fractional iteration of f(x)=1/(1+x)
Gottfried Helms
f(x) = 1/(1+x)
under iteration and a meaningful description for fractional
iteration. I thought, it should be possible to use
the above closed form, but then, when iterated, we get
the continued fraction [0,1,1,...,x] to a certain index
and I don't have an idea, how to interpolate here.
So I expressed f(x) in terms of a powerseries
f(x) = 1 - x + x^2 - ... + ...
and used fixedpoint-shift to generate another function
g(x) which has no constant term and gives
f(x) = g(x - x0) + x0
Fractional iterates are then computed using diagonalization.
In short, for the half-iterate from x=1 I got the complex value
f°0.5 (1) = 0.6 + 0.2*I
This was surprising (see also the reasoning in Matt's thread
"More tetration")
However, the next iterate, using the same function gives the
expected value
f°0.5 (0.6 + 0.2*I) = 0.5 = f°1(1)
Can someone find another expression for the iteration and
possibly calculate a concurring value?
Other values are
h real(f°h(1)) imag(f°h(1))
--------------------------------------
0 1 +0 *I
0.1 0.934866588 +0.119610522*I
0.2 0.844115725 +0.192964023*I
0.3 0.750413612 +0.223830162*I
0.4 0.667122962 +0.222594321*I
0.5 0.6 +0.2 *I
0.6 0.550328704 +0.164720804*I
0.7 0.517269014 +0.123111797*I
0.8 0.499154488 +0.07971447 *I
0.9 0.49406502 +0.037838766*I
1 0.5 +0 *I
1.1 0.514863934 -0.031828109*I
1.2 0.536392333 -0.056126858*I
1.3 0.562102367 -0.071877563*I
1.4 0.589329449 -0.07868729 *I
1.5 0.615384615 -0.076923077*I
1.6 0.637824206 -0.067768155*I
1.7 0.654768067 -0.053128135*I
1.8 0.665162007 -0.035368628*I
1.9 0.668885885 -0.016940237*I
2 0.666666667 -0 *I
A more explanative text and also a plot
for the fractional iterated can be found at
http://go.helms-net.de/math/tetdocs/FracIterAltGeom.htm
Regards -
Gottfried Helms
Matt
> Just for exercise I considered the function
>
> f(x) = 1/(1+x)
>
> under iteration and a meaningful description for fractional
> iteration.
Some while ago, when I was playing with this, I got, for the more
general f(x) = 1/(a*x + b),
f^n(x) = ((a*x + b - beta)*alpha^(n - 1) - (a*x + b - alpha)*beta^
(n - 1))/((a*x + b - beta)*alpha^n - (a*x + b - alpha)*beta^n)
where
alpha = (b + Sqr(b^2 + 4*a))/2
beta = (b - Sqr(b^2 + 4*a))/2
Gottfried Helms
Ah, interesting. Just tried some values and this gives the
same result, using a=b=1.
Short remark: isn't b - beta = alpha , and b - alpha = beta?
So we had
(a*x + alpha)* alpha^(n-1) - (a*x + beta) *beta^(n-1)
f^n(x) = --------------------------------------------------------
(a*x + alpha) *alpha^n - (a*x + beta) *beta^n
or is there a specific reason for the longer(or did I get something wrong) ?
Gottfried
Matt
No, you're right. I just bashed in the formula from some workings from
a while ago without thinking about it. Maybe there was some reason, to
do with something I was working on, why I originally wrote it as I
did, or maybe I just never properly tidied it up. Anyway, thanks for
pointing that out.
amy666
using moebius transforms :
iterate twice the following
(a + b x) / (c + d x)
denote that as M2(x)
then solve for a , b , c and d ;
M2(x) = 1/(1+x)
and then the solution is
(a + b x) / (c + d x)
..
not ?
regards
tommy1729
Gottfried Helms
> Just for exercise I considered the function
>
> f(x) = 1/(1+x)
>
> under iteration and a meaningful description for fractional
> iteration. I thought, it should be possible to use
For the function f°h(x) with x=1, depending on h I get
f°h(1) = phi + sqrt(5)*sum(k=1,inf,((2-3*phi)*(phi-1)^h) ^k)
where phi = (sqrt(5)-1)/2 ~ 0.61803...
Here the order of exponents h and k in the term
((2-3*phi)*(phi-1)^h) ^k
is critical, since phi-1 is negative and the order of
exponentiation is then relevant, if h is fractional.
(I think, the above can be simplified more...)
=== Not a powerseries ==========================
If I rewrite f°h(1) as f(h) to focus the dependency on the
height parameter, then actually we have a form like this
(using a,b,u for short)
f(h) = a + (b*u^h)^1 + (b*u^h)^2 + ...
which is not a powerseries in h. How would we call such a type
of series? (Similar series occur in fractional tetration if I
apply the diagonalization-method)
Gottfried Helms
(For few checked examples the results are identical to the results of
my first calculations)
Gottfried Helms
> Am 22.01.2009 22:19 schrieb Gottfried Helms:
>> Just for exercise I considered the function
>>
>> f(x) = 1/(1+x)
>>
>> under iteration and a meaningful description for fractional
>> iteration. I thought, it should be possible to use
>
> For the function f°h(x) with x=1, depending on h I get
>
> f°h(1) = phi + sqrt(5)*sum(k=1,inf,((2-3*phi)*(phi-1)^h) ^k)
>
> where phi = (sqrt(5)-1)/2 ~ 0.61803...
>
> Here the order of exponents h and k in the term
>
> ((2-3*phi)*(phi-1)^h) ^k
>
This is shorter expressed by the formula for geometric series:
In
f°h(1) = phi + sqrt(5)*sum(k=1,inf,((2-3*phi)*(phi-1)^h) ^k)
c = (2-3*phi)*(phi-1)^h
then write f1(h) for f°h(1) then with that c
f1(h) = phi + sqrt(5)*c/(1-c)
Hmm. Maybe, I'll arrive at Matt's formula this way...
--------------------------------------------------------
The previous question remains open: is there another
approach to the fractional iterates, which would give
different values? Because this function seems sufficient
simple so that the whereabouts of this "regular iteration
using fixpoint-shift" could be studied more easily than
in the tetration-discussion, where fractional iterates
(computed with this method) usually lead to strongly divergent
series, even with complex coefficients, for which we
then cannot find the values with arbitrary precision.
For instance, in the plot in [1] we notice, that the
spiral around the limit case f1(inf) = phi has nearly
equal angles when iterated with a fixed h. But this
is not exact the case (see pic 3).
Could an equi-angular interpolationmethod for fractional
iterates be found?
Gottfried Helms
[1] http://go.helms-net.de/math/tetdocs/FracIterAltGeom.htm
(updated)
Gottfried
alainv...@gmail.com
>
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Good morning,
The problem of continuous iteration (ax +b)/(cx +d)
is simply solve:
from a direct classification according to the number
of fixed points:zero, one, two,infinity and pole or not.
Here:
h(x) =1/(1+x) two fixed points x1 = (-1+sqrt(5))/2
x2 = (-1-sqrt(5))/2
and a pole x =-1
May be written (h(x)-x1)/(h(x)-x2)=a*(x-x1)/(x-x2) ,
for x = 0 ,h(0) = 1 then a = (1+sqrt(5)/(1-sqrt(5))
And also (h^[n](x)-x1)/(h^[n](x)-x2)=a^n *(x-x1)/(x-x2)
just mind 'a' is real negative!
Alain
A N Niel
<he...@uni-kassel.de> wrote:
> Just for exercise I considered the function
>
> f(x) = 1/(1+x)
>
> under iteration and a meaningful description for fractional
> iteration.
It is a linear fractional transformation, so do the fractional
iteration in terms of matrices. (The golden section appears in some of
the other answers, and that is because it is an eigenvalue of the
matrix involved).
dan73
In your first graph, "fractional iterations approaching
the golden ratio ", I noticed where x=1 and x=.25
and the difference above and below the x axis as h=.5.
Where my closed form method where n=1,2,3,4...
(sqrt(((n/2)^2)+1))+n/2 gives the golden ratio
when n = 1. For all higher values of (n) are all
the metal means.
