Taylor's theorem for multivariate functions

Taylor's theorem

As we have seen in a previous post, the essence of differentiation is a linear approximation. 

See also: Partial and total differentiation of multivariate functions

Given a totally differentiable function $z=f(x,y)$, we have in the neighbor of $P=(a,b)$,

$$f(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+o(\Vert X-P\Vert )$$

where $X=(x,y).$ By discarding the higher order terms $o(\Vert X-P\Vert )$, we have the equation of the tangent plane at $P$:

$$z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

that gives the linear approximation of $z=f(x,y)$ around the point $P$. But we may obtain better approximations by looking into the details of $o(\Vert X-P\Vert )$, that is, by incorporating higher-order derivatives. Just as in the univariate case, we have a multivariate version of Taylor's theorem, which can be stated concisely and proved easily by using differential operators.

See also: Differential operators

Theorem (Taylor's theorem)

Let $f(x,y)$ be a function of class $C^n$ on an open region $U\subset {\mathbb{R}}^2$, and $(a,b)\in U$. If the line segment between $(x,y)$ and $(a,b)$ is in $U$, the following holds:

$$\begin{array}{rcl}f(x,y)& =& \sum _{m=0}^{n-1}\frac{1}{m!}{(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y})}^{m}f(a,b)\\ \text{(Eq:Taylor)}& & & +\frac{1}{n!}{(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y})}^{n}f(a+\theta h,b+\theta k)\end{array}$$

where $h=x-a$, $k=y-b$, and $\theta \in (0,1)$. 

Remark. In (Eq:Taylor), $h$ and $k$ in the differential operators are regarded as constants. That is, we mean

$${(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y})}^{m}f(a,b)=\sum _{l=1}^m(\genfrac{}{}{0ex}{}{m}{l})[\frac{{\partial }^{m}f}{\partial x^l\partial y^{m-l}}(a,b)](x-a{)}^l(y-b{)}^{m-l}.$$

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Remark. Eq. (Eq:Taylor) is called the finite Taylor expansion of $f(x,y)$, and its last term (that involving $\theta$) is called the remainder or residual. In particular, when $(a,b)=(0,0)$, the finite Taylor expansion is also called the finite Maclaurin expansion. □

Proof. Let us define the function $g(t)=f(a+ht,b+kt)$. Since the line segment between $(x,y)$ and $(a,b)$ is in $U$, $g(t)$ is a function of class $C^n$ defined on an open interval containing $[0,1]$, and

$$\begin{array}{rcl}g(0)& =& f(a,b),\\ g(1)& =& f(x,y).\end{array}$$

By the finite (univariate) Maclaurin expansion of $g(t)$, we have

$$\begin{array}{}\text{(Eq:gMac)}& g(t)=\sum _{m=0}^{n-1}\frac{1}{m!}{g}^{(m)}(0)t^m+\frac{1}{n!}{g}^{(n)}(\theta t)t^n\end{array}$$

where $0<\theta <1$. $g^{(m)}(t)$ (the $m$-th derivative of $g(t)$) is given as

$$\begin{array}{rcl}{g}^{(m)}(t)& =& \frac{d^m}{d{t}^m}f(a+ht,b+kt)\\ & =& {(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y})}^{m}f(a+ht,b+kt).\end{array}$$

where we have used the "total differential operator" $\frac{d}{dt}$ defined by using the chain rule:

$$\frac{d}{dt}f(x(t),y(t))=(\frac{dx}{dt}\frac{\partial }{\partial x}+\frac{dy}{dt}\frac{\partial }{\partial y})f(x(t),y(t)).$$

By setting $t=0$, we have

$$g^{(m)}(0)={(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial y})}^{m}f(a,b).$$

Substituting these into (Eq:gMac) and setting $t=1$, we have (Eq:Taylor). ■

Example. Let's find the Maclaurin expansion of $f(x,y)=e^{2x-3y}$ at $(x,y)=(0,0)$ with $n=2$.

