This is the problem I am trying to solve:

Does there exist a polynomial $p(x)$ of degree $n$, with complex coefficients, such that $p(k)=1/k$ for each integer $k$ satisfying $1\le k\le n+2$?

I have come to the conclusion that no such polynomial exists but I don't know how to prove it? Perhaps by contradiction?

Thank you.

  • $\begingroup$ Not for $n=1$.. $\endgroup$ – amsmath Oct 11 '17 at 18:38

We know that $$p(z)-\frac{1}{z}=\frac{zp(z)-1}{z}$$ on $\mathbb{C}-\{0\}$. Now we see the polynomial $zp(z)-1$ of degree $n+1$ must have $n+2$ zeroes, namely $1,..,n+2$. This is only possible if $zp(z)-1=0$, so if $p(z)=\frac{1}{z}$, clearly a contradiction since any polynomial is continuous at $0$.


Proof by contradiction can certainly be an approach to this question. Here's one way to do it using Calculus. Assume such a polynomial $p$ exists, and define $\displaystyle F(x)=p(x)-\frac{1}{x}$. This function is continuous and differentiable on $\displaystyle \left[\frac{1}{n+2},1\right]$ and has $n+2$ roots on this interval because $F(1)=F(2)=\cdots=F(n+2)=0$ by assumption. By Rolle's Theorem, between each pair of roots there exists at least one point $c$ such that $F'(c)=0$. Thus we can conclude that

$$F'(x)=p'(x)+\frac{1}{x^2} \text{ has at least } n+1 \text{ roots on } \left[\frac{1}{n+2},1\right].$$

By the same token:

  • $F''(x)$ has at least $n$ roots in this interval;
  • $F'''(x)$ has at least $n-1$ roots in this interval;
  • $F^{(n+1)}(x)$ has at least $1$ root in this interval.

But since the $(n+1)$-st derivative of an $n$-th degree polynomial is identically zero, $\displaystyle F^{(n+1)}(x)=p^{(n+1)}(x)+\frac{d^{n+1}}{dx^{n+1}}\left(\frac{1}{x}\right)=\frac{(-1)^{n+1}(n+1)!}{x^{n+2}}$, and it does not have any roots — hence the contradiction.

I think a different proof can be constructed using Lagrange's interpolation formula, but I didn't work it out. The idea is that an $n$-th degree polynomial is completely determined by its values at $n+1$ distinct points. But here we've been prescribed $n+2$ values, which is one too many.


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