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Based on a previous question, I had the following conjecture and was wondering if anyone knew how to prove it or find a counterexample.

Consider the polynomial $$ p(t)=c\frac{(t-x_0)(t-x_1)\cdots(t-x_n)}{(x-x_0)(x-x_1)\cdots(x-x_n)}$$

where $x_0,x_1,\ldots,x_n$, and $x$ are distinct and all lie in the interval $[a,b]$, and $c\neq 0$. The polynomial $p(t)$ is the lagrange interpolation polynomial of degree $n+1$ satisfying $p(x_i) = 0$ for $i=0,\ldots,n$ and $p(x)=c$. Its $(n+1)$-th derivative is the constant $$p^{(n+1)}(t)=\frac{c(n+1)!}{(x-x_0)(x-x_1)\cdots(x-x_n)}$$

Conjecture: suppose $f(t)\in C^{(n+1)}[a,b]$ and satisfies $f(x_i) = 0$ for $i=0,\ldots,n$ and $f^{(n+1)}(t)>p^{(n+1)}(t)$ for all $t\in [a,b]$ or $f^{(n+1)}(t)<p^{(n+1)}(t)$ for all $t\in [a,b]$. Prove that $f(x) \neq p(x)=c$.

Or find a counter example to the conjecture

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1 Answer 1

up vote 0 down vote accepted

This is true. If $f(x)=p(x)$, then $f-p$ has $n+2$ distinct zeros, between which lie $n+1$ distinct zeros of $(f-p)'$, between which lie $n$ distinct zeros of $(f-p)''$, and so on, leading to a zero of $(f-p)^{(n+1)}$, contradicting the fact that $f^{(n+1)}\lessgtr p^{(n+1)}$.

Note that the proof uses neither the fact that $p$ is a polynomial, nor the fact that the function values are zero; it proves the more general fact that if two $C^n$ functions coincide at $n+1$ distinct points, their $n$-th derivatives must coincide at some point in between.

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