# Polynomial of degree $-\infty$?

I'm reading E.J Barbeau Polynomials. I'm in a page where he asks a polynomial of degree $-\infty$. Then I thought about $77x^{-\infty}+1$, but when I went for the answers, the answer to this question was zero.

Then I thought about making $n^{-\infty}$ on Mathematica and it outputed $Indeterminate$ as a result.

I thought the problem was in my understanding of exponantiation, then I tried to "algebrize" it. (I guess that's the name of the procedure)

Then I thought:

$2^3=\overbrace{2\cdot 2\cdot 2}^{\text{3 times}}$

$a^b=\overbrace{a\cdot a\cdot a\cdot ...}^{\text{b times}}$

And in this case:

$a^{-\infty}=\overbrace{a\cdot a\cdot a\cdot ...}^{{-\infty}\text{ times}}$

But this gave me no insight of what could be done to better understand this. I can't see why $n^{-\infty}=0$ so clearly.

With the last example, I'm thinking that there will be no $a$'s to multiply, can you help me?

I thought about some other thing:

$$2^{-8}=\frac{1}{256}=\frac{1}{2^8}$$

Then considering this example, I would get: $$a^{-\infty}=\frac{1}{\infty}=0$$ Right?

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Well, negative powers are defined as $x^{-k} := \frac{1}{x^k}$. Thus, you can define $$x^{-\infty} := \lim\limits_{k\to+\infty}\frac{1}{x^k}$$ and since for $|x|>1$ the absolute value of $x^k$ grows unboundedly with $k\to\infty$, you obtain that $x^{-\infty} = 0$ whenever $|x|>1$. –  Ilya Aug 21 '12 at 10:49
Consider $$\lim_{k\to\infty}\frac1{a^k}$$ –  Ｊ. Ｍ. Aug 21 '12 at 10:50
Surely the point is that we (some of us, rather) conventionally define the polynomial $0$ to have degree $-\infty$, so that formulae like $\deg(pq) = \deg(p)+\deg(q)$ continue to hold. –  Sean Eberhard Aug 21 '12 at 10:52
@Gustavo: would you please give a hint, are you looking for the reasons for $0$ to be a polynomial of the degree $-\infty$, or your question is different? It is not that clear from your post. –  Ilya Aug 21 '12 at 13:15
It sounds like the question in the book was "What polynomial should we define to have degree $-\infty$ ?" As it's not a mathematical question and has a "come up with the same opinion I already have" feel, it's a rather poor question in my opinion. –  MartianInvader Aug 21 '12 at 17:24

IMO it comes down to conventions. We say the zero polynomial has degree $-\infty$. Let's see why this is a good convention:

Usually the degree is the highest power with a non-vanishing coefficient. Following this logic it is not really clear what the degree of the zero-polynomial should be. We could just say it has no degree, or we could say it is just a special case of a degree $0$ polynomial (i.e. a constant polynomial), or maybe it's something different?

What properties does the degree have? More specifically what happens if I add or multiply two polynomials $P$ and $Q$ of degree, say, $n$ and $m$?

You can check that the degree of the sum of $P$ and $Q$ will be smaller or equal to the maximum of the degrees of $P$ and $Q$, while the product will have degree $m+n$.

In particular if we multiply any polynomial $P$ with the zero polynomial we want:

$$\deg 0=\deg P\cdot 0=\deg P+ \deg 0$$

To make sense of this equation $\deg 0$ has to be $\pm \infty$ but $+\infty$ doesn't agree with the property for sums. So $-\infty$ remains as the only sensible choice.

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Yes. I imagined this: $ax^2+bx^1+c^0$ which is a second degree polynomial if at least $a\neq 0$. If $(a=0)$ then it's a first degree polynomial, if $(a=b=0)$ it's a zero degree polynomial, if $(a=b=c=0)$ we have a polynomial of degree $-\infty$. –  Igäria Mnagarka Aug 21 '12 at 23:02
I got one doubt with this: If my suggested polynomial is going to be evaluated, I'm will meet some trouble on the $c\,$ term, since $c^0=1$ (and I know that the $c\,$ term could be $\neq 1$). I guess that $ax^2+bx^1+cd^0$ would be more adequate, isn't it? –  Igäria Mnagarka Aug 21 '12 at 23:37
It's $ax^2+bx^1+cx^0=ax^2+bx+c$ –  Simon Markett Aug 22 '12 at 8:20

In the division algorithm for polynomials you want to divide $f$ by a non-zero polynomial $g$ and get a remainder $r$ of smaller degree than tat of $g$: $f=qg+r$ where $q, r$ are polynomials and $\deg(r)<\deg(g)$. In case $\deg(g)=0$, i.e. $g$ is a non-zero constant then $r=0$, $\deg(r)=\deg(0)=-\infty$ makes this all work out nicely.

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You can write $n^{-\infty}=(\frac{1}{n})^{\infty}$ and if $|n|>1$ then you will get

$$\lim_{k\rightarrow -\infty}n^{k}=0$$ As $|n|<1$ then $$\lim_{k\rightarrow -\infty}n^{k}=\infty$$ thats why $77x^{-\infty}+1$ is undefined.

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You might mean $|n|\gt1$ instead of $|n|\lt1$, whatever the relevance of your remark to the question asked. –  Did Aug 21 '12 at 13:07
@did: it's not clear from the OP whether it asks about reasons for $0$ to be the polynomial of the degree $-\infty$, or why $n^{-\infty} = \mathsf{Indeterminate}$, or why $n^{-\infty} = 0$ (whatever the value of $n$ is meant in the latter expression). Moreover, I would say that the former meaning seems to be less probable given the text of OP (especially when updated). –  Ilya Aug 21 '12 at 13:12
@Seyhmus: you may want to mention that $n^{-\infty} = \infty$ for $n\in (0,1)$ which is one of the reasons, why $77x^{-\infty}$ is undefined. –  Ilya Aug 21 '12 at 13:13
@Ilya thanks for the comment. –  Seyhmus Güngören Aug 21 '12 at 13:21
Turn back to the beginning of the first chapter, a page or so before the problem you're attempting, and you should find Barbeau's definition of degree. It contains the words "a nonzero constant polynomial has degree $0$, but, by convention, the zero polynomial (all coefficients vanishing) has degree $-\infty$." The question then becomes rather easy.
@GustavoBandeira Maybe the question is, what is $0$? Is the zero polynomial the same thing as the natural number $0$, as the real number $0$,...? –  Simon Markett Aug 21 '12 at 11:07
Gustavo: Sorry but a polynomial is not a number (nor a function, by the way), but a sequence $(a_n)_{n\geqslant0}$ with values in a given ring, with the property that the set $\{n\mid a_n\ne0\}$ is finite. And this sequence, when viewed as a polynomial, is often denoted by $\sum\limits_na_nX^n$ by convention. –  Did Aug 21 '12 at 12:37
@SimonMarkett set-wise: $\Bbb{N} \subset \Bbb{N}[x].$ –  user2468 Aug 21 '12 at 13:35