Lemme explain this more
We used to take analyticality around a fixed point for granted and had almost no ideas about how to deal with non-analyticality, for example when it comes to the iterations of function\[f(z)=z+\sqrt{z}\]
This function only has a fixed point at 0 and has a derivative or multiplier as \(\infty\), our old methods totally invalidated
If you insist, that its inverse function is multivalued and thus one branch cut must preserved a fixed point with analyticality in nature, so there's still a chance: \(f_1^{-1}(z)=z+\frac{1-\sqrt{1+4z}}{2},f_2^{-1}(z)=z+\frac{1+\sqrt{1+4z}}{2}\)
Then the reality comes beat at u because after evaluation only \(f_1^{-1}(z)\) has a fixed point, also at 0. It's superattractive and thus only the Botcher equation would work.
If you do some substancial calculation you'll quickly realize that Botcher's function will not extend beyond a branch cut, after so many iterations, the function can only be defined on a subset of \(\mathbb{C}\)
So you can never get an iteration that's entirely-defined-on-C.
If you still try hard, you can build a superfunction that has infinitely many cuts and a discontinuity at a infinitely long curve, It's \(\C:Im(z)^2=Re(z)\), such computing can be superannoying: (in Mathematica)
where A[n] is the superfunction, slowly computing, low precisions and many branch cut issues.
While using the asymptotic method, we have very briefly:\[f^t(z)=z+t \sqrt{z}+\frac{1}{4} (t-1) t-\frac{(t-1) t}{16 \sqrt{z}}+\frac{(t+1) \left(t^2-t\right)}{96 z}+\frac{-2 t^4-2 t^3+5 t^2-t}{768 z^{3/2}}+\frac{(t+1) \left(12 t^4+13 t^3-53 t^2+28 t\right)}{15360 z^2}-\frac{\left(16 t^5+52 t^4-55 t^3-110 t^2+69 t+28\right) t}{61440 z^{5/2}}+\frac{(t+1) \left(240 t^6+838 t^5-1447 t^4-2053 t^3+3208 t^2-786 t\right)}{2580480 z^3}-\frac{t \left(720 t^7+4176 t^6-280 t^5-18102 t^4-385 t^3+21714 t^2-1630 t-6213\right)}{20643840 z^{7/2}}+\frac{(t+1) \left(1120 t^8+6908 t^7-3292 t^6-39338 t^5+19626 t^4+55911 t^3-40287 t^2-648 t\right)}{82575360 z^4}+O(z^{-9/2})\] at infinity
And our calculation can be balefully abbreviated as:
Here I\[Alpha][z_] and \[Alpha][z_] are superfunction and abelfunction of f(z)
It can be quickly computed and has very high precision (due to some convience I only saved 10 digits)
Such methods can be useful and also permit us to compute a precise series for any iterations (that in the compositional group)
We used to take analyticality around a fixed point for granted and had almost no ideas about how to deal with non-analyticality, for example when it comes to the iterations of function\[f(z)=z+\sqrt{z}\]
This function only has a fixed point at 0 and has a derivative or multiplier as \(\infty\), our old methods totally invalidated
If you insist, that its inverse function is multivalued and thus one branch cut must preserved a fixed point with analyticality in nature, so there's still a chance: \(f_1^{-1}(z)=z+\frac{1-\sqrt{1+4z}}{2},f_2^{-1}(z)=z+\frac{1+\sqrt{1+4z}}{2}\)
Then the reality comes beat at u because after evaluation only \(f_1^{-1}(z)\) has a fixed point, also at 0. It's superattractive and thus only the Botcher equation would work.
If you do some substancial calculation you'll quickly realize that Botcher's function will not extend beyond a branch cut, after so many iterations, the function can only be defined on a subset of \(\mathbb{C}\)
So you can never get an iteration that's entirely-defined-on-C.
