Why is $x = c$ an inflection point only if the lowest-order (above the second) non-zero derivative is of odd order (third, fifth, etc.).

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.

• Think about the Taylor polynomial. Nov 19 '20 at 15:34

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 $$1-x, 1-x+x^2,1-x+x^3,1-x+x^4,1-x+x^5$$: 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 $$f^{(k)}(0)=0\quad (2\leq k Then Taylor's theorem tells us that $$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$$.