Find $P(-\frac{1}{2})$ given a infinite points of a polynomial

Question

The problem

Given $P(x)$ a polynomial with real coefficients and satisfies

\begin{align} P(n)= 1^{2010} +2^{2010}+ \cdots + n^{2010}, \forall n \in \mathbb{N*} \end{align},

then calculate the value of $P(-\frac{1}{2})$.

Attempts

The first flash that came to my mind was that I have to use the Lagrange Interpolation, which I assume should apply for P(1) to P(2012). Note that using $\lim_{n \to + \infty}$ it's very easy to prove that $degP =2011$.

But then the normal binomial does not appear ( precisely, only the odd factors appear).

Then what should I do now? Any help is appreciated

What is $\Bbb N \ast$? Note that if two real polynomials agree on infinitely many points, then they must be equal. — Aryaman Maithani, Oct 09 '21 at 14:37
This wouldn't be from a contest held on year 2010 by any chance, would it? — Jyrki Lahtonen, Oct 09 '21 at 14:38
Mind you, I think $P$ has degree $2011$. See Faulhaber formula. Or, more simply, the sum $1+2+\cdots+n$ is a quadratic polynomial on $n$, the sum $1^2+2^2+\cdots+n^2$ is a cubic et cetera. — Jyrki Lahtonen, Oct 09 '21 at 14:40
One way would be to iterate the procedure from an old answer of mine 2010 times. I recall there being a shortcut in that method that comes to the fore here, but I may be very wrong :-) — Jyrki Lahtonen, Oct 09 '21 at 14:46
Consider this result by Donald Knuth: https://en.wikipedia.org/wiki/Faulhaber%27s_formula — Lazy, Oct 09 '21 at 14:53
I mean, with exponent $k=1$ we get $P_1(n)=n(n+1)/2$ and $P_1(-1/2)=-1/8$. With $k=2$ we get $P_2(n)=n(n+1)(2n+1)/6$ and $P_2(-1/2)=0$. The recollection I have is that $P_k(-1/2)$ vanishes for all even $k$, but I don't remember why :-( — Jyrki Lahtonen, Oct 09 '21 at 14:53
Consider $P(n,k)$ as general case, with arbitrary power $k$. Particular case $k=2010$. Prove that $P(n,2m)$ contains $(2n+1)$ as one of factors. Then just set $n=-1/2$. — Ivan Kaznacheyeu, Oct 09 '21 at 14:55
@Lazy The formula was found over 300 years before Knuth's time. — Jyrki Lahtonen, Oct 09 '21 at 15:02
@IvanKaznacheyeu That's a generic comment, can you elaborate on that? — Nikola Tolzsek, Oct 10 '21 at 07:23

Sil · Accepted Answer · 2021-10-18T07:39:21.457

The below is a simple argument which works under the assumption the polynomial $P(x)$ with given property $P(n)=1^{p} +2^{p}+ \cdots + n^{p}$ exists (here $p=2010$, but we really only need that $p$ is even). (Without such assumption we could show it exists using the argument similar to derive Faulhaber polynomials, but as currently stated we are given the polynomial $P(x)$)

Since $P(n)-P(n-1)=n^{p}$ for infinitely many values of $n$ (positive integers), and $P(x)$ is a polynomial, we must have equality of polynomials, i.e. $P(x)-P(x-1)=x^p$ for all $x$ real. This is because $P(x)-P(x-1)-x^p$ is a polynomial with infinitely many zeroes, hence must a be a zero polynomial.

Using $P(x-1)=P(x)-x^p$ we can get values of $P(x)$ at negative integers. For example $P(0)=P(1)-1=0, P(-1)=0, P(-2)=-1, P(-3)=-1-2^p$ and so on (note $(-n)^p=n^p$ since $p$ is even). It is not hard to extend the above by induction into \begin{align} P(-n)&=-1-2^p-\dots-(n-1)^p=-(1+2^p+\dots+(n-1)^p)=-P(n-1). \end{align} As before, this holds for infinitely many values so it must be equality of polynomials $P(-x)=-P(x-1)$ for all $x$ real. Now just set $x=\frac{1}{2}$ to get $$ P\Big(-\frac{1}{2}\Big)=-P\Big(\frac{1}{2}-1\Big)=-P\Big(-\frac{1}{2}\Big) $$ and hence $P(-\frac{1}{2})=0$.

Note: The above implies also $x(x+1)(2x+1) \mid P(x)$ for even $p$, and similarly $x(x+1) \mid P(x)$ for odd $p$.

Find $P(-\frac{1}{2})$ given a infinite points of a polynomial

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