I had just installed Sage, which includes PariGP, but still are stuck on excel, because I need to learn PariGP.
Thanks for your help.
I still chewing it. Since λ is negative, that means that his powers take complex values, then h(x) should be complex, and that's a mess with excel.
So, as practice, I tried with base b=2^⁽½⁾, and still trying to make it work.
The fixed point is 2, and λ=0,693147181
The blue line should be the tetration base b=1,414213562, and should be matching the cyan line (which I got with Excel).
![[Image: NQYymDY.png?1]](http://i.imgur.com/NQYymDY.png?1)
That way, it would match my own solution, and would give ⁰b≈1 and ⁻¹b≈0
![[Image: 7lC9PSh.jpg?1]](http://i.imgur.com/7lC9PSh.jpg?1)
![[Image: 8JBZF9Q.png?1]](http://i.imgur.com/8JBZF9Q.png?1)
I got that solution adjusting the coefficients with Excel's Solver, for a Taylor series with 12 coefficients, (expanded around 0), subject to these restrictions:
a₀=1
⁻¹b=0
⁰b=1
\( ^xb=b^{\,{^{x-1}b}} \)
I got these coefficients:
Thanks for your help.
I still chewing it. Since λ is negative, that means that his powers take complex values, then h(x) should be complex, and that's a mess with excel.
So, as practice, I tried with base b=2^⁽½⁾, and still trying to make it work.
The fixed point is 2, and λ=0,693147181
The blue line should be the tetration base b=1,414213562, and should be matching the cyan line (which I got with Excel).
![[Image: NQYymDY.png?1]](http://i.imgur.com/NQYymDY.png?1)
(04/14/2015, 04:48 PM)sheldonison Wrote: [/code]Maybe that curve is inverted?
\( \text{superfunction}(z)=S\left( (-\lambda)^z \cdot \sin(\pi z)\right) \)
And here is a graph, showing an analytic super-function for \( {0.01}^z \) the Op might be interested in, from -10 to 10, where it starts out oscillating around the primary fixed point, and then converges to the two cycle that the Op noted. Of course, this isn't tetration; Tet(-2) is by convention a logarithmic singularity. And this function has no uniqueness, I could have just as easily used sine or cosine, or an infinite number of other 1-cyclic functions.
That way, it would match my own solution, and would give ⁰b≈1 and ⁻¹b≈0
![[Image: 7lC9PSh.jpg?1]](http://i.imgur.com/7lC9PSh.jpg?1)
![[Image: 8JBZF9Q.png?1]](http://i.imgur.com/8JBZF9Q.png?1)
I got that solution adjusting the coefficients with Excel's Solver, for a Taylor series with 12 coefficients, (expanded around 0), subject to these restrictions:
a₀=1
⁻¹b=0
⁰b=1
\( ^xb=b^{\,{^{x-1}b}} \)
I got these coefficients:
Code:
a0 1
a1 -1,339425732123610E+00
a2 -3,822141746305190E+00
a3 4,850222934263920E+00
a4 5,354079775248650E+00
a5 -6,056488564613110E+00
a6 -2,938727974303330E+00
a7 2,550689528770280E+00
a8 4,116553795338360E-01
a9 1,989418887288240E-06
a10 1,291559624345350E-04
a11 -1,519921425061600E-07
a12 1,409996370575610E-08
a13 1,348473734742680E-10
a14 1,133187395821660E-11
a12 -3,860717643983480E-09
