How it looks (i.θ)ₐ - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: How it looks (i.θ)ₐ (/showthread.php?tid=987) Pages: 1 2 RE: How it looks (i.θ)ₐ - marraco - 04/23/2015 If the calculus are accurate, the extremes of the curve start to touch each other for some base at the right, and then the curves start running over themselves. Numerically, they seem to fit an ellipse very well. At some point, the tips start spiraling around a dot (it cannot be seen at this zoom level). I wonder if as $\vspace{10}{a \to \infty}$ the curve converges to a cardioid. RE: How it looks (i.θ)ₐ - marraco - 04/23/2015 (04/21/2015, 08:23 AM)sheldonison Wrote: and sexp(10) is a humongous number. The largest number pari-gp can represent is about $\text{sexp}_e\left(4.08\right)\approx 7.24676\cdot 10^{131475196{$That was my dumb mistake. I didn't needed sexp(10), but sexp(I*10). ¿Do you think that the tiny spirals at the end of the lines ͥ·ˣa (as x→∞) are accurate, or just a numerical artifact? ¿Did you attempted to find the base for which the lines turn into an "ellipse" (or something close)?. I guess a=e^(e^-1); that would "explain" the change of state. Now, if the bases$\vspace{15}{1really turn into ellipses, then it should be easy to find an algebraic expression for tetration to real exponents, or at least an important insight for $\vspace{20}{^{\frac{1}{x}}a}$ The base $\vspace{15}{a=e^{e^{-1}} }$ seems to match an ellipse with center near c=2.65599203615835 (Don't take that precision as accurate. I got it from Excel), relation of axis b=a, and radius $\vspace{20}{r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}}$, or $\left(imag(^{i.x}a)\right)^2 \,+\, \left({\frac{real(^{i.x}a)-c}{b}}\right)^2\,=\, r^2\\ \\ a=e^{e^{-1}} \\ b= e^{e^{-1}}\\ c= 2.655992036 \\ r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}\\ \\ W_{(x)}.e^{W_{(x)}}=x \,\Rightarrow \, e={\frac{1}{W_{(1)}}}^{\frac{1}{W_{(1)}}} \,\Rightarrow \, W_{(1)}=0,56714329 $ RE: How it looks (i.θ)ₐ - JmsNxn - 04/23/2015 (04/23/2015, 04:52 PM)marraco Wrote: Now, if the bases$\vspace{15}{1really turn into ellipses, then it should be easy to find an algebraic expression for tetration to real exponents, or at least an important insight for $\vspace{20}{^{\frac{1}{x}}a}$ The base $\vspace{15}{a=e^{e^{-1}} }$ seems to match an ellipse with center near c=2.65599203615835 (Don't take that precision as accurate. I got it from Excel), relation of axis b=a, and radius $\vspace{20}{r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}}$, or $\left(imag(^{i.x}a)\right)^2 \,+\, \left({\frac{real(^{i.x}a)-c}{b}}\right)^2\,=\, r^2\\ \\ a=e^{e^{-1}} \\ b= e^{e^{-1}}\\ c= 2.655992036 \\ r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}\\ \\ W_{(x)}.e^{W_{(x)}}=x \,\Rightarrow \, W_{(1)}=0,56714329 $ 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: $^^z a= \frac{1}{\G(1-z)} (\sum_{n=0}^\infty (^^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty (^^{n+1}a) \frac{(-w)^n}{n!}) w^{-z}\,dw)$ It is a periodic solution, except at eta. RE: How it looks (i.θ)ₐ - sheldonison - 04/23/2015 (04/23/2015, 04:52 PM)marraco Wrote: That was my dumb mistake. I didn't needed sexp(10), but sexp(I*10).Maracco, Beautiful graphs! Also, I*10 should work just fine! sexp(10*I)=0.318128684402243 + 1.33723930629532*I As I mentioned earlier, Kneser.gp doesn't implement tetration for baseseta, to bases 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 $\vspace{10}{^{-1}a \neq 0}$, $\vspace{10}{^{0}a \neq 1}$ RE: How it looks (i.θ)ₐ - marraco - 04/26/2015 (04/23/2015, 11:20 PM)sheldonison Wrote: I strongly discourage using sexp(z) for baseseta, and instead. The theta approximation for bases