Minimum of an approximated function

Question

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 !

Answer 1

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 $(1,1)$.

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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,

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

Answer 2

Define $x=1+u, y=1+v$

$$\begin{array}\\ f(1+u,1+v) &= 100((1+v)-(1+u)^2)^2 + (1-(1+u))^2\\ &= 100(1+v-1-2u-u^2)^2 + (-u)^2\\ &= 100(v-2u-u^2)^2 + u^2\\ &= 100(v^2+4u^2+u^4-4uv-2u^2v+4u^3)+u^2\\ &\approx 100(v^2-4uv)+401u^2\\ &= 100v(v-4u)+401u^2\\ &= 100(y-1)(y-1-4(x-1))+401(x-1)^2\\ &= 100(y-1)(y-4x+3)+401(x-1)^2\\ \end{array}$$

Answer 3

Since $f(x,y) = 100(y-x^2)^2 + (1-x)^2 $, around (1, 1) we have

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