Taylor Theorem for Two Variables

Consider approximating $z = f(x,y)$

by a quadratic polynomial of $x$

and

Aroximation
Note that the total differential is an approximation of the surface at by the tangent plane. If we approximate the surface by the quadratic polynomial, we expect better approximation.

Let

$f(x,y) = a_0 + a_1 x + a_2 y + a_3 x^2 + a_4 xy + a_5 y^2 + R_3$

, where

is an error term. Then find all 2nd order partial derivatives of $z = f(x,y)$

. If

is so small that we can neglect, then

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle a_0 + a_1 x + a_2 y + a_3 x^2 + a_4 xy + a_5 y^2$
$\displaystyle f_{x}(x,y)$	$\displaystyle =$	$\displaystyle a_1 + 2a_3x + a_4y, f_{y}(x,y) = a_2 + a_4 x + 2a_5y$
$\displaystyle f_{xx}(x,y)$	$\displaystyle =$	$\displaystyle a_3, f_{xy}(x,y) = a_4, f_{yy}(x,y) = a_5.$

Now put

. Then

$\displaystyle f(0,0)$	$\displaystyle =$	$\displaystyle a_0, f_x(0,0) = a_1, f_y(0,0) = a_2$
$\displaystyle f_{xx}(0,0)$	$\displaystyle =$	$\displaystyle a_3, f_{xy}(0,0) = a_4, f_{yy}(0,0) = a_5$

Thus we can express all coefficients of $f$

by the 2nd order partial derivatives. Therefore,

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle a_0 + a_1 x + a_2 y + a_3 x^2 + a_4 xy + a_5 y^2$
	$\displaystyle =$	$\displaystyle f(0,0) + f_{x}(0,0)x + f_{y}(0,0)y$
	$\displaystyle +$	$\displaystyle f_{xx}(0,0)x^2 + f_{xy}(0,0)xy + f_{yy}(0,0)y^2$

$\delta$ neighborhood
A $\delta$ neighborhood of is a set of such that $\vert(x-x_0,y-y_0)\vert < \delta$ .

Partial Differential Operator
Let be constants, We define $h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}$ by $(h\frac{\partial}{\partial x} + k\frac{\partial}{\partial y}) f(x_0,y_0) = h\frac{\partial f}{\partial x}(x_0,y_0) + k\frac{\partial f}{\partial y}(x_0,y_0)$ .

Theorem 4..7 If is the class $C^{n}$ in the $\delta$ neighborhood of $(x_{0},y_{0})$ , then for $(x_{0} + h,y_{0}+k)$ ,

$\displaystyle f(x_{0} + h,y_{0}+k)$	$\displaystyle =$	$\displaystyle f(x_{0},y_{0}) + \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)f(x_{0},y_{0})$
	$\displaystyle +$	$\displaystyle \frac{1}{2!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^2 f(x_{0},y_{0}) + \cdots$
	$\displaystyle +$	$\displaystyle \frac{1}{(n-1)!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^{n-1} f(x_{0},y_{0}) + R_{n}$

where,

$\displaystyle R_{n} = \frac{1}{n!}\left (h \frac{\partial}{\partial x} + k \fra... ...partial y} \right )^{n} f(x_{0} + \theta h,y_{0} + \theta k) (0 < \theta < 1)$

$\displaystyle \left (h \frac{\partial}{\partial x} + k \frac{\partial}{\partial... ...+ k \frac{\partial}{\partial y} \right )^{m-1} f(x,y), m = 2,3,\cdot \cdot$

NOTE Let $F(t) = f(x_{0} + h t, y_{0} + k t) (0 \leq t \leq 1)$ . Then $F(t)$ is the class $C^{n}$ in $t$ . Thus by Maclausin theorem,

$\displaystyle F(t) = F(0) + \frac{1}{1!}F'(0)t + \frac{1}{2!}F''(0)t^2 + \cdots + \frac{1}{(n-1)!}F^{(n-1)}(0)t^{n-1} + R_{n},$

$\displaystyle R_{n} = \frac{1}{n!}F^{(n)}(\theta t)t^n (0 < \theta < 1).$

Thus for

$\displaystyle F1.$	$\displaystyle =$	$\displaystyle f(x_{0} + h, y_{0} + k) = f(x_{0},y_{0}) + \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)f(x_{0},y_{0})$
	$\displaystyle +$	$\displaystyle \frac{1}{2!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^2 f(x_{0},y_{0}) + \cdots$
	$\displaystyle +$	$\displaystyle \frac{1}{(n-1)!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^{n-1} f(x_{0},y_{0}) + R_{n}$

Maclaurin Theorem
$f(x,y) = f(0,0) + xf_{x}(0,0) + yf_{y}(0,0) + \frac{1}{2}(x^2 f_{xx}(0,0) + 2xy... ... x} + y\frac{\partial}{\partial y})^{n-1}f(\theta x,\theta y) (0 < \theta < 1)$

Example 4..17 Given $f(x,y) = e^x \cos{y}$ . Find the Taylor polynomial of 2nd degree at $(0,0)$

By Maclaurin theorem, $f(x,y) = f(0,0) + xf_{x}(0,0) + yf_{y}(0,0) + \frac{1}{2}(x^2 f_{xx}(0,0) + 2xyf_{xy}(0,0) + y^2f_{yy}(0,0)) +R_3$ s.