So looking at your first graph of phi just add the
first x=.25 with second x=1 = 1.25 =(((n/2)^2)+n).
Then --
sqrt(1.25) + (h=.5)or(n/2) = the golden ratio.
Then if exploring these type of continued fractions
[1:1,1,1,1,...], [2:2,2,2,2,...],[3:3,3,3,3,...]
So given this formula for all continued fractions
of the above form will the both x values and h values
coincide for all the metal means?
Where n=2 which is the silver mean = sqrt(2)+ (h=1) or
(n/2) but by your graph, what is x=? and second x=??
Dan
alainv...@gmail.com
> In article <6ts692Fccn9...@mid.dfncis.de>, Gottfried Helms
To A.N Niel
Of course there are known solutions in term of matrices,
I only show there exist other ways, we might also use
Ad Hoc parametrization,
Example: x(t) = 1+2/t, t =>t +3 or h(x(t))= 1 +2/(t+3)
h^[r](x(t)=x(t+3*r) corresponds to h(x)=(5x-3)/(3x-1)
and directly we've got h^[r](x)=((3r+2)x-3r)/(3rx+2-r)
Alain
umu...@gmail.com
>
> Regards -
>
> Gottfried Helms
Something alike for the function f(x) = 1/(1-x) at:
http://hdebruijn.soo.dto.tudelft.nl/www/grondig/testable.htm
http://hdebruijn.soo.dto.tudelft.nl/jaar2005/italiaan.pdf
Han de Bruijn
beew...@hotmail.com
>
> Regards -
>
> Gottfried Helms
This problem is equivalent to finding the "fractional" Fibonacci
Number.
With
f_0(x) = x
f_(n+1)(x) = f_n( 1 / (1 + x) )
We have
f_n(x) = (F_n + F_(n-1)*x) / (F_(n+1) + F_n*x)
= ( 1 + (F_(n-1)/F_n)*x ) / ( (F_(n+1)/F_n) + x )
where F_n is the n-th Fibonacci Number. This fact is demonstrated as
follows:
f_0 = x
f_1 = 1 / (1 + x)
f_2 = (1 + x) / (2 + x)
f_3 = (2 + x) / (3 + 2x)
f_4 = (3 + 2x) / (5 + 3x)
f_5 = (5 + 3x) / (8 + 5x)
...
-MO
alainv...@gmail.com
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Bonsoir,
Ok,I do agree,
Alain
Gottfried Helms
Am 23.01.2009 11:58 schrieb alainv...@gmail.com:
> Good morning,
Good evening! :-)
>
> The problem of continuous iteration (ax +b)/(cx +d)
> is simply solve:
> from a direct classification according to the number
> of fixed points:zero, one, two,infinity and pole or not.
>
> Here:
> h(x) =1/(1+x) two fixed points x1 = (-1+sqrt(5))/2
> x2 = (-1-sqrt(5))/2
> and a pole x =-1
This looks as a very interesting classification. Can
I read a bit more about this systematic (online preferred;-))
> May be written (h(x)-x1)/(h(x)-x2)=a*(x-x1)/(x-x2) ,
> for x = 0 ,h(0) = 1 then a = (1+sqrt(5)/(1-sqrt(5))
>
> And also (h^[n](x)-x1)/(h^[n](x)-x2)=a^n *(x-x1)/(x-x2)
> just mind 'a' is real negative!
Yepp - I'll use this tomorrow. Thank you very much!
Kind regards-
Gottfried
Gottfried Helms
Yes, you see: all my functional iterations were done
in matrix-terms; so I did it here too (although I
hoped to find an alternative approach) The matrix-
approach is much versatile...
The problem, which I hoped to see more clearly, was
the translation of the problem using fixpoint-shift
with the associated powerseries/matrix-operator.
In the tetration-discussion we assume, that the
selection of different fixpoints for the shift
gives different results for the fractional iteration/
matrix-power - but we have no concrete computed
data (mostly because of the strong divergence of
the coefficients in the powerseries for fractional
iteration). Here the involved powerseries behave
more nicely; I had only convergent examples for
this iteration so far.
Thus, if there's something inherent in the regular
iteration (matrix-based) with fixpointshift, one
could possibly find it here under a magnifying
glass... (That's why I asked for concurring results)
Gottfried
Gottfried Helms
> Where my closed form method where n=1,2,3,4...
> (sqrt(((n/2)^2)+1))+n/2 gives the golden ratio
> when n = 1. For all higher values of (n) are all
> the metal means.
Well...
> So looking at your first graph of phi just add the
> first x=.25 with second x=1 = 1.25 =(((n/2)^2)+n).
> Then --
> sqrt(1.25) + (h=.5)or(n/2) = the golden ratio.
Hmm, difficult to parse...
>
> Then if exploring these type of continued fractions
> [1:1,1,1,1,...], [2:2,2,2,2,...],[3:3,3,3,3,...]
> So given this formula for all continued fractions
> of the above form will the both x values and h values
> coincide for all the metal means?
>
> Where n=2 which is the silver mean = sqrt(2)+ (h=1) or
> (n/2) but by your graph, what is x=? and second x=??
I didn't understand too much of this, I'm afraid...
Thanks anyway-
Gottfried
>
> Dan
Gottfried Helms
>
> Something alike for the function f(x) = 1/(1-x) at:
>
> http://hdebruijn.soo.dto.tudelft.nl/www/grondig/testable.htm
> http://hdebruijn.soo.dto.tudelft.nl/jaar2005/italiaan.pdf
>
> Han de Bruijn
Han, thanks!
Just as in your second source I found that fibonacci-
relation and tried to use this for the continuous
iteration here, but I didn't get a usable instrument
by this. I had the second fixpoint as well, but didn't
work it out, that it is repelling.
I'll have another look at your text later.
"for free ... no referee" :-) many of us have the same
obstacle...
Gottfried
Gottfried Helms
> This problem is equivalent to finding the "fractional" Fibonacci
> Number.
>
> With
> f_0(x) = x
> f_(n+1)(x) = f_n( 1 / (1 + x) )
>
> We have
> f_n(x) = (F_n + F_(n-1)*x) / (F_(n+1) + F_n*x)
> = ( 1 + (F_(n-1)/F_n)*x ) / ( (F_(n+1)/F_n) + x )
>
> where F_n is the n-th Fibonacci Number. This fact is demonstrated as
> follows:
>
> f_0 = x
> f_1 = 1 / (1 + x)
> f_2 = (1 + x) / (2 + x)
> f_3 = (2 + x) / (3 + 2x)
> f_4 = (3 + 2x) / (5 + 3x)
> f_5 = (5 + 3x) / (8 + 5x)
> ...
Yes, I got the same idea; it is the same idea as interpolation
of Bernoulli- and Bell-numbers. With the fibonacci-numbers
there is this old closed-from formula, in the same
spirit as Matt's nice closed-form in his first post.
My problem was to relate that fractional-fibonaccis
to the value/evaluation of the fractional f°h(x).
But your above formula looks very nice:
(F_n + F_(n-1)*x)
f_n(x) = -----------------------
(F_(n+1) + F_n *x)
----------------------
All the given formulae so far seem to give the same
results, including the regular iteration using matrix-
powers (and fixpoint-shift).
So what, for instance, if we try to involve the second
fixpoint. Will the results come out the same?
Hmmm ...
Gottfried
Gottfried Helms
Am 23.01.2009 18:43 schrieb Gottfried Helms:
> Am 23.01.2009 17:36 schrieb beew...@hotmail.com:
>> This problem is equivalent to finding the "fractional" Fibonacci
>> Number.
>>
>> With
>> f_0(x) = x
>> f_(n+1)(x) = f_n( 1 / (1 + x) )
>>
>> We have
>> f_n(x) = (F_n + F_(n-1)*x) / (F_(n+1) + F_n*x)
>> = ( 1 + (F_(n-1)/F_n)*x ) / ( (F_(n+1)/F_n) + x )
>>
>> where F_n is the n-th Fibonacci Number. This fact is demonstrated as
>> follows:
>>
>> f_0 = x
>> f_1 = 1 / (1 + x)
>> f_2 = (1 + x) / (2 + x)
>> f_3 = (2 + x) / (3 + 2x)
>> f_4 = (3 + 2x) / (5 + 3x)
>> f_5 = (5 + 3x) / (8 + 5x)
>> ...
