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

*To*: mathgroup at smc.vnet.net*Subject*: [mg71995] 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 06:26:02 -0500 (EST)*References*: <200612061104.GAA02807@smc.vnet.net>

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. >>>> >>>> >>>> >>> >>> >>> >> > >

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