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.
Best Answer
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 $1-x, 1-x+x^2,1-x+x^3,1-x+x^4,1-x+x^5$: