2017-06-17 11:25:48 +0100 | received badge | ● Popular Question (source) |
2017-06-09 23:33:25 +0100 | commented question | indicial equation Thank you dan_fulea. Typically N would be <20, and the rest of the constants would not be integers. It is a pity that there is not a readily available general term. I was hoping to find out some general properties about this new function: some derivatives and simple integrals. But 100 terms in the series is not so bad: it should provide a graph good enough. I am not sure if I should dare to ask you for help. In any event, thanks for your attention. Javier Garcia |
2017-06-09 13:38:47 +0100 | commented answer | indicial equation Thank you dan_fulea. Yes, you are right. My problem is related to the hypergeometric function, but it has a greater complexity. I have manually found the first 4 terms of the series, g(1) to g(4), and they look rather complicated. Do you think there is any hope that SageMath could solve this problem? |
2017-06-09 13:38:46 +0100 | answered a question | indicial equation Thank you dan_fulea. Yes, you are right. My problem is related to the hypergeometric function, but it has a greater complexity. I have manually found the first 4 terms of the series, g(1) to g(4), and they look rather complicated. Do you think there is any hope that SageMath could solve this problem? Thanks, Javier Garcia. |
2017-06-08 19:36:27 +0100 | asked a question | indicial equation Let N be an integer. Let a, b, A_1,...,A_N be constant real numbers. Let g(n) be a real function of integer variable n. g(n) satisfies the recurrence equation: (n + 1)(n + b)g(n+1) = (n + a)*g(n) + sum(i=1 to N)(A_i * g(n-i)) with g(n)=0 for n<0 . g(0) is obtained through boundary conditions, so it can be considered a given constant. I need to know the general functional form of the term g(n), as a function of the given constants of the problem: a,b,N,A_1,...,A_N and g(0). Could you help me? Could I programm in SAGE the code to provide the searched general term g(n)? Thanks for your attention. Javier Garcia |