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An alternate power series representation for ln(x)
This proof involves the use of a new operator:

and it's inverse:

and the little differential operator:

see here for more:

I'll use these specifically:


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


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.


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.

Messages In This Thread
An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 08:41 PM

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