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 How it looks (i.θ)ₐ marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 04/25/2015, 07:52 PM (This post was last modified: 04/25/2015, 08:00 PM by marraco.) (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 $\vspace{10}{^{-1}a \neq 0}$, $\vspace{10}{^{0}a \neq 1}$ Attached Files   JmsNxn solution.gp (Size: 3.23 KB / Downloads: 226) I have the result, but I do not yet know how to get it. « Next Oldest | Next Newest »

 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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