|Dec3-05, 12:55 PM||#1|
T(0) = 1
T(n) = T(n-1) + root(T(n-1))
how many recursion does T(n) need to grow to the number k?
can I get this? root(k) < m < c root(k)
c is constant and m is the times we need for T(n) goes to k.
Any help appreciated!!
|Dec3-05, 10:18 PM||#2|
You can bound it with a continous function.
Try solving dy/dx = sqrt(y) with y(0)=1, you can show that this function is bigger than T but gives a pretty good bound. Solving the above gives c=2 as the constant you're looking for.
BTW: Maybe I'm just dumb but I had to read you post about three times before I figured out that by "root" you meant square root or sqrt or [tex]\sqrt(.)[/tex].
|Similar Threads for: recursive question|
|Recursive Relations.||Set Theory, Logic, Probability, Statistics||0|
|recursive definition||Calculus & Beyond Homework||3|
|recursive algorithm||Calculus & Beyond Homework||6|
|recursive question||Calculus & Beyond Homework||1|
|non-recursive formula||General Math||3|