[math-fun] q-binomial-oids (ascii art included)
I can find the following in the literature:
___n-1 M-k | | (1-q ) ___M-1 n \~~ M | |k=0 n n (n-1)/2 | | (1+x q ) = > -------------- x q | |n=0 /__ n=0 ___n k | | (1-q ) | |k=1
Now I came up with
___n-1 M-k | | (1-q ) 1 \~~ M | |k=0 n n (n-1) -------------- = > ------------------------------- x q ___M-1 n /__ n=0 ___n-1 k ___n-1 k | | (1-x q ) | | (1-q q ) | | (1-x q ) | |n=0 | |k=0 | |k=0
This is certainly known. Can anyone point out where this is given? Herr Gosper? LaTeX sources are \prod_{n=0}^{M-1}{(1+x\,q^n)} & = & \sum_{n=0}^{M}{ \frac{\prod_{k=0}^{n-1}{(1-q^{M-k})}} {\prod_{k=1}^{n}{(1-q^k)}} \, x^n \, q^{n\,(n-1)/2} } and \frac{1}{\prod_{n=0}^{M-1}{(1-x\,q^n)}} & = & \sum_{n=0}^{M}{ \frac{\prod_{k=0}^{n-1}{(1-q^{M-k})}} {\prod_{k=0}^{n-1}{(1-q\,q^k)} \, \prod_{k=0}^{n-1}{(1-x\,q^k)}} \, x^n \, q^{n\,(n-1)} }
Jacobi! (this was hard to find)
The relations are respectively the special cases (a,b)=(-1,0) and (a,b)=(0,1) of an identity due to Jacobi:
___M-1 n ___n-1 M-k ___n-1 k | | (1-a x q ) | | (1-q ) | | (b q -a) | |n=0 \~~ M | |k=0 | |k=0 n n (n-1)/2 -----------------= > ----------------------------------- x q ___M-1 n /__ n=0 ___n-1 k ___n-1 k | | (1-b x q ) | | (1-q ) | | (1-b q ) | |n=0 | |k=0 | |k=0
To be found on p.795 of W. P. Johnson: {How Cauchy Missed Ramanujan's ${}_1\psi_1$ Summation}, American Mathematical Monthly, vol.111, no.9, pp.791-800, November-2004 * Joerg Arndt <arndt@jjj.de> [Mar 17. 2010 18:22]:
I can find the following in the literature:
___n-1 M-k | | (1-q ) ___M-1 n \~~ M | |k=0 n n (n-1)/2 | | (1+x q ) = > -------------- x q | |n=0 /__ n=0 ___n k | | (1-q ) | |k=1
Now I came up with
___n-1 M-k | | (1-q ) 1 \~~ M | |k=0 n n (n-1) -------------- = > ------------------------------- x q ___M-1 n /__ n=0 ___n-1 k ___n-1 k | | (1-x q ) | | (1-q q ) | | (1-x q ) | |n=0 | |k=0 | |k=0
This is certainly known. Can anyone point out where this is given? Herr Gosper?
LaTeX sources are
\prod_{n=0}^{M-1}{(1+x\,q^n)} & = & \sum_{n=0}^{M}{ \frac{\prod_{k=0}^{n-1}{(1-q^{M-k})}} {\prod_{k=1}^{n}{(1-q^k)}} \, x^n \, q^{n\,(n-1)/2} }
and
\frac{1}{\prod_{n=0}^{M-1}{(1-x\,q^n)}} & = & \sum_{n=0}^{M}{ \frac{\prod_{k=0}^{n-1}{(1-q^{M-k})}} {\prod_{k=0}^{n-1}{(1-q\,q^k)} \, \prod_{k=0}^{n-1}{(1-x\,q^k)}} \, x^n \, q^{n\,(n-1)} }
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Curious asymmetry --- (b q - a) on top, (1 - b q) on bottom --- have you checked there's no typo here? WFL On 3/19/10, Joerg Arndt <arndt@jjj.de> wrote:
Jacobi! (this was hard to find)
The relations are respectively the special cases (a,b)=(-1,0) and (a,b)=(0,1) of an identity due to Jacobi:
___M-1 n ___n-1 M-k ___n-1 k | | (1-a x q ) | | (1-q ) | | (b q -a) | |n=0 \~~ M | |k=0 | |k=0 n n (n-1)/2 -----------------= > ----------------------------------- x q
___M-1 n /__ n=0 ___n-1 k ___n-1 k
| | (1-b x q ) | | (1-q ) | | (1-b q )
| |n=0 | |k=0 | |k=0
To be found on p.795 of W. P. Johnson: {How Cauchy Missed Ramanujan's ${}_1\psi_1$ Summation}, American Mathematical Monthly, vol.111, no.9, pp.791-800, November-2004
* Fred lunnon <fred.lunnon@gmail.com> [Mar 19. 2010 15:20]:
Curious asymmetry --- (b q - a) on top, (1 - b q) on bottom --- have you checked there's no typo here? WFL
On 3/19/10, Joerg Arndt <arndt@jjj.de> wrote:
Jacobi! (this was hard to find)
The relations are respectively the special cases (a,b)=(-1,0) and (a,b)=(0,1) of an identity due to Jacobi:
___M-1 n ___n-1 M-k ___n-1 k | | (1-a x q ) | | (1-q ) | | (b q -a) | |n=0 \~~ M | |k=0 | |k=0 n n (n-1)/2 -----------------= > ----------------------------------- x q ___M-1 n /__ n=0 ___n-1 k ___n-1 k | | (1-b x q ) | | (1-q ) | | (1-b x q ) <--= HERE | |n=0 | |k=0 | |k=0
Thanks for spotting! An 'x' has to be inserted in the rightmost lower factor (done in the ascii art above). This was a error with TeX-ing up from pari/gp code: L(M,a,b)=prod(n=0,M-1,(1-a*x*q^n)/(1-b*x*q^n)) R(M,a,b)=sum(n=0,M, prod(k=0,n-1,1-q^(M-k)) / prod(k=1,n,1-q^k) * q^(n*(n-1)/2) * x^n * prod(k=0,n-1,b*q^k-a) / prod(k=0,n-1,1-b*x*q^k) ) \\ check first few: { for(M=0,5, f=L(M,a,b); g=R(M,a,b); print([M,f-g]); ); } \\ this prints [0, 0] [1, 0] [2, 0] [3, 0] [4, 0] [5, 0]
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