If you still try hard, you can build a superfunction that has infinitely many cuts and a discontinuity at a infinitely long curve, It's \(\C:Im(z)^2=Re(z)\), such computing can be superannoying: (in Mathematica)
Code:
Clear[F, B, B1, B2, B3, BI, IF]
F[z_] := z + Sqrt[z]
B[z_] := 1/z + 1 - z/2 + 3 z^2/2 - 35 z^3/8 + 101 z^4/8 -
599 z^5/16 + 1869/16 z^6 - 48733 z^7/128
B1[z_] :=
1.0640783082804308` - 0.02060483255029065` (-6.` + z) +
0.004348716183968336` (-6.` + z)^2 -
0.0007901451309530964` (-6.` + z)^3 +
0.00013565293405788404` (-6.` + z)^4 -
0.000022785630652999728` (-6.` + z)^5 +
3.750548741345441`*^-6 (-6.` + z)^6 -
6.07839369121214`*^-7 (-6.` + z)^7 +
1.0408560489666969`*^-7 (-6.` + z)^8 -
1.9589488351201014`*^-8 (-6.` + z)^9 +
3.3923628509269487`*^-9 (-6.` + z)^10 -
4.19175181489827`*^-10 (-6.` + z)^11 +
3.885382935067092`*^-11 (-6.` + z)^12 -
1.1779937148123413`*^-11 (-6.` + z)^13 +
6.536083288896178`*^-12 (-6.` + z)^14 -
2.9867841343622627`*^-12 (-6.` + z)^15 +
1.50119091703261`*^-12 (-6.` + z)^16 -
8.23562018916844`*^-13 (-6.` + z)^17 +
4.662696976292319`*^-13 (-6.` + z)^18 -
2.6108033074645186`*^-13 (-6.` + z)^19 +
1.434506597568819`*^-13 (-6.` + z)^20
B2[z_ /; Abs[z - 6] <= 3] := B1[z]
B2[z_ /; Abs[z - 6] > 3 && Re[z] > 10] := B2[IF[z]]^(1/2)
B2[z_ /; Abs[z - 6] > 3 && Re[z] <= 10] := B2[F[z]]^2
B3[z_ /; Im[z]^2 <= Re[z]] := B2[z]
B3[z_ /; Im[z]^2 > Re[z]] := Undefined
IF[z_] := 1/2 (1 + 2 z - Sqrt[1 + 4 z])
IF2[z_] := -(1/2) (1 + 2 (-z) + Sqrt[1 + 4 (-z)])
BI[z_] :=
z - z^2 + 3/2 z^3 - 7/2 z^4 + 81/8 z^5 - 249/8 z^6 + 1569/16 z^7 -
5077/16 z^8 + 135067/128 z^9
Clear[BI1, BI2, BI3, BI4, BI5, BI6, BI7, B4]
BI1[z_ /; Im[z]^2 >= -Re[z]] := Sqrt[BI[IF[z]]]
BI1[z_ /; Im[z]^2 < -Re[z]] := -Sqrt[BI[IF[z]]]
BI2[z_ /; Im[z]^2 >= -Re[z]] := Sqrt[BI1[IF[z]]]
BI2[z_ /; Im[z]^2 < -Re[z]] := -Sqrt[BI1[IF[z]]]
BI3[z_ /; Im[z]^2 >= -Re[z]] := Sqrt[BI2[IF[z]]]
BI3[z_ /; Im[z]^2 < -Re[z]] := -Sqrt[BI2[IF[z]]]
BI4[z_ /; Im[z]^2 >= -Re[z]] := Sqrt[BI3[IF[z]]]
BI4[z_ /; Im[z]^2 < -Re[z]] := -Sqrt[BI3[IF[z]]]
BI5[z_ /; Im[z]^2 >= -Re[z] && Abs[z] > 1] := Sqrt[BI4[IF[z]]]
BI5[z_ /; Im[z]^2 < -Re[z] && Abs[z] > 1] := -Sqrt[BI4[IF[z]]]
BI5[z_ /; Abs[z] <= 1] := BI4[z]
BI6[z_ /; Abs[z] <= 6/5] := BI5[z]
BI6[z_ /; Im[z]^2 >= -Re[z] && Abs[z] > 6/5] := Sqrt[BI6[IF[z]]]
BI6[z_ /; Im[z]^2 < -Re[z] && Abs[z] > 6/5] := -Sqrt[BI6[IF[z]]]
BI7[z_ /; 1/10 < Abs[z] < 2] := BI6[z]
BI7[z_ /; 1/10 >= Abs[z]] := BI6[F[z]]^2
BI7[z_ /; Im[z]^2 >= -Re[z] && Abs[z] >= 2] := Sqrt[BI7[IF[z]]]
BI7[z_ /; Im[z]^2 < -Re[z] && Abs[z] >= 2] := -Sqrt[BI7[IF[z]]]
B4[z_] := Exp[1/Log[BI7[z]]]
Clear[IBI, IBI1, IBI2, IBI3]
IBI[z_] :=
z + z^2 + z^3/2 + z^4 + z^5/8 + z^6/2 + (7 z^7)/16 + z^8 - (21 z^9)/
128 + z^10/8 + (71 z^11)/256 + z^12/2 + (5 z^13)/1024 + (7 z^14)/
