# §17.2 Calculus

## §17.2(i) $q$-Calculus

For $n=0,1,2,\dots$,

 17.2.1 $\left(a;q\right)_{n}=(1-a)(1-aq)\cdots(1-aq^{n-1}),$
 17.2.2 $\left(a;q\right)_{-n}=\frac{1}{\left(aq^{-n};q\right)_{n}}=\frac{(-q/a)^{n}q^{% {\left({{n}\atop{2}}\right)}}}{\left(q/a;q\right)_{n}}.$

For $\nu\in\Complex$

 17.2.3 $\left(a;q\right)_{\nu}=\prod_{j=0}^{\infty}\left(\frac{1-aq^{j}}{1-aq^{\nu+j}}% \right),$

when this product converges.

 17.2.4 $\displaystyle\left(a;q\right)_{\infty}$ $\displaystyle=\prod_{j=0}^{\infty}(1-aq^{j}),$ 17.2.5 $\displaystyle\left(a_{1},a_{2},\dots,a_{r};q\right)_{n}$ $\displaystyle=\prod_{j=1}^{r}\left(a_{j};q\right)_{n},$ 17.2.6 $\displaystyle\left(a_{1},a_{2},\dots,a_{r};q\right)_{\infty}$ $\displaystyle=\prod_{j=1}^{r}\left(a_{j};q\right)_{\infty}.$
 17.2.7 $\left(a;q^{-1}\right)_{n}=\left(a^{-1};q\right)_{n}(-a)^{n}q^{-{\left({{n}% \atop{2}}\right)}},$
 17.2.8 $\frac{\left(a;q^{-1}\right)_{n}}{\left(b;q^{-1}\right)_{n}}=\frac{\left(a^{-1}% ;q\right)_{n}}{\left(b^{-1};q\right)_{n}}\left(\frac{a}{b}\right)^{n},$
 17.2.9 $\left(a;q\right)_{n}=\left(q^{1-n}/a;q\right)_{n}(-a)^{n}q^{{\left({{n}\atop{2% }}\right)}},$
 17.2.10 $\frac{\left(a;q\right)_{n}}{\left(b;q\right)_{n}}=\frac{\left(q^{1-n}/a;q% \right)_{n}}{\left(q^{1-n}/b;q\right)_{n}}\left(\frac{a}{b}\right)^{n},$
 17.2.11 $\left(aq^{-n};q\right)_{n}=\left(q/a;q\right)_{n}\left(-\frac{a}{q}\right)^{n}% q^{-{\left({{n}\atop{2}}\right)}},$
 17.2.12 $\frac{\left(aq^{-n};q\right)_{n}}{\left(bq^{-n};q\right)_{n}}=\frac{\left(q/a;% q\right)_{n}}{\left(q/b;q\right)_{n}}\left(\frac{a}{b}\right)^{n}.$
 17.2.13 $\left(a;q\right)_{n-k}=\frac{\left(a;q\right)_{n}}{\left(q^{1-n}/a;q\right)_{k% }}\left(-\frac{q}{a}\right)^{k}q^{{\left({{k}\atop{2}}\right)}-nk},$
 17.2.14 $\frac{\left(a;q\right)_{n-k}}{\left(b;q\right)_{n-k}}=\frac{\left(a;q\right)_{% n}}{\left(b;q\right)_{n}}\frac{\left(q^{1-n}/b;q\right)_{k}}{\left(q^{1-n}/a;q% \right)_{k}}\left(\frac{b}{a}\right)^{k},$
 17.2.15 $\left(aq^{-n};q\right)_{k}=\frac{\left(a;q\right)_{k}\left(q/a;q\right)_{n}}{% \left(q^{1-k}/a;q\right)_{n}}q^{-nk},$
 17.2.16 $\left(aq^{-n};q\right)_{n-k}=\frac{\left(q/a;q\right)_{n}}{\left(q/a;q\right)_% {k}}\left(-\frac{a}{q}\right)^{n-k}q^{{\left({{k}\atop{2}}\right)}-{\left({{n}% \atop{2}}\right)}},$
 17.2.17 $\displaystyle\left(aq^{n};q\right)_{k}$ $\displaystyle=\frac{\left(a;q\right)_{k}\left(aq^{k};q\right)_{n}}{\left(a;q% \right)_{n}},$ 17.2.18 $\displaystyle\left(aq^{k};q\right)_{n-k}$ $\displaystyle=\frac{\left(a;q\right)_{n}}{\left(a;q\right)_{k}}.$
 17.2.19 $\left(a;q\right)_{2n}=\left(a,aq;q^{2}\right)_{n},$

more generally,

 17.2.20 $\left(a;q\right)_{kn}=\left(a,aq,\dots,aq^{k-1};q^{k}\right)_{n}.$
 17.2.21 $\left(a^{2};q^{2}\right)_{n}=\left(a;q\right)_{n}\left(-a;q\right)_{n},$
 17.2.22 $\frac{\left(qa^{\frac{1}{2}},-aq^{\frac{1}{2}};q\right)_{n}}{\left(a^{\frac{1}% {2}},-a^{\frac{1}{2}};q\right)_{n}}=\frac{\left(aq^{2};q^{2}\right)_{n}}{\left% (a;q^{2}\right)_{n}}=\frac{1-aq^{2n}}{1-a},$

