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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 :)

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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...


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

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  • $\begingroup$ Whoa. Firstly, thank you for the time taken to write this :) I understand your answers to my third and fourth questions, but for the second question, you only provided an example of how if the last non-zero derivative (below the second) is even, then it is still not a point of inflection there. However, how do we then conclude that it is as such for all even-numbered "last" derivatives? Lastly, for your answer to my first question, I am still having a hard time even beginning to understand the logic behind it... as it seems very complicated! Haha. $\endgroup$
    – Ethan Mark
    Sep 21 '20 at 12:21
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    $\begingroup$ My answer to number 2 was meant to show that you can't possible have a proof that works for all last non-zero-derivative possibilities, because here's one for which it's not true. But with this example, the claim that it's true in the odd case remains OK. But to be honest, the more general even "badness" is easy to show too: just look at $f(x) = x^{2n}$, for $n > 1$. These provide examples whose convexity behavior is the same on both sides of $x = 0$, and so they show that the theorem you might hope for does not work "in the even cases"... $\endgroup$ Sep 21 '20 at 13:07
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    $\begingroup$ ... (i.e, if the first nonzero derivative is an even derivative, then the origin is not necessarily an inflection). There's actually a deeper statement: if $f$ is differentiable, and the first nonzero derivative (not counting the "0th" derivative!) at the origin is an even derivative, then the origin is never an inflection point. The proof of that claim is basically a lot like the proof of the "odd" claim that you described in your "further research" paragraph. $\endgroup$ Sep 21 '20 at 13:10
  • $\begingroup$ I see. Thank you :) For the proof for my first question, we also have to do prove the other side right? Where the last derivative is negative? Then we combine the two to get non-zero derivative? $\endgroup$
    – Ethan Mark
    Sep 21 '20 at 13:34
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    $\begingroup$ If I understand your question, then the answer is "yes". It's typical in proofs like this that what you do to the left of 0 is really similar to what you do to the right of zero, so we write out one case and leave the other to the reader; similarly, what we do in the odd case is often very similar (but slightly different) to what we do in an even case, and one or the other may be 'left to the reader". Sometimes these "left to the reader" things can take hours, alas. But they do build up your skill-set, which is good. $\endgroup$ Sep 21 '20 at 13:59

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