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+--- Thread: Exploring Pentation - Base e (/showthread.php?tid=103)
RE: Exploring Pentation - Base e - Ivars - 02/15/2008
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?
RE: Exploring Pentation - Base e - quickfur - 02/22/2008
Ivars Wrote:[...]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.
I would like to "integrate" (or differentiate) e.g pentation to obtain "slower" operation (or faster).
[...]
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.
RE: Exploring Pentation - Base e - Ivars - 03/03/2008
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=1/137,0316766From 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