# Prove Taylor's Theorem by Cauchy Mean Value Theorem

When I try to prove Taylor Theorem by Cauchy mean value Theorem by instruction on my textbook, I have some problem with an equality.

Starting by letting $$R(x)=f(x)-[f(\alpha)+\frac{x-\alpha}{1!}f'(\alpha)+\dots+\frac{(x-\alpha)^{n-1}}{(n-1)!}f^{(n-1)}(\alpha)],$$ I have shown that $$R(\alpha)=R'(\alpha)=\dots=R^{(n-1)}(\alpha)=0$$ and $$R^{(n)}(x)=f^{(n)}(x)$$.

However, I fail to show that there exists $$\gamma_1$$ between $$\alpha$$ and $$\beta$$ such that $$\frac{R(\beta)}{(\beta-\alpha)}=\frac{R(\beta)-R(\alpha)}{(\beta-\alpha)^n-0^n}=\frac{R'(\gamma_1)}{n(\gamma_1-\alpha)^{n-1}}.$$

By applying $$\gamma_1 \in (\alpha,\beta)$$ or $$\gamma_1-\alpha \in (0,\beta-\alpha)$$, and a function $$g(x)=x^n$$ to Cauchy's Mean Value Theorem, I can get some similar equality, but I don't know how to move further.

Also, According to the text book, I should derive the Lagrange form of the remainder $$R(\beta)=\frac{(\beta-\alpha)^nf^{(n)}(\gamma_n)}{n!}$$ by the sequence $$\{\gamma_1,\gamma_2,...,\gamma_n\}$$.

Let $$g_n(x)=(x-\alpha)^n$$. Note that $$R(\alpha)=g_n(\alpha)=0$$. Then by the Cauchy's Mean Value Theorem applied to the functions $$R$$ and $$g_n$$ with respect to the interval $$(\alpha,\beta)$$, there is $$\gamma_1\in (\alpha,\beta)$$ such that $$\frac{R(\beta)}{(\beta-\alpha)^n}=\frac{R(\beta)-R(\alpha)}{g_n(\beta)-g_n(\alpha)}=\frac{R'(\gamma_1)}{g_n'(\gamma_1)}=\frac{R'(\gamma_1)}{n(\gamma_1-\alpha)^{n-1}}.$$ Now use the same argument for $$R'(x)$$ and $$g_{n-1}(x)=n(x-\alpha)^{n-1}$$, then we find $$\gamma_2\in (\alpha,\gamma_1)$$ such that $$\frac{R'(\gamma_1)}{n(\gamma_1-\alpha)^{n-1}}=\frac{R'(\gamma_1)-R'(\alpha)}{g_{n-1}(\gamma_1)-g_{n-1}(\alpha)}=\frac{R''(\gamma_2)}{g_{n-1}'(\gamma_2)}=\frac{R''(\gamma_2)}{n(n-1)(\gamma_2-\alpha)^{n-2}}.$$ Keep going and at the $$n$$-th step we obtain $$\gamma_n\in (\alpha,\gamma_{n-1})$$ such that $$\frac{R(\beta)}{(\beta-\alpha)^n}=\frac{R^{(n)}(\gamma_{n})}{n!} =\frac{f^{(n)}(\gamma_{n})}{n!}.$$