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 Just asking... bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/18/2008, 01:48 PM martin Wrote:Oy, I'm not so good at explaining things, but I'll try anyway. and exercise makes you better and better 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 $\lim_{n\to 0} \sqrt[n]{\frac{a_1^n+\dots+a_m^n}{m}} = \sqrt[m]{a_1 \dots a_m}$? 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 $\sqrt[c(2-x)]{x(2^{c(2-x)}-1)+1}$ is the generalized mean of what? *headscratch* « Next Oldest | Next Newest »