Question

expand e^xy in powers of (x-1) and (y-1) upto three degree terms by Taylor's series
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Answer 1

Step-by-step explanation:

Given: function $f(x,y)=e^{xy}$ and powers $(x-1)$ and $(y-1)$

To Find: Taylor's series expansion up to three-degree terms

Solution:

• Taylor's series expansion for a function of two variables

Taylor's series expansion of the function $f(x,y)$ in powers of $(x-a)$ and $(y-b)$ is given by -

$f(x,y) = f(a,b) + [(x-a)\frac{\partial f}{\partial x}+(y-b)\frac{\partial f}{\partial y}] + \frac{1}{2!}[(x-a)^{2} \frac{\partial^{2} f}{\partial x^{2}} + 2(x-a)(y-b) \frac{\partial^{2} f}{\partial x \partial y}+$

$(y-b)^{2} \frac{\partial^{2} f}{\partial y^{2}}] + \frac{1}{3!}[(x-a)^{3} \frac{\partial^{3} f}{\partial x^{3}} + 3(x-a)^{2} (y-b) \frac{\partial^{3} f}{\partial x^{2} \partial y} + 3(x-a) (y-b)^{2} \frac{\partial^{3} f}{\partial x \partial y^{2}} \\\\+(y-b)^{3} \frac{\partial^{3} f}{\partial y^{3}}]$

• Taylor's series expansion of $f(x,y)=e^{xy}$ in powers of $(x-1)$ and $(y-1)$

Taylor's series expansion of the function $f(x,y)=e^{xy}$ in powers of $(x-1)$ and $(y-1)$ is given by -

$f(x,y) = f(1,1) + [(x-1)\frac{\partial f}{\partial x}+(y-1)\frac{\partial f}{\partial y}] + \frac{1}{2!}[(x-1)^{2} \frac{\partial^{2} f}{\partial x^{2}} + 2(x-1)(y-1) \frac{\partial^{2} f}{\partial x \partial y}+$

$(y-1)^{2} \frac{\partial^{2} f}{\partial y^{2}}] + \frac{1}{3!}[(x-1)^{3} \frac{\partial^{3} f}{\partial x^{3}} + 3(x-1)^{2} (y-1) \frac{\partial^{3} f}{\partial x^{2} \partial y} + 3(x-1) (y-1)^{2} \frac{\partial^{3} f}{\partial x \partial y^{2}} \\\\+(y-1)^{3} \frac{\partial^{3} f}{\partial y^{3}}] \cdots \cdots \cdots (1)$

To get Taylor's series expansion, find partial derivatives of the functions as mentioned in the above expression.

Therefore, considering first-degree terms

$\frac{\partial e^{xy} }{\partial x} = ye^{xy} \ \& \ \frac{\partial e^{xy} }{\partial y} = xe^{xy} \ \cdots \cdots \cdots (2)$

second-degree terms are

$\frac{\partial^{2} e^{xy} }{\partial x^{2} } = y^{2} e^{xy}, \ \frac{\partial^{2} e^{xy} }{\partial y^{2}} = x^{2}e^{xy} \ \& \ \frac{\partial^{2} e^{xy} }{\partial x \partial y } = (xy +1) e^{xy} \cdots \cdots \cdots (3)$

and, three-degree terms are

$\frac{\partial^{3} e^{xy} }{\partial x^{3}} = y^{3} e^{xy}, \ \frac{\partial^{3} e^{xy} }{\partial y^{3}} = x^{3}e^{xy}, \ \frac{\partial^{3} e^{xy} }{\partial x^{2} \partial y } = y(xy +2) e^{xy} \ \& \ \frac{\partial^{3} e^{xy} }{\partial x \partial y^{2} } \\\\= x(xy +2) e^{xy} \cdots \cdots \cdots (4)$

Substituting (2), (3), and (4) in (1), we get -

$\Rightarrow f(x,y) = e + [e^{xy} (x-1)y+e^{xy} (y-1)x] + \frac{1}{2!}[(x-1)^{2} y^{2} e^{xy} + 2(x-1)(y-1)$

$(xy+1) e^{xy} + (y-1)^{2}x^{2} e^{xy}] + \frac{1}{3!}[(x-1)^{3} y^{3} e^{xy} + 3y(x-1)^{2} (y-1) (2+xy)e^{xy} + \\\\3x(x-1) (y-1)^{2} (2+xy) e^{xy} +(y-1)^{3} y^{3} e^{xy}]$

This is the required expression.

Hence, Taylor's series expansion of the function $f(x,y)=e^{xy}$ in powers of $(x-1)$ and $(y-1)$ is

$\Rightarrow f(x,y) = e + [e^{xy} (x-1)y+e^{xy} (y-1)x] + \frac{1}{2!}[(x-1)^{2} y^{2} e^{xy} + 2(x-1)(y-1)$

$(xy+1) e^{xy} + (y-1)^{2}x^{2} e^{xy}] + \frac{1}{3!}[(x-1)^{3} y^{3} e^{xy} + 3y(x-1)^{2} (y-1) (2+xy)e^{xy} + \\\\3x(x-1) (y-1)^{2} (2+xy) e^{xy} +(y-1)^{3} y^{3} e^{xy}]$