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 An alternate power series representation for ln(x) bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/08/2011, 01:38 PM (05/07/2011, 11:20 PM)JmsNxn Wrote: $\ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n$ But this follows directly from the definition of the logarithm, by the following equivalent transformations: $\begin{eqnarray} \ln(x) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n\\ &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x/e^a-1)^n\\ \ln(e^a y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\ a + \ln(y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\ \ln(y) &=& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n \end{eqnarray}$ Quote:sadly, doesn't converge for values less than 1 The $\ln$ series converges for $|y-1|<1$, hence your series converges for $|x/e^a-1|<1$, this should include all values $x\in (0,2e^a)$? « Next Oldest | Next Newest »

 Messages In This Thread An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 08:41 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 09:43 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/07/2011, 10:45 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 11:20 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/08/2011, 01:38 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/08/2011, 07:54 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/08/2011, 08:28 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/09/2011, 01:02 AM

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