Problem 2.1
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aaarbk commented
I think the solution to problem 2.1 is wrong.
I believe the crux of the issue is that f(n) = O(g(n))
doesn't imply lg f(n) = O(lg g(n))
.
A counterexample is f(n) = 2
and g(n) = 1
for all n, then f(n) lg f(n)
equals 2
and g(n) lg g(n)
equals 0
and we cannot bound f(n) lg f(n)
from above by g(n) lg g(n)
.
If taking constant functions feels funny, try f(n) = 2 - exp(-n)
and g(n) = 1 + exp(-n)
.