User per alexandersson - MathOverflow

Answer by Per Alexandersson for What mathematical field work with this kind of concept?

2013-05-17T09:02:22Z

It also sound a bit like neural networks; these are "trained" (adapted) to yield certain output for certain input.

Maybe you will also be interested in Markov processes, (which can also be trained to give certain output for some given input).

Both these concepts are related to something called genetic programming, where a large parameter space is sampled, evaluated, and mutated, so that the next iteration performs "better" than the previous.

Answer by Per Alexandersson for polynomial zero within a square

2013-05-12T17:12:33Z

Since $f$ is non-zero in a neighbourhood of the unit square, it follows that $1/f$ is holomorhic in the unit square. By the maximum modulus principle, $|1/f|$ attains its maximum on the boundary. If follows that the minimum of $|f|$ is attained at the boundary, of the unit square. This sort of hints that it should be very hard to find such a polynomial.

Answer by Per Alexandersson for Useful tricks in experimental mathematics

2013-05-05T21:26:41Z

The tricks I regularly use:

Create more examples. Always.
As a corollary of the above, time is well spent on making algorithms that presents examples nicely.
The Online Encyclopedia of Integers (OEIS), is your friend.
Or, if that does not work, put your sequence or constant into WolframAlpha.
If the numerical data looks strange, redo! Some software do not warn when the precision is lost. Some software (Mahtmematica for example), do not consider $1/2$ and $0.5$ to be equal.
Take time to learn your software! You are more tempted to try new stuff, if it is easy to code.

Elementary proof of K-saturation conjecture

2013-05-05T16:10:04Z

One variant of Fulton's K-saturation conjecture is as follows:

$K_{\lambda/\mu,w} > 0 \Leftrightarrow K_{n\lambda/n\mu,n w} > 0$ for any integer $n>0.$

Here $K_{\lambda/\mu,w}$ denotes the Kostka numbers (number of skew SSYT of shape $\lambda/\mu$ and weight $w.$

This has been proved in various ways, (Knutson, Tao), so it is no longe a conjecture, but to me the proofs are quite involved. The proof shows the similar statement for Littlewood-Richardson coefficients, using K-hives and the above follows as a corollary.

Question: Is there an elementary proof of the above statement? Could one expect a short proof of this?

Answer by Per Alexandersson for zeta(3) in terms of derivatives of zeta at 1/2 and pi

2013-05-05T14:15:09Z

~~This is just partial numerics, but the following Mathematica code strengthens the conjecture, with 500 decimal places:~~

Here is code for those that want to perform numerics in Mathematica:

prec = 500; (* Precision of calculations. *)
lhs = N[Zeta[3], prec]
rhs = N[(-Zeta'''[1/2]/Abs[Zeta[1/2]] -
 3 Zeta''[1/2] Zeta'[1/2]/Abs[Zeta[1/2]]^2 -
 2 Zeta'[1/2]^3/Abs[Zeta[1/2]]^3 - Pi^3/4)/7, prec]
lhs == rhs (* Gives true *)

Answer by Per Alexandersson for Probability of random (0,1) Toeplitz matrix being invertible

2013-04-30T08:32:08Z

This is just a partial answer:

Let $A$ be a $N \times N$ Toeplitz matrix, and consider the sequence of matrices $A_n$ for $n>N$ where we increase the size of $A$ to $n\times n$ and the new diagonals are filled with 0. Then the sequence of determinants $|A_n|$ will satisfy a linear recurrence. The zeros in a linear recurrence appears either in arithmetic progressions, or are sporadic. There is an upper bound on the number of sporadic zeros in linear recurrences, and these are "quite rare" in some sense. To have an infinite number of zeros in this sequence of determinants, we require that the symbol of the matrix have roots of unity, (this is sort of the same as the characteristic equation for the linear recurrence). Now, since the matrix is binary, the coefficients of the symbol (a polynomial) are also either 0 or 1.

