Taylor's Series For Two Variables Engineering Maths, Btech first year

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: $$f(x,y)=f(a,b)+\frac{1}{1!}[(x-a)f_x(a,b)+(y-b)f_y(a,b)]$$ $$+\frac{1}{2!}[(x-a{)}^2{f}_{xx}(a,b)+2(x-a)(y-b)f_{xy}(a,b)+(y-b{)}^2{f}_{yy}(a,b)]$$ $$+\frac{1}{3!}[(x-a{)}^3{f}_{xxx}(a,b)+(y-b{)}^3{f}_{yyy}(a,b)+3(x-a{)}^2(y-b)f_{xxy}(a,b)+3(x-a)(y-b{)}^2{f}_{xyy}(a,b)]$$ $$+.....$$ $$\text{where }{f}_x=\frac{\partial f}{\partial x};f_y=\frac{\partial f}{\partial y}$$ $$f_{xx}=\frac{{\partial }^{2}f}{\partial x^2};f_{xy}=\frac{{\partial }^{2}f}{\partial x\partial y};f_{yy}=\frac{{\partial }^{2}f}{\partial y^2}$$ $$f_{xxx}=\frac{{\partial }^{3}f}{\partial x^3};f_{yyy}=\frac{{\partial }^{3}f}{\partial y^3}$$ $$f_{xxy}=\frac{{\partial }^{3}f}{\partial x^2\partial y};f_{xyy}=\frac{{\partial }^{3}f}{\partial x{y}^2}$$ 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 $(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$ and 3 is the degree of $x$ and $y$ terms and order of partial derivatives.

Sample Problem 1 :

Expand $x^{2}y+3y-2$ in the powers of $(x-1)$ and $(y+2)$ upto three degree terms.

Solution :

Here degree implies derivative.
Given function is $f(x,y)=x^{2}y+3y-2$
By Taylor Series- $$f(x,y)=f(a,b)+\frac{1}{1!}[(x-a)f_x(a,b)+(y-b)f_y(a,b)]$$ $$+\frac{1}{2!}[(x-a{)}^2{f}_{xx}(a,b)+2(x-a)(y-b)f_{xy}(a,b)+(y-b{)}^2{f}_{yy}(a,b)]$$ $$+\frac{1}{3!}[(x-a{)}^3{f}_{xxx}(a,b)+(y-b{)}^3{f}_{yyy}(a,b)+3(x-a{)}^2(y-b)f_{xxy}(a,b)+3(x-a)(y-b{)}^2{f}_{xyy}(a,b)]$$ $$+.....$$ $$\text{Here, }x-a=x-1\Rightarrow a=1$$ $$y-b=y+2\Rightarrow b=-2$$ $$f(1,-2)=-2-6-2=-10$$ $$f_x=\frac{\partial f}{\partial x}=2xy,f_x(1,-2)=-4$$ $$f_y=\frac{\partial f}{\partial y}=x^2+3,f_y(1,-2)=1+3=4$$ $$f_{xx}=\frac{{\partial }^{2}f}{\partial x^2}=2y,f_{xx}(1,-2)=-4$$ $$f_{yy}=\frac{{\partial }^{2}f}{\partial y^2}=0,f_{yy}(1,-2)=0$$ $$f_{xy}=\frac{{\partial }^{2}f}{\partial x\partial y}=2x,f_{xy}(1,-2)=2$$ $$f_{xxx}=\frac{{\partial }^{3}f}{\partial x^3}=0,f_{xxx}(1,-2)=0$$ $$f_{yyy}=\frac{{\partial }^{3}f}{\partial y^3}=0,f_{yyy}(1,-2)=0$$ $$f_{xxy}=\frac{{\partial }^{3}f}{\partial x^{2}y}=2,f_{xxy}(1,-2)=2$$ $$f_{xyy}=\frac{{\partial }^{3}f}{\partial x{y}^2}=0,f_{xyy}(1,-2)=0$$ 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 $x{y}^2+\cos xy$ in the powers of $(x-1)$ and $(y-\pi /2)$ upto two degree terms.

Solution :

$$f(x,y)=x{y}^2+\cos xy$$ $$x-a=x-1\Rightarrow a=1$$ $$y-b=y-\pi /2\Rightarrow b=\pi /2$$ By Taylor Series, $$f(x,y)=f(a,b)+\frac{1}{1!}[(x-a)f_x(a,b)+(y-b)f_y(a,b)]$$ $$+\frac{1}{2!}[(x-a{)}^2{f}_{xx}(a,b)+2(x-a)(y-b)f_{xy}(a,b)+(y-b{)}^2{f}_{yy}(a,b)]$$ $$+\frac{1}{3!}[(x-a{)}^3{f}_{xxx}(a,b)+(y-b{)}^3{f}_{yyy}(a,b)+3(x-a{)}^2(y-b)f_{xxy}(a,b)+3(x-a)(y-b{)}^2{f}_{xyy}(a,b)]$$ $$+.....$$ $$\text{Here, }x-a=x-1\Rightarrow a=1$$ $$y-\pi /2=y-b\Rightarrow b=\pi /2$$ $$f(1,\pi /2)=(\pi /2{)}^2+0=\frac{{\pi }^2}{4}$$ $$f_x=\frac{\partial f}{\partial x}=y^2-y\sin xy,f_x(1,\pi /2)=\frac{{\pi }^2}{4}-\frac{\pi }{2}$$ $$f_y=\frac{\partial f}{\partial y}=2xy-x\sin xy,f_y(1,\pi /2)=\pi -1$$ $$f_{xx}=\frac{{\partial }^{2}f}{\partial x^2}=-y^2\cos xy,f_{xx}(1,\pi /2)=0$$ $$f_{yy}=\frac{{\partial }^{2}f}{\partial y^2}=2x-x^2\cos xy,f_{yy}(1,\pi /2)=2$$ $$f_{xy}=\frac{{\partial }^{2}f}{\partial x\partial y}=2y-x\cos xy,f_{xy}(1,\pi /2)=\pi$$ Putting values, $$=\frac{{\pi }^2}{4}+\frac{1}{1}[(x-1)(\frac{{\pi }^2}{4}-\frac{\pi }{2})+(y-\frac{\pi }{2})(\pi -1)]$$ $$+\frac{1}{2}[(x-1{)}^2(0)+(y-\pi /2{)}^2(2)+2(x-1)(y-\pi /2)(\pi -1)]+...$$ $$=\frac{{\pi }^2}{4}+(\frac{{\pi }^2}{4}-\frac{\pi }{2})(x-1)+(\pi -1)(y-\frac{\pi }{2})$$ $$+{(y-\frac{\pi }{2})}^2+(\pi -1)(x-1)(y-\frac{\pi }{2})+...$$