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Recursive question

by flying2000
Tags: recursive
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flying2000
#1
Dec3-05, 12:55 PM
P: 40
Suppoese

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!!
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uart
#2
Dec3-05, 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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