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+--- Thread: b^b^x with base 0<b<e^-e have three real fixpoints (/showthread.php?tid=1182)
b^b^x with base 0<b<e^-e have three real fixpoints - Gottfried - 10/29/2017
Triggered by a question in the math.stackexchange-forum I looked at real fixpoints of b^b^x and found that for the smallest b<e^-e have three real fixpoints (while b^x has only one).
See a short compilation in:
https://math.stackexchange.com/a/2494323/1714
I have not yet found a closed-form expression - for instance by the Lambert W as in the fixpoint-problem of the simple case b^x.
Gottfried
RE: b^b^x with base 0<b<e^-e have three real fixpoints - sheldonison - 11/07/2017
(10/29/2017, 11:14 AM)Gottfried Wrote: Triggered by a question in the math.stackexchange-forum I looked at real fixpoints of b^b^x and found that for the smallest b<e^-e have three real fixpoints (while b^x has only one).
See a short compilation in:
https://math.stackexchange.com/a/2494323/1714
I have not yet found a closed-form expression - for instance by the Lambert W as in the fixpoint-problem of the simple case b^x.
Gottfried
For the two-periodic fixed point pair of \( f(z,k)=-\exp(z)+1+k \)
The formal solution for the fixed point of f(x,k) has a power series where the two fixed points are
\( L=g(\pm\sqrt{6k});\;\;\;\;f(f(L))=L;\;\;\;\;f(g(\pm\sqrt{6k}),k)=g(\mp\sqrt{6k}); \)
Then this solutions for the two periodic fixed points L for f(z,k) can be converted to solutions for the two periodic fixed point pair of \( a^z \) as follows:
\( z=\frac{g(\pm\sqrt{6k})-\ln(-\ln(a))}{\ln(a)};\;\;\;\; k=\ln(-\ln(a))+1;\;\;\;\;a^{a^z}=z \)
The definition of the formal series solution for g boils down to this equation; with the x^2/6 term chosen so g x^1 term coefficient=1. The pari-gp formalperiod2 program returns "1/6" as the x2term.
Also see https://math.stackexchange.com/questions/2490769/solutions-of-aax-x-for-fixed-a0
\( -\exp(g(x))+1+\frac{x^2}{6}=g(-x) \)
Code:
{g=
+x^ 1* 1
+x^ 2* -1/6
+x^ 3* 1/20
+x^ 4* -1/90
+x^ 5* 523/151200
+x^ 6* -23/28350
+x^ 7* 239/1008000
+x^ 8* -19/340200
+x^ 9* 1471949/100590336000
+x^10* -6583/1964655000
+x^11* 94891697/130767436800000
+x^12* -49909/328378050000
+x^13* 18670028801/988601822208000000
+x^14* -520019/241357866750000
+x^15* -88448773393/67224923910144000000
+x^16* 254033333/492370048170000000
+x^17* -15331312862555281/60696039299990814720000000
+x^18* 1418708833351/19449109272763170000000
+x^19* -1799426008377623/80928052399987752960000000
+x^20* 1503421489421/265215126446770500000000
+x^21* -2299035792738061729699/2134218412281997023510528000000000
+x^22* 2825289970112783/13084388263251422617500000000
+x^23* 20135202374763121379/871109556033468172861440000000000
+x^24* -2544602861617837/181168452875788928550000000000
+x^25* 59429596033136172637237727/5070902947582024927861014528000000000000
+x^26* -528038311635637939/143170753680524776956750000000000
+x^27* 13010375215002577166435605861/8823371128792723374478165278720000000000000
+x^28* -38921031170396033459/94664502333562982523803100000000000
+x^29* 100402563034473871496430148044559/873634031969892267532834206929387520000000000000
+x^30* -15748316211640715812312013/550017324845909301985237676542500000000000}The pari-gp program to calculate the formal g-series for the two periodic fixed points of a generic function gf with a multiplier of (-1) at x=0, is as follows: Then for example, the error term for fixedaaz(0.04)=0.089600840934760930 is accurate to about about 17-18 decimal digits with a 30 term series. The other fixed point is fixedaaz(0.04,-1). For a>exp(-e), the two periodic fixed points are complex conjugate pairs.
Code:
/* formal period2 fixed point for generic function with slope=-1, 2-cyclic solution */
/* gs=formalperiod2(-exp(x)+1,30)[1]; */
/* returns [series,x2term] */
\ps 31
kf(a)=log(-log(a))-1;
fixedaaz(a,n) = {
if (n<>-1, n=1);
return((subst(gs,x,n*sqrt(kf(a)*6))-kf(a)-1)/log(a));
}
formalperiod2(gf,n) = {
local(i,gs,savegs,zs,z,zt,x2term,m2,s2,r2);
savegs=0;
gs = x+aecoeff*x^2+O(x^4);
z = subst(gf,x,gs)-subst(gs,x,-x)+acoeff*x^2;
zs = subst(polcoeff(z,3),aecoeff,x);
zt = -polcoeff(zs,0)/polcoeff(zs,1);
savegs = x + zt*x^2;
gs=savegs;
z = subst(gf,x,gs)-subst(gs,x,-x)+acoeff*x^2;
zs = subst(polcoeff(z,2),acoeff,x);
zt = -polcoeff(zs,0)/polcoeff(zs,1);
x2term = zt;
m2 = matrix(2,2);
s2 = matrix(2,1);
forstep (i=3,n,2,
gs = savegs + aocoeff*x^i + aecoeff*x^(i+1)+O(x^(i+3));
z = subst(gf,x,gs) - subst(gs,x,-x) + x2term*x^2;
zs = subst(polcoeff(z,i+1),aecoeff,x);
zs = subst(zs,aocoeff,0);
m2[1,1] = polcoeff(zs,1);
zs = subst(polcoeff(z,i+1),aocoeff,x);
zs = subst(zs,aecoeff,0);
m2[1,2] = polcoeff(zs,1);
s2[1,1] = -polcoeff(zs,0);
zs = subst(polcoeff(z,i+2),aecoeff,x);
zs = subst(zs,aocoeff,0);
m2[2,1] = polcoeff(zs,1);
zs = subst(polcoeff(z,i+2),aocoeff,x);
zs = subst(zs,aecoeff,0);
m2[2,2] = polcoeff(zs,1);
s2[2,1] = -polcoeff(zs,0);
r2 = matsolve(m2,s2);
savegs = savegs + r2[2,1]*x^i + r2[1,1]*x^(i+1);
);
return([savegs,x2term]);
}
gs = formalperiod2(-exp(x)+1,30)[1];