Recursive question

  1. 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!!
    Last edited: Dec 3, 2005
  2. jcsd
  3. uart

    uart 2,776
    Science Advisor

    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].
    Last edited: Dec 3, 2005
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