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 General question on function growth dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/08/2011, 07:37 AM 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? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/08/2011, 04:09 PM (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 dyitto Junior Fellow Posts: 13 Threads: 3 Joined: Mar 2011 03/08/2011, 04:41 PM 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. « Next Oldest | Next Newest »

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