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+--- Thread: [repost] A nowhere analytic infinite sum for tetration. (/showthread.php?tid=1188)
[repost] A nowhere analytic infinite sum for tetration. - tommy1729 - 03/20/2018
In the title I mentioned that this is a repost.
Indeed these ideas are not completely new and have been mentioned before.
However I Will reinvent my own wheel , by a different presentation of the/my ideas ...
The basic principle is very old : infinite Sum and telescoping.
We want to solve t(exp(x)) = e t(x).
Define
0 < a < 1.
f(x,a) = exp(x) + a x.
f^[n+1](x,a) = f( f^[n](x,a) ,a).
—-
Now consider
for Every real y there is a solution x to
f(x,a) = y.
Nomatter What a is , but still 0 < a < 1.
——
Now define the Sum from m = - Oo to Oo :
t(z,a) = ... + exp(z + 2)/f^[-2](1/2,a) + exp(z + 1)/f^[-1](1/2,a) + ...
So t(z,a) = sum_m exp(z - m)/f^[m](1/2,a).
Now let Lim a —> 0 :
z = x then :
Lim t(z,0) = t(z) = t(x)
t(exp(x)) = e t(x).
Qed
___
For practical reasons I suggest
Compute for x < 1 :
t(x) by computing t(exp(x))/e.
——
We can continue
since 1/z + 1/exp(z) + 1/exp(exp(z)) + ...
Is nowhere analytic since exp iterations are chaotic we have :
t(x) is nowhere analytic.
Also
t(exp(x)) = t(x) e.
ln t( exp(x) ) = ln t(x) + 1
Thus
ln t(x) = slog(x).
A nowhere analytic slog(x) !
——-
Due to the telescoping infinite Sum , this mzthod is probably the simplest nowhere analytic solution in the spirit of calculus.
I assume it was worthy of Being repeated.
Regards
Tommy1729