more generally,

 17.2.23 $\frac{\left(aq^{\frac{1}{k}},q\omega_{k}a^{\frac{1}{k}},\dots,q\omega_{k}^{k-1% }a^{\frac{1}{k}};q\right)_{n}}{\left(a^{\frac{1}{k}},\omega_{k}a^{\frac{1}{k}}% ,\dots,\omega_{k}^{k-1}a^{\frac{1}{k}};q\right)_{n}}=\frac{\left(aq^{k};q^{k}% \right)_{n}}{\left(a;q^{k}\right)_{n}}=\frac{1-aq^{kn}}{1-a},$

where $\omega_{k}=e^{2\pi i/k}$.

 17.2.24 $\lim_{\tau\to 0}\left(a/\tau;q\right)_{n}\tau^{n}=\lim_{\sigma\to\infty}\left(% a\sigma;q\right)_{n}\sigma^{-n}=(-a)^{n}q^{{\left({{n}\atop{2}}\right)}},$
 17.2.25 $\lim_{\tau\to 0}\frac{\left(a/\tau;q\right)_{n}}{\left(b/\tau;q\right)_{n}}=% \lim_{\sigma\to\infty}\frac{\left(a\sigma;q\right)_{n}}{\left(b\sigma;q\right)% _{n}}=\left(\frac{a}{b}\right)^{n},$
 17.2.26 $\lim_{\tau\to 0}\frac{\left(a/\tau;q\right)_{n}\left(b/\tau;q\right)_{n}}{% \left(c/\tau^{2};q\right)_{n}}=(-1)^{n}\left(\frac{ab}{c}\right)^{n}q^{{\left(% {{n}\atop{2}}\right)}}.$

## §17.2(ii) Binomial Coefficients

 17.2.27 $\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}=\frac{\left(q;q\right)_{n}}{\left(q;q\right)% _{m}\left(q;q\right)_{n-m}}\\ =\frac{\left(q^{-n};q\right)_{m}(-1)^{m}q^{nm-{\left({{m}\atop{2}}\right)}}}{% \left(q;q\right)_{m}},$ Defines: $\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}$: $q$-binomial coefficient (or Gaussian polynomial) Symbols: ${\left({{m}\atop{n}}\right)}$: binomial coefficient, $\left(a;q\right)_{n}$: $q$-Pochhammer (or $q$-shifted factorial), $q$: complex base, $m$: nonnegative integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/17.2.E27 Encodings: TeX, pMML, png
 17.2.28 $\lim_{q\to 1}\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}={\left({{n}\atop{m}}\right)}=% \frac{n!}{m!(n-m)!},$
 17.2.29 $\genfrac{[}{]}{0.0pt}{}{m+n}{m}_{q}=\frac{\left(q^{n+1};q\right)_{m}}{\left(q;% q\right)_{m}},$
 17.2.30 $\displaystyle\genfrac{[}{]}{0.0pt}{}{-n}{m}_{q}$ $\displaystyle=\genfrac{[}{]}{0.0pt}{}{m+n-1}{m}_{q}(-1)^{m}q^{-mn-{\left({{m}% \atop{2}}\right)}},$ 17.2.31 $\displaystyle\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}$ $\displaystyle=\genfrac{[}{]}{0.0pt}{}{n-1}{m-1}_{q}+q^{m}\genfrac{[}{]}{0.0pt}% {}{n-1}{m}_{q},$ 17.2.32 $\displaystyle\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}$ $\displaystyle=\genfrac{[}{]}{0.0pt}{}{n-1}{m}_{q}+q^{n-m}\genfrac{[}{]}{0.0pt}% {}{n-1}{m-1}_{q},$
 17.2.33 $\lim_{n\to\infty}\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}=\frac{1}{\left(q;q\right)_{% m}}=\frac{1}{(1-q)(1-q^{2})\cdots(1-q^{m})},$
 17.2.34 $\lim_{n\to\infty}\genfrac{[}{]}{0.0pt}{}{rn+u}{sn+t}_{q}=\frac{1}{\left(q;q% \right)_{\infty}}=\prod_{j=1}^{\infty}\frac{1}{(1-q^{j})},$

provided that $r>s$.