Yes, that's true. But actually I do not see, how this leads
to a notion of a fractional n here ...
Did I overlook something?
Gottfried
beew...@hotmail.com
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
My thought, which certainly may be incomplete, is that f_n(x) is of
the form:
f_n(x) = (a_n + b_n * x) / (c_n + d_n * x)
I made the intuitive leap that the form applies to fractional n. I
guess the real test would be if the iterative equation holds for
fraction n if one uses fractional Fibonacci numbers. It seems to.
f_(n+1) = (F_n + F_(n-1)*(1/(1+x)) ) / (F_(n+1) + F_n*(1/(1+x)) )
= (F_n + F_(n-1) + F_n * x) / (F_(n+1) + F_n + F_(n+1) * x)
= (F_(n+1) + F_n * x) / (F_(n+1) + F_(n+1) * x)
which is consistent with
f_n = ( F_n + F_(n-1)*x ) / ( F_(n+1) + F_n*x )
- MO
beew...@hotmail.com
>
> - Show quoted text -
Works if fractional F_n satisfies F_n = F_(n-1) + F_(n-2)
- MO
Matt
For f(x) = (a*x + b)/(c*x + d), I get
f^n(x) = a/c + (b*c - a*d)/c*((c*x - a + alpha)*alpha^(n - 1) -
(c*x - a + beta)*beta^(n - 1)) / ((c*x - a + alpha)*alpha^n - (c*x - a
+ beta)*beta^n)
with
alpha = (a + d + sqrt((a - d)^2 + 4*b*c)) / 2
beta = (a + d - sqrt((a - d)^2 + 4*b*c)) / 2
dan73
>> the golden ratio ", I noticed where x=1 and x=.25
>> and the difference above and below the x axis as h=.5.
>>
>Hmm. I don't see the focus of that thought...
>> Where my closed form method where n=1,2,3,4...
>> (sqrt(((n/2)^2)+1))+n/2 gives the golden ratio
>> when n = 1. For all higher values of (n) are all
>> the metal means.
>Well...
Could you run another graph but instead of the
golden ratio iterate the silver mean --[2:2,2,2,2..]
This way I could possibly explain it better after
observing the points of this new graph but then again
maybe not.
Dan
Gottfried Helms
>>> In your first graph, "fractional iterations approaching
>>> the golden ratio ", I noticed where x=1 and x=.25
>>> and the difference above and below the x axis as h=.5.
>>>
>> Hmm. I don't see the focus of that thought...
>
>>> Where my closed form method where n=1,2,3,4...
>>> (sqrt(((n/2)^2)+1))+n/2 gives the golden ratio
>>> when n = 1. For all higher values of (n) are all
>>> the metal means.
>> Well...
>
> Could you run another graph but instead of the
> golden ratio iterate the silver mean --[2:2,2,2,2..]
>
> This way I could possibly explain it better after
> observing the points of this new graph but then again
> maybe not.
>
> Dan
Hmmm. Cold... :-(
But if you use Matt's formula, then you can compute
as many examples as you need to arbitrary precision
and can then explain, what you mean/assume/want to
conjecture.
No?
Gottfried
alainv...@gmail.com
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Bonjour,
All continued fractions such as [2:2,2,2,2..] ,[1,2,1,2..]
correspond to functions h(x) = (a*x+b)/(c*x+d) .
Two notices:
g(x) =1/(1-x) has a cycle = 3 .
All homographic functions h(x) can be parametrized,
Ex. h(x)=1/(1+x) ,x =f(t) =(1+sqrt(5)-t*(1-sqrt(5))/2 ,
t=> a*t , a = ((1+sqrt(5))/(1-sqrt(5)) ,h(x)=h(f(t))=f(a*t)
And an interesting property : h^[r](x)=h^[r](f(t))=f(a^r*t),
Alain
amy666
>
> not ?
