How to prove log2(7 + 1/n) = Big Omega(1) I know that by definition to be...
How to prove log2(7 + 1/n) = Big Omega(1)
I know that by definition to be Big omega of 1 the f(n) is not bounded by n to infinity, but its bounded to a constant C.
Homework Answers
Big Omega Definition: ---------------------------- f(n)=Ω(g(n)) usually defined by saying for some constant c>0 and all large enough n, f(n)≥c g(n). log2(7 + 1/n) = Big Omega(1) Here, f(n) = log2(7 + 1/n) For all value of n > 0 the condition log2(7 + 1/n) >= 2.8 So, We can say that log2(7 + 1/n) >= 2.8*1 Where c = 2.8 and n > 0 and g(n) = 1 So, f(n)=Ω(g(n)) f(n)=Ω(1)
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How to prove log2(7 + 1/n) = Big Omega(1)
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