## §17.2(iii) Binomial Theorem

 17.2.35 $\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}(-z)^{j}q^{{\left({{j}\atop{2}}% \right)}}=\left(z;q\right)_{n}={(1-z)(1-zq)\cdots(1-zq^{n-1})}.$

In the limit as $q\to 1$, (17.2.35) reduces to the standard binomial theorem

 17.2.36 $\sum_{j=0}^{n}{\left({{n}\atop{j}}\right)}(-z)^{j}=(1-z)^{n}.$

Also,

 17.2.37 $\sum_{n=0}^{\infty}\frac{\left(a;q\right)_{n}}{\left(q;q\right)_{n}}z^{n}=% \frac{\left(az;q\right)_{\infty}}{\left(z;q\right)_{\infty}},$

provided that $|z|<1$. When $a=q^{m+1}$, where $m$ is a nonnegative integer, (17.2.37) reduces to the $q$-binomial series

 17.2.38 $\displaystyle\sum_{n=0}^{\infty}\genfrac{[}{]}{0.0pt}{}{n+m}{n}_{q}z^{n}$ $\displaystyle=\frac{1}{\left(z;q\right)_{m+1}}.$ 17.2.39 $\displaystyle\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q^{2}}q^{j}$ $\displaystyle=\left(-q;q\right)_{n},$ 17.2.40 $\displaystyle\sum_{j=0}^{2n}(-1)^{j}\genfrac{[}{]}{0.0pt}{}{2n}{j}_{q}$ $\displaystyle=\left(q;q^{2}\right)_{n}.$

When $n\to\infty$ in (17.2.35), and when $m\to\infty$ in (17.2.38), the results become convergent infinite series and infinite products (see (17.5.1) and (17.5.4)).

## §17.2(iv) Derivatives

The $q$-derivatives of $f(z)$ are defined by

 17.2.41 $\mathcal{D}_{q}f(z)=\begin{cases}\dfrac{f(z)-f(zq)}{(1-q)z},&z\neq 0,\\ f^{\prime}(0),&z=0,\end{cases}$ Defines: $\mathcal{D}_{q}$: $q$-differential operator (locally) Symbols: $q$: complex base and $z$: complex variable Referenced by: §17.2(iv), §18.27(i) Permalink: http://dlmf.nist.gov/17.2.E41 Encodings: TeX, pMML, png

and

 17.2.42 $f^{[n]}(z)=\mathcal{D}_{q}^{n}f(z)=\begin{cases}z^{-n}(1-q)^{-n}\sum_{j=0}^{n}% q^{-nj+{\left({{j+1}\atop{2}}\right)}}(-1)^{j}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q% }f(zq^{j}),&z\neq 0,\\ \dfrac{f^{(n)}(0)\left(q;q\right)_{n}}{n!(1-q)^{n}},&z=0.\end{cases}$

When $q\to 1$ the $q$-derivatives converge to the corresponding ordinary derivatives.

### Product Rule

 17.2.43 $\mathcal{D}_{q}(f(z)g(z))=g(z)f^{[1]}(z)+f(zq)g^{[1]}(z).$

### Leibniz Rule

 17.2.44 $\mathcal{D}_{q}^{n}(f(z)g(z))=\sum_{j=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{j}_{q}f% ^{[n-j]}(zq^{j})g^{[j]}(z).$

$q$-differential equations are considered in §17.6(iv).

## §17.2(v) Integrals

If $f(x)$ is continuous at $x=0$, then

 17.2.45 $\int_{0}^{1}f(x){d}_{q}x=(1-q)\sum_{j=0}^{\infty}f(q^{j})q^{j},$

and more generally,

 17.2.46 $\int_{0}^{a}f(x){d}_{q}x=a(1-q)\sum_{j=0}^{\infty}f(aq^{j})q^{j}.$

If $f(x)$ is continuous on $[0,a]$, then

 17.2.47 $\lim_{q\to 1-}\int_{0}^{a}f(x){d}_{q}x=\int_{0}^{a}f(x)dx.$

### Infinite Range

 17.2.48 $\int_{0}^{\infty}f(x){d}_{q}x=\lim_{n\to\infty}\int_{0}^{q^{-n}}f(x){d}_{q}x=(% 1-q)\sum_{j=-\infty}^{\infty}f(q^{j})q^{j},$

provided that $\sum_{j=-\infty}^{\infty}f(q^{j})q^{j}$ converges.

## §17.2(vi) Rogers–Ramanujan Identities

 17.2.49 $1+\sum_{n=1}^{\infty}\frac{q^{n^{2}}}{(1-q)(1-q^{2})\cdots(1-q^{n})}=\prod_{n=% 0}^{\infty}\frac{1}{(1-q^{5n+1})(1-q^{5n+4})},$ Symbols: $q$: complex base and $n$: nonnegative integer Referenced by: §17.14 Permalink: http://dlmf.nist.gov/17.2.E49 Encodings: TeX, pMML, png
 17.2.50 $1+\sum_{n=1}^{\infty}\frac{q^{n^{2}+n}}{(1-q)(1-q^{2})\cdots(1-q^{n})}=\prod_{% n=0}^{\infty}\frac{1}{(1-q^{5n+2})(1-q^{5n+3})}.$ Symbols: $q$: complex base and $n$: nonnegative integer Referenced by: §17.14 Permalink: http://dlmf.nist.gov/17.2.E50 Encodings: TeX, pMML, png

These identities are the first in a large collection of similar results. See §17.14.