Use the second-degree Taylor polynomial to determine if the point $(1,1)$ is a critical point.
$f(x,y)=xy-x-y-2\cos\left(x-y\right)$ around the point $(x,y)=(1,1)$
$$f_{x}=y+2\sin \left(x-y\right)-1\\
f_{xx}=2\cos \left(x-y\right)\\
f_{xy}=-2\cos \left(x-y\right)+1\\
f_{y}=x-2\sin \left(x-y\right)-1\\
f_{yy}=2\cos(x-y)\\
\ \\
f(1,1)=-3\\
f_{x}(1,1)=0\\
f_{xx}(1,1)=2\\
f_{xy}(1,1)=-1\\
f_{y}(1,1)=0\\
f_{yy}(1,1)=2\\
\ \\
-3+0(x-1)+0(y-1)+\frac{2}{2!}(x-1)^2+\frac{-1}{1!1!}(x-1)(y-1)+\frac{2}{2!}(y-1)^2=x^2-xy-x+y^2-y-2
$$
If a insert $(x,y)=(1,1)$ in $x^2-xy-x+y^2-y-2$ i get $=-3$.
How do i use the taylor polynomial to determine if $(1,1)$ is a critical point?
Best Answer
I don't know how to use the second degree Taylor polynomial in a sensible way to do this, but if you must, here it goes:
The 2nd degree Taylor polynomial of a scalar function expanded at $a$ is $$T^2 f(x;a)=f(a)+\nabla f(a)\cdot (x-a)+\frac{1}{2}(x-a)^T\mathrm H f(a)(x-a),$$ where $\nabla f$ is the gradient and $\mathrm Hf$ is the Hessian. Now $f$ has a critical point at $a$ if $\nabla f(a)=0$, in other words, if the second degree Taylor polynomial reduces to $$f(a)+(x-a)^T\mathrm Hf(a)(x-a).$$ But that's just a detour around simply calculating $\nabla f$, which is the actually important condition, which you have to do anyway to find the Taylor polynomial.
What you can do with the 2nd degree Taylor polynomial, though, is to determine the type of critical point: if the Hessian is positive [negative] definite, then we have a minimum [maximum], and if it is indefinite, then we have a saddle point. If it is semi-definite, then the second degree Taylor polynomial can't be used to find the type of critical point.