Question

Find the Taylor's series expansion of e x + y  in powers of x and y up to second degree terms.​

Answer 1

Answer:

The Taylor series expansion of $e^{x+y}$ up to second degree terms is $f(x,y)=1+x+y+\frac{x^2}{2}+xy+\frac{y^2}{2}$ in powers of $x$ and $y$.

Step-by-step explanation:

The $2^{nd}-$degree Taylor series expansion of $f(x,y)$ about the origin is

$f(x,y)=f(0,0)+f_x(0,0)(x-0)+f_y(0,0)(y-0)+\frac{f_{xx}(0,0)}{2} (x-0)^2$

              $+f_{xy}(0,0)(x-0)(y-0)+\frac{f_{yy}(0,0)}{2} (y-0)^2$

$f(x,y)=f(0,0)+xf_x(0,0)+yf_y(0,0)+x^2\frac{f_{xx}(0,0)}{2}+xyf_{xy}(0,0)$

              $+y^2\frac{f_{yy}(0,0)}{2}$  ...... (1)

Consider the function as follows:

$f(x,y)=e^{x+y}$

⇒ $f(x,y)=e^xe^y$

⇒ $f(0,0)=e^0e^0$

              $=1$

Compute $f_x(0,0)$ as follows:

$f_x(x,y)=\frac{\partial }{\partial x} (e^xe^y)$

            $=e^y\frac{\partial }{\partial x} (e^x)$

            $=e^ye^x$

$f_x(0,0)=e^0e^0$

            $=1$

Compute $f_y(0,0)$ as follows:

$f_y(x,y)=\frac{\partial }{\partial y} (e^xe^y)$

            $=e^xe^y$

$f_y(0,0)=e^0e^0$

            $=1$

Similarly,

Compute $f_{xx}(0,0), f_{xy}(0,0), f_{yy}(0,0)$ as follows:

$f_{xx}(x,y)=\frac{\partial^2 }{\partial x^2} (e^xe^y)$

            $=e^xe^y$

$f_{xx}(0,0)=1$

$f_{xy}(x,y)=\frac{\partial^2 }{\partial x \partial y} (e^xe^y)$

            $=e^xe^y$

$f_{xy}(0,0)=1$

$f_{yy}(x,y)=\frac{\partial^2 }{\partial y^2} (e^xe^y)$

            $=e^xe^y$

$f_{yy}(0,0)=1$

From (1), we get

⇒ $f(x,y)=1+x(1)+y(1)+x^2\frac{1}{2}+xy(1)+y^2\frac{1}{2}$

⇒ $f(x,y)=1+x+y+\frac{x^2}{2}+xy+\frac{y^2}{2}$

Therefore, the Taylor series expansion is $f(x,y)=1+x+y+\frac{x^2}{2}+xy+\frac{y^2}{2}$.

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