Timeline for The sum of same powers of all matrices modulo p
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Apr 8, 2013 at 0:16 | comment | added | ya-tayr | Since the formal series $f(x) = \sum_k \sum_A A^k x^k = \sum_A (I - Ax)^{-1}$ doesn't change if you replace the summand by $(I - (A+I)x)^{-1}$, the series satisfies $f(x) = (1-x) f(x/(1-x))$, which is a little bit like being a modular form of weight 1. | |
| Apr 7, 2013 at 22:14 | comment | added | paul garrett | Wonderfully clear! | |
| Apr 7, 2013 at 20:21 | history | answered | Sergei Ivanov | CC BY-SA 3.0 |