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