"2T (n/2)+ n/ log n" Code Answer's
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2T (n/2)+ n/ log n
Concerned Chinchilla
on Jan 05, 2021
t(n) = 2t(n/2) + n/log(n)
= 2(2t(n/4) + n/2/log(n/2)) + n/log(n)
= 4t(n/4) + n/log(n/2) + n/log(n)
= 4(2t(n/8) + n/4/log(n/4)) + n/log(n/2) + n/log(n)
= 8t(n/8) + n/log(n/4) + n/log(n/2) + n/log(n)
= 16t(n/16) + n/log(n/8) + n/log(n/4) + n/log(n/2) + n/log(n)
= n * t(1) + n/log(2) + n/log(4) + ... + n/log(n/2) + n/log(n)
= n(1 + Sum[i = 1 to log(n)](1/log(2^i)))
= n(1 + Sum[i = 1 to log(n)](1/i))
~= n(1 + log(log(n)))
= n + n*log(log(n)))
~= n*log(log(n)) [n -> infinity]
Source: stackoverflow.com
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