User:Matt Kwan/Cauchy

Lagrange form is proved here: https://docs.google.com/viewer?url=http://www.math.hawaii.edu/~jb/math216/lagrange.pdf&pli=1

To prove: $$R_n(x)=\frac{f^{(n+1)}(\xi)}{n!}x(x-\xi)^n$$ where $$f$$ is some $$n+1$$ times differentiable function, $$p_n$$ is the $$n$$th order taylor polynomial for $$f$$, and $$\xi \in (0, x)$$

set $$x$$, set $$n$$

$$ \begin{align} &g(t)=\sum_{k=0}^n \frac{f^{(k)}(t)}{k!}(x-t)^k\\ &g(x)=f(x)\\ &g(0)=p_n(x)\\ \end{align} $$

By the mean value theorem:

$$ \begin{align} g'(\xi)&=\frac{g(x)-g(0)}{x-0}\\ &=\frac{R_n(x)}x \end{align} $$

By the product rule:

$$ \begin{align} g'(t)&=f(t)+\sum_{k=1}^n \left( \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1} \right) \\ &=\sum_{k=0}^n \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \sum_{q=0}^{n-1} \frac{f^{(q+1)}(t)}{q!}(x-t)^q \\ &=\frac{f^{(n+1)}(t)}{n!}(x-t)^n\\ \end{align} $$

$$ \begin{align} R_n(x)=\frac{f^{(n+1)}(\xi)}{n!}x(x-\xi)^n\\ \end{align} $$