I have 2 doubts:-
1) Is (log* n)^n = O((logn)!) ?
2) Which is bigger, log(log* n) or log*(logn) ?
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templatetypedef
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Zephyr
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It is known as iterated logarithm ! You can check here :- https://en.wikipedia.org/wiki/Iterated_logarithm – Zephyr Aug 16 '17 at 10:03
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-.- it's log star and not a multiplication. I am sorry :) It was hard to notice. – Tony Aug 16 '17 at 10:05
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I know it's hard to notice. That's why I mentioned it in the title of the question. Can you help me with this question ? – Zephyr Aug 16 '17 at 10:06
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I am not sure why I am receiving down votes. Did I ask something wrong ? – Zephyr Aug 16 '17 at 10:21
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2You showed 0 efforts. (I didn't Down voted you). – Tony Aug 16 '17 at 10:22
1 Answers
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For 2), you have log*(log n) = log*(n)-1. Then let m = log*(n). You have m-1 > log(m) for sufficiently large m.
Hint:
For 1), let m = log*(n). Then the LHS is m^n, and the RHS is the factorial of the logarithm of an exponential tower of height m, i.e. the factorial of an exponential tower of height m-1.
Even disregarding the factorial, an exponential tower should grow much faster than a power.
Yves Daoust
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Just let me know if I am correct . In 1st case, we can consider log*n as approximately a constant (let it be k) . Now k^n = O(n!). So for very large n, k^n = O((logn)!). Hence (log *n)^n = O((log n)!). Am I correct? – Zephyr Aug 17 '17 at 08:21
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@Zephyr: in asymptotic computations, you may not consider log*n as constant. – Yves Daoust Aug 17 '17 at 09:18