Particular family of polynomials, proof of shape.

Question

I have found a particular family of polynomials seems to be nice with respect to some interests of mine.

The criteria I had:

1. $P(0) = 0$
2. $P(1) = 1$
3. $P'(x) > 0\hspace{1cm} \forall x\in\,\,]0,1[$
4. $P''(\xi) = 0 \text{ for 1 and only 1 } \xi \in[0,1]$

As far as I can tell by ocular inspection, the family of polynomials $$x\to P_{N,M}(x) = 1-(1-x^N)^M\\[0.5cm] M,N \in \mathbb N$$ has these properties for many pairs $(M,N)$, but can we prove (for which)?

Plot below is example for $N=4, M=8$:

As we can see, very smooth, strictly increasing, one inflection point around maybe $0.5-0.6$ somewhere.

Answer 1

I'm afraid you have to do the calculations....so since I've got a free evening here's the calculus joy:

\begin{equation} P'(x) = MN(1-x^N)^{M-1}x^{N-1}\\ P''(x) = MN(1-x^N)^{M-2}x^{N-2}[(1-M)Nx^N+(N-1)(1-x^N)] \end{equation}

The equation $P''(x)=0$ has clearly two solutions at $x=0$ and $x=1$ for all $M,N$ greater than $2$. It has also another unique nice solution in $[0,1]$:

\begin{equation} (1-M)Nx^N+(N-1)(1-x^N) = 0 \qquad\Longrightarrow\qquad x = \sqrt[N]{\frac{1-N}{1-MN}} \end{equation}