Just asking...

[bo198214]

Oy, I'm not so good at explaining things, but I'll try anyway.

and exercise makes you better and better

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
\( \lim_{n\to 0} \sqrt[n]{\frac{a_1^n+\dots+a_m^n}{m}} = \sqrt[m]{a_1 \dots a_m} \)?

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*