Why can't we conclude the extrema property of a function from its quadratic approximation when the discriminant is zero?
Suppose $f:\mathbf{R}^2\to\mathbf{R}$ is a smooth function and $P=(0,0)$ is a critical point of $f$. The second-derivative test is inconclusive when the discriminant at $P$ is zero: $$f_{xx}(0,0)f_{yy}(0,0)-(f_{xy}(0,0))^2=0\ .$$
For simplicity, assume further that $f(0,0)=0$ and $f_{xx}(0,0)\ne 0$. Then the quadratic approximation of $f$ at $P$ is of the form
\begin{align} g(x,y) =ax^2+bxy+cy^2=a\left[ (x+\frac{b}{2a}y)^2+(\frac{D}{4a^2})y^2\right], \end{align} where $D=4ac-b^2$. In the case when $D\ne 0$, $f$ and $g$ have the same property at $P$ by the second-derivative test.
But when $D=0$， we can no longer conclude the property of $f$ at $P$ from that of $g$ unless $f(x,y)=g(x,y)=a(x+\frac{b}{2a}y)^2$, where $g$ has a local min (resp. max) at $P$ when $a>0$ (resp. $a<0$). I would like to understand why.
Question: why can we no longer conclude the local min/max property for $f$ from $g$ when $D=0$?
1 answer
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Snoopy | (no comment) | Oct 7, 2022 at 17:25 |
You've basically walked right up to the explanation. The second-order approximation you describe, when $D = 0$, is
$$g(x, y) = a\left(x + \frac{b}{2a}y\right)^2$$
Define $z = x + \frac{b}{2a}y$, and we have $g(x, y) = g(z) = az^2$.
This means that $g$ describes a function that really only has one independent variable, a linear combination of $x$ and $y$. It looks like a parabola dragged along a line, and every point in the path of the vertex of the parabola is an extremum.
But since $g$ is only a second-order approximation to $f$, higher-order effects in $f$ may curve the path of the vertex with or against the curvature of the parabola, which will determine whether the critical point is an extremum or a saddle point. This isn't an issue when $D \neq 0$, because then the second-order effects will dominate the higher-order effects and knowing the behavior of $g$ near the critical point is sufficient.
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