16 + (1095 z^15)/2048 + z^16 - (15885 z^17)/32768 - (21 z^18)/
128 + (18443 z^19)/65536 + z^20/8 - (55841 z^21)/262144 + (71 z^22)/
256 + (324945 z^23)/524288 + z^24/2 - (2649857 z^25)/4194304 + (
5 z^26)/1024 + (6109987 z^27)/8388608 + (7 z^28)/16 - (
18206579 z^29)/33554432 + (1095 z^30)/2048 + (92290439 z^31)/
67108864 + z^32
IBI1[z_ /; Re[z] >= 0] := F[IBI[z^2]]
IBI1[z_ /; Re[z] < 0] := F2[IBI[z^2]]
IBI2[z_ /; Re[z] >= 0] := F[IBI1[z^2]]
IBI2[z_ /; Re[z] < 0] := F2[IBI1[z^2]]
IBI3[z_ /; Re[z] >= 0] := F[IBI2[z^2]]
IBI3[z_ /; Re[z] < 0] := F2[IBI2[z^2]]
A[n_] := InverseFunction[B4][B4[1.]^(2^n)]While using the asymptotic method, we have very briefly:\[f^t(z)=z+t \sqrt{z}+\frac{1}{4} (t-1) t-\frac{(t-1) t}{16 \sqrt{z}}+\frac{(t+1) \left(t^2-t\right)}{96 z}+\frac{-2 t^4-2 t^3+5 t^2-t}{768 z^{3/2}}+\frac{(t+1) \left(12 t^4+13 t^3-53 t^2+28 t\right)}{15360 z^2}-\frac{\left(16 t^5+52 t^4-55 t^3-110 t^2+69 t+28\right) t}{61440 z^{5/2}}+\frac{(t+1) \left(240 t^6+838 t^5-1447 t^4-2053 t^3+3208 t^2-786 t\right)}{2580480 z^3}-\frac{t \left(720 t^7+4176 t^6-280 t^5-18102 t^4-385 t^3+21714 t^2-1630 t-6213\right)}{20643840 z^{7/2}}+\frac{(t+1) \left(1120 t^8+6908 t^7-3292 t^6-39338 t^5+19626 t^4+55911 t^3-40287 t^2-648 t\right)}{82575360 z^4}+O(z^{-9/2})\] at infinity
And our calculation can be balefully abbreviated as:
Code:
Clear[L, A, \[Alpha], IA, I\[Alpha], F, IF, IF2]
F[z_] := z + z^(1/2)
IF[z_] := 1/2 (1 + 2 z - Sqrt[1 + 4 z])
IF2[z_] := 1/2 (1 + 2 z + Sqrt[1 + 4 z])
F[z_, t_] :=
z + t z^(1/2) + 1/4 (-1 + t) t - ((-1 + t) t)/(
16 Sqrt[z]) + ((1 + t) (-t + t^2))/(
96 z) + (-t + 5 t^2 - 2 t^3 - 2 t^4)/(
768 z^(3/2)) + ((1 + t) (28 t - 53 t^2 + 13 t^3 + 12 t^4))/(
15360 z^2) - (t (28 + 69 t - 110 t^2 - 55 t^3 + 52 t^4 + 16 t^5))/(
61440 z^(5/2)) +
1/(2580480 z^3) (1 + t) (-786 t + 3208 t^2 - 2053 t^3 - 1447 t^4 +
838 t^5 + 240 t^6) -
1/(20643840 z^(7/2))
t (-6213 - 1630 t + 21714 t^2 - 385 t^3 - 18102 t^4 - 280 t^5 +
4176 t^6 + 720 t^7) +
1/(82575360 z^4) (1 + t) (-648 t - 40287 t^2 + 55911 t^3 +
19626 t^4 - 39338 t^5 - 3292 t^6 + 6908 t^7 + 1120 t^8)
IA[z_] :=
z^2/4 + Log[z]/8 - 1/4 z Log[z] + Log[z]^2/16 - (
1/96 + Log[z]/16 + Log[z]^2/32)/z
I\[Alpha][z_] :=
N[Nest[IF, IA[z + 10000 + 2.350354884909475`100], 10000], 10]
A[z_] := 2 Sqrt[z] + 1/4 Log[z] + Log[2]/2 +
1/192/z + (-1/6144 Log[z]^3 -
Log[2]/1024 Log[z]^2 + (5 - 3 Log[2]^2)/1536 Log[z] +
1/768 (1 + 5 Log[2] - Log[2]^3))/z^(3/2)
\[Alpha][z_] :=
N[A[Nest[F, N[z, 100], 10000]] - 10000 - 2.350354884909475, 10]It can be quickly computed and has very high precision (due to some convience I only saved 10 digits)
Such methods can be useful and also permit us to compute a precise series for any iterations (that in the compositional group)
Regards, Leo 