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 Exploring Pentation - Base e Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 02/15/2008, 08:24 PM (This post was last modified: 02/15/2008, 09:49 PM by 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 Junior Fellow  Posts: 22 Threads: 1 Joined: Feb 2008 02/22/2008, 12:21 AM Ivars Wrote:[...] 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 , the derivative f'(x) must necessarily grow faster than f itself. The converse is also true: if f(x) grows asymptotically slower than , then f'(x) will grow asymptotically slower than f itself. When this difference is precisely by a factor of x (i.e., 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 Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 03/03/2008, 08:04 PM (This post was last modified: 03/03/2008, 08:09 PM by Ivars.) I took I added asymtotic values of negative infinite of base e heptation and  to the sum I mentioned before: Sum  =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-infinity = -3,751 and e-infinity = -5,693. Then I put them in the same sum, obtaining: Sum  =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 = 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-Sum+Sum = 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-infinity and e-infinity and more. Then we could see how does the sum Sum-Sum+Sum-Sum+Sum-Sum+.........converge. Ivars « Next Oldest | Next Newest »

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