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How it looks (i.θ)ₐ
#15
(04/23/2015, 11:15 PM)JmsNxn Wrote: I've been trying to follow this thread, and finally I have something to contribute. For the bases it might be easier to use my expansion of this function. See http://arxiv.org/pdf/1503.07555v1.pdf

I came up with a holomorphic expression for for these bases. It's a fast converging expression as well. It is not as messy as a Taylor series expansion of this same function. It's also a single holomorphic expression for all , greatly reducing computational time.
This is for the periodic/pseudoperiodic extension of tetration (regular koenigs iteration) for bases

The expression isn't so easy to write out:


Unfortunately, I cannot make it work.

I made this code in PariGP (I attached the file)
Code:
\p 64

\\base
a=1.2

\\Delta w, for integration step.
\\To do: iterate until converging precision.
Dw=.01

\\Value of w where the integration/sum will be truncated
\\To do: use only the necessary number of terms to achieve desired precision
w_mx=14

\\Value of n where the summation (inside the integral) will be truncated
\\To do: use only the necessary number of terms to achieve desired precision
n_mx=100

\\Value of n where the summation (outside the integral) will be truncated
\\To do: use only the necessary number of terms to achieve desired precision
m_mx=100

\\n° w values to be evaluated.
\\depends on the integration step Dw
n_w=truncate(1+(w_mx-1)/Dw)

\\Value of w, as function of vector index
\\Integration from 1≤w≤∞ will be truncated at w_mx
w(i)=1+(i-1)*Dw

\\n° rows
\\used as index for the summation
\\rows=n_mx

\\To do: check if some z cause errors.
Invgamma(z)=1/gamma(z)

\\precalculation of ⁿ⁺¹a
\\xa=˟a=ⁿ⁺¹a
\\xa[1]=°⁺¹a=a
Size_xa=max(n_mx,m_mx)
xa=vector(Size_xa,n,a);
for (n=2,Size_xa, xa[n]=a^xa[n-1]);


\\precalculation of factorial
InvFact_n=vector(Size_xa,n,1/factorial(n-1));


\\Integrating Factor, function of w
\\∫ IntSum.w⁻z dw
IntSum=vector(n_w,i,{
                   sum(n=1, n_mx,
                       xa[n]*InvFact_n[n]*(-w(i))^(n-1), 0.)
                   })
Integral(z)=sum(n=1,n_w-1, (IntSum[n]*w(n)^(-z)+IntSum[n+1]*w(n+1)^(-z))/2*Dw )


\\To do: check division by zero.
\\Tetration ^za=za(z)
za(z)={Invgamma(1-z)
       *(sum(m=1,m_mx,
             xa[m]*(-1)^(m+1)*InvFact_n[m]/(m-z))          
         +Integral(z))
      }

There is a problem with the integral. for values w>14, it start growing very fast, and cannot be computed.




Anyways, I truncated the summations and the integral limits, before it starts diverging, to see what I get. Here is the same expression with the variables I used in the PariGP code:




These are the names of the functions in the code:




It produces something close to the right answer, but ,

[Image: 6M8jnhY.jpg?1]


Attached Files
.gp   JmsNxn solution.gp (Size: 3.23 KB / Downloads: 226)
I have the result, but I do not yet know how to get it.
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Messages In This Thread
How it looks (i.θ)ₐ - by marraco - 04/18/2015, 11:20 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 02:40 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/19/2015, 08:40 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 11:19 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 02:35 AM
RE: How it looks (i.θ)ₐ - by Gottfried - 04/20/2015, 07:53 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 02:34 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 03:20 AM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/21/2015, 08:23 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 04:52 PM
RE: How it looks (i.θ)ₐ - by JmsNxn - 04/23/2015, 11:15 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/23/2015, 11:20 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/26/2015, 12:50 AM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/26/2015, 05:08 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 03:46 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 02:21 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/25/2015, 07:52 PM



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