Taylor Series
For two variables :
Before diving into Taylor Series for two variables, you should know Taylor Series for one variable, so if you don't know Taylor Series for one variable Click Here for a quick look.
A function of two variables \(x,y\) can be expanded using Taylor Series using following formula:
\[ f(x,y) = f(a,b) + \frac{1}{1!}[(x - a)f_x(a,b) + (y - b)f_y(a,b)] \]
\[ + \frac{1}{2!}[(x - a)^2f_{xx}(a,b) + 2(x - a)(y - b)f_{xy}(a,b) + (y - b)^2f_{yy}(a,b)] \]
\[ + \frac{1}{3!}[(x - a)^3f_{xxx}(a,b) + (y - b)^3f_{yyy}(a,b) + 3(x - a)^2(y - b)f_{xxy}(a,b) + 3(x - a)(y - b)^2f_{xyy}(a,b) ] \]
\[ + \hspace{10pt} ..... \]
\[ \text{where } f_x = \frac{\partial f}{\partial x}; \hspace{10pt} f_y = \frac{\partial f}{\partial y} \]
\[ f_{xx} = \frac{\partial ^2f}{\partial x^2}; \hspace{10pt} f_{xy} = \frac{\partial ^2f}{\partial x
\partial y}; \hspace{10pt} f_{yy} = \frac{\partial ^2f}{\partial y^2} \]
\[ f_{xxx} = \frac{\partial ^3f}{\partial x^3}; \hspace{10pt} f_{yyy} = \frac{\partial ^3f}{\partial y^3} \]
\[ f_{xxy} = \frac{\partial ^3f}{\partial x^2\partial y}; \hspace{10pt} f_{xyy} = \frac{\partial ^3f}{\partial xy^2} \]
Yeah it's scary, but if you know Taylor Series for one variable, then it is easy to remember. In taylor Series for one variable, you had \(f(a)\), in two variable you have \(f(a,b) \); in one variable, you had derivatives of the function, in two variables, you have partial derivatives of the function w.r.t. \(x,y\); in one variable you had \((x - a)\), in two variables, you have \((x - a)\) and \((y - b) \).
All the terms in Taylor Series for two variables follow a pattern known as identities of \(a,b\). For example, the second term is simply \((a + b)^1\), where 1 is both the degree of \( (x - a) \) and \((y - b)\) and the order of partial derivatives of the function w.r.t. \(x, y\), and the \(x\) terms are with partial derivative w.r.t. \(x\) and the \(y\) terms are with partial derivatives w.r.t. \(y\).
The third term resembles with \((a + b)^2 = a^2 + 2ab + b^2\), here 2 is the degree of \(x\) and \(y\) terms and order of partial derivatives. Similarly the third term resembles \((a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2\) and 3 is the degree of \(x\) and \(y\) terms and order of partial derivatives.
Sample Problem 1 :
Expand \(x^2y + 3y - 2\) in the powers of \((x - 1)\) and \((y + 2)\) upto three degree terms.
Solution :
Here degree implies derivative.
Given function is \( f(x,y) = x^2y + 3y -2 \)
By Taylor Series-
\[ f(x,y) = f(a,b) + \frac{1}{1!}[(x - a)f_x(a,b) + (y - b)f_y(a,b)] \]
\[ + \frac{1}{2!}[(x - a)^2f_{xx}(a,b) + 2(x - a)(y - b)f_{xy}(a,b) + (y - b)^2f_{yy}(a,b)] \]
\[ + \frac{1}{3!}[(x - a)^3f_{xxx}(a,b) + (y - b)^3f_{yyy}(a,b) + 3(x - a)^2(y - b)f_{xxy}(a,b) + 3(x - a)(y - b)^2f_{xyy}(a,b) ] \]
\[ + \hspace{10pt} ..... \]
\[ \text{Here, } x - a = x - 1 \Rightarrow a = 1 \]
\[ y - b = y + 2 \Rightarrow b = -2 \]