So, in some sense, the question is related to the probability that a polynomial with coefficients either 0 or 1 has a root of unity.

Answer by Per Alexandersson for L-systems and Sierpinski Triangle

2013-04-27T17:12:47Z

Once you get the grips around constructing the Hilbert curve: https://en.wikipedia.org/wiki/Hilbert_curve then this is not really all that hard to construct.

Any fractal that consists of $n$ identical, smaller copies of itself, and where the pieces are connected "start to finish", could probably be generated by a suitable l-system.

As an exercise: * The Koch kurve consists of 4 smaller copies of itself, connected in the end-points. * n-flakes: https://en.wikipedia.org/wiki/N-flake#Hexaflake etc...

Sequence of semi-standard Young tableaux, counting

2012-08-31T14:37:31Z

Let $\lambda$ be a partition and define $T_\lambda^n(k)$ to be the number of semi-standard young tableaux of shape $(k\lambda_1,k\lambda_2,\dots,k \lambda_n).$ Now, one can prove that $T_\lambda^n(k)$ is a polynomial in $k.$

Is this well-known? Is there a known formula for $T_\lambda^n(k)$? Would it be interesting to find a general formula for this polynomial?

(I know about the hook-length formula, but since it is a product with $|\lambda|$ factors, this is not really useful for studying asymptotics of $T_\lambda^n(k)$.)

(The asymptotics for the corresponding Schur polynomials are also quite interesting. The roots of $S_{k \lambda}(t,1,1,\dots,1) = 0$ will for example accumulate close to the unit circle as $k$ grows.)

EDIT: It seems that finding Kostka numbers, $K_{k\lambda, k\beta}$ is a refinement of the above problem, and it is known that these grows polynomially. An article from 2007, Degrees of Stretched Kostka coefficients, by B. McAllister finds the degree of these, but the exact formulas for the polynomials seems to be unknown.

Answer by Per Alexandersson for Learning stochastic calculus, want to know what the notation of this function means

2013-03-13T21:20:45Z

I would strongly suspect $1_{[a_i,b_i)}$ is the function which is one on the interval $[a_i,b_i)$ and zero elsewhere.

Answer by Per Alexandersson for All possible linear combinations of positive half-integers with coefficients +/- 1

2013-03-05T22:07:17Z

Hm, maybe rewriting this by adding $p_1+p_2+\dots$ on both sides, and dividing by two, then this is isomorphic to $P' = f_1 p_1 + \dotsc$ where $f_i$ is either 0 or 1. Thus, it is the number of ways to construct $P'$ as a sum using 0 or one $p_1.$ Thus, expand $\prod_i (1+x^{p_i})$ and look for the coefficient of $x^{P'}$ to find the number you are looking for.

I don't know if this helps you, but maybe this interpretation helps a bit. What do you know of the numbers $p_i$?

Answer by Per Alexandersson for Eigenvalues of principle minors Vs. eigenvalues of the matrix

2013-03-05T10:48:41Z

No, it is not true; Consider the matrix

$$A= \begin{pmatrix} \frac{9}{2} & \frac{9}{20} & \frac{21}{20} & -\frac{3}{2}\\ -\frac{79}{11} & -\frac{3}{110} & \frac{23}{110} & \frac{31}{11}\\ \frac{6}{11} & \frac{21}{55} & \frac{114}{55} & \frac{6}{11}\\ \frac{16}{11} & \frac{12}{55} & \frac{128}{55} & \frac{5}{11} \end{pmatrix} $$ which has eigenvalues 0,1,2,3 (so it is positive semi-definite, but not definite.) The four principal minors are $$\frac{27}{22},\frac{189}{110},\frac{153}{55},\frac{36}{11}$$ sorted in increasing order. This should give a definite negative answer to your question.

