# Proving stationary points of inflection

Edit

For the purposes of proving the statement below, a stationary point of inflection of a curve shall be defined as a point on the curve where the curve changes concavity.

Problem

Suppose $$f(x)$$ is $$k$$ times differentiable with $$k \mod 2 \equiv 1$$ and $$k \geq 3$$. Then, if $$f^{(n)}(c) = 0$$ for $$n = 1, ..., k - 1$$ and $$f^{(k)}(c) \neq 0$$, prove that $$c$$ is a stationary point of inflection of $$f$$.

I have successfully proven the cases where $$k = 3$$ and $$k = 5$$ (or so I think) and I am currently trying to devise a proof for the general case above. I am trying to use the ideas from my two proofs (they are largely based on the second derivative test) and am thinking along the lines of induction, but I am not sure if that is wise. Any suggestions/hints/help will be greatly appreciated!

As I am not so well-versed in mathematical proof-writing as I would like to be, I am also providing my proofs for the $$k = 3$$ and $$k = 5$$, so that the community may critique them for me!

Proof for $$k = 3$$

Suppose $$f^{(3)}(c) > 0$$

$$\because f^{(3)}(c) = \lim \limits_{x \to c} \frac {f^{(2)}(x) - f^{(2)}(c)} {x - c} = \lim \limits_{x \to c} \frac {f^{(2)}(x)} {x - c}$$

$$\therefore \lim \limits_{x \to c} \frac {f^{(2)}(x)} {x - c} > 0$$

When $$x \rightarrow c^+$$, $$x > c$$

For $$\lim \limits_{x \to c^+} \frac {f^{(2)}(x)} {x - c} > 0$$, $$f^{(2)}(x) > 0$$

When $$x \rightarrow c^-$$, $$x < c$$

For $$\lim \limits_{x \to c^-} \frac {f^{(2)}(x)} {x - c} > 0$$, $$f^{(2)}(x) < 0$$

$$\because f^{(2)}(x)$$ changes sign at $$c$$

$$\therefore f$$ changes concavity at $$c$$

$$\implies$$ By definition, $$c$$ is a stationary point of inflection of $$f(x)$$

Similarly, if $$f^{(3)}(x) < 0$$, then $$\lim \limits_{x \to c} \frac {f^{(2)}(x)} {x - c} < 0$$

When $$x \rightarrow c^+$$

For $$\lim \limits_{x \to c^+} \frac {f^{(2)}(x)} {x - c} < 0$$, $$f^{(2)}(x) < 0$$

When $$x \rightarrow c^-$$

For $$\lim \limits_{x \to c^-} \frac {f^{(2)}(x)} {x - c} < 0$$, $$f^{(2)}(x) > 0$$

$$\because f^{(2)}(x)$$ changes sign at $$c$$

$$\therefore f$$ changes concavity at $$c$$

$$\implies$$ By definition, $$c$$ is a stationary point of inflection of $$f(x)$$

To conclude, suppose $$f(x)$$ is $$3$$ times differentiable. If $$f^{(n)}(c) = 0$$ for $$n = 1, 2$$ and $$f^{(3)}(c) \neq 0$$, then $$c$$ is a stationary point of inflection of $$f$$.

Proof for $$k = 5$$

Suppose $$f^{(5)}(c) > 0$$

Let $$g(x) = f^{(3)}(x)$$

$$\because g^{(1)}(c) = 0$$ and $$g^{(2)}(c) > 0$$

$$\therefore g(x)$$ has a minimum at $$c$$

$$\because g(c) = 0$$

$$\therefore$$ for all $$x$$ near $$c$$, $$g(x) > 0$$

$$\implies f^{(2)}(x)$$ is an increasing function near $$c$$

In particular, when $$x \rightarrow c^-$$, $$f^{(2)}(x) < f^{(2)}(c)$$ and when $$x \rightarrow c^+$$, $$f^{(2)}(x) > f^{(2)}(c)$$

$$\because f^{(2)}(c) = 0$$

$$\therefore f^{(2)}(x)$$ changes sign at $$c$$

$$\implies f(x)$$ changes concavity at $$c$$

$$\therefore$$ By definition, $$c$$ is a stationary point of inflection of $$f(x)$$

