# Characterising the number of turning points of a 'generalised' polynomial

Is it possible to find the number of turning points of a power function whose largest exponent is some real number known to lie between $$(n,n+1)$$ for some $$n\in\mathbb{Z}$$?

To give an example

Consider the function $$f:(0,1)\rightarrow\mathbb{R}$$ with: $$f(z)=A(1-z)^{\gamma+1}+Bz(1-z)^{\gamma}+Cz^{\gamma+1}$$

where A,B,C are real constants and $$\gamma\in(0,1)$$. This implies that the highest power of $$z$$ lies in (1,2), and my initial intuition was that this should imply that the function has at most 1 turning point. Any suggetsions on how to go about proving this?

• I think this should clearly be false. Let $P(x)$ be a polynomial with $k$ turning points, all strictly positive. Then $P(x^\alpha)$ is a generalized polynomial with $k$ turning points, no matter how small $\alpha$ is. . – user44191 Jan 20 at 16:37
• What is the definition of "turning point"? – Alexandre Eremenko Jan 20 at 18:27

## 1 Answer

If by a turning point you mean a point where the function switches from increasing to decreasing or vice versa, then the function $$f$$ given by $$f(z)=A(1-z)^{\gamma+1}+Bz(1-z)^{\gamma}+Cz^{\gamma+1}$$ with $$A=27,B=4,C=18,\gamma=2/5$$ has (not one but) two turning points, near $$0.74$$ and near $$0.98$$.

Here is the graph of $$f$$:

And here are relevant values of $$f$$:

$$f(0)=27>f(8/10)\approx17.7f(1)=18.$$

• Unless I'm missing something, the function with the given values only has one turning point close to 0.78. The function is then increasing and approaches 18 from below as z approaches 1. – Jon Jan 20 at 17:25
• @Jon : Oops! I had copied the value of $B$ from a wrong place. This is now corrected, and everything else holds. – Iosif Pinelis Jan 20 at 18:47