Polynomial of degree $-\infty$?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.