RE: Re: RE: RE: Functional decomposition (solving f[f[x]] = g[x] for given g)

*To*: mathgroup at smc.vnet.net*Subject*: [mg71972] RE: [mg71885] Re: [mg71872] RE: [mg71839] RE: [mg71764] Functional decomposition (solving f[f[x]] = g[x] for given g)*From*: "Ingolf Dahl" <ingolf.dahl at telia.com>*Date*: Wed, 6 Dec 2006 06:04:05 -0500 (EST)*Organization*: Göteborg University*Reply-to*: <ingolf.dahl at telia.com>

Hi again dear MathGroup, I have checked the suggested procedure below for solving the equation f(f(x))=exp(x), and it works smoothly. Maybe this is already well-known. Maybe it is known, but not implemented in Mathematica code. Maybe it is included in the book by Aczel, I have not been able to check that yet, since I have not the book available. If you know where to find an already published solution, please inform me. But anyway, I have not spent too many hours on it, and it is a nice demonstration of the powers of Mathematica. Thus it is relevant to publish it here. Moreover, the solution I obtain is characterized by some mathematical constants, which should be thumbprints for uniqueness. I have not been able to localize these constants anywhere else. But what I have found is that the solution is related to half-integer values of "tetration" in Wikipedia and PowerTower in MathWorld (naturally extending into other fractional values). Wikipedia states; "At this time there is no commonly accepted solution to the general problem of extending tetration to the real or complex numbers, although it is an active area of research." What I claim is that (in the Mathematica notation from FurtherExamples to "Power" in Help) PowerTower[E,-1/2] = "HalfwayFromZero" = 0.4987433645316708791828373750416860523968424`43. PowerTower[E,-3/2] = "HalfwayToZero" = -0.6956636150637165750917227913152630811483903`42. (PowerTower[E,-2] = -Infinity, PowerTower[E,-1] = 0, PowerTower[E,0] = 1 by natural extension) Here is my solution: I started by doing a routine for decomposition of some functions g(x) into q equal functions. Thus f(f(f...(f(x))...))=g(x), with q f:s. First the series expansion to the power n is derived. Then the function is composed p times with itself. We thus get the series expansion of the function to the decompositional power p/q. The requirement on g is that g[0)=0, with positive derivative at origin and with a Maclaurin expansion around origin. Attributes[FractionalSeries]= {HoldFirst}; FractionalSeries[ func_/; And[ func[0.]==0., SeriesCoefficient[ Series[ func[var], { var,0,1}],1]>0],x_, { p_Integer/; p>0, q_Integer/; q>0}, n_Integer/; n>0]:= Module[ { a,f1,fn,f1s}, f1[var_]= Sum[ var^i* a[i], { i,1,n}]; fn[var_]:= Nest[ f1, var,q]; a[1]= a[1] /. Solve[ Reduce[ { SeriesCoefficient[ Series[ fn[var], { var,0,1}],1]== SeriesCoefficient[ Series[ func[var], { var,0,1}],1], a[1]>0}, a[1]], a[1]][ [1]]; Do[ a[i]= a[i] /. Solve[ { SeriesCoefficient[ Series[ fn[var], { var,0,i}],i]== SeriesCoefficient[ Series[ func[var], { var,0,i}],i]}, a[i]][ [1]], { i,2,n}]; Series[ f1[x], { x,0,n}]; ComposeSeries@@ Table[ Series[ f1[x], { x,0,n}], {p}] Example of use: Sinh is such a function, and let us take it to the decompositional power 2/3. Define ftwothirds[y_] := Normal[FractionalSeries[Sinh[#1] & , var, {2, 3}, 29]] /. {var -> y}; Then ftwothirds[ftwothirds[ftwothirds[0.3]]] - Sinh[Sinh[0.3]] evaluates to 5.551115123125783*^-17 and ftwothirds[ftwothirds[ftwothirds[SetAccuracy[0.3, 60]]]] - Sinh[Sinh[SetAccuracy[0.3, 60]]] evaluates approx to 1.65*^-24 The problem is that Exp is not such a function, since Exp[0} == 1. But Exp behaves asymptotically as 2*Sinh. Let us find the series expansion of 2*Sinh to the decompositional power 1/2. (There is a more complicated relation between the series expansions for 