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 True or False Logarithm bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/27/2010, 04:53 AM (This post was last modified: 07/27/2010, 08:02 AM by bo198214.) Hey, the following function $f_n$ iapproaches - if the limit $n\to\infty$ exists - the intuitive logarithm to base $b$, i.e. the intuitive Abel function of $bx$ developed at 1: $f_n(x)=-\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-x^k}{1-b^k}$ The question is whether this is indeed the logarithm, i.e. if $\lim_{n\to\infty} f_n(x) = \log_b(x)$ for $\left|1-\frac{x}{b}\right|<1$, provided that the limit exists at all. It has a certain similarity to Euler's false logarithm series (pointed out by Gottfried here) as it can indeed be proven that $f(b^m) = m$ for natural numbers $m$ (even for $m=0$ in difference to Euler's series): $f_n(b^m) = -\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-b^{mk}}{1-b^k}$ if we now utilize that $\frac{1-y^m}{1-y}=\sum_{i=0}^{m-1} y^i$ for $y=b^k$ then we get $ f_n(b^m) = -\sum_{k=1}^n \left( n \\ k \right) (-1)^{k}\sum_{i=0}^{m-1} b^{ki} = \sum_{i=0}^{m-1}\left(1-\sum_{k=0}^n \left(n\\k\right)(-1)^{k} b^{ki}\right)$ $f_n(b^m)=\sum_{i=0}^{m-1} 1-(1-b^i)^n$ Hence $\lim_{n\to\infty} f_n(b^m) = m$ But is this true also for non-integer $m$? Do we have some rules like $\lim_{n\to\infty} f_n(x^n)=n \lim_{n\to\infty} f_n(x)$, or even $\lim_{n\to\infty} f_n(xy)=\lim_{n\to\infty} f_n(x) + f_n(y)$? « Next Oldest | Next Newest »

 Messages In This Thread True or False Logarithm - by bo198214 - 07/27/2010, 04:53 AM RE: True or False Logarithm - by Gottfried - 07/27/2010, 01:12 PM RE: True or False Logarithm - by Gottfried - 07/27/2010, 02:41 PM RE: True or False Logarithm - by bo198214 - 08/11/2010, 02:37 AM RE: True or False Logarithm - by andydude - 04/25/2012, 09:37 PM

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