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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, 09:43 PM (This post was last modified: 05/07/2011, 10:05 PM by JmsNxn.) Doing some tests it converges for 1, 2, 4, 5 after about 10 terms, it fails to converge for 6. Still, I'm enthralled this works. Doing some more tests, if I do the same process over again, but take the little taylor series of ln(x) centered about 0 instead of 1 (we can do this because negative infinity is the identity of $\bigtriangleup$) I find the normal taylor series of ln(x): $ln(x) = \sum_{n=1}^{\infty} (x-1)^n \frac{(-1)^{n+1}}{n}$ so this leads me to believe that if we take the little taylor series centered about K and then do the same process to get a normal power series expression, and we let K tend to infinity, we will have an ever growing radius of convergence. That is since doing this method centered about 1 has a bigger radius of convergence than doing this method centered about 0. « 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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