= Solution
At the origin the needed <partial derivatives> are
$$
f(0,0)=1,\quad f_x(0,0)=0,\quad f_y(0,0)=-3,\quad f_{xx}(0,0)=2,\quad f_{xy}(0,0)=f_{yy}(0,0)=0.
$$
The <Taylor series> through total degree two is therefore
$$
f(0,0)+f_x(0,0)x+f_y(0,0)y+\tfrac12\bigl(f_{xx}(0,0)x^2+2f_{xy}(0,0)xy+f_{yy}(0,0)y^2\bigr),
$$
which gives
$$
\boxed{f(x,y)=1-3y+x^2+O(\|(x,y)\|^3)\quad\text{as }(x,y)\to(0,0).}
$$
Here total degree is what matters: $x^2y^2$ is quartic, not quadratic. The exact omitted <polynomial> is $y^3+x^2y^2+x^2y^4$, which verifies the stated <Big O notation> remainder directly.
Back to article page