$$\begin{array}{rcl}{f}_x(x,y)& =& 2e^{2x-3y},\\ f_y(x,y)& =& -3e^{2x-3y},\\ f_{xx}(x,y)& =& 4e^{2x-3y},\\ f_{xy}(x,y)& =& -6e^{2x-3y},\\ f_{yy}(x,y)& =& 9e^{2x-3y}.\end{array}$$ 

The finite Maclaurin expansion with $n=2$ is given by 

 $$\begin{array}{rcl}f(x,y)& =& f(0,0)+f_x(0,0)x+f_y(0,0)y\\ & & +\frac{1}{2}\{f_{xx}(\theta x,\theta y)x^2+2f_{xy}(\theta x,\theta y)xy+f_{yy}(\theta x,\theta y)y^2\}\end{array}$$

where $0<\theta <1$. Thus, we have

$$e^{2x-3y}=1+2x-3y+\frac{1}{2}(4x^2-12xy+9y^2)e^{2\theta x-3\theta y}.$$

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Asymptotic expansion

Corollary (Second-order asymptotic expansion)

Let $f(x,y)$ be a function of class $C^2$on an open region $U\subset {\mathbb{R}}^2$, and $P=(a,b),X=(x,y)\in U$. Suppose that the line segment between $X$ and $P$ is in $U$. Then,

$$\begin{array}{rcl}f(x,y)& =& f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\\ & & +\frac{1}{2}[f_{xx}(a,b)(x-a{)}^2+2f_{xy}(a,b)(x-a)(y-b)+f_{yy}(a,b)(y-b{)}^2]\\ & & +o(\Vert X-P{\Vert }^2)\end{array}$$

as $X\to P$.

Proof. Let $h=x-a$ , $k=y-b$, then $\Vert X-P{\Vert }^2=h^2+k^2$. Let $A=f_{xx}(a,b)$, $B=f_{xy}(a,b)$, $C=f_{yy}(a,b)$. By Taylor's theorem, 

$$\begin{array}{rcl}f(x,y)& =& f(a,b)+f_x(a,b)h+f_y(a,b)k\\ & & +\frac{1}{2}[A{h}^2+2Bhk+C{k}^2]+\frac{1}{2}r(h,k)\end{array}$$

where

$$\begin{array}{rcl}r(h,k)& =& (A^{\prime }-A)h^2+2(B^{\prime }-B)hk+(C^{\prime }-C)k^2\\ A^{\prime }& =& f_{xx}(a+\theta h,b+\theta k),\\ B^{\prime }& =& f_{xy}(a+\theta h,b+\theta k),\\ C^{\prime }& =& f_{yy}(a+\theta h,b+\theta k)\end{array}$$

for some $\theta \in (0,1)$. Since $f(x,y)$ is of class $C^2$, all the second derivatives are continuous. Therefore, $A^{\prime }-A$, $B^{\prime }-B$ and $C^{\prime }-C$ all converge to 0 as $X\to P$. Hence

$$(A^{\prime }-A)\frac{h^2}{h^2+k^2},(B^{\prime }-B)\frac{hk}{h^2+k^2},(C^{\prime }-C)\frac{k^2}{h^2+k^2}$$

all converge to 0 as $X\to P$. In fact,

$$\begin{array}{rcl}|(A^{\prime }-A)\frac{h^2}{h^2+k^2}|& \le & |A^{\prime }-A|\to 0,\\ |(B^{\prime }-B)\frac{hk}{h^2+k^2}|& \le & |B^{\prime }-B|\to 0,\\ |(C^{\prime }-C)\frac{k^2}{h^2+k^2}|& \le & |C^{\prime }-C|\to 0.\end{array}$$

Thus, $r(h,k)=o(\Vert X-P{\Vert }^2)$. ■

Example. The second-order asymptotic expansion of $f(x,y)=e^{2x-3y}$ at $(x,y)=(0,0)$ is

$$e^{2x-3y}=1+2x-3y+\frac{1}{2}(4x^2-12xy+9y^2)+o(x^2+y^2).$$

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