Intuition for sign of third derivative

[economicsnerd]

[I asked this question over a year ago, but I thought I'd try again.]

Let $I\subseteq \mathbb R$ be an interval and $f:I\to\mathbb R$ be a $C^\infty$ function.

I have the following characterizations:
1) $f'\geq 0$ everywhere iff $f$ is increasing.
2) $f''\geq 0$ everywhere iff $f$ is convex.

The underlined properties above are very nice for a couple reasons:
- They're easy to interpret/visualize. e.g. An increasing function is one with all secant lines having slope $\geq 0$; a convex function is one with all secants lying above its graph.
- They're both global properties.
- They're both easy to state without having defined a derivative. That is, I can define an increasing function or a convex function, even if my audience doesn't understand what a derivative is.

Is there a nice interpretable condition which is equivalent to $f'''\geq0$? Ideally, I'd be interested in a condition which (like monotonicity and convexity) is global and makes no reference to differentiation.

 

[FactChecker]

I guess f''' > 0 means the function is getting tightly more curved upward as the independent variable increases.

 

[mfb]

I guess f''' > 0 means the function is getting tightly more curved upward as the independent variable increases.

It is hard to visualize this with functions like f(x)=x^3 in the interval [-1,0] as the function is not curved upwards. "More upwards" in the sense of a larger second derviative, sure, but it's hard to understand that in an intuitive way.

 

[mfb]

Huh?
The third derivative of f(x)=x^3 is 6 everywhere, and 6>0.

 

[FactChecker]

Huh?
The third derivative of f(x)=x^3 is 6 everywhere, and 6>0.

I stand corrected. I deleted my incorrect answer.

 

[mathwonk]

well x^3 is going from curved downwards to curved upwards, so it is getting more curved upwards in a sense as you move to the right. unfortunately the curvature is not exactly proportional to the second derivative, so although the second derivative is increasing the curve itself is apparently straightening out to the right. what is the formula for the curvature of y = x^3, anyway?

it seems to be |f''|/{1 + (f')^2}^(3/2). good grief. so we get |6x|/(1+9x^4)^(3/2), hmmmm. well anyway that's why it is straightening out.

 

[davidmoore63@y]

How about:

Given two points A and B where B is to the right of A : a function with f"'>0 is one which at B lies above any parabola which is osculating at A. [by osculating I mean it matches f in respect of ordinate, slope and second derivative].

Not particularly nice I suppose

 

[economicsnerd]

I think I found one that works okay.

For some fixed $d>0$ and $\lambda\in(0,1)$, let $g(x)$ denote the "gap" between the graph of $f$ and its secant between $x$ and $x+d$, measured proportion $\lambda$ of the way along. That is, let $g(x):= [\lambda f(x+d) + (1-\lambda) f(x)] - f(x+\lambda d)$. What a positive third derivative requires is that this gap be increasing in $x$.

One could explain that to a person with no calculus training, right?

 

[mfb]

I guess it is possible to prove that statement with a taylor approximation to second order and a remainder which uses the third derivative. Then use that this derivative is positive everywhere.
Didn't check it in detail, but I also did not find a counterexample and the proposed formula looks reasonable.