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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?

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    $\begingroup$ 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. . $\endgroup$ – user44191 Jan 20 at 16:37
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    $\begingroup$ What is the definition of "turning point"? $\endgroup$ – Alexandre Eremenko Jan 20 at 18:27
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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 $f$:

enter image description here

And here are relevant values of $f$:

$$f(0)=27>f(8/10)\approx17.7<f(95/100)\approx18.3>f(1)=18.$$

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  • $\begingroup$ 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. $\endgroup$ – Jon Jan 20 at 17:25
  • $\begingroup$ @Jon : Oops! I had copied the value of $B$ from a wrong place. This is now corrected, and everything else holds. $\endgroup$ – Iosif Pinelis Jan 20 at 18:47

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