2

$t^2 + t$ is a polynomial.

What is

$t^{-2} + t^{-1}$ called?

More generally,

$t^{-n} + t^{-(n-1)} + \cdots + t^{-2} + t^{-1} + c$.

Sin Nombre
  • 622
  • 4
    If $P(t)=t^2+t$ then $t^{-2}+t^{-1}$ is called $P(\frac{1}{t})$. – N. S. Nov 03 '20 at 02:08
  • 4
    It’s equal to $\frac{1+t}{t^2}$, a quotient of polynomials, so it’s a rational function. – Brian M. Scott Nov 03 '20 at 02:09
  • I know of no more specific term than "rational function" of $t$. This previous question and answer(s) may shed more light: https://math.stackexchange.com/questions/574001/why-cant-polynomials-have-negative-exponents-or-division-by-a-variable But note that your expression is a polynomial when expressed in the transformed variable $x = \frac 1t$ – Deepak Nov 03 '20 at 02:09
  • @N.S. Is there a name to it? Like "hypergeometric polymonomial reciprocated gamma" functions? – Sin Nombre Nov 03 '20 at 02:09
  • 3
    https://en.wikipedia.org/wiki/Laurent_polynomial – Torsten Schoeneberg Nov 03 '20 at 02:10
  • 3
    The question in the title does not match the question in the body of the post. The polynomial function $f(t)=t^2+t$ is not one-to-one and thus does not have an inverse function. Even if by "inverse" you meant "reciprocal" rather than "inverse function," the post still doesn't make sense, because $t^{-2}+t^{-1}$ is not the reciprocal of $t^2+t$. – symplectomorphic Nov 03 '20 at 02:18
  • You could also say it is the principal part of a Laurent series. Except that this would not include the constant term, strictly speaking. – Tob Ernack Nov 03 '20 at 02:37

3 Answers3

2

For the sake of completeness, I summarize the comments above:

As pointer out by Torsten and manthanomen, there are specific names (Laurent polynomial) for such functions. However, these are not widely known, so I agree with Brian and Deepak and suggest using a more general concept called rational function:

A function $f(x)$ is called a rational function if and only if it can be written in the form $f(x) = \frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomial functions of $x$ and $Q$ is not the zero function.

If you do not want to call it rational function and want to define these functions merely through the regular polynomials then saying $f(t) = P\left(\tfrac{1}{t}\right)$ as suggested by N. S. is the way to go.

Finally, as pointed out by David, $$(t+t^2)^{-1} = \frac{1}{t + t^2} \ne \frac{1 + t}{t^2} = t^{-1} + t^{-2},$$ hence not an inverse number in the usual meaning, and as noticed by symplectomorphic, $f(t) = t + t^2$ as a function does not have an inverse function in the usual meaning because it is not injective: $f(0) = f(-1) = 0$.

Nikita Skybytskyi
  • 403
1

Let $k$ be a field. The linear map $\overline{\cdot}: k[t, t^{-1}] \to k[t, t^{-1}]$ which sends $t$ to $t^{-1}$ is often called the "bar involution'' in certain subjects in algebra (for example, Hecke algebras and quantum groups). So $t^{-2} + t^{-1}$ could be called "the image of $t^2 + t$ under the bar involution", i.e., $\overline{t^2 + t} = t^{-2} + t^{-1}$. Of course, as one of the commenters has noted, the image under the bar involution is not the multiplicative inverse of the polynomial.

manthanomen
  • 3,186
  • You want Laurent polynomials as the domain and codomain. As wild speculation, I imagine the terminology "bar" comes from physics using $t=\exp(is)$, since then $t^{-1} = \exp(-is) = \exp(\overline{is}) = \overline{\exp(is)} = \overline{t}$. I personally have not heard it. – Joshua P. Swanson Nov 03 '20 at 02:31
  • Yes, thanks for catching that mistake. – manthanomen Nov 03 '20 at 02:37
1

In some areas of algebraic combinatorics, the related operation $f(q) \mapsto q^{\mathrm{deg}(f)} f(q^{-1})$ is called "reversal". Here one usually has $f(0) \neq 0$, making it an involution.

Joshua P. Swanson
  • 9,634