Timeline for Evaluation of q-Pochhammer series
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 19, 2018 at 13:26 | comment | added | Zhou | @ darij grinberg, Thank you very much! | |
| Aug 19, 2018 at 13:24 | history | edited | Zhou | CC BY-SA 4.0 |
improve detail!
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| Aug 19, 2018 at 13:04 | history | edited | Zhou | CC BY-SA 4.0 |
added 71 characters in body
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| Aug 19, 2018 at 13:02 | comment | added | darij grinberg | ... you can compute its $x$-derivative in the latter ring (since the $x$-derivative operators on $\mathbb{Q}\left[\left[q,x\right]\right]$ and on $\mathbb{Q}\left(x\right)\left[\left[q\right]\right]$ agree on $\mathbb{Q}\left[x\right]\left[\left[q\right]\right]$). I'm fascinated by how much goes on behind the scenes in seemingly simple Laurent-series arguments. | |
| Aug 19, 2018 at 13:01 | comment | added | darij grinberg | ... more formally, to the image of the latter injection). Thus, its inverse $g_q\left(x\right)$ belongs to $\mathbb{Q}\left(x\right)\left[\left[q\right]\right]$ as well. This allows us to do things like substituting $1+x$ or $1-x$ for $x$ or take derivatives at $x = 1$. In your second-to-last displayed equation, you observe that $\dfrac{xf_q\left(1+x\right)}{\left(q;q\right)_\infty}$ actually lies in $\mathbb{Q}\left[x\right]\left[\left[q\right]\right]$ (that is, it uses only polynomials, not rational functions in $x$), whence it embeds into $\mathbb{Q}\left[\left[q,x\right]\right]$, and ... | |
| Aug 19, 2018 at 12:58 | comment | added | darij grinberg | Oh, I see. The canonical injection $\mathbb{Q}\left(x\right) \to \mathbb{Q}\left(\left(x\right)\right)$ (sending $x$ to $x$) of the ring of rational functions in $x$ into the ring of Laurent series in $x$ gives rise to a canonical injection $\mathbb{Q}\left(x\right)\left[\left[q\right]\right] \to \mathbb{Q}\left(\left(x\right)\right)\left[\left[q\right]\right]$ (sending $q$ to $q$). Consider these two injections as inclusions. Then, your first displayed equation shows that $f_q \left(x\right)$ belongs to $\mathbb{Q}\left(x\right)\left[\left[q\right]\right]$ (or, ... | |
| Aug 19, 2018 at 12:45 | comment | added | darij grinberg | What ring are we working in? Formal power series in $q$ over the rational functions in $x$ ? | |
| Aug 19, 2018 at 12:42 | comment | added | Zhou | @ darij grinberg $1/x$ is from $\sum_{k\ge 0}\frac{q^k}{(q;q)_k}\frac{1}{1-xq^k}$ with $k=0$ and substitute $1−x$ to $x$. | |
| Aug 19, 2018 at 11:38 | comment | added | darij grinberg | From what I understand, $f_q\left(1-x\right)$ is not well-defined. But $\dfrac{f_q\left(1-x\right)}{\left(q;q\right)_\infty}$ is well-defined (as a formal power series in $q$ and $x$), because your equality $\dfrac{f_q\left(x\right)}{\left(q;q\right)_\infty} = \sum\limits_{k\geq 0} \dfrac{q^k}{\left(q;q\right)_k} \cdot \dfrac{1}{1-xq^k}$ shows that $\dfrac{f_q\left(x\right)}{\left(q;q\right)_\infty} \in \mathbb{Q}\left[x\right]\left[\left[q\right]\right]$ and we can substitute $1-x$ for the polynomial indeterminate $x$. But how do you get a $\dfrac{1}{x}$ in the result? | |
| Aug 19, 2018 at 11:28 | comment | added | darij grinberg | ... side counts the partitions of $n$ such that each parts is $\leq m$. But there is a bijection between the former and the latter partitions, given by conjugation (= transposition). | |
| Aug 19, 2018 at 11:27 | comment | added | darij grinberg | Ah, I see. Your second equality sign follows from the identity $\dfrac{1}{\left(x;q\right)_\infty} = \sum\limits_{k\geq 0} \dfrac{x^k}{\left(q;q\right)_k}$, which appears on en.wikipedia.org/wiki/Q-Pochhammer_symbol#Identities . I still don't see what this identity has to do with the $q$-binomial identity, but at least I see how to prove it: For any $m \geq 0$ and $n \geq 0$, the coefficient of $x^m q^n$ on the left hand side counts the partitions of $n$ into at most $m$ parts, while the coefficient of $x^m q^n$ on the right ... | |
| Aug 19, 2018 at 11:22 | comment | added | darij grinberg | Care to elaborate how you're applying the $q$-binomial theorem? I don't see any $q$-binomial coefficients around... | |
| Aug 19, 2018 at 7:02 | history | answered | Zhou | CC BY-SA 4.0 |