The Recurrence Relation from the function we get is:
$T(n) = T(n^{1/2}) + C$
$C$: constant time so say ‘$1$’
then,
$T(n) = T(n^{1/2}) + 1$
Now,
Lets Assume $n = 2^{t}$,
Therefore,
$n^{1/2}$ $=$ $2^{t/2}$
Therefore the equation becomes,
$T(2^{t}) = T(2^{t/2}) + 1$
Now, Say there is a function $P(t)$ such that $P(t) = T(2^{t}),$ $….(i)$
So the equation becomes,
$P(t) = P(t/2) + 1$
Now,
Using masters theorem,
$P(t)=\Theta (\log_{2}t)$
and $P(t) = T(2^{t}) = T(n)$ $from(i)$
So, $T(n) = \Theta (\log_{2}t)$
Now,
$\because n = 2^{t},$
$ \therefore t=\log_{2}n$
$ \therefore log_{2}t=\log_{2}\log_{2}n$
$Hence, T(n)=\Theta (\log_{2}\log_{2}n)$