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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 $f$.

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

end of post-commment addition

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.

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  • $\begingroup$ 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 $c$? $\endgroup$
    – Ethan Mark
    Sep 25 '20 at 4:31
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    $\begingroup$ 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.) $\endgroup$ Sep 25 '20 at 15:53
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    $\begingroup$ 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. $\endgroup$ Sep 25 '20 at 15:55
  • $\begingroup$ 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? $\endgroup$
    – Ethan Mark
    Sep 25 '20 at 16:35
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    $\begingroup$ 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. " $\endgroup$ Sep 26 '20 at 15:54

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