Yuko
Reputation
Next privilege 250 Rep.
 Nov 18 comment Solution to a recurrence relation @BrianM.Scott Can you please help me with this problem? math.stackexchange.com/questions/1534243/… Nov 18 comment Solving the recurrence $a_{n+1}=a_n+(n+1)^2, a_0 = 1$ using generating functions @GEdgar Derivative of $\frac{z}{1-z}$ is $\frac{1}{(1-z)^2}$, and what happens to the $n$, should I just delete them and put $\frac{1}{1-z}$? Nov 18 comment Solving the recurrence $a_{n+1}=a_n+(n+1)^2, a_0 = 1$ using generating functions @GEdgar Check my edit. Nov 18 comment Solving the recurrence $a_{n+1}=a_n+(n+1)^2, a_0 = 1$ using generating functions @CalvinKhor Check my edit. I guess that's how it is going to be. But, now how can I apply partial decomposition to this? Nov 18 revised Solving the recurrence $a_{n+1}=a_n+(n+1)^2, a_0 = 1$ using generating functions added 170 characters in body Nov 18 accepted What's the relation between Stirling numbers and the generating functions? Nov 18 accepted Recurrence relation about square of Fibonacci number Nov 18 accepted Is there any English version of Récoltes et Semailles? Nov 18 accepted Prove the identity $x^n = \sum^{n}_{k=0}S_{n,k}(x)_k$ Nov 18 asked Validity in Kripke frame Nov 18 revised Recurrence relation about square of Fibonacci number deleted 323 characters in body Nov 17 revised Solving the recurrence $a_{n+1}=a_n+(n+1)^2, a_0 = 1$ using generating functions added 82 characters in body Nov 17 asked Solving the recurrence $a_{n+1}=a_n+(n+1)^2, a_0 = 1$ using generating functions Nov 16 comment Recurrence relation about square of Fibonacci number I just thought that we should be consistent, that's why when I was trying the substitution, I substituted everywhere. Nov 16 comment Recurrence relation about square of Fibonacci number I see, I thought, we should substitute everywhere. Nov 16 comment Recurrence relation about square of Fibonacci number At the beginning, you use $F_{n+2} = F_{n+1} + F_n$, in the equation at $F_{n+3}^2 = F_{n+2}^2 + F_{n+1}^2 + 2F_{n+2}F_{n+1}$, so you substitute, right? But, how did you end up with $F_{n+3}^2 = F_{n+2}^2 + F_{n+1}^2 + 2F_{n+1}F_n + 2F_{n+1}^2$? I got something completely different. Nov 16 accepted Solution to a recurrence relation Nov 16 comment What's the relation between Stirling numbers and the generating functions? And how do they relate with Stirling numbers? Nov 16 comment Solution to a recurrence relation @BrianM.Scott Yes, and I already did. math.stackexchange.com/questions/1532420/… Nov 16 asked What's the relation between Stirling numbers and the generating functions?