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 Just asking... Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 07/19/2008, 03:48 AM bo198214 Wrote: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 True! I tried to prove it, too, but the without your first element I stuck. Amazing! Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »