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Just asking...
#4
martin Wrote:I know, I said this formula works for 0 < x < 1, the other values are naturally obtained
- for x<0 by iterated log[2^^(x+1)]/log(2)
- for x>1 by iterated 2^(x-1)

So whats your requirement then? You can use any function on 0..1.
Of course, perhaps it should be continuous. And perhaps also differentiable, and perhaps also infinitely often differentiable and perhaps also analytic.

Hm, I didnt try it yet but I guess the formula is not infinitely often differentiable, did you test it? You only need to consider one patch point, for example 0 or 1. If it is infinitely differentiable there then also on all the other points. By this method by the way Andrew (which is also here on the forum) constructed an analytic solution. This is one thread that deals with this construction.
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Messages In This Thread
Just asking... - by martin - 07/14/2008, 09:49 PM
RE: Just asking... - by bo198214 - 07/14/2008, 10:53 PM
RE: Just asking... - by martin - 07/15/2008, 11:40 AM
RE: Just asking... - by bo198214 - 07/16/2008, 08:00 PM
RE: Just asking... - by martin - 07/16/2008, 09:52 PM
RE: Just asking... - by bo198214 - 07/16/2008, 10:36 PM
RE: Just asking... - by bo198214 - 07/16/2008, 10:52 PM
RE: Just asking... - by martin - 07/17/2008, 11:28 AM
RE: Just asking... - by bo198214 - 07/18/2008, 01:48 PM
RE: Just asking... - by martin - 07/18/2008, 06:37 PM
RE: Just asking... - by bo198214 - 07/18/2008, 10:02 PM
RE: Just asking... - by Gottfried - 07/19/2008, 03:48 AM
RE: Just asking... - by Ivars - 07/19/2008, 07:04 AM
RE: Just asking... - by martin - 07/19/2008, 09:03 AM
RE: Just asking... - by Ivars - 07/19/2008, 07:45 PM
RE: Just asking... - by andydude - 07/21/2008, 04:05 AM
RE: Just asking... - by martin - 07/21/2008, 06:31 PM
RE: Just asking... - by Ivars - 07/22/2008, 06:07 AM



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