0
$\begingroup$

So according to wikipedia inflection point,

If the second derivative, $f''(x)$ exists at $x_0$, and $x_0$ is an inflection point for $f$, then $f''(x_0) = 0$, but this condition is not sufficient for having a point of inflection, even if derivatives of any order exist. In this case, one also needs the lowest-order (above the second) non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection, but an undulation point.

It is also mentioned that:

the condition that the first nonzero derivative has an odd order implies that the sign of $f'(x)$ is the same on either side of $x$ in a neighborhood of $x$

My question is why is it the case? For example, if $f^{3}(c)$ is nonzero, while $f''(c) = 0$. How does the first condition contribute to the fact that $(x, f(x))$ is not an undulation point?. I found this non-obvious but I couldn't find any explanation online. So I am seeking a proof (or any kind of intuition) for this. Thanks in advance.

$\endgroup$
1
  • $\begingroup$ Think about the Taylor polynomial. $\endgroup$
    – saulspatz
    Nov 19 '20 at 15:34
1
$\begingroup$

Following the Taylor development, assuming WLOG the inflection point at $x=0$, we have

$$f(x)\approx f(0)+f'(0)x+f^{(n)}(0)\frac{x^n}{n!}$$

where $n$ is the order of the first nonzero derivative ($n>1$). For $x$ small enough, the higher order terms have no influence. To get an inflection, the last term must be an odd function, so that the curve can cross the straight line $y=f(0)+f'(0)x$.

Below, plots of $1-x, 1-x+x^2,1-x+x^3,1-x+x^4,1-x+x^5$:

enter image description here

$\endgroup$
1
$\begingroup$

Assume that $f$ is $C^\infty$ in a neighborhood of $x=0$. Then the the tangent to the graph of $f$ at $x=0$ is given by $y=t(x):=f(0)+f'(0) x$. We now have to study the function $g(x):=f(x)-t(x)$ in the neighborhood of $x=0$. In any case $g(0)=g'(0)=0$. When all derivative values $f^{(n)}(0)=0$ $\>(n\geq2)$ these derivative values cannot tell us anything about the behavior of $g$ near $x=0$, since $g$ has all derivatives $=0$ at $x=0$.

Now let us assume that there is some $n\geq2$ with $$f^{(k)}(0)=0\quad (2\leq k<n),\qquad f^{(n)}(0)=:c\ne 0\ .$$ Then Taylor's theorem tells us that $$g(x)=c{x^n\over n!}+o(x^n)=x^n\left({c\over n!}+o(1)\right)\qquad(x\to0)\ .$$ This says that $g(x)$ is of a single sign in a punctured neighborhood of $x=0$, when $n$ is even, and $g$ (having the $x$-axis as tangent at $x=0$) changes sign when $x$ passes through $0$, when $n$ is odd. The latter behavior you would describe as an inflection point of $f$ at $x=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.