On locating inflection points

Question

From what I have learnt, a point of inflection of a curve is, by definition, a point where the curve changes concavity.

The Simple Case

Thus, if, for a point, $c$, on a given function, $f(x)$, $f'(c) = f''(c) = 0$ and $f'''(c) \neq 0$, then $c$ is a point of inflection. I believe I understand the explanation for this, as, by definition, the second derivative describes concavity, so the third would necessarily describe the rate of change of concavity. Then, since $f'(c) = f''(c) = 0$ and $f'''(c) \neq 0$, we can conclude that the rate of change of the second derivative is non-zero, so concavity is changing and $c$ is a point of inflection. Please feel free to correct me if my explanation for this is wrong!

The General Case

However, on doing a little more research, I found out that this phenomenon can actually be generalised as follows: If $f(x)$ is $k$ times continuously differentiable in a certain neighborhood of a point $x$ with $k$ odd and $k ≥ 3$, while $f^{(n)}(x_0) = 0$ for $n = 2, …, k − 1$ and $f^{(k)}(x_0) ≠ 0$, then $f(x)$ has a point of inflection at $x_0$.

Questions

I do not understand how to explain this, since I thought that only the third derivative (and not other higher-order derivatives) describes rate of change of concavity, so I have the following four questions:

1. How do we generalise my observation about the feature of the third derivative to any odd-numbered derivative (below the second)?

2. Why does this generalisation only apply to odd-numbered derivatives (below the second)? In other words, why does it not apply to even-numbered derivatives (below the second)?

3. I also know that there can be inflection points where the second derivative is undefined. How, then, do we confirm that there is an inflection point there? Is the fact that the second derivative is undefined a sufficient condition?

4. As an extension to my third question, what if the second derivative is defined and equals zero at the particular point, but the third derivative is undefined? How, then, do we confirm that there is an inflection point there?

Background

Perhaps I might add that I am currently taking a introductory module in calculus at university level, so my level of knowledge about calculus at present may not be in-depth enough to understand the sophisticated explanations that I expect would be coming my way. I have learnt IVT, EVT, Rolle's Theorem, MVT, Cauchy's MVT and L'Hopital's Rule, but that is about it as far as theorems are concerned, so I would greatly appreciate it if there are any intuitive/"lower-level" explanations to this :)

Answer 1

That's a lot of questions. To answer number 2, look at the examples $$ f(x) = x^4 $$ and $$ f(x) = x^5. $$

The first has three zero derivatives at $x = 0$, but the fourth is nonzero. Nonetheless, there's no inflexion, because we go from concave up, left of $x = 0$, to concave-up (right of $x = 0). Hence, no inflection.

The second is an example of your "little more research" -- by having its first nonzero derivative be of odd order, it actually does constitute a point of inflection.

For #3, #4: Not every problem is easy. When you have an undefined second derivative...you often have to explicitly compute the concavity to the left and right of the inflexion point to confirm that it really is an inflection. There's no "general shortcut." What's cool is that in so many cases, mere differentiability actually does work, which sort of justifies all the work we put in learning calculus: it often pays off. In the cases where it doesn't, you can hardly get upset at it!

For #1: The proof that a function with these properties is convex up on the right, and convex down on the left (at least if the first nonzero derivative is an odd one, and positive), at least in a neighborhood of the point being considered, is basically a repeated application of the mean value theorem. I started writing out the proof, but ... I have to go do other stuff, so I'm abandoning my start partway. The gist is that if the first nonzero derivative $f^{(k+1)}(0) = M$, then in a small enough region just to the right of zero, $f(x)$ looks a lot like $q(x) = M x^{k+1}$ (like, "so similar that all their derivatives have the same signs everywhere on that interval"), which is enough to show that $f'$ is increasing on one side, and decreasing on the other, and you're done. I'm leaving behind the start of my top-of-the-head proof, but it's only the start of the trail...

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One proof goes something like this:

First, if we're looking at $x = a$, then replace $f$ with $$ g(x) = f(x - a) - f(a) $$ and then we've got the same hypotheses for $g$, but at the origin, and with $g(0) = 0$. So now we're proving this:

If $g(0) = g'(0) = \ldots = g^{(k)}(0) = 0$, but $g^{(k+1)}(0) > 0$, and $k$ is even, then $g$ has a point of inflection at $0$.

To show this, we'll show that for some $b > 0$,we have that $$ g': [-b, b] \to \Bbb R $$ is (i) increasing on $[0, b]$ but (ii) decreasing on $[-b, 0]$.

Actually, I'll only show item (i), and leave $(ii)$ to you.

