Applications of Calculus

Let \(P(x)=(x-x_0)(x-x_1)\cdots(x-x_n)(x-x_{n+1})\). We have \[P^{\prime}(x)=\sum_{j=0}^{n+1}\frac{P(x)}{x-x_j}\quad\mbox{i}\quad P^{\prime\prime}(x)=\sum_{j=0}^{n+1}\sum_{k\neq j}\frac{P(x)} {(x-x_j)(x-x_k)}\;.\] Therefore \[P^{\prime\prime}(x_i)=2P^{\prime}(x_i) \sum_{j\neq i}\frac1{(x_i-x_j)}\] for \(i=0,1,\dots,n+1\). Thus the condition of the problem is equivalent to \(P^{\prime\prime}(x_i)=0\) for \(i=1,2,\dots,n\). Therefore \[x(x-1)P^{\prime\prime}(x)=(n+2)(n+1)P(x).\] It is easy to see that there is a unique monic polynomial of degree \(n+2\) satisfying the above differential equation. On the other hand, the monic polynomial \(Q(x)=(-1)^nP(1-x)\) satisfies the same equation and has degree \(n+2\), so we must have \((-1)^nP(1-x)=P(x)\), which implies the statement.

If \(P(x)=(x-\alpha)^kQ(x)\), then \(P^{\prime}(x)=(x-\alpha)^kQ^{\prime}(x)+k(x-\alpha)^{k-1}Q(x)\)

If \(P(x)+1\) has a triple zero at point 1, then its derivative \(P^{\prime}(x)\) has a double zero at that point. Similarly, \(P^{\prime}(x)\) has a double zero at point \(-1\) too. It follows that \(P^{\prime}(x)\) is divisible by the polynomial \((x-1)^2(x+1)^2\). Since \(P^{\prime}(x)\) is of degree at most 4, it follows that \[P^{\prime}(x)=c(x-1)^2(x+1)^2=c(x^4-2x^2+1)\] for some constant \(c\). Now \(P(x)=c(\frac15x^5-\frac23x^3+x)+d\) for some real numbers \(c\) and \(d\). The conditions \(P(-1)=1\) and \(P(1)=-1\) now give us \(c=-15/8\), \(d=0\) and \[P(x)=-\frac38x^5+\frac54x^3- \frac{15}8x. \]

Let us assume w.l.o.g. that \(n=\mathrm{deg}P\geq\mathrm{deg}Q\). Let \(P_0=\{z_1,z_2,\dots,z_k\}\) and \(P_1\) \(=\{z_{k+1}\), \(z_{k+2},\dots, z_{k+m}\}\). Polynomials \(P\) and \(Q\) coincide at \(k+m\) points \(z_1,z_2,\dots,z_{k+m}\). The result will follow if we show that \(k+m > n\).

We have \[P(x)=(x-z_1)^{\alpha_1}\cdots (x-z_k)^{\alpha_k}=(x-z_{k+1})^{\alpha_{k+1}}\cdots (x-z_{k+m})^{\alpha_{k+m}}+1\] for some natural numbers \(\alpha_1,\dots,\alpha_{k+m}\). Let us consider \(P^{\prime}(x)\). We know that it is divisible by \((x-z_i)^{\alpha_i-1}\) for \(i=1,2,\dots,k+m\); hence, \[\prod_{i=1}^{k+m}(x-z_i)^{\alpha_i-1}\mid P^{\prime}(x).\] Therefore, \(2n-k-m=\deg\prod_{i=1}^{k+m}(x-z_i)^{\alpha_i-1}\leq \mathrm{deg}P^{\prime}=n-1\), i.e. \(k+m\geq n+1\), as desired.

Let \(a < b\) be two zeros of polynomial \(P\). Assume w.l.o.g. that \(P^{\prime}(a) > 0\) and consider the point \(c\) in the interval \([a,b]\) in which \(P\) attains a local maximum (such a point exists since the interval \([a,b]\) is compact). We know that \(P(x)=P(c)+(x-c)[P^{\prime}(c)+o(1)]\). If for example \(P^{\prime}(c) > 0\) (the case \(P^{\prime}(c) < 0\) leads to a similar contradiction), then \(P(x) > P(c)\) would hold in a small neighborhood of \(c\), a contradiction. It is only possible that \(P^{\prime}(c)=0\), so \(c\) is a root of \(P^{\prime}(x)\) between \(a\) and \(b\).