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Just asking...
#3
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)

I just wanted to know if this is a way to obtain the values for non-integer x.

Thanks,
Martin
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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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