Exploring Pentation - Base e

[Ivars]

BTW, the value:

1/(-1,850354529) = -0,540436972651802

while cos(-1) which is the real part of -I*e^(-I) in complex plane is

cos (-1) = -0,54030230586814

The difference between infinite negative pentation of e and cos(-1) being 0,025%.

I have a feeling something is being rather closely approximated by infinite negative pentation of e (e.g. alpha, (-I*e^(-I )), ) . Where would the next steps of approximation hide?

[quickfur]

[...]
I would like to "integrate" (or differentiate) e.g pentation to obtain "slower" operation (or faster).
[...]

Given any function f(x) that grows asymptotically faster than \( e^x \), the derivative f'(x) must necessarily grow faster than f itself. The converse is also true: if f(x) grows asymptotically slower than \( e^x \), then f'(x) will grow asymptotically slower than f itself. When this difference is precisely by a factor of x (i.e., \( \lim_{x\rightarrow\infty}f(x)/f'(x)=Cx+D \) for constants C≠0, D), then f(x) is a polynomial. There exist functions whose asymptotic ratio with their derivatives lie between 0 and Cx+D... I'll leave this as an exercise for the reader.

Since tetration, pentation, and higher grow a lot faster than exponentiation, their derivative must necessarily be faster. Moreover, the difference between the derivative of a pentation and pentation will be much greater than the difference between the derivative of tetration and tetration.

[Ivars]

I took

I added asymtotic values of negative infinite of base e heptation and [9] to the sum I mentioned before:

Sum [5] =1/1,85035452902718^8-1/(2*1,8503545290271^7+1/(3*1,8503545290271^6-1/(4*1,8503545290271^5+1/(5*1,8503545290271^4=0,007297583=1/137,0316766

From Andrew's graph, I found the values to be roughly e[7]-infinity = -3,751 and e[9]-infinity = -5,693.

Then I put them in the same sum, obtaining:

Sum [7] =1/3,751^8-1/(2*3,751)^7+1/(3*3,751)^6-1/(4*3,751)^5+1/(5*3,751)^4=3,20285E-05

Sum[9] = 1/5,693^8-1/(2*5,693)^7+1/(3*5,693)^6-1/(4*5,693)^5+1/(5*5,693)^4=3,15992E-05

Then I made Sum [5,7,9] = Sum[5]-Sum[7]+Sum[9] = 0,007297583-3,20285E-05+3,15992E-05=0,0072971534=1/137,039738252

So after this, approximation of alpha =0.07297352570(5) got even better, as I expected, but of course I do not know the exact values of e[7]-infinity and e[9]-infinity and more.

Then we could see how does the sum Sum[5]-Sum[7]+Sum[9]-Sum[11]+Sum[13]-Sum[15]+.........converge.

Ivars