cormen chapter 3

Question

Consider the two functions (logn)^k  and  n^ϵ, where k>1 and ϵ>0. 

The solution is (logn)^k= O(n^ϵ). 

My doubt is that if we take ϵ as0.000000000000000000000000000000.......000000000001 that is very small wont this result be wrong??

Can someone check?

Comments

(logn)^k= (n^ϵ)

Take log both side 

k. loglogn = ϵ logn 

so k. loglogn = O(ϵ.logn)  means  k. loglogn$\leqslant$ C ϵ.logn [ here c is some constant]

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but ϵ  can be fraction lie 0.00000000000000000000000000000000000000...........1. will it not make ϵ log n very small?

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constant C will take care of it. C can be 10000000........000000000000000000

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oh yeah wow awesome. thanku so much for the explanation.