>
>
> regards
>
> tommy1729
that seems the most logical way for a half-iterate to me ...
are your results consistant with this ?
analogue for 1/3 iterate ...
not ??
galathaea
so there are two matrix spaces
which both generalise continuous iteration notions
and agree on their answers
there is the modular group SL(2, Z)
and it's continuous real SL(2, R) for orbit generation
(or even it's complexification SL(2, C))
this has all the magic number theoriness
that fascinates the ellipticists and their ilk
an element x e SL(2, K)
is an application of a corresponding modular transform
over the linear fractional function space
so x^2 is two applications
and so on
continuous iteration is then
exp(ln(x) z)
where z is in R or C
(and thus central for lack of left-right ambiguity)
for any element of SL(2, ?) that has a logarithm
like the one given
this makes sense
then there is the comtet-helms construction
over the infinite matrix operator space M_oo(R or C)
(or more rigorously over some hilbert subalgebra H_oo)
this space has operator that correspond
to the application of some function
with local power series representation
and also operators like the derivative
that are not themselves functions
the same approach is taken for continuous operator orbit
y e H_oo(K)
y^z = exp(ln(y) z)
for z e K
now
through the classic taylor decomposition
of the linear fractional forms
there is an embedding
i_2 : SL(2, Z) -> H_oo(Q)
where the construction is exponentially consistent
e.g.
i_2(x^z) = (i_2(x))^z in H_oo(C) on some branch
i'm sure this is an easy application
of the properties of direct limits..
what is also interesting
is that the solution is simplified
using ramanujan's similarity trick
in his third quarterly report
he looks at continuous iteration
using the prototypical
p
f(x) = a x
n n
(n) (p - 1)/(p - 1) p
so f (x) = a x
then he uses his similarity rule
-1
if F = Q f Q
z -1 z
then F = Q f Q
to reduce problems to the prototypical form
this seems to be the "natural" approach
for this type of continuous iteration
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
Gottfried Helms
> for the fractional iterated can be found at
> http://go.helms-net.de/math/tetdocs/FracIterAltGeom.htm
>
it would be possible, to find an interpolation-scheme
based on the requirement of an equiangular trajectory.
That means, that with
f°h(x) the h'th iterate of f(x)
finf(x) the limit for h->inf (if this converges)
then for a fixed (possibly fractional d)
arg(f°(h+d)(x) - finf(x) ) - arg(f°h(x) - finf(x) )
= <const dependend on d>
= Pi/d ( for our function here)
the angles for a fixed stepwidth of iteration are equal.
(modulo 2 Pi, actually).
--------------------
While I did not yet find such a concurring method for interpolation
I experimented with the representation of f(x) as iteration-series
instead of powerseries and came across a rather strange
iteration-trajectory.
First - because f(x) has an oscillating trajectory while
its iterates converge to the limit-value, I used the twice iterated
ff(x) = f(f(x))
here.
x+1
ff(x) = ---- = 1/2*(1 + x/2 - (x/2)^2 + (x/2)^3 - (x/2)^4 ... )
x+2
The inverse is
2x -1
ff°-1(x) = ------- = -1 + x + x^2 + x^3 + ...
1 - x
Second, then I asked for an iterable function ft(x) such that
const + x + ft(x) + ft°2(x) + ft°3(x) + ... = ff(x)
This gives, using the idea that
const+ x + ft(x) + ft°2(x) + ft°3(x) + ... = ff(x)
const+ ft(x) + ft°2(x) + ft°3(x) + ... = ff( ft(x) )
----------------------------------------------------
x = ff(x) - ff( ft(x) )
and thus
ff( ft(x) ) = ff(x) - x
ft(x) = ff°-1( ff(x) - x)
2x^2 + 3x
ft(x) = - -----------
(x + 1)^2
The taylor-expansion of ft(x) is
ft(x) = -3*x + 4*x^2 - 5*x^3 + 6*x^4 - ... + ...
This gives the "strange" iteration-trajectory (lines wrapped when positive values occur)
Table: for x=1; then ft°0(x), ft°1(x), ...
