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

*To*: mathgroup at smc.vnet.net*Subject*: [mg71996] Re: [mg71972] RE: [mg71885] Re: [mg71872] RE: [mg71839] RE: [mg71764] Functional decomposition (solving f[f[x]] = g[x] for given g)*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 7 Dec 2006 06:26:08 -0500 (EST)*References*: <200612061104.GAA02807@smc.vnet.net> <B2493AAB-0463-49C0-9E90-118BB276E47F@mimuw.edu.pl>

I meant "class C1", of course: differentiable, with continuous derivative. Andrzej Kozlowski On 7 Dec 2006, at 17:58, Andrzej Kozlowski wrote: > To my mind an equally interesting problem (though admittedly not if > one wishes to apply Mathematica) is to find the widest class of > functions g for which the equation admits no solutions. There is > one case which is immediately obvious: the functional equation f(f > (x))==g(x) cannot have any solutions f (of class C2) for smooth > functions g whose derivative is g'(x) <0 everywhere; g(x) = exp(-x) > is such a function. The proof is completely trivial, but as far as > I can tell no-one seems to have mentioned it explicitly so here it is: > if f(f(x)) = g(x) then f'(f(x)) f'(x)) = g'(x) is negative. But > that means that f' has opposite sign at x and f(x). That in turn > implies (if f is of class C2) that f'(t) = 0 for some t between x > and f(x). But that would mean that g'(t) = 0, contradicting the > fact that g'(x)<0 everywhere. > I have given very little thought to this but it seems to me that > the assumption g'(x)>0 for all x should suffice for the existence > of f (though may not suffice for the existence of a nice enough f, > e.g. analytic). I would expect also that examples can be > constructed where g'(x)>0 for some x and g'(y)< 0 for some why and > where no f of class C2 can exists, but I can't find any obvious ones. > > Andrzej Kozlowski > > On 6 Dec 2006, at 20:04, Ingolf Dahl wrote: > >> 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: RE: Functional decomposition (solving f[f[x]] = g[x] for given g)***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**References**:**RE: Re: RE: RE: Functional decomposition (solving f[f[x]] = g[x] for given g)***From:*"Ingolf Dahl" <ingolf.dahl@telia.com>

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

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

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