Homogenous polynomials as sum or differences of squares and symmetric polynomials

2013-02-26T12:33:55Z

I seem to recall that a general homogenous real polynomial $P$ of even degree in $n$ variables, $n\geq 3,$ cannot always be expressed as $P(x_1,\dotsc,x_n)=\sum_j a_j Q_j^2(x_1,\dotsc,x_n)$ where $a_j \in \mathbb{R},$ and the $Q_j^2$ are homogenous of the same degree as $P.$ (Please, correct me if I am wrong).

Now, what if we know that $P$ is symmetric? Is it still true that a general polynomial cannot be expressed as above? And if it can, what if we require that the $Q_j$ themselves are symmetric?

Question: If $P$ is symmetric and homogenous of even degree, can it be expressed as a sum/difference of squares of homogenous and symmetric polynomials?

I know that if $P$ is symmetric, and $P(x_1+t,x_2+t,\dotsc,x_n+t)=P(x_1,\dotsc,x_n),$ then there is always such a representation as the one above, but can one lose the translation-invariant condition and still have sum-and-difference of squares representation?

Answer by Per Alexandersson for Homogenous polynomials as sum or differences of squares and symmetric polynomials

2013-02-27T20:39:21Z

My college provided me with a simple example: $$x^2 + xy + y^2$$ cannot be expressed as a sum/difference of symmetric, homogenous polynomials of degree 1, for obvious reasons.

(Each homogenous, symmetric polynomial of degree 1 in two variables are of the form a(x+y). Thus, all possible $Q_j$ are of the form $a'(x+y)^2$ which, of course, cannot give the polynomial above.)

Answer by Per Alexandersson for The right conformal map to make a certain picture

2013-02-26T22:03:35Z

Well, if $z \mapsto z^2$ gives circles, then $(x,y) \mapsto (x^2-y^2,xy)$ should give you ellipses with the horizontal axis twice as large as the vertical. Note, this map is not conformal.

It therefore "feels" like you cannot find such a map that you are looking for, ellipses are not really conformal objects in some sense, maybe someone else can give a rigorous proof on why there cannot be such a map?

Periodicity of a specific non-linear ODE of second order

2013-02-08T15:40:00Z

Consider the second-order ODE: $$\ddot{x} + x+x^2=0,$$ here $\ddot{x}$ is the second derivative w.r.t. $t$. Take initial values $x(0)=0.5$ and $\dot{x}(0)=0.$

Question: is the solution periodic or not?

Comment: Numerical experiments seems to show that the solution is periodic when $x(0)<0.5$ and if $x(0)>0.5$ then the solution fails to be periodic, in fact, $x(t)\rightarrow -\infty.$

(Asked by Prof. J.E. Björk, Stockholm Univ.)

Answer by Per Alexandersson for To what extent should a pure mathematician care about the meaning of things?

2013-02-07T21:15:18Z

To me, it is a bit like asking the meaning of music; it does not have to be "applied", but may exist and be beautiful in it self.

Now, beautiful mathematics is usually simpler to understand, than ugly mathematics, so sometimes, it might be necessary to do beautiful, non-meaningful things, to simplify mathematics which has a meaning. This was sort of the story when the complex numbers were invented; these were strange, and did not have a meaning, but simplified the mathematics. Now, we cannot live without them.

Answer by Per Alexandersson for Elementary question: distinct elements in a set

2013-02-03T17:17:05Z

"Let $x,y,z,$ be pairwise distinct", is perfectly fine.

Discriminant on boundary of semi-algebraic surface

2011-11-22T13:32:17Z

Let $P(t)$ be a polynomial in $t$ of degree $n$, with some contiguous coefficients (not the first or last) being $x_1,\dots,x_k$ and the rest of the coefficients are fixed.

(E.g. $p(t)=1+2t+x_1t^2+x_2t^3+(5i+7)t^6$ is ok).

Consider the set $S$ defined as $(x_1,\dots,x_k) \in \mathbb{C}^k$ with the property that the roots $t_1,\dots,t_{n}$ of $P(t)=0$ may be ordered increasingly w.r.t modulus, and such that $$|t_j|=|t_{j+1}| = \dots = |t_{k+j}|,$$ for some fixed $j.$

This set is real $k$-dimensional.

