+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3)
+--- Thread: General question on function growth (/showthread.php?tid=606)
General question on function growth - dyitto - 03/08/2011
I've been reading this explanation.
Now take functions
f(x) = x^x
g(x) = (x + 1)^(x + 1)
According to the definition of "little-oh", I'd conclude that f(x) = o(g(x)).
Am I right?
RE: General question on function growth - bo198214 - 03/08/2011
(03/08/2011, 07:37 AM)dyitto Wrote: I've been reading this explanation.
Now take functions
f(x) = x^x
g(x) = (x + 1)^(x + 1)
According to the definition of "little-oh", I'd conclude that f(x) = o(g(x)).
Am I right?
Yes, because f is not \( \Omega(g) \): if you would chose any constant C>0 (> 0 is essential though omitted in that text, better look at wikipedia), then you always find
x^x < C (x+1)^(x+1) for large enough x
because
x^x / (x+1)^(x+1) < (x+1)^x / (x+1)^(x+1) = 1/(x+1) < C
RE: General question on function growth - dyitto - 03/08/2011
Intuitively I would say that the above functions f and g have about the same growth rate, since f simply stays one step behind g.
A function with a REAL different growth rate would be:
h(x) = x^(x^x)
So if I wanted to look into the relative growth of hyperoperational functions, then these Bachmann–Landau notation apparently wouldn't be of much use in this context.