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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.

Vishu
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    very remotely related https://math.stackexchange.com/questions/446130/quickest-way-to-determine-a-polynomial-with-positive-integer-coefficients – gt6989b Dec 31 '20 at 13:22
  • Suppose you use your definition. When will you know you can stop adding the $\text{sgn}...$? Let's say $i=1000000$ and you haven't found a non-zero $\text{sgn}...$ for the last 10000 $i$'s. You still don't know if you should go on to $i=1000001$ . In other words, your definition might be right, but to use it, you must add an infinity of $\text{sgn}...$. – MasB Dec 31 '20 at 20:28
  • @BernardMassé After a certain point, all the terms will be zero. You keep going until that happens. Yes, there have to be an infinite amount of terms, as the degree of a polynomial can be arbitrarily large. – Vishu Dec 31 '20 at 20:32
  • @Tavish How do you know there aren't anymore non-zero terms? Sums to infinite are okay if they converge. – MasB Dec 31 '20 at 22:31
  • @BernardMassé Would I have claimed that this sum returns a finite degree if this was divergent? Consider what happens when $i+j \ge d+1$. – Vishu Jan 01 '21 at 07:52

3 Answers3

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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.

Hagen von Eitzen
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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

miracle173
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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.

Korn Kruaykitanon
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    The challenge is finding a purely mathematical expression for the degree, without using any words. :) – Vishu Dec 31 '20 at 13:31
  • Well, the condition we do an unbounded search here is also a predicate of variable $P(x)$. This can be viewed as a function equally good as what you’re looking for. – Korn Kruaykitanon Dec 31 '20 at 13:34
  • No, this is not a function of any sorts. How would you write that mathematically? – Vishu Dec 31 '20 at 13:35
  • $\inf{,n\in\Bbb N\mid \sup_{x>1}x^{-1}|P(x)|<\infty,}$ – Hagen von Eitzen Dec 31 '20 at 13:39
  • @Tavish That IS a function. It describes unambiguously a number for each polynomial. May be not written the way you are looking for, but its definitely a function. – jjagmath Dec 31 '20 at 14:39
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    @jjagmath Seems too strange of a function to me. Definitely not what I was looking for. – Vishu Dec 31 '20 at 14:56
  • In Mathematics often the functions are defined that way. For example, the important function $g(k)$ in Additive Number Theory is defined as he minimum number of $k$-th powers of naturals needed to represent all positive integers. – jjagmath Dec 31 '20 at 15:07
  • @jjagmath They might be. What’s your point? – Vishu Dec 31 '20 at 15:09
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    @Tavish My point is that " the smallest non-negative integer $n$ such that $\lim_{x\to \infty} \frac{P(x)}{x^n}$ exists." does define a function, and this is a common way to define functions in Mathematics. So your comment saying that is not a function of any sorts or asking to writing it "mathematically" its out of place. It IS a function and is written as its usual in Mathematics (its written mathematically) – jjagmath Dec 31 '20 at 16:28
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    @jjagmath in the context of question i dont think this answer is what the OP is looking for/book of author is expecting ,in that case i can make it simpler d(p(x))='the smallest natural $j$ such that $\frac{d^j}{dx^j}P(x)$$ is a constant ,and we can make more ,the question loses its elegance with that idea. – Albus Dumbledore Dec 31 '20 at 16:42
  • @jjagmath My comment is very much in its place. Although it might not be literally true, I think I still convey my point. That sort of strange function isn’t what I’m looking for. – Vishu Dec 31 '20 at 16:43
  • @AlbusDumbledore Yes, might as well say $d(p(x)$= the degree of $p(x)$. – Vishu Dec 31 '20 at 16:52
  • @AlbusDumbledore I understand what Tavish is looking for and I know the Korn's answer is not a good one. I just disagree with the the comment saying that "is not a function of any sort" (is a perfectly valid definition of a function) and saying it's not "mathematically ". I think that Tavish comment saying "Definitely not what I was looking for" is what he should have said from the beginning, instead of discarding the answer as nonsense – jjagmath Dec 31 '20 at 17:07
  • @jjagmath I said that because I didn’t know this could be called a function. Now I do. But this answer is still not relevant. – Vishu Dec 31 '20 at 17:12
  • I agree with you that the answer is not a good one. And I'm glad to see that I help you to break the false impression many people have that a function has to be something that needs to be defined by a formula. – jjagmath Dec 31 '20 at 17:17
  • @jjagmath Sure, you get to learn something everyday am I right. – Vishu Dec 31 '20 at 17:46