$$f(x, y) = e^{2x+xy+y^2}$$
Find the 2nd order taylor polynomial to the above function about (0,0)
The formula is:
$$P(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\frac 12[f_{xx}(x-a)^2+2f_{xy}(x-a)(y-b)+f_{yy}(y-b)^2]$$
$$f_x=e^{2x+xy+y^2}(2+y)$$
$$f_y=e^{2x+xy+y^2}(x+2y)$$
$$f_{xx}=e^{2x+xy+y^2}(2+y)^2$$
$$f_{yy}=e^{2x+xy+y^2}(x+2y)^2+2e^{2x+xy+y^2}$$
$$f_{xy}=e^{2x+xy+y^2}(2+y)(x+2y)+e^{2x+xy+y^2}$$
But I still get the wrong answer. What I am doing wrong?
Best Answer
We have:
$$f(x, y) = e^{2x+xy+y^2}$$
Finding partials and evaluating them at the point $(a, b) = (0,0)$, yields:
Next, we have:
$P(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\dfrac 12\left[f_{xx}(a,b)(x-a)^2+2f_{xy}(a,b)(x-a)(y-b)+f_{yy}(a,b)(y-b)^2\right]$
So, we get:
$$P(x, y) = 1 + 2 x + 0 +\frac 12[4 x^2 + 2(1) x y + 2 y^2] = 1 + 2x + 2x^2 + xy + y^2$$
Notice that the book forgot to divide $\frac 12 (4)$ and that is the source of their error. Your answer is correct.