Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
An alternate power series representation for ln(x)
#1
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/tetrationforu...40#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.
Reply
#2
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.
Reply
#3
(05/07/2011, 08:41 PM)JmsNxn Wrote: This proof involves the use of a new operator:


and it's inverse:


and the little differential operator:

(The notation is more unambiguous than in your previous thread Smile )

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]
you get a better ln-typesetting.
Reply
#4
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
Reply
#5
(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 1
The series converges for , hence your series converges for , this should include all values ?
Reply
#6
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.
Reply
#7
(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:
Reply
#8
(05/08/2011, 08:28 PM)bo198214 Wrote: Oh thats just:


Oh that's so simple! Thanks for the help.
Reply


Possibly Related Threads...
Thread Author Replies Views Last Post
  A Notation Question (raising the highest value in pow-tower to a different power) Micah 8 1,186 02/18/2019, 10:34 PM
Last Post: Micah
Question Taylor series of i[x] Xorter 12 9,694 02/20/2018, 09:55 PM
Last Post: Xorter
  Functional power Xorter 0 1,084 03/11/2017, 10:22 AM
Last Post: Xorter
  2 fixpoints related by power ? tommy1729 0 1,200 12/07/2016, 01:29 PM
Last Post: tommy1729
  Taylor series of cheta Xorter 13 10,290 08/28/2016, 08:52 PM
Last Post: sheldonison
  Inverse power tower functions tommy1729 0 1,640 01/04/2016, 12:03 PM
Last Post: tommy1729
  Remark on Gottfried's "problem with an infinite product" power tower variation tommy1729 4 4,589 05/06/2014, 09:47 PM
Last Post: tommy1729
  [integral] How to integrate a fourier series ? tommy1729 1 2,213 05/04/2014, 03:19 PM
Last Post: tommy1729
  about power towers and base change tommy1729 7 7,162 05/04/2014, 08:30 AM
Last Post: tommy1729
  Iteration series: Series of powertowers - "T- geometric series" Gottfried 10 15,278 02/04/2012, 05:02 AM
Last Post: Kouznetsov



Users browsing this thread: 1 Guest(s)