Is it true, (in general), that the discriminant $D(x_1,\dots,x_n) = Discr_t(P)$ satisfy $D(x_1,\dots,x_n)=0$ on the $k-1$-dimensional boundary of $S$?

Answer by Per Alexandersson for Picture of a 3 dimensional amoeba.

2013-01-30T12:30:37Z

There is now a 3-dimensional amoeba on wikipedia.

You are welcome. The main reason for the difficulty of making such amoebas is that it is the projection of a 4-dimensional surface (the zero set of the polynomial). Finding a parametrization of the zeros set of an arbitrary polynomial in 2 variables is highly non-trivial. The case on wikipedia is linear, thus we may easily parametrize it. Now, this set is projected with $Log|\cdot|$, thus one has to be a bit careful which points to include to make the picture "pretty", since there may be zeros of the polynomial "far" away, and close to 0.

I did not draw the projection of the parametrized 4-dimensional surface, but rather the projection of a lot of points on the surface, chosen in a manner that makes them somewhat evenly distributed.

EDIT: Also, rumour has it that Frank Sottile has some 3-dimensional amoebas somewhere on his web page, but I have not been able to find them.

EDIT 2: Now, amoebas never look like soap bubbles, but the connected components of their complements do. (These components are always convex, aka. soap bubbles)

Answer by Per Alexandersson for Asymptotic Behavior of Non-Analytic Function of the Eigenvalues

2013-01-06T08:52:17Z

To create a partial answer, eigenvalues of banded Toeplitz matrices accumulate on some real algebraic curves in $\mathbb{C}$, (this has been proved by Schmidt and Spitzer around 1970.). Now, if we also attach a point-mass at each eigenvalue, (for fixed matrix size) with equal mass at each point, and with total mass one, we get a probability measure. These measures converge in a certain sense to some limit measure, see for example the book by Bender and Böttcher .

Now, the sum $c_n = \frac{1}{n} \sum_{k=1}^n |\lambda_{n,k}|$ can then be interpreted as the ~~center of mass~~ mean distance to 0 of all the eigenvalues from matrix of size $n \times n.$ Say that the associated point measure is called $\mu_n$ (mass $1/n$ at each eigenvalue), then we know that $\mu_n \to \mu$ for some $\mu$ in some sense.

Note that $c_n = \int_{\mathbb{C}} |z| d\mu_n(z)$ and then what you are looking for is $c = \int_{\mathbb{C}} |z| d\mu(z)$.

As an explicit example, taking your matrix to be tridiagonal, will give rise to characteristic polynomials which are also a family of orthogonal polynomials, w.r.t some measure (the limit of point measures, actually), see e.g. this paper i just googled.

Answer by Per Alexandersson for sequences with a fractal dimension

2012-12-22T18:46:48Z

Taking a truncated integer sequence is essentially the same as defining the "heights" of the line endpoints of a 2-dimensional curve, similar to http://www.gameprogrammer.com/fractal.html my guess is that this is not really new, but rather a cumbersome way to define self-similar curves.

Making a reasonable scaling will in some examples, as above, give a well-defined limit curve. Maybe it is possible to compute the Hausdorff dimension by doing a discrete Fourier transform of the sequence. The spectrum should be quite interesting, I think.

Littlewood-Richardson Coefficients from Kostka coefficients

2012-12-12T13:46:13Z

I read on page 4 here that the Kostka coefficients $K_{\lambda,\mu}$ are specializations of the Littlewood-Richardson coefficients $c^\tau_{\sigma,\lambda}$ by specializing $\sigma,\tau$ depending on $\mu$ in a simple manner (certain sums of parts of $\mu$).

I have two questions: Is there a similar specialization/translation for Kostka coefficients obtained from skew shapes, $K_{\lambda,\mu}^\nu$ where $\lambda/\nu$ is a skew shape?

