Finding the Degree of a Polynomial, as a function of the Polynomial

Question

Recently, I was wondering about the following question:

Given a polynomial $P(x)$ with real coefficients, express its degree $d$ as a function of only $P(x)$, i.e. $d(P(x))$. Only elementary functions/operations and values of $P$ (or its derivatives/anti-derivatives) at any value may be used.

There must be several creative ways to do this. After some experimentation, I came up with the following:

$$d(P(x)) = \sum_{i=1}^{\infty} \text{sgn}\left(\sum_{j=0}^{\infty} \left| \frac{d^{i+j} P}{dx^{i+j}} (0) \right| \right)$$

Are there simpler functions that work? Any inputs are welcome.

Answer 1

You can define the degree of a polynomial $P(x)$ to be the smallest non-negative integer $n$ such that $\lim\limits_{x \to +\infty} \frac{P(x)}{x^n}$ exists.

Answer 2

We have $$ \deg P=\lim_{n\to\infty}\frac{\ln (P(n)^2+n^{-n})}{2\ln n}.$$ The strange addition in the numerator is to ensure that all terms are defined and finite (for $n>1$) and also to correctly lead to $\deg 0=-\infty$.

This may be considered "simpler" than your expression because it does not use any derivatives and uses only a single limit instead of nested series (though your series turn out to be sums after all). Since it is clear that infinitely many evaluations are needed, it seems that this is about as elementary as you can get.

Answer 3

If $$P(x)=a_nx^n+\cdots+a_0$$ then $$d(x):=\frac{P'(\frac 1 x)}{xP(\frac 1 x)}=\frac{na_n+\cdots+a_1x^{n-1}}{a_n+\cdots+a_0x^{n}}$$ and $$d(0)=n$$, at least if degree>0