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 An alternate power series representation for ln(x) JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/07/2011, 11:20 PM (This post was last modified: 05/08/2011, 12:20 AM by JmsNxn.) Yeah, I was aware of a direct relation to differentiation: $\frac{d}{dx} f(x) = e^{(\bigtriangleup \frac{d}{dx} f(x)) + x - f(x)}$ And I've extended the definition of the series to: $\ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n$ And I've found a beautiful return: $\ln(x) = e + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{ne}}(x-e^e)^n$ Converges for values as high as 30. My computer overflows before it stops converging. I think my assumptions were correct. If this is true, I might find an infinite convergent power series for ln(x) edit: sadly, doesn't converge for values less than 1 « 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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