New tetration method based on continuum sum and exp-series
#5
(04/20/2010, 07:45 PM)tommy1729 Wrote: \( \log_b\left(\frac{\mathrm{tet}'_b(z)}{\mathrm{tet}'_b(0) \log(b)^z}\right) = \sum_{n=0}^{z-1} \mathrm{tet}_b(n) \)

somewhat off topic but i was thinking about :

for some 'suitable' positive real z :

\( \mathrm{tet}'_b0(0) \log(b0)^z=\mathrm{tet}'_b1(0) \log(b1)^z \)

suppose we find a relation (function) between b0 and b1 ( function from b0 to b1 ) that holds for b0 , b1 < eta but could be extended (coo [apart from some poles perhaps (at e.g. eta ) ] ) to b0 or b1 > eta then Ansus and mike might have a better prospect with their equations...

regards

tommy1729

however note that the function mapping b0 to b1 will 'turn' from mapping b1 to b0.

so potential usefullness is only in the interval base [eta,'turning point']

( hoping turning point > eta , but recall we get to choose parameter z ! )

to visualize , consider f = x^2 and map f(-x) onto f(x) and then it will " turn " at ' turning point 0 ' and do the opposite ; 'f(-x) to f(x)' as x grows from -oo to + oo.

let the function mapping real b0 to real b1 be called m.

the reason for this turning is intuitive , if b0 < m(b0) and log(b0) = 1

then b0 + epsilon > m(b0 + epison) because log(b0 + epsilon) > 1 and the equation \( \mathrm{tet}'_b0(0) \log(b0)^z=\mathrm{tet}'_b1(0) \log(b1)^z \) cannot be solved for real b0 and real b1 with b1 > b0.

note that this gives the upper bound of (b0 or b1) = e.

if im not mistaken ... i know i know ... this isnt formal.

( UFO gottfried ? :p )

regards

tommy1729
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RE: New tetration method based on continuum sum and exp-series - by tommy1729 - 04/20/2010, 08:12 PM

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