1.00000 -1.25000
10.0000 -1.90083 -1.87779 -1.84139 -1.77595 -1.62789 -1.05615
297.390 -1.99664 -1.99661 -1.99659 (...) -1.81339 -1.71794 -1.45277
0.669339 -1.04211
538.141 -1.99814 -1.99813 -1.99813 (...) -1.84197 -1.77708 -1.63084 -1.07238
175.062 -1.99429 -1.99422 -1.99415 (...) -1.85002 -1.79242 -1.66943 -1.26234
8.71840 -1.88651 -1.85560 -1.80275 -1.69389 -1.36425
2.79166 -1.66671 -1.25018
9.98010 -1.90063 -1.87750 -1.84090 -1.77501 -1.62540 -1.04227
533.998 -1.99813 -1.99812 -1.99811 (...) -1.87604 -1.83848 -1.77026 -1.61277 -0.968715
1051.70 -1.99905 -1.99905 -1.99905 (...)
--------------------------------------
We have, that for x=1
ff(x) = 2/3 = 0.6666... and ff(x)-1/2 = 0.16666....
The sequence of partial sums of the iterationseries
x + ft°1(x) + ft°2(x) + ... at x=1 begins with
1.00000
-0.250000 9.75000 7.84917 5.97139 4.13000 2.35404 0.726155
-0.329993 297.060 295.063 293.066 (...) 6.88689 5.02539 3.21200 1.49407
0.041291 0.71063
-0.331482 537.810 535.812 533.813 (...) 11.6906 9.77379 7.87273 5.99459 4.15262 2.37554 0.744695
-0.327686 174.734 172.740 170.746 (...) 10.1216 8.21722 6.33417 4.48416 2.69173 1.02230
-0.240036 8.47837 6.59185 4.73625 2.93351 1.23961
-0.124637 2.66702 1.00032
-0.249862 9.73024 7.82961 5.95211 4.11121 2.33621 0.710803
-0.331468 533.667 531.668 529.670 (...) 9.63354 7.73388 5.85784 4.01936 2.24910 0.636331
-0.332384 1051.36 1049.36 1047.37 (...)
and the plot is that of a sawtooth-function with variable frequency and amplitude.
Adding the const 0.5 and inverting the value 1/(0.5 + ps_n) of the n'th partial sum
should converge to 1/ff(x) = 6 . What we get is this:
2.00000 0.666667
4.00000 0.0975610 0.119772 0.154526 0.215983 0.350380 0.815558
5.88212 0.00336067 0.00338337 0.00340638 0.00342971 0.00345336 0.00347733 (...)
1.84743 0.826016
5.93408 0.00185767 0.00186459 0.00187156 0.00187859 0.00188566 0.00189280 (...)
5.80335 0.00570665 0.00577235 0.00583957 0.00590837 0.00597881 0.00605095 (...)
3.84668 0.111379 0.141007 0.190976 0.291247 0.574840
2.66409 0.315754 0.666526
3.99779 0.0977494 0.120054 0.154988 0.216863 0.352584 0.825898
5.93358 0.00187208 0.00187910 0.00188619 0.00189332 0.00190051 0.00190776 (...)
5.96603 0.000950694 0.000952504 0.000954321 0.000956145 0.000957976 0.000959814 (...)
5.99930 0.0000193199 0.0000193206 0.0000193214 0.0000193221 0.0000193229 0.0000193236 (...)
5.20115 0.0268407 0.0283416 0.0300189 0.0319056 0.0340431 0.0364847 (...)
4.61776 0.0547609 0.0612878 0.0695489 0.0803294 0.0949653 0.115920 (...)
1.68799
1.06248
4.30465 0.0742561 0.0866394 0.103842 0.129271 0.170422 0.247329 (...)
1.12729
3.63414 0.133483 0.177679 0.262382 0.479475
1.64817
(...)
5.80446 0.00567281 0.00573772 0.00580414 0.00587210 0.00594168 0.00601292 (...)
4.69775 0.0503595 0.0558409 0.0626394 0.0712876 0.0826458 0.0981969 (...)
5.99766 0.0000650337 0.0000650421 0.0000650506 0.0000650591 0.0000650675 0.0000650760 (...)
5.73916 0.00768118 0.00780056 0.00792369 0.00805077 0.00818198 0.00831753 (...)
