Prove or find a counterexample for the following Assume that

Prove or find a counterexample for the following. Assume that f(n) and g(n) are monotonically increasing functions that are always larger than 1.

Prove or find a counterexample for the following. Assume that f(n) and g(n) are monotonically increasing functions that are always larger than 1. (1) f (m) o(g(n)) log (f (m)) o(log (g(ra))) o(29 (n)) (3) f(n)

Solution

(1) f(n) = o(g(n)) log(f(n)) = o(log(g(n)))

given that

f(n) is o(g(n)) is true then log(f(n)) = o(log(g(n))) also true... this statement is true..

explanation..

lets take f(n) = o(g(n))

f(n) is o(g(n)), means for every constant k>0, the function g(n) is greater than the f(n)

now apply both sides log, means we are decreasing growth rate of f(n) and g(n), both by log, equally..

since we are decreasing their growth rate equally, for every contant k>0 , log(g(n)) is greater than log(f(n))

means log(f(n)) = o(log(g(n)))..hence proved..

(2) f(n) = O(g(n)) log(f(n)) = O(log(g(n)))

f(n) is O(g(n)) is true then log(f(n)) = O(log(g(n))) also true... this statement is true..

explanation:

lets take f(n) = O(g(n))

f(n) is O(g(n)), means their is a constant k>0,where the function g(n) is greater than or equal to the f(n),tightly bounded by g(n)

now apply both sides log, means we are decreasing growth rate of f(n) and g(n), both by log, equally..

since we are decreasing their growth rate equally, for the same contant k>0 , log(g(n)) is greater than or equal to log(f(n))

means log(f(n)) = O(log(g(n)))..hence proved..

(3) f(n) = o(g(n)) 2^f(n) = o(2^g(n))

(4) f(n) = O(g(n)) 2^f(n) = O(2^g(n))

similary 3,4 are also true.. like 1 and 2

difference is

in 3,4 we increasing growth rate f(n) and g(n), exponentially, equally on both...sides

hence they are also true..