Similarly, if $$f^{(5)}(c) < 0$$, then $$g(x)$$ has a maximum at $$c$$

$$\because g(c) = 0$$

$$\therefore$$ for all $$x$$ near $$c$$, $$g(x) < 0$$

$$\implies f^{(2)}(x)$$ is a decreasing function near $$c$$

In particular, when $$x \rightarrow c^-$$, $$f^{(2)}(x) > f^{(2)}(c)$$ and when $$x \rightarrow c^+$$, $$f^{(2)}(x) < f^{(2)}(c)$$

$$\because f^{(2)}(c) = 0$$

$$\therefore f^{(2)}(x)$$ changes sign at $$c$$

$$\implies f(x)$$ changes concavity at $$c$$

$$\therefore$$ By definition, $$c$$ is a stationary point of inflection of $$f(x)$$

To conclude, suppose $$f(x)$$ is $$5$$ times differentiable. If $$f^{(n)}(c) = 0$$ for $$n = 1, ..., 4$$ and $$f^{(5)}(c) \neq 0$$, then $$c$$ is a stationary point of inflection of $$f$$.

Having been able to come up with these two proofs largely by myself, with some help from my professor, I am actually quite excited on trying a proof for the general case where I am leaning towards induction (actually, it is the only form I can think of), but as my ideas for $$k = 3$$ and $$k = 5$$ are not exactly identical, I am not sure if induction is the way to go.

I am also trying to stick to second derivative tests (or something of similar difficulty) as I am currently only taking an introductory calculus module at university, so I do not have such "high-powered" tools at my disposal, such as Taylor's Series/Theorem and the likes of it.

Also, apologies for the lengthy post!

Edit 2

Proof for the general case (Many thanks to John Hughes for the guidance)

Let $$g(x) = f(x + c) - f(c)$$

$$\implies g(0) = 0$$ and $$g^{(k)}(0) = f^{(k)}(c)$$

Then, it suffices to prove that, if $$0$$ is a stationary point of inflection of $$g$$, $$c$$ will be a stationary point of inflection of $$f$$.

Suppose $$g^{(3)}(0) > 0$$

$$\because g^{(3)}(c) = \lim \limits_{x \to 0} \frac {g^{(2)}(x) - g^{(2)}(0)} {x - 0} = \lim \limits_{x \to 0} \frac {g^{(2)}(x)} {x}$$

$$\therefore \lim \limits_{x \to 0} \frac {g^{(2)}(0)} {x} > 0$$

When $$x \rightarrow 0^+$$, $$x > 0$$

For $$\lim \limits_{x \to 0^+} \frac {g^{(2)}(x)} {x} > 0$$, $$g^{(2)}(x) > 0$$

$$\because g^{(2)}(x) > 0$$ for some $$x \in (0, b)$$ and $$f^{(2)}(x) = g^{(2)}(x - c)$$,

$$\therefore f^{(2)}(x) > 0$$ for some $$x \in (c, b + c)$$

When $$x \rightarrow 0^-$$, $$x < 0$$

For $$\lim \limits_{x \to 0^-} \frac {g^{(2)}(x)} {x} > 0$$, $$g^{(2)}(x) < 0$$

$$\because g^{(2)}(x) < 0$$ for some $$x \in (-b, 0)$$ and $$f^{(2)}(x) = g^{(2)}(x - c)$$,

$$\therefore f^{(2)}(x) < 0$$ for some $$x \in (-b + c, c)$$

$$\implies f^{(2)}$$ changes sign near $$c$$

$$\implies f$$ changes concavity at $$c$$

$$\therefore c$$ is a stationary point of inflection of $$f$$

Similarly, if $$g^{(3)}(0) < 0$$, then $$\lim \limits_{x \to 0} \frac {g^{(2)}(x)} {x} < 0$$

When $$x \rightarrow 0^+$$

For $$\lim \limits_{x \to 0^+} \frac {g^{(2)}(x)} {x} < 0$$, $$g^{(2)}(x) < 0$$

$$\because g^{(2)}(x) < 0$$ for some $$x \in (0, b)$$ and $$f^{(2)}(x) = g^{(2)}(x - c)$$,

$$\therefore f^{(2)}(x) < 0$$ for some $$x \in (c, b + c)$$

When $$x \rightarrow 0^-$$

For $$\lim \limits_{x \to 0^-} \frac {g^{(2)}(x)} {x} < 0$$, $$g^{(2)}(x) > 0$$

$$\because g^{(2)}(x) > 0$$ for some $$x \in (-b, 0)$$ and $$f^{(2)}(x) = g^{(2)}(x - c)$$,

$$\therefore f^{(2)}(x) > 0$$ for some $$x \in (-b + c, c)$$

$$\implies f^{(2)}$$ changes sign near $$c$$

$$\implies f$$ changes concavity at $$c$$

$$\therefore c$$ is a stationary point of inflection of $$f$$

To conclude, suppose $$f(x)$$ is $$k$$ times differentiable with $$k \mod 2 \equiv 1$$ and $$k \geq 3$$. If $$f^{(n)}(c) = 0$$ for $$n = 1, ..., k - 1$$ and $$f^{(k)}(c) \neq 0$$, then $$c$$ is a stationary point of inflection of $$f$$.

