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 \( 1 < a \le e^{1/e} \) 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 \( ^^z a \) 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 \( \Re(z) > 0 \), greatly reducing computational time.
This is for the periodic/pseudoperiodic extension of tetration (regular koenigs iteration) for bases \( 1 < a \le e^{1/e} \)

The expression isn't so easy to write out:

\( ^^z a= \frac{1}{\G(1-z)} (\sum_{n=0}^\infty (^^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty (^^{n+1}a) \frac{(-w)^n}{n!}) w^{-z}\,dw) \)

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.


\( ^^z a= \frac{1}{\G(1-z)} (\sum_{n=0}^\infty (^^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty (^^{n+1}a) \frac{(-w)^n}{n!}) w^{-z}\,dw) \)

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:


\( ^z a= \frac{1}{\G(1-z)} (\sum_{m=0}^{m_{mx}} (^{m+1} a)\frac{(-1)^m}{m!(m+1-z)} + \int_1^{n_{w}} (\sum_{n=0}^{n_{mx}} (^{n+1}a) \frac{(-w)^n}{n!}) w^{-z}\,dw) \)

These are the names of the functions in the code:


\( za(z) \,=\, ^z a \\

Invgamma(1-z) \,=\, \frac{1}{\G(1-z)} \,\,\,(crashes \, at \, integer \, z)\\

IntSum[w] \,=\, (\sum_{n=0}^{n_{mx}} (^{n+1}a) \frac{(-w)^n}{n!}) \\
Integral(z) \,=\, \int_1^{n_{w}} IntSum[w] w^{-z}\,dw) \)

It produces something close to the right answer, but \( \\[10pt]

{^{-1}a \neq 0} \), \( \\[10pt]

{^{0}a \neq 1} \)

[Image: 6M8jnhY.jpg?1]


Attached Files
.gp   JmsNxn solution.gp (Size: 3.23 KB / Downloads: 869)
I have the result, but I do not yet know how to get it.
Reply


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



Users browsing this thread: 1 Guest(s)