SOLUTION We first find all 2nd partial derivatives of $f(x,y) = e^x \cos y$ . $f_{x} = e^x \cos{y}, f_{y} = e^x(-\sin{y}) = -e^x \sin{y}$ , $f_{xx} = e^x \cos{y}$ , $f_{xy} = -e^{x}\sin{y}$ , $f_{yy} = -e^x \cos{y}$ . Theorem4.7, let $x_0 = 0, y_0 = 0, x = h, y = k$ . Then $f(0,0) = 1, f_{x}(0,0) = 1, f_{y}(0,0) = 0, f_{xx}(0,0) = 1, f_{xy}(0,0) = 0, f_{yy}(0,0) = -1$ . Thus

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle f(0,0) + xf_{x}(0,0) + yf_{y}(0,0) + \frac{1}{2!}\left\{x^{2}f_{xx}(0,0)\right.$
	$\displaystyle +$	$\displaystyle \left.2xyf_{xy}(0,0) + y^{2}f_{yy}(0,0) \right\} + R_{3}$
	$\displaystyle =$	$\displaystyle 1 + x + \frac{1}{2!}\left\{x^2 - y^2\right\} + R_{3} \ensuremath{ \blacksquare}$

Exercise4-16
Note that Taylor polynomial of 2nd degree of a function at $(\frac{\pi}{2},1)$ means expressing the function using $(x-\frac{\pi}{2})$ and .

Exercise 4..17 Find the 2nd degree Taylor polynomial of $f(x,y) = \sin{xy}$ at $(\frac{\pi}{2},1)$ .

SOLUTION $f_{x} = y\cos{xy}, f_{y} = x\cos{xy}, f_{xx} = -y^{2}\sin{xy}, f_{xy} = \cos{xy} -xy\sin{xy}, f_{yy} = -x^{2}\sin{xy}$ . Thus in Theorem4.7, let $x = x_{0} + h = \frac{\pi}{2} + h$ , $y = y_{0} + k = 1 + k$ . Then

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle f(\frac{\pi}{2},1) + (x - \frac{\pi}{2})f_{x}(\frac{\pi}{2},1) + ... ...},1) + \frac{1}{2!}\left\{(x - \frac{\pi}{2})^{2}f_{xx}(\frac{\pi}{2},1)\right.$
	$\displaystyle +$	$\displaystyle \left.2(x - \frac{\pi}{2})(y-1)f_{xy}(\frac{\pi}{2},1) + (y-1)^{2}f_{yy}(\frac{\pi}{2},1) \right\} + R_{3}$
	$\displaystyle =$	$\displaystyle -\frac{1}{2!}\left\{(x - \frac{\pi}{2})^{2} - (x - \frac{\pi}{2})(y-1) - \frac{\pi^{2}}{4}(y-1)^{2}\right\} + R_{3} \ensuremath{ \blacksquare}$

Proof By Taylor theorem, for $(x_{0} + h, y_{0} + k) \in D$ , we have

$\displaystyle f(x_{0} + h,y_{0}+k)$	$\displaystyle =$	$\displaystyle f(x_{0},y_{0}) + hf_{x}(x_{0},y_{0}) + kf_{y}(x_{0} + y_{0})$
	$\displaystyle +$	$\displaystyle \frac{1}{2}[h^2 f_{xx}(x_{0}+\theta h,y_{0}+\theta k) + 2hkf_{xy}(x_{0}+\theta h,y_{0}+\theta k)$
	$\displaystyle +$	$\displaystyle k^2 f_{yy} (x_{0}+\theta h,y_{0}+\theta k)], (0 < \theta < 1)$

Now let $Q = f(x_{0} + h, y_{0} + k) - f(x_{0},y_{0})$ . Then $f_{x}(x_{0},y_{0}) = f_{y}(x_{0},y_{0}) = 0$ and

$\displaystyle Q = \frac{1}{2}[h^2 f_{xx}(x_{0}+\theta h,y_{0}+\theta k) + 2hkf_... ...x_{0}+\theta h,y_{0}+\theta k) + k^2 f_{yy} (x_{0}+\theta h,y_{0}+\theta k) ].$

