• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Infinite tetration of the imaginary unit tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/22/2011, 03:15 PM (This post was last modified: 06/22/2011, 03:25 PM by tommy1729.) (06/21/2011, 11:19 PM)tommy1729 Wrote: @ sheldon : with superfunction , do you mean the regular superfunction expanded at the fixpoint ? @ everyone : euh , why does 1.7129 i ^ 1.7129 i ^ ... ( 0.5 + 0.5 i) give a circle and what is its radius ? well , the radius can be easily computed from two iterations by basic trigonometry ... but i wonder if there is something special about it ... we have a new fundamental constant 1.7129... relating e , pi , i just like eulers e^pi i + 1 = 0 tommy1729 sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 06/22/2011, 03:23 PM (This post was last modified: 06/23/2011, 01:50 PM by sheldonison.) (06/21/2011, 12:13 PM)Gottfried Wrote: ....I made a picture, how the fixpoint, the log of the fixpoint, and the base according to the Shell-Thron-description are connected. (I've also put it in the hyperop-wiki). In my notation I always used u (for the log of the fixpoint), t =exp(u) for the fixpoint and b=exp(u/t) for the base....My conjecture is that as long as the period is an irrational real number, then I suspect the superfunction is analytic. As we have seen, for the base with a real period=3, starting with a point near L, and iterating the function x=b^x three times, doesn't get you back to the initial starting point, which seems to be a big problem, which prevents the definition of the superfunction as being real periodic, with a period=3. The function f(z)=b^(L+z) for a base on the Shell-Thron region, has a Taylor series developed in the neighborhood of the fixed point L with the following form: $b^{(L+z)}=L+\sum_{n=1}^{\infty}a_n z^n$ $a1=\log(L)$, which is also number on the unit circle, since abs(log(L))=1. In Gottfried's notation, a1=u. $\text{period}=2\pi i /\log(\log(L)) = 2\pi i /\log(a1)$, where the period is a real number The goal is to show that if the period is irrational, then for good rational approximations of the period (developed from the continued fraction), where: $\text{period}\approx\frac{m}{n}$ $\exp_b^{o m}(L+z)\approx L+z$ and that as the approximation for the period becomes more and more exact, then iterating b^(L+z) m times also becomes more and more exact, so that in the limit, for a small enough z (to avoid fractal chaos), then $\lim_{m\to\infty} \exp_b^{o m}(L+z) = L+z$, where m is the numerator of a continued fraction approximation of the period. In that case, Then the period "works". For all integer values of m, we can define the superfunction for x=(m mod period), starting from initial value L+z: $\text{SuperFunction}_{L}(x) = \exp_b^{o m}(L+z)$ where $x = (m \bmod \text{period})$. And for any real x, a sequence of more and more accurate approximations can be used, $x \approx (m \bmod \text{period})$ (edit not sure how to define this sequence). $\text{SuperFunction}_{L}(x) = \exp_b^{o x}(L+z) = \lim_{m\to\infty} \exp_b^{o m}(L+z)$, and then there is justification for developing a superfunction of b^z, which has a real period, and decays to the constant L as $\Im(x) \to i\infty$. $\text{SuperFunction}_{L}(x) = L + \sum_{n=1}^{\infty} c_n\exp(2n\pi ix/\text{period})$ Going back to the series for $b^{(L+z)}=L+\sum_{n=1}^{\infty}a_n z^n$ I conjecture that there aren't restrictions on a2..an, for b^(L+z), so that real period works, so long as the period is irrational, and the superfuntion is developed with z small enough to avoid fractal chaos. But if the period is a rational number, there are more strict restrictions on a2..an, since for a small integer number of iterations, then iterating $\exp_b^{o \text{m}}(L+z)=L+z$ must be exact if period=m/n. For example, a2..an=0 always works, and many other examples also work for a rational period, but in the general case for arbitrary a2..an, the rational period doesn't work. That means that in the general case, the superfunction on the Shell Thron boundary with a rational period doesn't work, and we can't develop the superfunction with that rational period. Whereas, the conjecture is that we can develop the superfunction whenever the period is an irrational number on the Shell-Thron boundary. It will be