An alternate power series representation for ln(x) - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: An alternate power series representation for ln(x) ( /showthread.php?tid=628) |

An alternate power series representation for ln(x) - JmsNxn - 05/07/2011
This proof involves the use of a new operator: and it's inverse: and the little differential operator: see here for more: http://math.eretrandre.org/tetrationforum/showthread.php?tid=627&pid=5740#pid5740 I'll use these specifically: and The proof starts out by first proving: first give the power series representation of e^x And given: We take the ln of e^x to get an infinite series of deltations, if: represents a series of deltations then and therefore if we let x = e^x and now we have an infinite lowered polynomial which when little differentiated equals itself. It's the little polynomial equivalent of e^x's perfect taylor series. therefore: and using the chain rule: where f'(x) is taken to mean the little derivative of f(x). we get the result: and now we can solve for the k'th little derivative of ln(x) using the little power rule, k E N and so if the little derivative Taylor series is given by: where is the n'th little derivative of . we can take the little derivative taylor series of ln(x) centered about 1. The first term is equal to 0, so I'll start the series from n = 1. Here it is in two steps: now let's plug this in our formula for ; it works for an infinite sum because is commutative and associative. now since: now take the lns away and and now if we let x = ln(x) we get: And there, that's it. I haven't been able to check if this series converges, I don't have many convergence tests. I'm just sure that the algebra is right. I haven't tried computing it. In my gut it doesn't look like it will converge, but I have faith. RE: An alternate power series representation for ln(x) - JmsNxn - 05/07/2011
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 ) I find the normal taylor series of ln(x): 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. RE: An alternate power series representation for ln(x) - bo198214 - 05/07/2011
(05/07/2011, 08:41 PM)JmsNxn Wrote: This proof involves the use of a new operator: (The notation is more unambiguous than in your previous thread ) But your operator can be expressed with the classical differentiation, see: Or purely functional with the composition operation : PS: when you write ln with backslash in front: Code: `[tex]\ln(x)[/tex]` RE: An alternate power series representation for ln(x) - JmsNxn - 05/07/2011
Yeah, I was aware of a direct relation to differentiation: And I've extended the definition of the series to: And I've found a beautiful return: 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 RE: An alternate power series representation for ln(x) - bo198214 - 05/08/2011
(05/07/2011, 11:20 PM)JmsNxn Wrote: But this follows directly from the definition of the logarithm, by the following equivalent transformations: Quote:sadly, doesn't converge for values less than 1The series converges for , hence your series converges for , this should include all values ? RE: An alternate power series representation for ln(x) - JmsNxn - 05/08/2011
I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the went to in the denominator. Otherwise, though, that's a nice proof. RE: An alternate power series representation for ln(x) - bo198214 - 05/08/2011
(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 went to in the denominator. Otherwise, though, that's a nice proof. Oh thats just: RE: An alternate power series representation for ln(x) - JmsNxn - 05/09/2011
(05/08/2011, 08:28 PM)bo198214 Wrote: Oh thats just: Oh that's so simple! Thanks for the help. |