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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?

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    @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!2012-08-21

4 Answers 4

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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.

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    It's $ax^2+bx^1+cx^0=ax^2+bx+c$2012-08-22
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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.

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    @JenniferDylan, thx but the question was more of rethorical nature. It was supposed to say: Yes, $0$ is a polynomial, but not necessarily the "same" $0$.2012-08-21
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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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    @Ilya thanks for the comment.2012-08-21
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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.