recursion - Calculating the Recurrence Relation T(n)=T(n-1)+logn -


we solve recurrence relation through repeating substitution:

t(n)=t(n-1)+logn

i started substitution , got following.

t(n)=t(n-2)+log(n)+log(n-1)

by logarithm product rule, log(mn)=logm+logn,

t(n)=t(n-2)+log[n*(n-1)]

continuing this, get

t(n)=t(n-k)+log[n*(n-1)*...*(n-k)]

we know base case t(1), n-1=k -> k=n+1, , substituting in get

t(n)=t(1)+log[n*(n-1)*...*1]

clearly n*(n-1)*...*1 = n! so,

t(n)=t(1)+log(n!)

i not know how solve beyond point. answer o(log(n!))? have read other explanations saying Θ(nlogn) , follows o(nlogn) , Ω(nlogn) upper , lower bounds respectively.

this expands out log (n!). can see because

t(n) = t(n - 1) + log n

= t(n - 2) + log (n - 1) + log n

= t(n - 3) + log (n - 2) + log (n - 1) + log n

= ...

= t(0) + log 1 + log 2 + ... + log (n - 1) + log n

= t(0) + log n!

the exact answer depends on t(0) is, θ(log n!) fixed constant value of t(0).

a note - using stirling's approximation, θ(log n!) = θ(n log n). might relate existing complexity classes.

hope helps!


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