interesting to see if the conjecture turns out to be true, and how it can be proven (or disproven). - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/22/2011, 07:05 PM nice post sheldon. i think a period of 3 might works if we use other numbers then the complex ones , but thats off-topic and needs more investigation before i continue to talk about that. however i noticed another thing needs proof , well at least i did not see one yet , it might be trivial but im doubtful : prove that the cycles are neutral and do not self-intersect : neither attracting nor repelling nor self-crossing. note that if all the cyles are of finite length => not attracting = not repelling <=> not self-crossing. tommy1729 Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 06/23/2011, 08:55 AM (This post was last modified: 06/23/2011, 09:20 AM by Gottfried.) Hi Sheldon - likely a typo in the formula: (06/22/2011, 03:23 PM)sheldonison Wrote: case, Then the period "works". For all integers, we can define the superfunction for x=(m mod period), starting from initial value L+z: $\text{SuperFunction}_{L}(x) = \exp_B^{o m}(L+z)$. And forFor the parameter of the superfunction I think you want to write m (or for exp_b the iterator/height x ?) I'm still thinking about your post and probably need some more time. (I experimented with some plots so far). However, since you introduce the continued fractions here - that made me thinking, whether periods of quadratic roots are different (and the curves look different) from other irrational periods since the continued fraction is periodic; especially whether the golden ratio for the period-parameter gives a distinctive curve. Then we have the difference between algebraic irrationals and transcendents; for the latter the liouville-numbers might be a special class in regard of the approximation to a dense line. That numbers have the property to have irrationality measure of infinity and arbitrarily "close to" rational numbers - much "closer" than any algebraic irrational number. Thus it might be possible, that they are some bridge between the rational periods with their distinct accumulation points and the seemingly smooth curve of the golden-ratio period on the other extreme. What I'm looking at at the moment is your problem of the analycity of the superfunction -I just have to get familiar with the understanding of that general problem first. The only thing what I know beforehand is, that I can only create a formal powerseries for the superfunction if u is an irrational root of the complex unity - the standard approach to the diagonalization of the associated Bell-matrix does not work for repeated eigenvalues (which occur, if u is a rational such root). But I think I need one or two days to chew all this to more clarity. (Here in Hassia/Germany I have a nice four-day "short holiday" with a free (christian) day today and my half-term allows me to stay away from university tomorrow... so I'll invite the sun for that very welcome break :-) ) Also, there was a nice graph accidentally, when I forgot to remove the connecting lines of the excel-plot of some trajectories - just an artistic impression: (The header is not correct/was not finished when the data were added to the graph) Gottfried Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 06/23/2011, 01:13 PM (This post was last modified: 06/23/2011, 04:53 PM by sheldonison.) (06/23/2011, 08:55 AM)Gottfried Wrote: Hi Sheldon - likely a typo in the formula: (06/22/2011, 03:23 PM)sheldonison Wrote: case, Then the period "works". For all integers, we can define the superfunction for x=(m mod period), starting from initial value L+z: $\text{SuperFunction}_{L}(x) = \exp_B^{o m}(L+z)$. And forFor the parameter of the superfunction I think you want to write m (or for exp_b the iterator/height x ?) ....Hey Gottfried, I don't think its a typo, but I tried to make the original post slightly clearer. x=(m mod period), m would be an integer iteration. The "o m" notation is iteration m times. Lower case b works better; here from your previous example, $b\approx 1.7129i$, $\text{period}\approx 2.9883$ $\text{SuperFunction}_{L}(0) = (L+z)$ $\text{SuperFunction}_{L}(1) = b^{(L+z)}$ $\text{SuperFunction}_{L}(2) = \exp_b^{o 2}(L+z)$ $\text{SuperFunction}_{L}(0.0117) = \exp_b^{o 3}(L+z)$ $\text{SuperFunction}_{L}(1.0117) = \exp_b^{o 4}(L+z)$ $\text{SuperFunction}_{L}(2.0117) = \exp_b^{o 5}(L+z)$ $\text{SuperFunction}_{L}(0.0234) = \exp_b^{o 6}(L+z)$ $\text{SuperFunction}_{L}(x) = \exp_b^{o m}(L+z)$ where $x=(m \bmod \text{period})$ This allows you to use integer iterations to gradually fill in all of the points for the superfunction on the real axis between 0 and the period. Since the function is real periodic, this means we can define the superfunction anywhere on the real axis. Then what I did, was to take the Fourier analysis of the function, using polynomial interpolations from several nearby points on either side to get a fairly exact value for a sampling of evenly spaced points; the results were posted here. I'm going to read up on what's out there on Fatou sets. Anybody have a suggested reference? Specifically, we're interested in a Periodic Fatou set generated from a complex parabolic fixed point. I'm also trying to formulate a question for math overflow, to verify the existence of real periodic Fatou sets, generated with a superfunction of a function with a periodic parabolic fixed point, similar to what is seen on the Shell-Thron boundary. - Sheldon sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 06/23/2011, 06:10 PM (This post was last modified: 06/23/2011, 07:50 PM by sheldonison.) (06/23/2011, 01:13 PM)sheldonison Wrote: I'm going to read up on what's out there on Fatou sets. Anybody have a suggested reference? Specifically, we're interested in a Periodic Fatou set generated from a complex parabolic fixed point. I'm also trying to formulate a question for math overflow, to verify the existence of real periodic Fatou sets, generated with a superfunction of a function with a periodic parabolic fixed point, similar to what is seen on the Shell-Thron boundary. - SheldonLooks like my intuition is correct. They're called Siegel discs, which are part of the Classification of Fatou components. In particular, Siegel discs occur around "occur around irrationally indifferent periodic points", of exactly the type that we have been describing in this thread. Siegel proved their existence. I'll do some more reading -- there are requirements on the Brjuno Number of the convergents of the continued fraction expansion of α, which would be 1/period. The requirements don't seem that stringent, which implies that on the Shell-Thron boundary, for many/most cases if the period is irrational, $\text{period}=2 \pi i/\log(\log(L))$, then there is a real periodic superfunction. Then the real periodic superfunction can be mapped to a unit disc, where the boundary of the unit disc would be generating the superfunction from an initial starting point of 0, where the fractal behavior caused by $\log_b(0)$, or $\exp_b^{-n}(0)$. The interior of the Siegel disc would correspond to where the superfunction is analytic, with a real period, converging to the fixed point as $\Im(x)\to i\infty$, which would correspond to the center of the disc. - Sheldon Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 06/24/2011, 08:07 AM (This post was last modified: 06/24/2011, 09:24 AM by Gottfried.) Hi Sheldon - I've fiddled a bit with the problem of rational periods to see how far, and why it differs from the irrational-period orbits. I don't know whether this is of any use at all, at least it has been an interesting exercise. For an example let's take the period of 4, so the parameter phi=Pi/2, then u=cos(phi)+I*sin(phi)=1*I, t=exp(u)=e^I, b=exp(u/t)=e^(I/e^I). Let's call our base-function f(x)=b^x, and if h-fold iterated f(x,h). I use a slightly different shift for the recentering of the series for the regular-tetration base-function, but found it convenient in my tetration-analyses: x'= (x+1)*t and x"=x/t -1 so g(x,h) = f(x',h)" and conversely f(x,h)=g(x",h)' . The Bell-matrix for g(x) = f(x')" = (b^x')" = b^((x+1)*t)/t-1 is triangular and just the matrix of Stirlingnumbers 2nd kind rowwise rescaled by powers of u and reciprocals of factorials (its generating function g(z)=exp(u*z)-1 ). However, having repeated eigenvalues here the standard computation of diagonalization fails, so we cannot compute the schroeder-function which is needed for the regular superfunction. On the other hand, when the diagonal has all units then we have the alternative of the matrix-logarithm