Timeline for Expressions involving $q$-binomial coefficients?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 18, 2015 at 12:19 | comment | added | Lev Borisov | Yup, got the sign wrong. | |
| Jan 18, 2015 at 6:03 | comment | added | Hjalmar Rosengren | You have a minus sign wrong. My answer can be simplified to $q^{2a^2-a}(1-q^{2b+2-2a})(q^{2b-4a+4};q^2)_{2a-1}/(q;q)_{2a}$. To see this, rewrite the three factors involving $(q;q)_k$ using $(q;q)_{2k}=(q;q^2)_k(q^2;q^2)_k$, $(q;q)_{2k+1}=(q;q^2)_{k+1}(q^2;q^2)_k$ and the other two factors using $(a;q^2)_k=(-1)^kq^{k^2-k}(q^{2-2k}/a;q^2)_k$. | |
| Jan 18, 2015 at 2:48 | comment | added | Lev Borisov | I am getting $q^{2a^2-a}(q^{2b+2-2a}-1)(q^{2b-4a+4};q^2)_{2a-1}/(q;q)_{2a}$... | |
| Jan 18, 2015 at 1:28 | comment | added | Lev Borisov | I think your formula is a bit off, but I will surely figure it out now. | |
| Jan 18, 2015 at 1:16 | vote | accept | Lev Borisov | ||
| Jan 17, 2015 at 15:44 | comment | added | Hjalmar Rosengren | You could add that Askey's advice "whenever you see a sum involving binomial coefficients, write it in hypergeometric notation", is equally valid in the $q$-case. | |
| Jan 17, 2015 at 15:43 | history | answered | Hjalmar Rosengren | CC BY-SA 3.0 |