2*Sinh and Sinh than I thought from the beginning.) Simplify[FractionalSeries[2*Sinh[#1] & , y, {1, 2}, 29]] (returns long expression) If we use this expression, we might build a function f2sinh to extend the series expansion to whole real axis. f2sinh should now be THE smooth solution to the equation f2sinh[f2sinh[x]] == Sinh[x] . f2sinh[x_Real] := Module[{y, steps}, y = If[MachineNumberQ[x], SetAccuracy[x, 24], x]; steps = 0; While[Abs[y] > 0.07, steps = steps + 1; y = ArcSinh[y/2]]; y = fpoly[y]; (If[MachineNumberQ[x], N[#1], #1] & )[Nest[2*Sinh[#1] & , y, steps]]] where fpoly[y_] := Sqrt[2]*y + y^3/(9*Sqrt[2]) - (11*y^5)/(2700*Sqrt[2]) + (479*y^7)/(1020600*Sqrt[2]) - (426679*y^9)/(6246072000*Sqrt[2]) + (85589473*y^11)/(7557747120000*Sqrt[2]) - (1403953837837*y^13)/(689720002171200000*Sqrt[2]) + (1031743413846247*y^15)/(2669216408402544000000*Sqrt[2]) - (1496209724379058102331*y^17)/(19591834900362000756480000000*Sqrt[2]) + (17812978070670318115902941*y^19)/(1145769688642970528240463360000000*Sqrt[2 ]) - (7533260191446785723514332836231*y^21)/(232533958310090868706402038912000000 0000*Sqrt[2]) + (136561976266545063524556174223350449*y^23)/(1985445516947137017537762080849 52576000000000*Sqrt[2]) - (719728529960035731908565151209614126425839277*y^25)/(4853778747825775229320 514450035010550004224000000000000*Sqrt[2]) + (100427683629367623630378641852188624798406912708777*y^27)/(3101826723913052 963398192041352673632022399364096000000000000*Sqrt[2]) - (230559003784630013844879523283194881704293155047113459273*y^29)/(3226456204 1869955579438812968736894041769959385166794752000000000000*Sqrt[2]) + SetAccuracy[0., 44]*(y/SetAccuracy[0.07, 60])^29; The limit 0.07 in this definition is a bit arbitrarily chosen, and you might decrease it to increase accuracy, which anyway is good along the whole real axis. I have estimated the truncation error to be as large as the last term in the series. Now we can assume that Exp to a fractional decompositional power asymptotically should be identical to 2*Sinh to the same fractional decompositional power, with a relative error that is less than the relative error of Exp compared to 2*Sinh, if the decompositional power is less than one. Thus for any x, we might climb up to a large value of Exp...[Exp[x]] (say x>60), fetch a value using f2sinh[x], and then climb down again. Thus we might calculate the (smooth) solution to the equation fexp[fexp[x]]== Exp[x] as fexp[x_Real] := Module[{y, steps}, y = If[MachineNumberQ[x], SetAccuracy[x, 24], x]; steps = 0; While[y < 60, steps = steps + 1; y = Exp[y]]; y = f2sinh[y]*(1 + SetAccuracy[0., y/1.15]); (If[MachineNumberQ[x], N[#1], #1] & )[Nest[Log[#1] & , y, steps]]] The limit of fexp[x] when x->-Infinity is the constant HalfwayToZero defined above. And fexp[HalfwayToZero] = 0, fexp[0]= HalfwayFromZero, fexp[HalfwayFromZero] = 1, fexp[1] = Exp[HalfwayFromZero], fexp[Exp[HalfwayFromZero] = E und so weiter. Any devils hiding anywhere? One little devil is that fexp[x] approaches HalfwayToZero fast when z -> -Infinity, so there easily is a loss of accuracy. For large negative values of x, fexp will approximately follow the formula fexp[x] ~ HalfwayToZero + Exp[x]/f´[HalfwayToZero] where f´[HalfwayToZero] = 0.5688037784984972069377219276490924736217`40. For calculations of derivatives etc. of f(x) in the region around x = 0, a slightly simpler formula might be used instead of f(x): fexpsimpl[x_Real] := Nest[Log, Nest[2Sinh[#] &, fpoly[Nest[ArcSinh[#/2] &, Nest[Exp, x, 4], 8]], 8], 4] Thus we can obtain the zeroth, first second and third derivatives of fexp(x) at x=0: {0.498743364531670879182837375041686052396842401697695, 0.87682850111901739234161348908444926737095344685410, 0.489578037438247862041903165991088781934758253721, 