General question on function growth - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/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. Yes, because f is not : 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 Bachmannâ€“Landau notation apparently wouldn't be of much use in this context. |