Minimum of an approximated function I have the following function
$$ f(x,y) = 100(y-x^2)^2 + (1-x)^2$$
which has minimum at point $(1,1)$. I was asked to find it's quadratic approximation by using Taylor's theorem. By using the derivatives
$$f_x=400(x^3-xy)+2x-2$$
$$f_y=200(y-x^2)$$
$$f_{xx}=400(3x^2-y)+2$$
$$f_{yy}=200$$
$$f_{xy}=-400x$$

I found Taylor's expansion around $(0,0)$ as follows:
$$Q(x, y) = f(0,0)+f_x (0,0)x +f_y(0,0)y+\frac{1}{2}f_{xx}(0,0)x^2+f_{xy}xy + \frac{1}{2} f_{yy}(0,0)y^2 $$

$$=1 -2x +0y +\frac{1}{2}2x^2+0xy+\frac{1}{2}200y^2$$
$$=x^2-2x+100y^2+1$$

However, this new function has a minimum at point $(1,0)$. My profesor told me that this can't be right and that the new approximated function $Q$ should have min at the same point $(1,1)$.

Im not sure why is this true and I cant find any mistakes in my solution. Am I wrong in thinking that it makes sense that the new fucntion $Q$ doesnt have the same minima with the old?

Does anyone have any insight? 

Thanks in advance !
Note that the Taylor expansion is to be performed around the minimum $(1,1)$. Denote $\epsilon = O( y-1)=O(x-1)$. Then, the quadratic approximation can be obtained as follows,

$$ f(x,y) = 100(y-x^2)^2 + (1-x)^2$$
$$= 100[y-1-(x^2-1)]^2 + (1-x)^2$$
$$= 100[(y-1)^2-2(y-1)(x^2-1)+(x^2-1)^2] + (1-x)^2$$
$$= 100[(y-1)^2-2(y-1)(x-1)(x+1)+(x-1)^2(x+1)^2] + (1-x)^2$$
$$= 100[(y-1)^2-4(y-1)(x-1)+4(x-1)^2] + (1-x)^2+O(\epsilon^3)$$
$$= 100(y-2x+1)^2 + (1-x)^2+O(\epsilon^3)$$

Note, the approximate function still has its minimum at $(1,1)$.



Edit:

$$f_x=400(x^3-xy)+2x-2 = 0$$
$$f_y=200(y-x^2)=0$$
$$f_{xx}=400(3x^2-y)+2 = 800+2$$
$$f_{yy}=200$$
$$f_{xy}=-400x=-400$$

Put it together,

$$f(x, y) = 0+f_x (x-1) +f_y(y-1)+\frac{1}{2}f_{xx}(x-1)^2+f_{xy}(x-1)(y-1) + \frac{1}{2} f_{yy}(y-1)^2 $$
$$ = \frac{1}{2}(800+2)(x-1)^2-400(x-1)(y-1) + 100 (y-1)^2 $$
$$ =(x-1)^2+ 400(x-1)^2-400(x-1)(y-1) + 100 (y-1)^2 $$
$$ =(x-1)^2+ 100(y-2x+1)^2 $$
Since
$f(x,y) = 100(y-x^2)^2 + (1-x)^2
$,
around (1, 1) we have

$\begin{array}\\
f(1+x,1+y) 
&amp;= 100((1+y)-(1+x)^2)^2 + (1-(1+x))^2\\
&amp;= 100((1+y)-(1+2x+x^2))^2 + x^2\\
&amp;= 100(y-2x-x^2)^2 + x^2\\
&amp;= 100(y^2+4x^2+x^4-4xy-2x^2y+4x^3)+x^2\\
&amp;\approx 100(y^2-4xy)+x^2\\
\end{array}
$