Given a Taylor polynomial of degree $2$, study if there is a saddle point at a point

Question

True or false? "The Taylor polynomial of order $2$ of a field $f$ at point $(-2,1)$ is $$p(x,y)=1+\dfrac 12(^2+4xy+y^2),$$ so $f$ has in $(-2, 1)$ a saddle point."

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Let's take a look at the critical points of $f$: $$\nabla f(x,y)=\nabla p(x,y)=(x+2y,2x+y)\quad\Rightarrow\quad(x+2y,2x+y)=(0,0)\quad\Rightarrow\quad\begin{cases}x+2y&=0\\2x+y&=0\end{cases}\quad\Rightarrow\quad\text{The critical points are }\{(0,0)\},$$

hence $(-2,1)$ is not a critical point, hence $$\boxed{(-2,1)\quad\text{is not a saddle point}}.$$

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I do not know if it is correct to say that $ f $ "is equal" to $ p $. I did exercises where the partial derivatives coincide with those of the Taylor Polynomial of order 2, but here I do not know if it is like that.

Are my reasoning correct?

Thank you!

Answer 1

Taylor expansion describes the local behavior of a function. As a result, a more rigorous way to state the result is

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Since

$$ \left.\nabla f(x, y)\right|_{(x, y) = (-2, 1)} = \left.\nabla p(x, y)\right|_{(x, y) = (-2, 1)} = (0, -3) \neq (0, 0), $$

$(-2, 1)$ is not a critical point of $f(x, y)$, hence $(-2, 1)$ is not a saddle point.

Answer 2

No, you may not say that $f$ is equal to $p$, though you can tell them equal to the second order.

From the definition of the Taylor development, you can conclude that the gradients and Hessians of $f$ and $p$ are equal at $(-2,1)$.

The gradient is indeed $0$ and the determinant of the Hessian

$$\begin{vmatrix}2&4\\4&2\end{vmatrix}$$ is negative, hence you do have a saddle point.

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This hold iff $x$ and $y$ denote the displacements from $(-2,1)$ and not absolute coordinates.