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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:

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

That would lead me to:

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

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?

Addendum:

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?

Chris Eagle
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Red Banana
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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 $|x|>1$. – SBF Aug 21 '12 at 10:49
  • Consider $$\lim_{k\to\infty}\frac1{a^k}$$ – J. M. ain't a mathematician Aug 21 '12 at 10:50
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    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. – SBF Aug 21 '12 at 13:15
  • @Ilya It's preciselly what yoy suggested. – Red Banana Aug 21 '12 at 15:03
  • @GustavoBandeira: please, would you clarify it in OP? – SBF Aug 21 '12 at 15:21
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    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
  • @MartianInvader See here and here. I tried to show some effort to answer it and as I predicted here, MSE users could provide me with valuable advices as you can see below. I do agree with you that it's a poor question, but unfortunately, these are the only questions I can elaborate. – Red Banana Aug 21 '12 at 19:19
  • @Gustavo I think you misunderstood me... I meant the question you refer to in the book, asking for a degree $-\infty$ polynomial, was a poor question for a math book. Asking for clarification on this site about a poor book question is indeed appropriate! – MartianInvader Aug 21 '12 at 19:29

4 Answers4

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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.

Simon Markett
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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 $-\infty$. – Red Banana 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? – Red Banana Aug 21 '12 at 23:37
  • It's $ax^2+bx^1+cx^0=ax^2+bx+c$ – Simon Markett Aug 22 '12 at 08:20
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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.

Chris Eagle
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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.

Seyhmus Güngören
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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). – SBF 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. – SBF Aug 21 '12 at 13:13
  • @Ilya thanks for the comment. – Seyhmus Güngören Aug 21 '12 at 13:21
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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.

i. m. soloveichik
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