Quote:

Originally Posted by **hotmail590**

Can someone please help me with this problem?

Suppose that the function f satisfies the recurrence relation

2 * f ( sqrt(n) ) + log(n)

whenever n is a perfect square greater than 1 and f(2) = 1

As this is it is not a recurrence relation as it stands.

It should look something like:

$\displaystyle

f(n)=2 f(\sqrt{n})+\log(n)

$

then as $\displaystyle f(2)=1$ we would have:

$\displaystyle

f(16)=2f(4) +\log(16)=2 (2f(2)+\log(4))+\log(16)

$

$\displaystyle

=2(2+\log(4))+\log(16)=4+2\log(4)+\log(16)

$

RonL