An alternate power series representation for ln(x)
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An alternate power series representation for ln(x)
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05/08/2011, 08:28 PM
(05/08/2011, 07:54 PM)JmsNxn Wrote: I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the \( e^{an} \) went to in the denominator. Otherwise, though, that's a nice proof. Oh thats just: \( \frac{1}{e^{an}}(x-e^a)^n=\frac{(x-e^a)^n}{(e^a)^n} = \left(\frac{x-e^a}{e^a}\right)^n = \left(\frac{x}{e^a} - 1\right)^n \) |
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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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