0.1222616742183662064640316453590024071635142693} It should be possible to calculate Exp to other fractional decompositional powers, and to calculate PowerTower for other first arguments than E and for fractional second argument, following the general scheme here. That all for today! Ingolf Dahl > -----Original Message----- > From: Gregory Duncan [mailto:gmduncan at gmail.com] > Sent: den 2 december 2006 11:11 > To: mathgroup at smc.vnet.net > Subject: [mg71885] Re: [mg71872] RE: [mg71839] RE: [mg71764] > Functional decomposition (solving f[f[x]] = g[x] for given g) > > The problem falls generally into the the area of functional equations. > See J. Aczel > Functional Equations: History, Applications and Theory or any > of his other books. One can also try to find a fixed point in > a function space. > > > On 12/1/06, Ingolf Dahl <ingolf.dahl at telia.com> wrote: > > Some further comments on the equation f(f(x)) = g(x): > > For the case g(x) = exp(x) some work have been done, see > > http://www.math.niu.edu/~rusin/known-math/index/39-XX.html , > > http://www.math.niu.edu/~rusin/known-math/99/sqrt_exp > > and > http://www.newton.dep.anl.gov/newton/askasci/1993/math/MATH023.HTM . > > (found by a Google search for "f(f(x))") > > > > However, I have not penetrated deeply enough into all references to > > really see if there are good methods to find (precise) numerical > > values for at least one solution of the equation. Or a plot... > > > > But some thoughts further, somewhat fishing on deep water: > If we have > > a function g(x) that is such that g(0)=0 (or g(x0)=x0 for > some x0) and > > where > > g(x) also has a Maclaurin expansion with a positive first > coefficient, > > it should be easy to find a Maclaurin series which composed > to itself > > is identical to that of g(x). Examples of such functions g(x) are > > sin(x) and sinh(x). Then there is the question if this series > > converges in some range, I have not tested yet. (In the links above > > they refer to a theorem by Polya, stating that convergence > range must > > be limited.) For some other functions, we could also think > of expansions in x^(Sqrt(2)). > > Suppose that we find such a function f(x) for g(x) = > sinh(x), name it > > fsinh(x), with convergence in some region near origo. Then > it might be > > possible to extend that to the whole axis if we for any x > apply ginv > > (the inverse of g) a number of times until we reach the convergence > > region. (we might have to use extended precision.) There we > apply f, > > and then apply g the same number of times as we did with > ginv. Then we > > should have obtained f(x). > > > > If we now look at the case g(x) = exp(x), we can note that > if x > 20, > > exp and 2*sinh are essentially identical (to machine > precision). Thus > > we could assume that fsinh(x)*Sqrt[2] is a good > approximation of the > > solution f(x) for g(x)=exp(x) and x>20, and then we might > use the same > > procedure as for > > g(x)=x^2+1 to extend this to zero. With the power of Mathematica > > available, just a few lines of code maybe are all that is > needed. But > > maybe some devils are hiding in the details? > > > > Is there anyone who knows more about this? > > > > Ingolf Dahl > > > > > > > -----Original Message----- > > > From: Ingolf Dahl [mailto:ingolf.dahl at telia.com] > > > Sent: den 30 november 2006 12:06 > > > To: mathgroup at smc.vnet.net > > > Subject: [mg71839] RE: [mg71764] Functional decomposition > (solving > > > f[f[x]] = g[x] for given g) > > > > > > Hi Kelly, > > > Here is a rapid amateur answer (but without scary rhymes, > as in my > > > last answer). If you are satisfied with the numerical > values of the > > > function, you can investigate > > > > > > g[x_]:=1+x^2; > > > ginv[x_]:=Sqrt[x-1]; > > > f[x_Real]:=Nest[ginv,Power[Nest[g,x,9],Sqrt[2]],9] > > > > > > f[x_Real] appears to be a good solution to f[f[x]] == g[x] to > > > approximate machine precision. The solution is based on two > > > observations: > > > 1) Nest[g,x,9] is equivalent to Nest[f,x,18], and we reach very > > > large values when this is calculated. There g[x] is approximately > > > equal to x^2 (to machine precision), and f[x] can be > approximated by > > > x^Sqrt[2]. Thus we might reach a good approximation of > Nest[f,x,19] > > > by Power[Nest[g,x,9],Sqrt[2]]. > > > 2) We can then apply the inverse function ginv of g 9 > times to reach > > > f[x]. > > > The relative error is reduced by the application of the inverse > > > function. > > > > > > We can note that f[0.] is 0.6420945043908285` which do > not appear to > > > be a known mathematical constant. But now it is. > > > (If someone can check... I cannot reach the "The Inverse Symbolic > > > Calculator" at CESM. Is it down?) > > > > > > Generalize and write f[x] as DecompositionPower[x,1/2] and the > > > solution to f[f[f[x]]]\[Equal]g[x] as > DecompositionPower[x,1/3] For > > > the given function g then we can try > > > > > > > DecompositionPower[x_,p_]:=Nest[ginv,Power[Nest[g,x,9],Power[2,p]],9 > > > ] > > > > > > and we might for instance plot > > > > > > Plot[DecompositionPower[0.,p],{p,0,1}] > > > > > > The function DecompositionPower[x_,p_] should be quite easy to > > > generalize to a large set of polynomial functions g. > Maybe it then > > > should have the name FractionalNest? > > > > > > Best regards > > > > > > Ingolf Dahl > > > > > > -----Original Message----- > > > From: Kelly Jones [mailto:kelly.terry.jones at gmail.com] > > > Sent: 28 November 2006 12:04 > > > To: mathgroup at smc.vnet.net > > > Subject: [mg71839] [mg71764] Functional decomposition (solving > > > f[f[x]] = g[x] for given g) > > > > > > Given a function g[x], is there a standard methodology to find a > > > function f[x] such that: > > > > > > f[f[x]] == g[x] > > > > > > A "simple" example (that doesn't appear to have a closed form > > > solution): > > > > > > f[f[x]] == x^2+1 > > > > > > There are probably several (approximable? > > > power-series-expressible?) answers, but the most natural answer > > > would be everywhere positive and monotone increasing for x>0. > > > > > > For x^2, there are at least two continuous solutions: > > > x^Sqrt[2] and x^-Sqrt[2], the former being more "natural". > > > It's somewhat amazing that x^2+1 is so much harder. > > > > > > Of course, the next step would be to find f[x] such that: > > > > > > f[f[f[x]]] == g[x] > > > > > > for given g[x], and so on. > > > > > > Thought: is there any operator that converts functional > composition > > > to multiplication or something similar? That would reduce this > > > problem to find nth roots and applying the inverse operator? > > > > > > Other thought: For some reason, taking the geometric > average of the > > > identity function and g, and then iterating, seems like a > good idea. > > > EG, Sqrt[x*g[x]], Sqrt[x*Sqrt[x*g[x]]], and so on. > > > > > > -- > > > We're just a Bunch Of Regular Guys, a collective group > that's trying > > > to understand and assimilate technology. We feel that > resistance to > > > new ideas and technology is unwise and ultimately futile. > > > > > > > > > > > > > > > >

**Follow-Ups**:**Re: RE: Re: RE: RE: Functional decomposition (solving f[f[x]] = g[x] for given g)***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Vandermonde Matrix/Optimization question**

**need help defining a MakeBoxes rule for Piecewise**

**Re: RE: RE: Functional decomposition (solving f[f[x]] = g[x] for given g)**

**Re: RE: Re: RE: RE: Functional decomposition (solving f[f[x]] = g[x] for given g)**