• Sorry, but could you define "stationary point of inflection" please? Sep 22 '20 at 16:01
• @K.defaoite I have edited the post :) Sep 22 '20 at 16:28

This is great. I want to make a first suggestion for shortening/simplifying your proof. Observe that if you prove the theorem in the case where $$c = 0$$ and $$f(0) = 0$$, then you've also proved it in the general case, for if $$g$$ is a function that satisfies your general hypotheses, you can define $$f(x) = g(x+c) - g(c).$$ Now $$f(0) = 0$$ as required, and by applying basic differentiation rules, you have $$f^{(k)}(0) = g^{(k)}(c),$$ so your "special case" theorem tells you that $$f$$ has an inflection at $$0$$, so $$g$$ has an inflection at $$c$$. So now you can change the start of your proof to this:

Suppose $$f(x)$$ is $$k$$ times differentiable with $$k \mod 2 \equiv 1$$ and $$k \geq 3$$. Then, if $$f^{(n)}({\color{red} 0}) = 0$$ for $$n = {\color{red} 0},1, ..., k - 1$$ and $$f^{(k)}({\color{red} 0}) \neq 0$$, prove that $${\color{red} 0}$$ is a stationary point of inflection.

Proof for $$k = 3$$.

Suppose $$f^{(3)}({\color{red} 0}) > 0$$

$$\because f^{(3)}({\color{red} 0}) = \lim \limits_{x \to {\color{red} 0}} \frac {f^{(2)}(x) - > f^{(2)}({\color{red} 0})} {x - {\color{red} 0}} = \lim \limits_{x \to {\color{red} 0}} \frac {f^{(2)}(x)} {x}$$

$$\therefore \lim \limits_{x \to {\color{red} 0}} \frac {f^{(2)}(x)} {x } > 0$$

When $$x \rightarrow {\color{red} 0}^+$$, $$x > {\color{red} 0}$$

For $$\lim \limits_{x \to {\color{red} 0}^+} \frac {f^{(2)}(x)} {x } > 0$$, $$f^{(2)}(x) > 0$$ $${\color{blue} {questionable}}$$

When $$x \rightarrow {\color{red} 0}^-$$, $$x < {\color{red} 0}$$

For $$\lim \limits_{x \to {\color{red} 0}^-} \frac {f^{(2)}(x)} {x} > 0$$, $$f^{(2)}(x) < 0$$

$$\because f^{(2)}(x)$$ changes sign at $${\color{red} 0}$$

$$\therefore f$$ changes concavity at $${\color{red} 0}$$

$$\implies$$ By definition, $${\color{red} 0}$$ is a stationary point of inflection of $$f(x).$$

Similarly, $${\color{red} \ldots}$$

The claim that because the limit is positive, the function is positive at $$x$$ doesn't quite make sense, because $$x$$ doesn't mean anything outside the context of the limit.

Added post-comments What you've written is that "Because $$\lim_{x \to 0} \frac{f(x)}{x} > 0$$, $$x > 0$$. Let me try to explain why that sentence is meaningless, even though the underlying idea -- that if $$\lim_{x \to 0} q(x) > 0$$, then $$q$$ is positive in some neighborhood of zero -- is indeed correct. Suppose I told you that $$\sum_{i = 0}^\infty a_i$$ is an odd integer. Can you say anything about $$a_i$$? Of course not, because $$i$$ here doesn't mean anything. It only meant something when you were performing the sum, where you said "first take $$i = 0$$, so that's $$a_0$$; then take $$i = 1$$ and add it, so that's $$a_0 + a_1$$ so far. Now take $$i = 2$$ and get $$a_2$$, and add that to the sum-so-far, to get $$a_0 + a_1 + a_2$$, and so on.

In just the same way, when you say $$\lim_{t\to 0} r(t)$$, the "t" has no meaning outside the limit. As an example, $$\lim_{t \to 0} \cos(t) = 1 > 0$$. Does that mean $$\cos(t) > 0$$? Well, it implies that fact for some values of $$t$$, but $$\cos(\pi) = -1 < 0$$, so it doesn't imply it for all $$t$$, does it? For which values of $$t$$ is it true? Answer: for all values of $$t$$ that are near enough to zero, or ... expressed more formally, there's some number $$s > 0$$ such that for $$-s< t < s$$, we have $$\cos(t) > 0$$. That's the statement you wanted to assert when you simply said "$$f(x) > 0$$".

