2.7 Taylor formula

Given $f$ of $C^r$ where $f(x_1,...,x_n)=f(\mathbf{x})$, how to estimate $f$ near ${\mathbf{x}}_0$?

For any $\mathbf{v}\in {\mathbb{R}}^n$, let $g(t)=f({\mathbf{x}}_0+\mathbf{v}t)$ is of $C^r$. We have $$\begin{array}{rl}{g}^{\prime }(t)& =\partial f({\mathbf{x}}_0+t\mathbf{v})\mathbf{v}=\sum _{i=1}^n\frac{\partial f}{\partial x_i}({\mathbf{x}}_0+t\mathbf{v})v_i\\ g^{″}(t)& =\sum _{i=1}^n\sum _{j=1}^n\frac{{\partial }^{2}f}{\partial x_j\partial x_i}({\mathbf{x}}_0+t\mathbf{v})v_i{v}_j\\ g^{(k)}(t)& =\sum _{i_1,...,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathbf{x}}_0+t\mathbf{v})v_{i_1}\cdots v_{i_k}\\ g^{(k)}(0)& =\sum _{i_1,...,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathbf{x}}_0)v_{i_1}\cdots v_{i_k}\\ f({\mathbf{x}}_0+t\mathbf{v})& =f({\mathbf{x}}_0)+\sum _{k=1}^r\frac{t^k}{k!}\sum _{i_1,...,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathbf{x}}_0)v_{i_1}\cdots v_{i_k}+o(t^r),t\to 0\end{array}$$

Assuming ${\mathbf{v}}^{\ast }$ is the identity vector along $\mathbf{v}$, let $t=\parallel \mathbf{v}\parallel$, then $$\begin{array}{rl}f({\mathbf{x}}_0+\mathbf{v})& =f({\mathbf{x}}_0)+\sum _{k=1}^r\frac{\parallel \mathbf{v}{\parallel }^k}{k!}\sum _{i_1,...,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathbf{x}}_0)v_{i_1}^{\ast }\cdots v_{i_k}^{\ast }+o(\parallel \mathbf{v}{\parallel }^r)\\ & =f({\mathbf{x}}_0)+\sum _{k=1}^r\frac{1}{k!}\sum _{i_1,...,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathbf{x}}_0)v_{i_1}\cdots v_{i_k}+o(\parallel \mathbf{v}{\parallel }^r),\mathbf{v}\to \mathbf{0}\end{array}$$

This is not true!! Since in the process above, $o(t)$ may have something to do with $\mathbf{v}$!!

So we can use the Lagrange remainder to deal with this problem. $$\begin{array}{rl}f({\mathbf{x}}_0+\mathbf{v})=f({\mathbf{x}}_0)& +\sum _{k=1}^{r-1}\frac{1}{k!}\sum _{i_1,...,i_k=1}^n\frac{{\partial }^{k}f}{\partial x_{i_k}\cdots \partial x_{i_1}}({\mathbf{x}}_0)v_{i_1}\cdots v_{i_k}\\ & +\frac{1}{r!}\sum _{i_1,...,i_r=1}^n\frac{{\partial }^{r}f}{\partial x_{i_r}\cdots \partial x_{i_1}}({\mathbf{x}}_0+\theta \mathbf{v})v_{i_1}\cdots v_{i_r}\end{array}$$

where $\mathbf{v}\to \mathbf{0}$ and $\theta \in (0,1)$. According to the continuity of the $r^{\text{th}}$ order derivative, we have

Terminally,

Here we can find that for multiple variables function, even expanding the Taylor series with Peano remainder requires the $r^{\text{th}}$ order derivative to be continuous.

The uniqueness of Taylor formula.

Theorem 2.7.1 Assuming $f\in C^r$, $P$ is a polynomial where $\mathrm{deg}P\le r$ satisfying that

Then $P$ is the Taylor polynomial of degree $r$ of $f$ at ${\mathbf{x}}_0$.

Proof To simplify, assuming ${\mathbf{x}}_0=\mathbf{0}$, then we have

where $T$ is the Taylor polynomial of degree $r$ of $f$. Since $f(\mathbf{x})-P(\mathbf{x})=o(\parallel \mathbf{x}{\parallel }^r)$, we have

where $Q_k(\mathbf{x})$ is a homogeneous mapping of degree $k$.

Then it’s proved by mathematical induction.

• $\mathbf{x}\to \mathbf{0}\Rightarrow Q_0=\mathbf{0}$.

• Assuming $Q_s(\mathbf{x})=0$ for any $0\le s<k$, then

  

  meaning $Q_k(\mathbf{x})=o(\parallel \mathbf{x}{\parallel }^k)$. Consider

  

  Therefore, $Q_k(\mathbf{x})=\mathbf{0}$.

Terminally, $P(\mathbf{x})-T(\mathbf{x})=\mathbf{0}$. $◻$

Example 2.7.1 Seek Taylor polynomial of degree $3$ of $\ln (1+x+y+z)$ at $(x,y,z)=(0,0,0)$. Since

So $$\begin{array}{rl}\ln (1+x+y+z)& =(x+y+z)-\frac{(x+y+z{)}^2}{2}+\frac{(x+y+z{)}^3}{3}+o((x+y+z{)}^3)\\ & =\underset{\text{Taylor polynomial of degree }3}{\underbrace{(x+y+z)-\frac{(x+y+z{)}^2}{2}+\frac{(x+y+z{)}^3}{3}}}+o(\parallel (x,y,z){\parallel }^3)\end{array}$$