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?
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
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 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.
Prove the following using the following definition of O,Big-omega,Theta, small omega Σki=1 ?i ?i = ?(nk )??? ? > 1.
How would I prove that log2(32n) = O(n) (Big Oh of N) I got: 2 log2(3) n <= c * n , however, I do not know how to continue from this part. Thanks
Prove that if f (n) = O (g (n)) and g (n) = Ohm (h (n)), it is not necessarily true that f(n) = O (h (n)). You may assume that low degree (i.e., low-exponent) polynomials do not dominate higher degree polynomials, while higher degree polynomials dominate lower ones. For example, n^3 notequalto O (n^2), but n^2 = O (n^3). Prove that if f (n) = O (g (n)) and g (n) = Ohm (h (n)), it is not necessarily...
1 question) Arrange the following in the order of their growth rates, from least to greatest: (5 pts) n3 n2 nn lg n n! n lg n 2n n 2 question)Show that 3n3 + n2 is big-Oh of n3. You can use either the definition of big-Oh (formal) or the limit approach. Show your work! (5 pts.) 3 question)Show that 6n2 + 20n is big-Oh of n3, but not big-Omega of n3. You can use either the definition of big-Omega...
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,...
please be clear with the steps taken and understandable 1. Prove that if f(n) = Θ(n2) for all f(n), then ΣΑ(n)-6(n3). i=1 2. Prove that if f.(n) are linear functions - i.e., that f(n)-Θ(n) for all Tn A(n) then Σ if.(n) = Θ(n3). Y definition of Big-Oh. ou are not required to use the formal i1
What are the Big-Oh and Omega orders of the following code fragment? What is Tilde approximation? The fragment is prameterized on the variable n. Assume that you are measuring the number of swap calls. for(int j=0;j<n-1;j++){ int z = j; for (int i=j+1; i<n; i++){ if(a[i] < a[z]){ z=i;} } if(z!= j){ swap(a[j], a[z]); //count these } }
Please Help. Where did I go wrong? Results for this submission Entered Answer Preview sqrt{{[4*sqrt{(x^2)+1}]/3)+75) 4.12 +1 +75 3 The answer above is NOT correct. (1 point) Solve the separable differential equation 4x – byvx2 +1 By/2 +1 = Subject to the initial condition: y(0) = 10. y= sqrt( (4* sqrt(x^2+1}}/3 + 75)