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I am having trouble solving a proof. Where t(n) <= cn^1.6, c being a constant. In general, Big Omega is the opposite of Big O in that it is the best case scenerio and looks for the lower bound. So there exists a c and and n0 such that n >= n0. But I am uncertain how to apply this to the proof and how to manipulate the constants in the equation to find c and n0 and to prove that t(n) is Omega(n^1.6).

t(n) = (n-3logn)^1.6 + 5n^1.5 + 7 is Omega(n^1.6)

Could anyone offer some insight on how to do this type of problem? Thanks in advance!

Also so I dont get any criticism as was received from the comment below me, this is not a homework problem but an example taken from a set of exercises so that it is easier for someone to explain the general concept behind this type of problem.

deedex11
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  • You basically need to prove (and that's quite trivial) that omega-wise log ^ 1.6 n and n ^ 1.5 is "insignificant" compared to n ^ 1.6. – chx Feb 24 '11 at 22:48

1 Answers1

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Definition of Big-Omega: f(n)=Omega(g(n)) iff constants C and K exist such that for all n > K, f(n) > C * g(n)

In other words, you need to be able to say something like this: "I choose C = 5, and now for all n > 1000, f(n) > 5 * g(n), so there."

Let's look at your problem now.

t(n) = (n-3logn)^1.6 + 5n^1.5 + 7 is Omega(n^1.6)

C * n^1.6 < (n-3logn)^1.6 + 5n^1.5 + 7

Divide by n^1.6

C < ((n-3logn)/n)^1.6 + 5n^-0.1 + 7/n^1.6
C < (1 - 3logn/n)^1.6 + 5n^-0.1 + 7/n^1.6

So let's look at these three terms one by one (more formal proof would be needed, of course, but these are straightforward).

1. (1 - 3logn/n)^1.6 = (1 - 0.smth)^1.6 = (0.smth)^1.6 < 1 for n > 2
2. 5n^-0.1 = 5/n^0.1 = 5/smth greater than 1 < 5 for n > 2
3. 7/n^1.6 = 7/smth large < 1 for n > 2

So we see that for any n > 2, C < 1 + 5 + 1 = 7

Now you get to say "I choose C=7, and for any n > 2, C*n^1.6 < ..."

taddy hoops
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iluxa
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