How Many Recursions Does T(n) Need for k?

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The discussion focuses on determining the number of recursions required for the function T(n) defined as T(0) = 1 and T(n) = T(n-1) + √(T(n-1)) to reach a value k. It is established that the growth of T(n) can be bounded by a continuous function, specifically through the differential equation dy/dx = √y with the initial condition y(0) = 1. The constant c is determined to be 2, providing a useful approximation for the number of recursions needed, denoted as m, where √k < m < 2√k.

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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:
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You can bound it with a continuous 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:

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