Can curvature ever be greater than at relative extremum?

[WK95]

Homework Statement

For a generic function y=f(x) which is twice-differentiaable, is it possible for there to be a curvature on the curve of that function that is greater than the curvature at its relative extremum?

Homework Equations

The Attempt at a Solution

From visualization and a sketch, I would say yes. but I'd like to be able to explain this mathematically.

At the maximum,

K = (d2y/dx2) since dy/dx=0 at the extremum making the denominator equal to 0.

 

[Dr. Courtney]

I think so, but no non-parametric examples come to mind.

 

[Ray Vickson]

Homework Statement

For a generic function y=f(x) which is twice-differentiaable, is it possible for there to be a curvature on the curve of that function that is greater than the curvature at its relative extremum?

Homework Equations

The Attempt at a Solution

From visualization and a sketch, I would say yes. but I'd like to be able to explain this mathematically.

At the maximum,

K = (d2y/dx2) since dy/dx=0 at the extremum making the denominator equal to 0.

The denominator had better not be zero! You mean that the denominator = 1.

Anyway, the answer to your question is yes, we can devise a function with maximal curvature at a non-stationary point. I will explain how we can do that.

First, start with a base function, such as $f_1(x) = c_0 + c_1(x-a)(b-x)$ whose maximum occurs at the average of $a$ and $b$; that is, $x_1 = (a+b)/2$. Imagine that the gap $b-a$ is large and the coefficient $c_1 > 0$ is small; that will make the maximum at $x_1$ occur where the graph $y = f(x)$ has "small" curvature. Suppose that $x_1 > 2$, say, and that $f_1(1) = c_0 + c_1 (1-a)(b-1) = 1$ by choice of $c_0$. Now consider the new function
f_2(x) = \begin{cases} x &amp; x &lt; 1 \\<br /> f_1(x) &amp; x \geq 1<br /> \end{cases}
The function $f_2(x)$ is maximized at $x = x_1$, which is a stationary point. It is continuous, but has a discontinuous derivative at $x = 1$. Now imagine "smoothing out" the break at $x = 1$ by a nearby infinitely-differentiable smooth function, to get a new function $f_3(x)$ that has everywhere continuous derivatives of all orders, but closely resembles the function $f_2(x)$. Its curvature will be very large and maximal near the point $x = 1$, but its only stationary point will be near $x = x_1$, which is $> 2$.

Such smoothing functions occur, for example, in constrained optimization, where they are sometimes used to approximate absolute barrier functions. For example, we can approximate the function $|x|$ by a $C^{\infty}$ function such as $\sqrt{x^2 + \epsilon^2}$, which is close to $|x|$ for small $|\epsilon|$, and for $x$ values away from 0. In a similar way, you can approximate a "ramp" function such as $f(x) = x, \: x < 0$ and $f(x) = 0, \; x \geq 0$ by an infinitely smooth nearby version. You can do the same type of thing to the non-smooth function $f_2$ constructed above.