\[ f(1,-2) = -2 - 6 - 2 = -10 \]
\[ f_{x} = \frac{\partial f}{\partial x} = 2xy, \hspace{10pt} f_{x}(1,-2) = -4 \]
\[ f_{y} = \frac{\partial f}{\partial y} = x^2 + 3, \hspace{10pt} f_{y}(1,-2) = 1 + 3 = 4 \]
\[ f_{xx} = \frac{\partial ^2f}{\partial x^2} = 2y, \hspace{10pt} f_{xx}(1,-2) = -4 \]
\[ f_{yy} = \frac{\partial ^2f}{\partial y^2} = 0 , \hspace{10pt} f_{yy}(1,-2) = 0 \]
\[ f_{xy} = \frac{\partial ^2f}{\partial x \partial y} = 2x, \hspace{10pt} f_{xy}(1,-2) = 2 \]
\[ f_{xxx} = \frac{\partial ^3f}{\partial x^3} = 0 , \hspace{10pt} f_{xxx}(1,-2) = 0 \]
\[ f_{yyy} = \frac{\partial ^3f}{\partial y^3} = 0, \hspace{10pt} f_{yyy}(1,-2) = 0 \]
\[ f_{xxy} = \frac{\partial ^3f}{\partial x^2y} = 2, \hspace{10pt} f_{xxy}(1,-2) = 2 \]
\[ f_{xyy} = \frac{\partial ^3f}{\partial xy^2} = 0, \hspace{10pt} f_{xyy}(1,-2) = 0 \]
Putting values,
\[ f(x,y) = -10 + \frac{1}{1}[(x - 1)(-y) + (y + 2)(4)] \]
\[ + \frac{1}{2}[(x - 1)^2(-4) + (y + 2)^2(0) + 2(x - 1)(y + 2)(2)] \]
\[ + \frac{1}{3\times 2}[(x - 1)^3(0) + (y + 2)^3(0) + 3(x - 1)^2(y + 2)(2) + 2(x - 1)(y + 2)^2(0)] + ...\]
\[ = -10 -4(x - 1) + 4(y + 2) - 2(x - 1)^2 + 2(x - 1)(y + 2) + (x - 1)^2(y + 2) + ... \]
Sample Problem 2 :
Expand \( xy^2 + \cos{xy} \) in the powers of \((x - 1)\) and \((y - \pi /2)\) upto two degree terms.
Solution :
\[ f(x,y) = xy^2 + \cos{xy} \] \[ x - a = x - 1 \Rightarrow a = 1 \] \[ y - b = y - \pi /2 \Rightarrow b = \pi /2 \] By Taylor Series, \[ f(x,y) = f(a,b) + \frac{1}{1!}[(x - a)f_x(a,b) + (y - b)f_y(a,b)] \] \[ + \frac{1}{2!}[(x - a)^2f_{xx}(a,b) + 2(x - a)(y - b)f_{xy}(a,b) + (y - b)^2f_{yy}(a,b)] \] \[ + \frac{1}{3!}[(x - a)^3f_{xxx}(a,b) + (y - b)^3f_{yyy}(a,b) + 3(x - a)^2(y - b)f_{xxy}(a,b) + 3(x - a)(y - b)^2f_{xyy}(a,b) ] \] \[ + \hspace{10pt} ..... \] \[ \text{Here, } x - a = x - 1 \Rightarrow a = 1 \] \[ y - \pi /2 = y - b \Rightarrow b = \pi /2 \] \[ f(1, \pi /2) = (\pi /2)^2 + 0 = \frac{\pi^2}{4} \] \[ f_{x} = \frac{\partial f}{\partial x} = y^2 - y\sin{xy}, \hspace{10pt} f_{x}(1, \pi/2) = \frac{\pi ^2}{4} - \frac{\pi}{2} \] \[ f_{y} = \frac{\partial f}{\partial y} = 2xy - x\sin{xy}, \hspace{10pt} f_{y}(1, \pi/2) = \pi - 1 \] \[ f_{xx} = \frac{\partial ^2f}{\partial x^2} = -y^2\cos{xy}, \hspace{10pt} f_{xx}(1, \pi/2) = 0 \] \[ f_{yy} = \frac{\partial ^2f}{\partial y^2} = 2x - x^2\cos{xy}, \hspace{10pt} f_{yy}(1, \pi/2) = 2 \] \[ f_{xy} = \frac{\partial ^2f}{\partial x \partial y} = 2y - x\cos{xy}, \hspace{10pt} f_{xy}(1, \pi/2) = \pi \] Putting values, \[ = \frac{\pi ^2}{4} + \frac{1}{1}[(x - 1)(\frac{\pi^2}{4} - \frac{\pi}{2}) + (y - \frac{\pi}{2})(\pi - 1)] \] \[ + \frac{1}{2}[(x - 1)^2(0) + (y - \pi /2)^2(2) + 2(x - 1)(y - \pi /2)(\pi - 1)] + ... \] \[ = \frac{\pi ^2}{4} + \left(\frac{\pi ^2}{4} - \frac{\pi}{2} \right)(x - 1) + (\pi - 1)\left(y - \frac{\pi}{2} \right) \] \[ + \left(y - \frac{\pi}{2} \right)^2 + (\pi - 1)(x - 1)\left(y - \frac{\pi}{2} \right) + ... \]