to define a formal (bivariate) powerseries for the superfunction in terms of x and the height-parameter h. But although the powers of u on the diagonal have absolute values of the unit, it is again not possible to do the matrix-logarithm. (If u were a complex unit-root of *irrational* order, then all integer powers of u, and thus eigenvalues, are different, especially different from 1, and the diagonalization can be used to build up the matrices/formal powerseries for regular iteration) So looking again at the formal expression for the superfunction in x and h (by diagonalization, and where the parameters u and h are still kept symbolic and not yet evaluated) we find, that in the denominators occur expressions (1-u^4), (1-u^8), (1-u^12) and so forth (which is well known). These become zero if evaluated and produce singularities. Here I remembered an analoguous case: if we remove singularities from the zeta-function zeta*(s)=zeta(s)-1/(s-1) we get an entire function, and the same is possible with the gamma-function as well; if we remove all the singularities by a similar scheme, we get the "incomplete gamma" which is again entire, and nicely converging (and has also apparently interesting and useful properties) So I looked what it would give, if I remove the singularites here as well, thus creating a somehow "incomplete" exponentiation/tetration g*(x) resp g*(x,h). I cancelled all occurences of (1-u^4) in the denominators and could now evaluate the formal powerseries for our selected parameter u=1*I. So it is possible to define a schroederfunction and also a superfunction, and we talk about g*(x,h) and f*(x,h)=g*(x",h)' in the following. Because the explicite schroeder-function is only needed intermediately I do not show the coefficients here. More concise is the matrix POLY of coefficients, which have to be premultiplied by the consecutive powers of the x-parameter of g*(x,h) (or let's say z0 instead of x because we use it as complex-valued initial value b^^0 for the iteration) and postmultiplied by consecutive powers of u^h . Code:Poly               0  u^(0h)*        u^(1h)*       u^(2h)*       u^(3h)*  u^(4h)*       u^(5h)*        u^(6h)*      u^(7h)* -------------------------------------------------------------------------------------------------------------------------   (z0/2)^0/0!)*        .              .             .             .        .             .              .            .   (z0/2)^1/1!)*        .              2             .             .        .             .              .            .   (z0/2)^2/2!)*        .         -2+2*I         2-2*I             .        .             .              .            .   (z0/2)^3/3!)*        .          2-6*I          12*I        -2-6*I        .             .              .            .   (z0/2)^4/4!)*        .           16*I      -28-44*I       48+24*I  -20+4*I             .              .            .   (z0/2)^5/5!)*        .        -8-40*I             .             .        .        8+40*I              .            .   (z0/2)^6/6!)*        .      720+480*I    -288-192*I             .        .    -720-480*I      288+192*I            .   (z0/2)^7/7!)*        .  -12880+1120*I  10080-2016*I  -2352+1344*I        .  12880-1120*I  -10080+2016*I  2352-1344*I      ...               ...     ...            ...           ...         ...        ...           ...           ... ------------------------------------------------------------------------------------------------------------------------- Here the singularities are removed in the way, that at the coefficient z0^{4k+1},z0^{4k+2},z0^{4k+3},z0^{4k+4} in the denominator the (1-u^4)^k was cancelled. Reconstruction of the original g(x,s) by this means to divide the rows of POLY by that multiplicities of zeros (but I'm not good in application and explanation of the L'Hospital-rule here, and under which circumstances I could make use of it for our case here) Now in our case it also comes out to be more convenient to evaluate the postmultiplication first, so I got the formal powerseries for g*(x,h) for the first few h: g*(x,0) = x, g*(x,1), g*(x,2) , g*(x,3), g*(x,4), g*(x,5)=g*(x,1) where it becomes periodic. Also only the first powers up to x^4 have non-null coefficients: Code:Integer heights ; periodic; only the coefficients up to z0^4 are not null          0  h=0  h=1     h=2   h=3  h=4  h=5   ...   ---------+----+-----+------+-----+----+-----+----   z0^0/0!*    .    .       .     .    .    .   ...   z0^1/1!*    1    I      -1    -I    1    I   ...   z0^2/2!*    .   -1     1-I     I    .   -1   z0^3/3!*    .   -I     3*I  -2*I    .   -I   z0^4/4!*    .    1  -6-5*I   6*I    .    1   z0^5/5!*    .    .       .     .    .    .     ...       ... ...     ...   ...  ...   ...  ...   ---------+----+-----+------+-----+----+-----+---- example: g*(z0,2) = - z0 + (1-I)/2*z0^2 - 3 I/6*z0^3 - (6+5I)/24*z0^4 The coefficients-matrix POLY allows fractional heights, however taking the complex u to fractional powers is not unambiguous, so the following is just for an impression: Code:Fractional heights (meaningful ?); h=0.5 g*(z0,0.5)=                  +0.70710678+0.70710678*I *z0                  -0.10355339 .            *z0^2                 -0.014297740 .            *z0^3               0.0046213626-0.0096763770*I *z0^4                 0.011785113-0.017677670*I *z0^5                 0.015699029+0.024328478*I *z0^6               -0.022230690-0.0026321999*I *z0^7                0.010378519-0.0080516725*I *z0^8             -0.0010434735-0.00042966558*I *z0^9               0.0024291464-0.0017456513*I *z0^10           ... The function g*(x,h) can also be converted from the expression using u^(k*h) here, which is Dirichlet-like in terms of h, into a taylor-series in terms of that parameter; we get then another version of the matrix POLY. I can send this coefficients later. So, what is this all good for? Hmm, really don't know, I'd say it reduces to a nice exercise. Perhaps it gives another impulse for the comparision with the whereabouts of the bases b which allow periods of irrational length... Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/24/2011, 12:25 PM sheldon , we dont know if the period is a brjuno number , which is why we are not certain if its a siegel disk. ( else i would have posted it myself ) i dont know much about brjuno numbers but it seems deciding if a algebraic number of degree 3 is a brjuno number is already partially unresolved , hence i doubt period 2.98... is easy. ( i also dont know much about herman rings , feel free to inform me ) although you are correct with probability 1 , we know how " tricky and weird " tetration can be. i would also like to point out that 1.7129i^z is periodic with 2pi i /ln(1.7129 i) = 3.57977339 + 1.22650561 i ( another mysterious number ) , such that 1.7129i^(z + 3.57977339 + 1.22650561 i) ends up in the same orbit as 1.7129i^z and this probably defines the boundary of the siegel disk ( because of the univalent riemann mapping ) and settles my question about large initial values like 1729 + 1729 i. so we have at least 2 types of periods : 2.98... and 3.57977339 + 1.22650561 i that dominate the bahaviour of the superfunction sexp_1.7129i(z) on the main branch. sexp_1.7129i(z) will have its fixpoints (of 1.7129i^z) at +/- oo i. i dare to consider ( not yet conjecture :p ) e§ = 1.7129i^z ln§ = ln(z) / ln(1.7129i) 2§ = e§ - 1/e§ for real x and some suitable real constant c sexp_1.7129i(x) = ln§ ln§ ln§ ... 2§^[x]( e§ e§ e§ ... © ) - with the correct branches - in analogue to my 2sinh method. ( plz dont copy the same arguments as in the sinh thread if irrelevant ) regards tommy1729 sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 06/24/2011, 04:36 PM (This post was last modified: 06/24/2011, 08:03 PM by sheldonison.) (06/24/2011, 12:25 PM)tommy1729 Wrote: sheldon , we dont know if the period is a brjuno number , which is why we are not certain if its a siegel disk. .... although you are correct with probability 1 , we know how " tricky and weird " tetration can be. i would also like to point out that 1.7129i^z is periodic with 2pi i /ln(1.7129 i) = 3.57977339 + 1.22650561 i ...Hmmm, we don't even know if the period is irrational either, although a random base on the Shell-Thron is going to be irrational, with probability 1. I wasn't able to follow the other part of your post. 3.57977339 + 1.22650561i would be a period of 1.7129^z, but I'm not sure how that effects the superfunction. The superfunction has a period of 2.9883, and 2.9883 is not a period of 1.7129^z. Presumably, the coefficients of the series for the superfunction can be calculated