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tetration base > exp(2/5)
#3
It appears that f(x) = 0 + 5/2 x + 5/2 x^5/5! + 5/2 x^6/6! + 5/2 x^10/10!
+ 5/2 x^11/11! + ...

is the only multisection type function that satisfies for all x > 0 :

x =< f(x) < exp(x).

The other candidates a/b seem to violate already within the bounds

0 < x < 11b.

SO assuming this is correct , probably the method based on base exp(2/5) converges the best.
As expected in post 1 and this answers part 1 of post 2.

More investigation is desired but probably the fact that the only number b such that 2/b is a good appr to 1/e is b = 5 is fundamental.

Large a in a/b lead to problems for x >> 1 because of the many powersums needed.
This makes some sense because 1/e < 1/2.

***

As for the cousin question , proving true is as hard as the cousin prime conjecture and probably equally hard as the twin prime conjecture.
Both unsolved as of today.

However proving false might be easier.

///

Notice fake function theory can in a way be an approximate inverse multisection.

a1 x^2 + a2 x^4 + ... = multisect

fake ( a1 x^2 + a2 x^4 + ... ) ~ Original function.

Nice.

///

regards

tommy1729
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Messages In This Thread
tetration base > exp(2/5) - by tommy1729 - 02/09/2015, 11:59 PM
RE: tetration base > exp(2/5) - by tommy1729 - 02/10/2015, 10:43 PM
RE: tetration base > exp(2/5) - by tommy1729 - 02/11/2015, 12:29 AM

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