waiting for you
No, if f(n) o(g(n)), it does not necessarily imply that g(n) o(f(n)).
If f(n) o(g(n)), it means that the growth rate of f(n) is smaller than the growth rate of g(n).
A whole number, f, is a factor of another whole number, n, if f goes into n some exact number of times - that is, it leaves no remainder. Another way of putting it is: if you have a number n, then f is a factor of n if you can find a number, g (not necessarily different), such that f*g = n
this is not correct F F F F F F F G A G F F F G F A N D A A J B E D J I N G
∫ f(x)/[f(x) + g(x)]n dx = ∫ 1/[f(x) + g(x)]n - 1 dx - ∫ g(x)/[f(x) + g(x)]n dx
G. F. N. has written: 'Tit for tat for the Lord knows what'
S-U-F-F-E-R-I-N-G
F. G. M. N. Poelhekke has written: 'Guinee-Bissau'
To prove that the function f(n) is big theta of g(n), you need to show that there exist positive constants c1, c2, and n0 such that for all n greater than or equal to n0, c1g(n) f(n) c2g(n). This means that f(n) grows at the same rate as g(n) within a constant factor for sufficiently large n.
It's spelled F-R-I-G-H-T-E-N-I-N-G.
W t f n i g g a
Yes, it is true that the function f(n)2 is asymptotically smaller than the function g(n)2.