Remainder form of Taylor polynomial at $x_0$: $ \frac{1}{n!}f^{(n)}(x_0 +\theta(x-x_0))(x-x_0)^n$ with $\theta \to \frac{1}{n+1}$ as $ x \to x_0$

Question

If the function $f: \mathbb R \to \mathbb R$ is $n+1$ times differentiable at $x_0$ and $f^{(n+1)}(x_0) \neq 0$, then a form of the remainder in Taylor's Formula is supposedly

$$r_n(x_0;x) = \frac{f^{(n)}\big(x_0 +\theta(x-x_0)\big)}{n!}(x-x_0)^n,$$ where $ 0< \theta < 1$ and $\theta = \theta(x)$ approaches $\frac{1}{n+1}$ as $x$ approaches $x_0.$

How would one derive this?

This seems quite strange to me.

For example, taking $f(x) =e^x$, the normal Taylor expansion for $n=2$ at $0$ would have the form (Lagrange Remainder): $$e^x = 1 + x + \frac12x^2 +\frac{1}{3!}e^{\xi}x^3,$$ but in this case it is

$$e^x = 1 + x + \frac12x^2 +\frac{1}{2!}e^{\theta(x)}x^2$$.

The properties of $\theta$ do not make much sense. Am I missing something?

Answer 1

Using Taylor-Lagrange formula we have, for $x$ in the neighborhood of $x_0$: $$ f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+\frac{f^{(n)}(x_0+\theta_x(x-x_0))}{n!}(x-x_0)^n\tag{1} $$ with $\theta_x\in(0,1)$, and we are interested in $\lim_{x\to x_0}\theta_x$. Taking one step further in Taylor-Lagrange formula we have also $$\eqalign{ f(x)&=\sum_{k=0}^{n-1}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\cr&\phantom{=}+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+\frac{f^{(n+1)}(x_0+\phi_x(x-x_0))}{(n+1)!}(x-x_0)^{n+1}\tag{2}} $$ with $\phi_x\in(0,1)$. Comparing $(1)$ and $(2)$ we get $$ \frac{f^{(n)}(x_0+\theta_x(x-x_0))}{n!} =\frac{f^{(n)}(x_0)}{n!} +\frac{f^{(n+1)}(x_0+\phi_x(x-x_0))}{(n+1)!}(x-x_0) $$ or $$ \frac{f^{(n)}(x_0+\theta_x(x-x_0))-f^{(n)}(x_0)}{x-x_0}=\frac{f^{(n+1)}(x_0+\phi_x(x-x_0))}{ n+1 } $$ that is $$ \theta_x=\frac{1}{n+1}\cdot\frac{f^{(n+1)}(x_0+\phi_x(x-x_0))}{ \dfrac{f^{(n)}(x_0+h_x)-f^{(n)}(x_0)}{h_x} } $$ where $h_x=\theta_x(x-x_0)$. Letting $x$ tend to $x_0$ and noting that $f^{(n+1)}(x_0)\ne0$, we get $$\lim_{x\to x_0} \theta_x=\frac{1}{n+1}.$$ which is the desired conclusion.$\qquad\square$

Answer 2

In the following I have a slightly different notation of the Lagrange form of the remainder than the OP's. If the OP insists on his notation please comment.

Let the function $f: \mathbb R \to \mathbb R$ be $k+1$ times differentiable at $x_0$. Denote its $k$-th order Taylor polynomial by $P_k(t)$ and define the function $F(t)$ by $$F(t) = f(t) + f'(t)(x-t)+\frac{f''(t)}{2!}(x-t)^2 + \ldots + \frac{f^{(k)}}{k!}(x-t)^k. $$ We show a way to arrive at the Lagrange- and Cauchy remainder forms.

Note that our remainder term is given by \begin{align*} R_k(x) =& f(x) - P_k(x) \\ =& F(x) - F(x_0) . \end{align*}

Let $G$ be any real-valued function, continuous on the closed interval $[x_0,x]$ and differentiable with a non-vanishing derivate on the open interval $(x_0,x)$.

By Cauchy's mean value theorem we then have $$\frac{F'(\xi)}{G'(\xi)} = \frac{F(x)-F(x_0)}{G(x)-G(x_0)}, \quad (*) $$ for some $\xi \in (x_0,x)$.

Now consider the derivative of $F(t)$, \begin{align*} F'(t) =& f'(t) + \big(f^{(2)}(t)(x-t) - f'(t)\big) + \Bigg(\frac{f^{(3)}(t)}{2!}(x-t)^2-\frac{f^{(2)}}{1!}(x-t)\Bigg) + \ldots + \Bigg( \frac{f^{(k+1)}(t)}{k!}(x-t)^k-\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\Bigg) \\ =& \frac{f^{(k+1)}(t)}{k!}(x-t)^k. \end{align*}

Plugging this into $(*)$ and rearranging terms we find an expession for the remainder $$ R_k(x) = F'(\xi)\frac{G(x)-G(x_0)}{G'(x)}. $$ Choosing $G(t) = (x-t)^{k+1}$ yields the Lagrange form of the remainder $$ R_k(x) = \frac{f^{k+1}(\xi)}{(k+1)!}(x-x_0)^k. $$ Choosing $G(t) = t-a$ yields the Cauchy form of the remainder $$ R_k(x) = \frac{f^{k+1}(\xi)}{k!}(x-\xi)^k (x-x_0). $$

PS: The last step of choosing $G(t)$ seems a bit obscure to me, although it works out obviously. Maybe someone can elaborate on what choices of $G(t)$ are valid?