0
$\begingroup$

The tag description:

The Pochhammer symbol is the notation used for rising and falling factorials. The $q$-Pochhammer symbol is the $q$-analog.

To be a $q$-analog, the $q$-Pochhammer symbol must equal the original expression (rising or falling factorial) as $q \rightarrow 1$ (or at least $q \rightarrow 1^-$). This is not the case, as is very easy to check. Details in my earlier question here. As I point out there, Wolfram MathWorld and Wikipedia both commit the same error. I'm not sure what the description should be instead, but we can't just have false statements lying around.

$\endgroup$
3
  • $\begingroup$ Note the Wikipedia article on q-analog which explains: "q-analogues are most frequently studied in the mathematical fields of combinatorics and special functions. In these settings, the limit q → 1 is often formal, as q is often discrete-valued (for example, it may represent a prime power)." $\endgroup$
    – hardmath
    Sep 1, 2019 at 19:05
  • $\begingroup$ Yet nowhere does it say that the limit criterion is not necessary for something to be a $q$-analog. $\endgroup$
    – H.v.M.
    Sep 1, 2019 at 22:29
  • $\begingroup$ As you've posted a Question on the main Math.SE site about the justification for the terminology (or lack thereof), I'll try to address your presumed lack of justification there. $\endgroup$
    – hardmath
    Sep 1, 2019 at 22:43

1 Answer 1

Reset to default
1
$\begingroup$

I've edited the tag info for to avoid the problematic claim that $q$-Pochhammer symbol is a $q$-analog of the Pochhammer symbol.

I'll post a discussion of the history of this "convention" on the Main Math.SE Question, including a link to a previous Question there that discusses the infelicity of such terminology.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .