<s>[INST] Multivariable taylor polynomial $$f(x, y) = e^{2x+xy+y^2}$$
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?
 [/INST] We have:
$$f(x, y) = e^{2x+xy+y^2}$$
Finding partials and evaluating them at the point $(a, b) = (0,0)$, yields:


*

*$f_x(x, y) = (y+2) e^{x y+2 x+y^2} \implies f_x(0,0) = 2$

*$f_y(x, y) = e^{x y+2 x+y^2} (x+2 y) \implies f_y(0,0) = 0$

*$f_{xx}(x, y) = (y+2)^2 e^{x y+2 x+y^2} \implies f_{xx}(0,0) = 4$

*$f_{xy}(x, y) = (y+2) e^{x y+2 x+y^2} (x+2 y)+e^{x y+2 x+y^2} \implies f_{xy}(0,0) = 1$

*$f_{yy}(x, y) = e^{x y+2 x+y^2} (x+2 y)^2+2 e^{x y+2 x+y^2} \implies f_{yy}(0,0) = 2 $


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.
</s>