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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    $\begingroup$ A polynomial only contains positive, integral indices. So $x^{-4}+x^{3}$ is not a polynomial. $\endgroup$ Nov 6 '12 at 13:25
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    $\begingroup$ A polynomial has only positive exponents. The degree of a rational function, that is a quotient of two polynomials, in your case $(x^7 + 1)/x^4$ is usually defined as the difference of the degrees of the involved polynomials. $\endgroup$
    – martini
    Nov 6 '12 at 13:26
  • $\begingroup$ @Shaktal : Thanks for adding the proper editing tags. $\endgroup$
    – bonCodigo
    Nov 6 '12 at 13:43
  • $\begingroup$ @martini : Thank you for guiding to the correct definition. So the expression I have is a rational function. If the resulting expression of the difference of a rational function is polynomial then degree can be found using polynomial exponential rule. In my case, the difference is resulting in a rational function. Can I say the degree is 3 in that case? Because upper degree of numerator is 7, upper degree of denominator is 4, the difference is 3... $\endgroup$
    – bonCodigo
    Nov 6 '12 at 14:19
  • $\begingroup$ @bonCodigo Yes. $\endgroup$
    – martini
    Nov 6 '12 at 14:20

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 $-4$.

  • $\begingroup$ Then what is the degree of $ \dfrac 1x$, as a polynomial (or monomial) in $ x$? Again if this is not a polynomial in $ x$, then why do we write $ f(x)= \dfrac 1x$? $\endgroup$
    – Manjoy Das
    May 4 '20 at 7:46

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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    $\begingroup$ Thanks, this is surely a more natural setting than a Laurent series. $\endgroup$
    – Alexander Gruber
    Feb 5 '15 at 23:44

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.


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