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In theory, we define the degree of a polynomial as the highest exponent it holds.

However when there are negative and positive exponents are present in the function, I want to know the basis that we define the degree. Is the order of a polynomial degree expression defined by the highest magnitude of available exponents?

For example in $x^{-4} + x^{3}$, is the degree $4$ or $3$?

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    @BillDubuque appreciate your input. Before posting this question I did a search and found out that Laurent polynomials have both positive, negative exponents. However it wasn't the level I was searching for. The indepth analysis of Laurent is a little beyond the 'freedom' allowed in my question at this point. But as I am interested in reading, I will do in due time.2012-11-07

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In abstract algebra, we write the set of all polynomials with coefficients in a ring $R$ as $R[x]$.
Here "polynomials" means expressions of the form $a_0+a_1x+a_2x^2+\cdots +a_nx^n$ where $a_0,\ldots,a_n\in R$ and $n$ is finite (Note: if you don't know what a ring is, just think of the $a$'s as numbers). So, in this context, your expression $x^{-4}+x^3$ isn't in a polynomial, because all the powers of $x$ have to be non-negative.

We can generalize this construction, though. The first thing we can do is drop the requirement that $n$ must be finite. If we do this, we get $R[[x]]$, the set of formal power series in $R$.

A futher generalization is the set of formal Laurent series in $R$, denoted $R(\hspace{-0.5pt}(x)\hspace{-0.5pt})$, and this is a setting in which we can answer your question. Formal Laurent series have the form $\sum_{n\in \mathbb{Z}}a_nx^n$ where $a_n=0$ for all but finitely many negative $n$. In other words, formal Laurent series are formal power series which are allowed to have a finite number of negative exponents too.

The order of a formal Laurent series is defined as the smallest $n$ such that $a_n\not= 0$. This is kind of like the degree of a polynomial, but for negative integers. The degree of a formal Laurent series is defined in the same way as the degree of a polynomial, though the degree may not exist (since all of the $a_n$ for $n>0$ are still allowed to be nonzero).

So, considered as a formal Laurent series, we would say that $x^{-4}+x^3$ has degree $3$ and order $-4$.

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For the sake of completeness, I would like to add that this generalization of polynomials is called a Laurent polynomial. This set is denoted $R[x,x^{-1}]$.

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    Thanks, this is surely a more natural setting than a Laurent series.2015-02-05
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Your question is essentially: how does one define the degree of a Laurent polynomial?

My guess is that the best of of doing this is to move from degrees to "degree pairs." Let me explain this for polynomials first.

Given a polynomial $P(x) = \sum_{n:\mathbb{N}} a_n x^n,$ define the degree pair of $P$ to be the pair $(i,j)$ where $i$ is the least natural number such that $a_i$ is non-zero, and $a_j$ is the greatest natural number such that $a_j$ is non-zero. Denote this $\mathrm{dpa}(P)$. For example:

$\mathrm{dpa}(-3x^5+x^2+x) = (1,5)$

This extends straightforwardly to Laurent polynomials, and seems to be relatively well-behaved.