Asymptotically tight bounds on lg(n!)

Related BrainMass Solutions

• Stirling's Approximations and Growth Rate of Functions : Limits, Infimum, Supremum, Asymptotic Upper Bound, Asymptotically Negligible, Asymptotic Lower Bound, Asymptotically Dominant and Asymptotically Tight Bound

  Stirling's approximations and growth rate of functions are investigated with regard to limits, infimum, supremum, asymptotic upper bound, asymptotically negligible, asymptotic lower bound, asymptotically dominant and asymptotically tight bound.

• Growth Rate of Functions : Limits, Infimum, Supremum, Asymptotic Upper Bound, Asymptotically Negligible, Asymptotic Lower Bound, Asymptotically Dominant and Asymptotically Tight Bound

  Growth rate of function pairs is analyzed with respect to asymptotic upper bound, asymptotically negligible, asymptotic lower bound, asymptotically dominant and asymptotically tight bound. The solution is detailed and well presented.

• Combinational Arithmetic Circuit : Sum of n k-bit numbers

  [b] Give a tight asymptotic upper (big-Oh) bound in terms of n and k, on the size of this circuit (definition below). Briefly explain your answer.

• Recurrence Running Time

  63537 Recurrence Running Time Give asymptotic upper and lower running time bounds for T(n) for each of the recurrences. Assume that T(n) is constant for n <= 2. Make bounds as tight as possible, and justify solutions.

• Recurrences

  Use a recursive tree to give an asymptotically tight solution to the recurrence T(n) = T(n-a) + T(a) + cn, where n > = 1 and c > 0 are constants.

• Definition of small O-notation

  A weak upper bound on n! is n n , since each of the n terms in the factorial product is at most n. The o-notation is used to denote an upper bound that is NOT asymptotically tight. Hence, intuitively !

• Smooth functions

  Again, f(k*n) = (k*n)^{lg(k*n)} = k^{lg(k*n)} * n^{lg(k*n)} => f(k*n) = k^{lg(k)} * n^{lg(n)} * k^{lg(n)} * n^{lg(k)} which belongs to O(f(n)) Hence proved.

• Solving a recurrence

  =2^k*T(2^2^0)+k*2^k =2^k*T(2)+k*2^k =2^k+k*2^k =(k+1)*2^k Since n=2^2^k, then 2^k=lg n and k=lg lg n Thus T(n)=theta((lg n)*(lg lg n)) This shows how to solve a recurrence using theta notation.

• Proof Limit Solved

  =2^k*T(2^2^0)+k*2^k =2^k*T(2)+k*2^k =2^k+k*2^k =(k+1)*2^k Since n=2^2^k, then 2^k=lg n and k=lg lg n Thus T(n)=theta((lg n)*(lg lg n)) ***Please show intermediate steps Please see the attached file. The expert solves proofs.