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 Just asking... bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/18/2008, 10:02 PM martin Wrote:bo198214 Wrote:Thats interesting! Can you give a proof for n=0, i.e that $\lim_{n\to 0} \sqrt[n]{\frac{a_1^n+\dots+a_m^n}{m}} = \sqrt[m]{a_1 \dots a_m}$? Erm... no. Hasn't this already been proven somewhere? Yes, but knowing an believing are two different things When I considered that formula I realized that I never learned the logarithm formula during studying: $\log(x) = \lim_{n\to\infty} (\sqrt[n]{x} - 1) n$ This is just what you get when you invert the famous Euler formula $e^x = \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}$. Knowing that formula, we can do much more with the limit $\lim_{\eps\to 0} \left(\frac{1^\eps + a^\eps}{2}\right)^{1/\epsilon} =\lim_{n\to\infty} \left(\frac{1+\sqrt[n]{a}}{2}\right)^n = \lim_{n\to\infty} \left(\frac{2+\sqrt[n]{a}-1}{2}\right)^n = \lim_{n\to\infty} \left(1+\frac{(\sqrt[n]{a}-1)n}{2n}\right)^n = e^{\ln(x)/2} = \sqrt[2]{x}$ This is just a particular case with $m=2$, $a_1=a$ and $a_2=1$ in the formula $\lim_{\epsilon\to 0} \sqrt[\epsilon]{\frac{a_1^\epsilon +\dots+ a_m^\epsilon}{m}}=\sqrt[m]{a_1\dots a_m}$. with the above base idea it is then no more difficult to prove the general case. This formula is really amazing Quote:It should merely interpolate 2^^x between x=0 and x=1 - the graph seemed smooth enough (together with the other iterated values beyond) to assume it represented the actual values to at least four decimal digits, but in other tables I've seen, only three decimal digits match. Aha and there is the answer why you considered your formula to be tetration. Because you compared it with some table values! Now that you are catched, you have to tell where you found this table! However, to all my knowledge, there is no *the* tetration. There are different approaches which result in different tetrations (though some only differ at the 10th or higher significant digit) that all satisfy those properties like being analytic strictly increasing and so on. « Next Oldest | Next Newest »