convex open set \(\Omega \subset {\mathbb R}^n\). Then for any \(a,b \in \Omega\), there is a point \(\xi \neq a,b\) on the line segment joining \(a\) and \(b\) such that
\[{}f(b){}\]
\[{}= \sum_{|\alpha| < k} \frac{\partial^\alpha f(a)}{\alpha!} (b-a)^\alpha{}\]
\[{}+ \sum_{|\beta| = k} \frac{\partial^\beta f(\xi)}{\beta!} (b-a)^\beta. {}\]
The hard part of the theorem follows immediately from the 1D theory applied to the function \(g(t) := f(t b + (1-t)a)\) on the interval \([0,1]\):
\[ g(1) = \sum_{m=0}^{k-1} \frac{g^{(m)}(0)}{m!} + \frac{g^{(k)}(\xi)}{k!}\]
where the one-dimensional theory applies because \(g\) will automatically be \(C^k\) on a neighborhood of \([0,1]\) because the line segment joining \(a\) to \(b\) is entirely contained in the domain \(\Omega\).
Proposition
If \(g\) is as above, then for any \(t \in [0,1]\) and any \(m \leq k\),
\[{}g^{(m)}(t){}\]
\[{}= \sum_{|\alpha| = m} \partial^\alpha f(t b + (1-t)a ){}\]
\[{}\cdot \frac{m!}{\alpha!} (b-a)^\alpha.{}\]
Proof
By induction on \(m\): The case \(m=0\) is vacuously true. Otherwise, assume the identity has been proved for \(m-1\):
\[{}g^{(m-1)}(t){}\]
\[{}= \sum_{|\alpha| = m-1} \partial^\alpha f(t b + (1-t)a ){}\]
\[{}\cdot \frac{(m-1)!}{\alpha!} (b-a)^\alpha.{}\]
Differentiate both sides with respect to \(t\) and apply the chain rule:
\[{}g^{(m)}(t){}\]
\[{}= \sum_{|\alpha| = m-1} \sum_{j=1}^n (b_j - a_j) {}\]
\[{}\cdot \partial_j \partial^\alpha f(t b + (1-t)a ){}\]
\[{}\cdot \frac{(m-1)!}{\alpha!} (b-a)^\alpha.{}\]
Letting \(e_j\) be the multiindex in \({\mathbb R}^n\) whose \(j\)-th entry is one (with all other entries being zero), this means
\[{}g^{(m)}(t){}\]
\[{}= \sum_{j=1}^n \sum_{|\alpha| = m-1} \partial^{\alpha+e_j} f(t b + (1-t)a ){}\]
\[{}\cdot \frac{(m-1)!}{\alpha!} (b-a)^{\alpha+e_j}.{}\]
\[{}= \sum_{j=1}^n \sum_{|\alpha| = m-1} \partial^{\alpha+e_j} f(t b + (1-t)a ){}\]
\[{}\cdot (\alpha_j+1) \frac{(m-1)!}{(\alpha+e_j)!} (b-a)^{\alpha+e_j}.{}\]
Now express the entire sum as a sum over multiindices \(\beta\) with \(|\beta| = m\). The question becomes: in what ways may the multiindex \(\beta\) be expressed as \(\alpha + e_j\)? The answer is the following. For any index \(j\) such that \(\beta_j \neq 0\), one may write \(\beta = (\beta-e_j) + e_j\) and \((\beta-e_j)\) is a valid multiindex (because its entries are not zero). Every pair \((\alpha,j)\) occurs as a solution to one of these equations (namely, \(\beta = \alpha + e_j\) has \(\beta - e_j = \alpha)\) and no pair \((\alpha,j)\) solves the equation for more than one \(\beta\). Thus
\[{}g^{(m)}(t){}\]
\[{}= \sum_{|\beta|=m}\sum_{\substack{j=1 \\ \beta_j \neq 0}}^n \partial^{\beta} f(t b + (1-t)a ) \beta_j{}\]
\[{}\cdot \frac{(m-1)!}{\beta!} (b-a)^{\beta}.{}\]
Including terms on the right-hand side for which \(\beta_j = 0\) does not change the right-hand side because of the factor of \(\beta_j\) that each term is multiplied by. Upon doing this, the sum \(\sum_j \beta_j\) simpy equals \(m\), and this completes the induction argument.
2. Exercise
Exercise
Suppose that \(f\) instead takes values in \({\mathbb R}^m\) but that all the other hypotheses of Taylor's Theorem above continue to hold. If \(\varphi\) is a convex function on \({\mathbb R}^m\), show that there is some \(\xi\) on the line segment joining \(a\) to \(b\) such that
\[{}\varphi \left( f(b) - \sum_{|\alpha| < k} \frac{\partial^\alpha f(a)}{\alpha!} (b-a)^\alpha \right){}\]
\[{}\leq \varphi \left( \sum_{|\beta| = k} \frac{\partial^\beta f(\xi)}{\beta!} (b-a)^\beta \right). {}\]
Hint
Reduce to the real-valued case by following the same procedure that was used for the Mean Value Theorem.