with some sort of formula from the series for B^(z-L). The Fourier series coefficients are equivalent to the Taylor Series coefficients of the superfunction function wrapped around the Siegel disc. The algorithm I used seems to works, but it isn't very elegant, and it only works a little bit inside the Siegel disc, where convergence is much better. Update, one Thesis paper I started to read, written by Edgar Arturo Saenz Maldonado on the Brjuno number seems to have the formulas. $\lambda=\exp(2\pi i \alpha)$ $f(z)=\lambda z + \sum_{n>=2}a_n z^n$. And .... the formal power series of h {the Seigel disc function} is given by $h(z)=\sum_{i>=1}h_i z^n$ If h is the solution of the functional equation ... $f(h(z))=h(\lambda z)$, the coefficients of the series must satisfy (formally) the following recursive relation: $h_n$=1, for n=1, and for n>=2, $h_n = \frac{1}{\lambda^n-\lambda}\sum_{n=2}^{n}a_m \sum_{n1+...+n_m=n} h_{n1}h_{n2}...h_{n_m}$ where in the second summation, $n_i>=1$ "... By the formulas in question, it is possible to determine the coefficients of the formal power series of $h_f$; the denominators of these coefficients can be written as products of the form $\lambda^n-\lambda$, for n>=2, since $\alpha$ is an irrational number these products could be very small....", which is where the Brjuno number comes from. So this would be a closed form equation for the superfunction for bases on the Shell-Thron boundary, where $\lambda=\exp(2\pi i \alpha)$, and $\alpha$ is an irrational Brjuno number. - Sheldon Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 06/24/2011, 09:44 PM (06/24/2011, 04:36 PM)sheldonison Wrote: [ "... By the formulas in question, it is possible to determine the coefficients of the formal power series of $h_f$; the denominators of these coefficients can be written as products of the form $\lambda^n-\lambda$, for n>=2, since $\alpha$ is an irrational number these products could be very small....", which is where the Brjuno number comes from.Hi Sheldon - in the thesis-article on page 2 and a bit more explicite on page 6 there is that formula for the rational approximation/irrationality measure of irrational numbers. Here the parameter mu is used and if I understand the argument correctly, then it says, if it is greater than 2 then there is some solution. Since all algebraic irrationals (roots of polynomials) have exactly the degree mu=2 the numbers mu>2 are all transcendental. That's also the background, on which Liouville-numbers may be of special interest: they have mu=infinite and are near the rational numbers again (just from "the other side of the globe", so-to-speak). Unfortunately I am far too unfamiliar with all the higher concepts involved (even of the fourier-decomposition). So I cannot say much more than that the coefficients of h seem simply to satisfy the formula for the eigenmatrix/schroeder-function computed by diagonalization of a triangular Bell-matrix (for functions satisfying f(0)=0, and f'(0)<>0,1 ) - but that's only assumed by the appearance of the formula, I didn't look at it really deep. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post [repost] A nowhere analytic infinite sum for tetration. tommy1729 0 1,298 03/20/2018, 12:16 AM Last Post: tommy1729 [MO] Is there a tetration for infinite cardinalities? (Question in MO) Gottfried 10 12,719 12/28/2014, 10:22 PM Last Post: MphLee Remark on Gottfried's "problem with an infinite product" power tower variation tommy1729 4 5,419 05/06/2014, 09:47 PM Last Post: tommy1729 Problem with infinite product of a function: exp(x) = x * f(x)*f(f(x))*... Gottfried 5 7,242 07/17/2013, 09:46 AM Last Post: Gottfried Wonderful new form of infinite series; easy solve tetration JmsNxn 1 4,534 09/06/2012, 02:01 AM Last Post: JmsNxn The imaginary tetration unit? ssroot of -1 JmsNxn 2 5,624 07/15/2011, 05:12 PM Last Post: JmsNxn the infinite operator, is there any research into this? JmsNxn 2 5,680 07/15/2011, 02:23 AM Last Post: JmsNxn Tetration and imaginary numbers. robo37 2 5,606 07/13/2011, 03:25 PM Last Post: robo37 Infinite Pentation (and x-srt-x) andydude 20 27,324 05/31/2011, 10:29 PM Last Post: bo198214 Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Gottfried 91 97,684 03/03/2011, 03:16 PM Last Post: Gottfried

Users browsing this thread: 1 Guest(s)