I need a lemma:

Lemma: If

1. $h(0) = 0$, and
2. $h$ is differentiable at $0$, and
3. the sequence of positive numbers $x_1, x_2, \ldots, x_n$ converges to $0$, and
4. $h(x_i) \le 0$ for all $i$, then

$$h'(0) \le 0$$. The proof is fairly easy: We know \begin{align} h'(0) &= \lim_{s \to 0}\frac{h(0+s) - h(0)}{s} \\ &= \lim_{s \to 0}\frac{h(0+s) - 0}{s} & \text{hypothesis 1}\\ &= \lim_{s \to 0}\frac{h(s)}{s}. \end{align} Now suppose that $h'(0) = M > 0$. Then for every $\epsilon > 0$, there's a $\delta > 0$ such that $|x - 0| < \delta$ implies that $|\frac{h(x)}{x} - M| < \epsilon$. Well, since that's true for every epsilon, it must be true for $\epsilon = \frac{M}{2}$. So let's choose $\delta$ to be a number that makes this difference less that $M/2$. Then for every $x$ with $$ |x - 0| < \delta $$ we have $$ | \frac{h(x)}{x} - M | < M/2 $$ which tells us that $\frac{h(x)}{x} \ge M / 2 > 0. In words: near $0$, the ratio $h(x)/x$ is at least $M/2$ away from zero.

Now pick a number $n$ so large that $$ |x_n - 0| < \delta, $$ which you can do because the $x_i$s converge to $0$. We can conclude that $$ \frac{h(x_n) / x_n} \ge M/2 $$ But we know $x_n > 0$ (by hypothesis 3) and $h(x_n) \le 0$ (hypothesis 4), so this is impossible. QED

A corresponding theorem where $h(x_i) \ge 0$ also works, with essentially the same proof, as does one where the $x_i$ converge to $0$ from below rather than from above. You should probably write out the theorems in each of these cases to get them right.

Let's look at the case where $k = 4$ to keep things simple. I think once you see that, you can work out the details for $k = 6, 8, $ and then (by induction?) do the general case.

So we have $f(0) = f'(0) = \ldots = f^{(4)} = 0$, and $f^{(5)}(0) > 0$, and we're claiming that for some small distance $b$ to the right of $x = 0$, the function $f'$ is increasing. I'll show this by showing that (**) for some number $b$, we have $f''(x) > 0$ for $0 < x \le b$. You can then apply the mean value theorem to show that $f$ is increasing.

Suppose (**) is false. Then (picking $b = 1/n$), for every positive integer $n$, we have that on the interval $(0, 1/n]$, there's a point, which I'll call $x_n$, where $f''(x) \le 0$. These points $x_n$ are all positive, and they clearly converge to zero. Our lemma lets us conclude that

Answer 2

As suggested by @John, I will provide an answer to OP's question using the Taylor's theorem. It will make for a much shorter answer. We want to prove the following:

If $f$ is $k$ times differentiable in a neighborhood of $x_0$, with $k$ odd and $k \geq 3$. Also, $f^{(n)}(x_0) = 0$ for $2 \leq n < k$ and $f^{(k)}(x_0) \neq 0$. Then $f$ has an inflection point at $x_0$. An inflection point being where the $f^{(2)}(x)$ changes sign.

By Taylor's theorem: $$\begin{align} f^{(2)}(x) &= f^{(2)}(x_0) \; + \; \sum_{j = 3}^k \frac{f^{(j)}(x_0)}{(j-2)!}(x -x_0)^{j-2}\; + h(x)(x-x_0)^{(k-2)}\\ &= \frac{f^{(k)}(x_0)}{(k-2)!}(x -x_0)^{k-2}\; + h(x)(x-x_0)^{(k-2)}\\ &= \left(\frac{f^{(k)}(x_0)}{(k-2)!} + h(x)\right)(x-x_0)^{(k-2)}\\ &= g(x)(x-x_0)^{(k-2)}\text{ say}\\ \text{where }&\lim_{x \rightarrow x_0} h(x) = 0\\ \end{align}$$

It follows that $\exists \delta>0,x \in (x_0-\delta,x_0 + \delta) \implies |h(x)| < |\frac{f^{(k)}(x_0)}{(k-2)!}|$. Therefore sign of $g(x)$ does not change for $x \in (x_0-\delta,x_0 + \delta)$. However since $k-2$ is odd, the sign of $(x-x_0)^{(k-2)}$ changes at $x_0$, which makes the sign of the whole expression $g(x)(x-x_0)^{(k-2)}$ change. Which is the same as saying that sign of $f^{(2)}(x)$ changes sign at $x_0$. Hence $x_0$ is an inflection point of $f$.