Next let $A = f_{xx}(x_{0}+\theta h,y_{0}+\theta k), B = f_{xy}(x_{0}+\theta h,y_{0}+\theta k), C = f_{yy} (x_{0}+\theta h,y_{0}+\theta k)$ . Then

$\displaystyle Q = \frac{1}{2}\left(Ah^2 + 2Bhk + Ck^2\right) = \frac{A}{2}\left[(h + \frac{Bk}{A})^2 + \frac{(AC - B^2)k^2}{A^2}\right].$

Thus the sign of $Q$

is determinde by the sign of $A$

and the sign of $AC-B^2$

1. If $\Delta = AC - B^2 > 0, A > 0$ , then since $f(x,y)$ is the class $C^2$ function, for any $h,k$ such that $\vert h\vert,\vert k\vert$ is sufficiently small and never 0 simulteneously, we have $Q > 0$ . Thus, $f(x_{0},y_{0})$ is a local minimum.

Check
positive $\big[$ positive $+\frac{(\mbox{positive})\mbox{positive}}{\mbox{positive}}\big]$ . Thus .

2. If $\Delta = AC - B^2 > 0, A < 0$ , then since $f(x,y)$ is the class $C^2$ func, for any $h,k$ such that $\vert h\vert,\vert k\vert$ is sufficiently small and never 0 simulteneously, we have $Q < 0$ . Thus $f(x_{0},y_{0})$ is a local maximum.

Check
negative $\big[$ positive $+\frac{(\mbox{positive})\mbox{positive}}{\mbox{positive}}\big]$ . Thus .

3. If $\Delta = AC - B^2 < 0$ and $A > 0$ , then $C < 0$ which gives a saddle point. Similarly, if $\Delta < 0$ and $A < 0$ , then $C > 0$ which give a saddle point. $\blacksquare$

Example 4..18 Find the local extrema of the following function.

$\displaystyle f(x,y) = x^2 + xy + 3y^2 + x +y$

SOLUTION In Example4.16, we found the critical point. Now we check to see whether the function takes a local extremum at the critical point. Now by the 2nd derivative test,

$\displaystyle f_{xx}(x,y) = 2, f_{xy}(x,y) = 1, f_{yy}(x,y) = 6, \Delta = 11$

Thus, $f(-\frac{5}{11}, -\frac{1}{11}) = -\frac{3}{11}$ is a local minimum. $\blacksquare$

Exercise 4..18 Find the local extrema of th following function.

$\displaystyle f(x,y) = x^3 + y^3 - 3xy$

Check
Multiply the equation 4.1 by and multiply the equation 4.2 by . Then subtract the latter one from the former one to obtain . From this, we get and put this back to the equation 4.1, then . Thus, .

SOLUTION Let $f(x,y) = x^3 + y^3 - 3xy$ . Then we have $f_{x} = 3x^2 - 3y, f_{y} = 3y^2 - 3x$ . If $f(x,y)$ takes the local maxima at $(x_{0},y_{0})$ , then

$\displaystyle f_{x}(x_{0},y_{0})$	$\displaystyle =$	$\displaystyle 3x_{0}^2 - 3y_{0} = 0$	(4.1)
$\displaystyle f_{y}(x_{0},y_{0})$	$\displaystyle =$	$\displaystyle 3y_{0}^2 - 3x_{0} = 0$	(4.2)

Solve this for $x_{0},y_{0}$ . Then substitute the equation $y_{0} = x_{0}^2$ , which is derived from the equation 4.1, to the equation 4.2. Then

$\displaystyle 3x_{0}^4 - 3x_{0} = 3x_{0}(x_{0}^3 - 1) = 0.$

Thus $x_{0} = 0, 1$ . Hence, $(x_{0}=0,y_{0}=0), (x_{0}=1, y_{0} =1)$ .

Now we apply the 2nd derivative test. Since $f_{xx}(x,y) = 6x, f_{xy}(x,y) = -3, f_{yy}(x,y) = 6y$ , at $(0,0)$ we have

$\displaystyle \Delta = f_{xx}(0,0)f_{yy}(0,0) - (f_{xy}(0,0))^2 = -9 < 0.$

This shows that $f(0,0)$

is not extrema. Now at $(1,1)$

, we have

$\displaystyle \Delta$	$\displaystyle =$	$\displaystyle f_{xx}(1,1)f_{yy}(1,1) - (f_{xy}(1,1))^2 = 6\cdot6 - (-3)^2 = 27 > 0$
$\displaystyle A$	$\displaystyle =$	$\displaystyle f_{xx}(1,1) = 6 > 0.$