Let be n=2m therefore n/2=2m−1 and we have
T(2m)=2T(2m−1)+m2m
Let S(m)=T(2m) , then S(m)=2S(m−1)+m2m
Let we prove by induction that S(m)≤c⋅m2⋅2m ;
Assume it is true for k<m,then
S(m)≤2c(m−1)2⋅2m−1+m⋅2m=2m⋅c⋅(m2+1−2m+cm))≤m2⋅2m
if c>1. Base of induction will be fulfilled if C would be taken as S(1)≤2c,c≥1
It s always possble.
Now we return to variable n=log(m) and have T(n)≤c⋅n⋅(log(n))2
This is upper bound of increasing means T(n)=O(n⋅(log(n))2) this is solution og given reccurence.
Comments