Secondly: Is there a way to "go back", i.e. can I express $c^\tau_{\sigma,\lambda}$ as some linear combination of some skew Kostka coefficients $K_{\lambda,\mu}^\nu$?

I would like to see if polynomiality of the map $n \mapsto K_{n \lambda, nw}^{n \nu}$ implies polynomiality for a similar map with LW-coefficients.

Monic polynomial with integer coefficients with roots on unit circle, not roots root of unity?

2012-11-28T09:07:31Z

There are certainly non-monic polynomials of degree 4 with all roots on the unit circle, but no roots are roots of unity; $5 - 6 x^2 + 5 x^4$ for example.

Now, for a monic polynomial of degree $n$, this is impossible (I think).

So, my question is, given a monic polynomial with integer coefficients of degree $n$, what is the maximal number of roots that can lie on the unit circle, and not be roots of unity?

For example, $1 + 3 x + 3 x^2 + 3 x^3 + x^4$ has two roots on the unit circle, and two real roots.

Answer by Per Alexandersson for A question about the Axiom of Choice and straight lines in the Euclidean plane.

2012-11-23T06:54:12Z

How about tangents to a deltoid, http://mathworld.wolfram.com/Deltoid.html ?

It do look like the tangents really do cover the entire plane, and all of those are non-parallel.

http://imageshack.us/photo/my-images/42/deltoidtangents.png

Answer by Per Alexandersson for How many matrices are possible for the given arrangement?

2012-11-11T17:48:42Z

Let $a_n$ be the number of $2 \times n$ -matrices avoiding constant 2*2-submatrices. Then

$$a_n = \frac{2^{-n} \left(4 \left(17+4 \sqrt{17}\right) \left(3+\sqrt{17}\right)^n+\left(\sqrt{17}-17\right) \left(\sqrt{17}-3\right)^n e^{i \pi n}\right)}{17 \left(3+\sqrt{17}\right)}$$

This should be fairly straightforward to prove, let $v(n)=(e_{01}(n),e_{10}(n),e_{00}(n),e_{11}(n))$ be the vector of number of $2\times n$-matrices ending with column 01, 10, 00 resp. 11.

We then have the recursion $$v(n+1)=\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{pmatrix} v(n)$$

Since this is symmetric, we may diagonalize this and from here, it should be straightforward to find the formula above. (I cheated a bit in Mathematica).

EDIT: Of course, $e_{01}(n)=e_{10}(n)$ and $e_{00}(n)=e_{11}(n)$ by symmetry, so one can of course reduce the above to a 2 by 2 matrix recursion instead, with entries 2,2 and 2,1. Eigenvalues of this matrix are $1/2 (3 + \sqrt{17}), 1/2 (3 - \sqrt{17})$ which explains the strange formula above.

Answer by Per Alexandersson for Are there any non-planar graphs containing only K(3,3) as a subgraph that are not 4-colourable?

2012-10-27T22:24:51Z

Wikipedia:

As Wagner showed, every graph that has no K5 minor can be decomposed via clique-sums into pieces that are either planar or an 8-vertex Möbius ladder, and each of these pieces can be 4-colored independently of each other, so the 4-colorability of a K5-minor-free graph follows from the 4-colorability of each of the planar pieces.

From http://en.wikipedia.org/wiki/Hadwiger_conjecture_%28graph_theory%29

Answer by Per Alexandersson for Prove that a particular polynomial sequence is a Sturm sequence

2012-10-17T20:14:53Z

This is just a partial answer, but anyway: ( I use $s[n,k]$ for $s^n_k$ since I just copy-pasted from Mathematica:

I do not know if this helps, or if you already know, but you have the recurrence $$s[n- 1,k + 1] = s'[n, k] - x \cdot s'[n - 1, k]$$

Note, by induction, we assume $s[n,k]$ and $s[n-1,k]$ to have interlacing roots. By thm 1.47 in http://arxiv.org/abs/math/0612833 (look into this work), we know that the derivatives have interlacing roots as well.

