Taylor Theorem for Two Variables

Consider approximating $z = f(x,y)$ by a quadratic polynomial of $x$ and $y$.

Aroximation

Note that the total differential is an approximation of the surface $z = f(x,y)$ at $(x_0, y_0)$ 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 $R_3$ is an error term. Then find all 2nd order partial derivatives of $z = f(x,y)$. If $R_3$ is so small that we can neglect, then

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

$$f_{x}(x,y) = a_1 + 2a_3x + a_4y, f_{y}(x,y) = a_2 + a_4 x + 2a_5y$$

$$f_{xx}(x,y) = a_3, f_{xy}(x,y) = a_4, f_{yy}(x,y) = a_5.$$

Now put $x = 0, y = 0$. Then

$$f(0,0) = a_0, f_x(0,0) = a_1, f_y(0,0) = a_2$$

$$f_{xx}(0,0) = a_3, f_{xy}(0,0) = a_4, f_{yy}(0,0) = a_5$$

Thus we can express all coefficients of $f$ by the 2nd order partial derivatives. Therefore,

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

$$= f(0,0) + f_{x}(0,0)x + f_{y}(0,0)y$$

$$+ f_{xx}(0,0)x^2 + f_{xy}(0,0)xy + f_{yy}(0,0)y^2$$

$\delta$ neighborhood

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

Partial Differential Operator

Let $h,k$ 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 $z = f(x,y)$ is the class $C^{n}$ in the $\delta$ neighborhood of $(x_{0},y_{0})$, then for $(x_{0} + h,y_{0}+k)$,

$$f(x_{0} + h,y_{0}+k) = f(x_{0},y_{0}) + \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)f(x_{0},y_{0})$$

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

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

where,

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

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

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,

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

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

Thus for $t=1$,

$$F1. = f(x_{0} + h, y_{0} + k) = f(x_{0},y_{0}) + \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)f(x_{0},y_{0})$$

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

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

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

$$f(x,y) = f(0,0) + xf_{x}(0,0) + yf_{y}(0,0) + \frac{1}{2!}\left\{x^{2}f_{xx}(0,0)\right.$$

$$+ \left.2xyf_{xy}(0,0) + y^{2}f_{yy}(0,0) \right\} + R_{3}$$

$$= 1 + x + \frac{1}{2!}\left\{x^2 - y^2\right\} + R_{3} \ensuremath{ \blacksquare}$$

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 $(y-1)$.

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

$$f(x,y) = f(\frac{\pi}{2},1) + (x - \frac{\pi}{2})f_{x}(\frac{\pi}{2},1) + ... ...},1) + \frac{1}{2!}\left\{(x - \frac{\pi}{2})^{2}f_{xx}(\frac{\pi}{2},1)\right.$$

$$+ \left.2(x - \frac{\pi}{2})(y-1)f_{xy}(\frac{\pi}{2},1) + (y-1)^{2}f_{yy}(\frac{\pi}{2},1) \right\} + R_{3}$$

$$= -\frac{1}{2!}\left\{(x - \frac{\pi}{2})^{2} - (x - \frac{\pi}{2})(y-1) - \frac{\pi^{2}}{4}(y-1)^{2}\right\} + R_{3} \ensuremath{ \blacksquare}$$

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

$$f(x_{0} + h,y_{0}+k) = f(x_{0},y_{0}) + hf_{x}(x_{0},y_{0}) + kf_{y}(x_{0} + y_{0})$$

$$+ \frac{1}{2}[h^2 f_{xx}(x_{0}+\theta h,y_{0}+\theta k) + 2hkf_{xy}(x_{0}+\theta h,y_{0}+\theta k)$$

$$+ k^2 f_{yy} (x_{0}+\theta h,y_{0}+\theta k)], (0 < \theta < 1)$$

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

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

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

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

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

$Q =$   positive$\big[$positive$+\frac{(\mbox{positive})\mbox{positive}}{\mbox{positive}}\big]$. Thus $Q > 0$.

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

$Q =$   negative$\big[$positive$+\frac{(\mbox{positive})\mbox{positive}}{\mbox{positive}}\big]$. Thus $Q < 0$.

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.

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

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,

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

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.

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

Check

Multiply the equation 4.1 by $y_0$ and multiply the equation 4.2 by $x_0$. Then subtract the latter one from the former one to obtain $3x_0^3 - 3y_0^3 = 0$. From this, we get $x_0 = y_0$ and put this back to the equation 4.1, then $3x_0^2 - 3x_0 = 3x_0(x_0 - 1) = 0$. Thus, $x_0 = 0,1$.

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

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

$$f_{y}(x_{0},y_{0}) = 3y_{0}^2 - 3x_{0} = 0$$ (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

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

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

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

This shows that $f(0,0)$ is not extrema. Now at $(1,1)$, we have

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

$$A = f_{xx}(1,1) = 6 > 0.$$

Thus, $f(1,1) = 1 + 1 - 3 = -1$ is the local minimum $\blacksquare$

Exercise A

1.

Find the 2nd degree Taylor polynomial of $f$ at $(a,b)$.．

(a) $${f(x,y) = x^{2}y, (a,b) = (1,1)}$$ (b) $${f(x,y) = \cos{xy}, (a,b) = (1, \frac{\pi}{2}) }$$

(c) $${f(x,y) = \log{(1 - x + y)}, (a,b) = (1,1)}$$ (d) $${f(x,y) = xe^{2x+y}, (a,b) = (0,0)}$$

2.

Find the local extrema of the following functions.

(a) $${f(x,y) = 2x - x^{2} - y^{2}}$$ (b) $${f(x,y) = x^{2} - 6y^{2} + y^{3}}$$

(c) $${f(x,y) = x^{3} - 3x + y}$$ (d) $${f(x,y) = x^{2} + xy + y^{2} - 3x - 3y}$$

Exercise B

1.

Find the local extrema of the following functions．

(a) $${f(x,y) = 2x^2 + y^2 -xy - 7y}$$ (b) $${f(x,y) = \frac{x}{y^2} + xy }$$

(c) $${f(x,y) = x^3 + y^3 - 3xy}$$ (d) $${f(x,y) = (x^2 + y^2)^2 - 2(x^2 - y^2)}$$

2.

Find the 2nd degree Taylor polynomial of $f$ at $(a,b)$．

(a) $${f(x,y) = e^x \cos{y}, (a,b) = (0,0)}$$

(b) $${f(x,y) = \log{(x + y^2)}, (a,b) = (2,1)}$$