The opposite form of that assertion about intervals is that if EVERY interval around zero, no matter how small, contains a zero of some function $$f$$, and the limit as $$x \to 0$$ exists, then that limit must be zero, rather than being strictly positive. That's the content of the "little lemma" I prove below.

Here's a little lemma:

Suppose $$f$$ is continuous, so that $$\lim{x \to 0} f(x) = L$$ exists, and for every number $$b>0$$, there's a number $$-b < x_b < b$$ with $$f(x_b) = 0$$. Then the limit must be zero.

That's not hard to prove (you do have to use epsilons and deltas, and the triangle inequality --- proof by contradiction works best here). From that lemma, we can say the following:

If $$\lim{x \to 0} f(x) = L \ne 0$$, then there's some number $$b$$ such that for all $$x$$ with $$-b < x < b$$, $$f(x) \ne 0$$.

In fact, with just a little more work (you need the intermediate value theorem), you can show

If $$\lim{x \to 0} f(x) = L > 0$$, then there's some number $$b$$ such that for all $$x$$ with $$-b < x < b$$, $$f(x) > 0$$.

...and a similar result holds for the case $$L < 0$$. Now you can replace the "questionable" line with this:

In the case $$\lim \limits_{x \to {\color{red} 0}^+} \frac {f^{(2)}(x)} {x } > 0$$, the lemma tells us there's some interval $$-b < x < b$$ such that $$\frac {f^{(2)}(x)} {x } > 0$$; for positive values of $$x$$, this implies that in the interval $$0 < x < b$$, $$f^{(2)}(x) > 0$$; for $$-b < x < 0$$, we can conclude that $$f^{(2)}(x) < 0$$.

To handle the inductive case, ... you're right. The pattern isn't obvious. You'd like to use the theorem in the $$n-2$$ case to prove the $$n$$ case, but what function would you apply it to?

I suspect that you can actually make this work by looking at the function $$h(x) = \begin{cases} \frac{f(x)}{x^2} & x \ne 0 \\ 0 & x = 0 \end{cases},$$ which (by your hypotheses) is continuous and differentiable (although both these need proving). I think that the theorem's result for $$h(x)$$ in case $$n-2$$ will prove it for $$f(x)$$ in case $$n$$. But to be honest, that's just a guess right now.

Still...nice work on working through the challenges of proving a new theorem. If it feels really good, even if it took a long time, then you've taken a first step towards being a mathematician.

• Thank you for your comments! I have two questions: Firstly, how is it we can just prove the case where $c=0$? I thought that we have to prove for all $c$? Secondly, correct me if I am wrong here, but I thought I learnt/proved somewhere in the course that when $\lim \limits_{x \to c} f(x) > 0$ for $x$ near $c$? Sep 25 '20 at 4:31
• Suppose we know the theorem is true at $0$ for any function $f$, i.e., if $f$ has $k$ derivatives (where $k$ is even, $\ge 2$) being $0$ at $0$, then $f$ has an inflection at $0$. Suppose that $g$ has $k$ (again even, at least 2) derivatives being $0$ at $c$. I want to show $g$ has an inflection at $c$. I define $f(x) = g(x + c)$. You can check $f(0) = g(0+c) = g(c) = 0; f'(0) = g'(0+c) = g'(c) = 0$, and so on. So our baby theorem applies to $f$, and we conclude that for $x$ near, but below $0$, $f''(x)$ has one side, and for $x$ near, but above $o$, $f'(x)$ has the opposite sign.(cont.) Sep 25 '20 at 15:53
• Now $f''(x) = g''(x+c)$, so for numbers near but a little below $c$, $g''(x) = f''(x-c)$, which is $f''(u)$ where $u$ is near but a little below $0$, so it has some sign; for numbers near but a little above $c$, $g''(c) = f''(v)$ where $v$ is near but a little above $0$, hence has the opposite sign. So we're done. Sep 25 '20 at 15:55
• Ah. I see what you mean now. So just to clarify, we are indeed proving the hypothesis for all $c$, it is just that in my hypothesis I used $f(x)$ and my $f(x)$ is actually the $g(x)$ in the proof you are suggesting, am I right? Sep 25 '20 at 16:35
• The jump from "0 is a stationary point of inflection of g" to "c is a stationary point of inflection of f" probably deserves a one-sentence argument, saying that if $g''(x)$ is negative for $-b <x < 0$, then $f''(x) = g''(x + c)$ must be negative for $-b+c < x < c$, or something like that. Otherwise fine. " Sep 26 '20 at 15:54