5.64481 0.0106915 0.0109238 0.0111665 0.0114201 0.0116854 0.0119634 (...)
5.87217 0.00365244 0.00367927 0.00370649 0.00373412 0.00376217 0.00379063 (...)
5.92669 0.00206932 0.00207791 0.00208658 0.00209531 0.00210412 0.00211300 (...)
4.17310 0.0837157 0.0997000 0.122979 0.159820 0.226182 0.376468 (...)
5.55768 0.0135927 0.0139699 0.0143685 0.0147905 0.0152379 0.0157131 (...)
4.38765 0.0687022 0.0792075 0.0934113 0.113633 0.144591 0.197464 (...)
5.22757 0.0257749 0.0271566 0.0286938 0.0304141 0.0323519 0.0345511 (...)
4.29480 0.0749362 0.0875614 0.105159 0.131295 0.173892 0.254468 (...)
5.96608 0.000949225 0.000951030 0.000952841 0.000954660 0.000956485 0.000958317 (...)
5.59218 0.0124296 0.0127444 0.0130756 0.0134243 0.0137922 0.0141806 (...)
4.48268 0.0626992 0.0713647 0.0827489 0.0983414 0.120939 0.156444 (...)
5.99958 0.0000117495 0.0000117497 0.0000117500 0.0000117503 0.0000117506 0.0000117508 (...)
5.37835 0.0199616 0.0207830 0.0216745 0.0226455 0.0237070 0.0248723 (...)
5.21334 0.0263470 0.0277921 0.0294036 0.0312120 0.0332554 0.0355824 (...)
4.06486 0.0921716 0.111820 0.141706 0.192233 0.294064 0.584852
5.12661 0.0299276 0.0318025 0.0339259 0.0363502 0.0391440 0.0423978 (...)
5.40192 0.0190921 0.0198425 0.0206540 0.0215343 0.0224926 0.0235396 (...)
4.80192 0.0449308 0.0492570 0.0544918 0.0609520 0.0691184 0.0797584 (...)
5.98475 0.000425022 0.000425384 0.000425746 0.000426109 0.000426472 0.000426836 (...)
5.61235 0.0117583 0.0120398 0.0123350 0.0126450 0.0129710 0.0133141 (...)
5.87645 0.00352659 0.00355160 0.00357696 0.00360269 0.00362879 0.00365527 (...)
5.50078 0.0155534 0.0160488 0.0165766 0.0171401 0.0177431 0.0183899 (...)
5.35644 0.0207788 0.0216700 0.0226406 0.0237016 0.0248663 0.0261506 (...)
5.99989 0.00000312033 0.00000312035 0.00000312037 0.00000312039 0.00000312041 0.00000312043 (...)
Hmmm ...
Gottfried Helms
Gottfried Helms
Am 25.01.2009 11:52 schrieb alainv...@gmail.com:
(...)
> All homographic functions h(x) can be parametrized,
> Ex. h(x)=1/(1+x) ,x =f(t) =(1+sqrt(5)-t*(1-sqrt(5))/2 ,
> t=> a*t , a = ((1+sqrt(5))/(1-sqrt(5)) ,h(x)=h(f(t))=f(a*t)
> And an interesting property : h^[r](x)=h^[r](f(t))=f(a^r*t),
-----------------------------------------------------^^^^^^^^
I surely like irony and fine sarcasm, especially if cleverly
hidden in symbols, but don't you think, this criticism of
long mathematical derivations goes a bit too far... :-)))
<duck&run>...
<G>
alainv...@gmail.com
Bonjour,
Neither hidden irony nor sarcasm ,
I do appreciate your threads on this forum;
sure we , times and again, go into "long mathematical derivations"
and lose feet,
Alain
Matt
It occurred to me later that this method fails when (a - d)^2 + 4*b*c
= 0, but I only just got round to trying to figure it out. In this
case we have a function of the form
f(x) = (a*x + b^2)/(-c^2*x + a + 2*b*c)
with a solution
f^n(x) = ((a + b*c - b*c*n)*x + b^2*n) / (-c^2*n*x + a + b*c +
b*c*n)