Originally Posted by
jzellt I need find a closed form for this recurrence relation:
b sub n = b sub n-1 + n^2 for n >0 and b sub 0 = 3
Anybody know how to do this?
There is probably a more elegant way than this:
1. Calculate the first five values
2. Calculate the differences between consecutive values
3. Repeat this process until you get a constant sequence.
4. Determine the number of repetitions. That is the degree of the closed form: Code:
degree 3: 3 4 8 17 33 ...
degree 2: 1 4 9 16 ...
degree 1: 3 5 7 ...
degree 0: 2 2
That means you are looking for a term with degree 3:
$\displaystyle s_n = a\cdot n^3+b\cdot n^2+c\cdot n + d$
From the initial row you know:
s_0 = 3
s_1 = 4
s_2 = 8
s_3 = 17
Thus you have to solve a system of simultaneous equations:
$\displaystyle \left|\begin{array}{r}d=3\\a+b+c+d = 4\\8a+4b+2c+d=8\\27a+9b+3c+d=17\end{array}\right.$
After a few steps you'll get:
$\displaystyle s_n = \frac13 n^3 + \frac12 n^2 + \frac16n+3$