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.

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.