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Just asking...
#9
martin Wrote:Oy, I'm not so good at explaining things, but I'll try anyway.

and exercise makes you better and better Smile

Quote:all the "common" mean values (arithmetic, geometric, quadratic and harmonic mean) can be expressed in the form [(a(1)^n+a(2)^n+...+a(m)^n)/m]^(1/n). For n=1, this is the arithmetic mean, for n=2 the quadratic, for n=-1 the harmonic, and, by analytic continuation or whatever you may call it, for n=0 the geometric mean.

Thats interesting! Can you give a proof for n=0, i.e that
?

Quote: I assumed that, with such a flexible mean value calculation, it just had to work. All I had to do was figure out the appropriate parameter n for a given interval [x ... x+1]. But lately I started doubting this assumption as well.

So the formula is the generalized mean of what? *headscratch*
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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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