Taylor's Theorem

Taylor's Theorem Let $k\ge 1$ be an integer and let the function $f:\mathbf{R}\to \mathbf{R}$ be $k$ times differentiable at the point $a\in \mathbf{R}$. Then there exists a function $h_k:\mathbf{R}\to \mathbf{R}$ such that $$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{{}^{″}}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(k)}(a)}{k!}(x-a{)}^k+h_k(x)(x-a{)}^k,$$ and $$\underset{x\to a}{lim}{h}_k(x)=0$$

• Note: $p_n(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{{‘}^{\prime }}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(k)}(a)}{k!}(x-a{)}^k+h_k(x)(x-a{)}^k$

Mean-value forms of the reminder Let $f:\mathbf{R}\to \mathbf{R}$ be $k+1$ times differentiable on the open interval with $f^{(k)}$ continuous on the closed interval between $a$ and $x$. Then $$R_k(x)=\frac{f^{(k+1)}(c)}{(k+1)!}(x-a{)}^{k+1}$$ for some $c$ between $a$ and $x$. Moreover, if there exists a real number $M$ such that $f^{(k+1)}\le M$ for all $x$, then $$R_k(x)\le \frac{M}{(k+1)!}|x-a{|}^{k+1}$$

• Note : This means that we can bound the magnitude of residual. See the Hoeffding’s lemma.

Proof Le $$g(t)$$ be the difference between $$(f-p)-R_n$$ where $$p$$ is $$k$$-th-order Taylor polynomial of $$f(x)$$ at $$x=t$$. $$g(t)=f(x)-(f(t)+f^{\prime }(t)(x-t)+\frac{f^{(2)}(t)}{2!}(x-t{)}^2+\cdots +\frac{f^{(k)}(t)}{k!}(x-t{)}^k-R_k(x)\frac{(x-t{)}^{k+1}}{(x-a{)}^{k+1}}$$ We have 1. $$g(x)=f(x)-f(x)+0+\cdots +0=0$$ 2. $$g(a)=f(x)-(f(a)+f^{\prime }(t)(x-a)+\frac{f^{(2)}(t)}{2!}(x-a{)}^2+\cdots +\frac{f^{(k)}(t)}{k!}(x-a{)}^k-R_k(x)\frac{(x-a{)}^{k+1}}{(x-a{)}^{k+1}}$$. Then $g(a)=f(x)-p_k(x)-R_k(x)=0$ As $g(t)$ satisfies Rolle's theorem, there exists $c$ between $a$ and $x$ such that $g^{\prime }(c)=0$. With some derivations, we have $$R_k(x)=\frac{f^{k+1}(c)}{(k+1)!}(x-a{)}^{k+1}$$.