Post History
#3: Post edited
by
Snoopy
·
2022-10-05T20:59:53Z (about 2 years ago)
- 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](https://en.wikipedia.org/wiki/Second_partial_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 $f$ 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 $D=0$?
- 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](https://en.wikipedia.org/wiki/Second_partial_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 $D=0$?
#2: Post edited
by
Snoopy
·
2022-10-05T20:58:25Z (about 2 years ago)
Why can't we conclude extrema property of a function from its quadratic approximation when the discriminant is zero?
- Why can't we conclude the extrema property of a function from its quadratic approximation when the discriminant is zero?
#1: Initial revision
by
Snoopy
·
2022-10-05T20:58:08Z (about 2 years ago)
Why can't we conclude 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](https://en.wikipedia.org/wiki/Second_partial_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 $f$ 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 $D=0$?