
#1
Dec305, 12:55 PM

P: 40

Suppoese
T(0) = 1 T(n) = T(n1) + root(T(n1)) 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!! 



#2
Dec305, 10:18 PM

Sci Advisor
P: 2,751

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]. 


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