Question

Expand f(x,y)= e^xy in Taylor’s series at (1, 1) up to first degree term​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Step-by-step explanation:

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

To Find: Taylor's series expansion up to first-degree term

Solution:

• Taylor's series expansion for two variables

For the function $f (x,y)$, Taylor's series expansion can be written as

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

where the expansion  $f (x,y) = f(a,b) + [(x-a)\frac{\partial f}{\partial x} + (y-b)\frac{\partial f}{\partial y}]$ is up to the first-degree term.

• Expansion of $f(x,y) = e^{xy}$ up to first-degree term

For $f (x,y) = e^{xy} \ \text{at} \ (1,1)$, Taylor's series expansion up to the first-degree term would be;

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

$\Rightarrow f (x,y) = e + (x-1)ye^{xy} + (y-1)xe^{xy}$

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

Hence, Taylor's series expansion of $f(x,y) = e^{xy}$ is $f (x,y) = e + e^{xy}[(x-1)y + (y-1)x$