How can I go prove ( sqrt( (n+1)^3 ) ) is Big Omega (n * sqrt(n))...
How can I go prove ( sqrt( (n+1)^3 ) ) is Big Omega (n * sqrt(n)) using the formal definitions of Big Oh, Big Theta, and Big Omega?
Homework Answers
f(n) = Omega(g(n)) means there are positive constants c and n0, such that f(n) >= cg(n) for all n ≥ n0 sqrt((n+1)^3) => (n+1) sqrt(n+1) (n+1) sqrt(n+1) = Omega(n * sqrt(n)) => (n+1) sqrt(n+1) >= cn * sqrt(n) Let's assume c = 1 => (n+1) sqrt(n+1) >= n * sqrt(n) => (n+1) >= n (because sqrt(n+1) is always > sqrt(n)) (n+1) >= n for all positive values of n so, (n+1) sqrt(n+1) = Omega(n * sqrt(n)) for c = 1 and n0 = 0
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How can I go prove ( sqrt( (n+1)^3 ) ) is Big Omega (n *
sqrt(n))...
Formal Definitions of Big-Oh, Big-Theta and Big-Omega: 1. Use the formal definition of Big-Oh to prove that if f(n) is...
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Use the properties of Big - Oh, Big - Omega, and Big - Theta to prove...
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I understand how it was simplified to n^(∈/(sqrt(logn))), but I'm trying to understand how to prove...
I understand how it was simplified to n^(∈/(sqrt(logn))), but
I'm trying to understand how to prove that logn grows faster for
0<∈<1. The derivative seems too complicated to prove this via
Lhopital's Rule, so I tried using WolframAlpha to compare the two
with logn as the numerator:
http://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+(logn)%2F(n%5E(0.5%2F(sqrt(logn))))&rawformassumption=%7B%22FunClash%22,+%22log%22%7D+-%3E+%7B%22Log10%22%7D
However, this gives me a result of 0 for any value above 0,
which would mean that n^(∈/(sqrt(logn))) grows at a faster rate,
even when 0<∈<1.
When I try to graph it,...
Formal Definitions of Big-Oh, Big-Theta and Big-Omega:
1. Use the formal definition of Big-Oh to prove that if f(n) is a decreasing function, then f(n) = 0(1). A decreasing function is one in which f(x1) f(r2) if and only if xi 5 r2. You may assume that f(n) is positive evervwhere Hint: drawing a picture might make the proof for this problem more obvious 2. Use the formal definition of Big-Oh to prove that if f(n) = 0(g(n)) and g(n)...
Use the properties of Big - Oh, Big - Omega, and Big - Theta to prove that if f (n) = theta (3 Squareroot n) and g (n) = Ohm (f (n) + 7 f (n)^2 + 49 Squareroot n), then g (n)^3 = Ohm (n^2). You may use the fact that n^a = 0 (n^b) if and only if a lessthanorequalto b, where a and b are constants.
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I understand how it was simplified to n^(∈/(sqrt(logn))), but
I'm trying to understand how to prove that logn grows faster for
0<∈<1. The derivative seems too complicated to prove this via
Lhopital's Rule, so I tried using WolframAlpha to compare the two
with logn as the numerator:
http://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+(logn)%2F(n%5E(0.5%2F(sqrt(logn))))&rawformassumption=%7B%22FunClash%22,+%22log%22%7D+-%3E+%7B%22Log10%22%7D
However, this gives me a result of 0 for any value above 0,
which would mean that n^(∈/(sqrt(logn))) grows at a faster rate,
even when 0<∈<1.
When I try to graph it,...
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