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 $\{\gamma_1,\gamma_2,...,\gamma_n\}$.


1 Answer 1


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 $$\frac{R(\beta)}{(\beta-\alpha)^n}=\frac{R^{(n)}(\gamma_{n})}{n!} =\frac{f^{(n)}(\gamma_{n})}{n!}.$$


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