Can curvature ever be greater than at relative extremum?

WK95
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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


curvature_equation.jpg


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.
 
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I think so, but no non-parametric examples come to mind.
 
WK95 said:

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


curvature_equation.jpg


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.
 
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