Now, the recursion is quite similar to many other that appear in the link above, so it might not be too hard to prove stability.

EDIT: I show/sketch below that the operator $f \mapsto A f - x f'$ always produce roots interlaced (and to the right) of the roots of $f$, provided $A>\deg f$ and that all roots of $f$ are positive. (We may assume leading term of $f$ has positive coefficient).

Clearly if f has a root of multiplicity m, then the result will have the same root with multiplicity m-1, so the problem is essentially to ensure that no new multiple roots may appear.

Assume now we have two consecutive roots $0\le a \le b$ of $f$ and that $f$ is positive between these. The derivative of $f$ this first positive, and then negative in $[a,b]$, thus $-x f'$ is first negative and then positive. Hence, $A f - x f'$ is negative in $a$, and positive in $b.$ Thus, we have a root of $A f - x f'$ in the interval $[a,b]$. (Mutatis mutandis for the case when f is negative between $a$ and $b$).

Now, if the degree of $f$ is even, $f$ is positive to the right of its largest root. The derivative is also positive here (since even degree poly), so $x f'$ is positive. Hence, $A f - x f'$ is negative in this point. However, notice this is a polynomial with even degree, and with a leading coefficient! Hence, it must eventually cross the real line and grow to infinity. (Mutatis mutandis for the case when f is odd).

This proves that $f \mapsto A f - x f'$ produces interlacing roots, and eventual multiplicities are decreased, no new multiple roots can be introduced.

Answer by Per Alexandersson for Eigenvalues of infinite matrices

2012-10-14T11:32:59Z

I have done some research on banded Toeplitz matrices, (where we consider a sequence of finite matrices, and show where the eigenvalues accumulate). There is also quite old literature on this subject, see references in this paper I shamelessly advertise: http://arxiv.org/abs/1208.5607

Answer by Per Alexandersson for Collisions between rooks taking random flights on an $N$ by $M$ chessboard

2012-09-23T19:46:58Z

So, I did some numerical experiments on 4 rooks on a k times k board. Each data point is the mean of 500 runs.

The x axis is the width/height of the board, the y axis is the number of iterations needed for it to be only one rook left.

EDITED: So, I did some changes and now my data conforms with the others:

1) Choose rook. 2) Choose direction. 3) If direction does not allow moving in that direction, goto 2. 4) Move 1,2,.. or k steps in direction chosen, where k gives a boundary square.

I.e. this does not count non-moves (which the image ABOVE do).

The image below shows mean of 500 runs, k rooks on a k*k board, starting at k=1.

New formula for counting SSYTs?

2012-09-07T13:54:01Z

I have a proof that given a partition $\lambda=(\lambda_1,\dots,\lambda_l)$ then the number of semi-standard Young tableaux of shape $\lambda$ with entries in $1,2,\dots, n$ is given by

$$\frac{1}{1!2!\cdots (n-1)!} \prod_{1\leq i\lt j\leq n} (\lambda_i-i)-(\lambda_j-j).$$ (We define $\lambda_j:=0$ if $j \gt l.$)

The product is also recognized as a Vandermonde determinant.

There are plenty of product formulas (over boxes in the tableau) and determinant formulas (but not in Vandermonde form, as far as I can tell) for the number of such SSYTs, but I have not seen a this particular one in the literature or in any article I've come across.

Is this formula known? Is this formula of any interest?

Comment by Per Alexandersson

2013-05-15T16:00:02Z

Is this maybe some type of Eulerian numbers?

Comment by Per Alexandersson

2013-05-13T21:21:11Z

This is a really messy question, and I think you better ask over at math.stackexchange or programmers.stackexchange since this is not research and really more about programming.

Comment by Per Alexandersson

2013-05-13T21:19:32Z

There should be some generalization of the ham sandwich theorem that takes care of more than two people (I think). It is trivial to generalize to a power of two people, for example (by iterating ham sandwich thm). However, a related, and mostly unsolved problem, is to divide a polygon into n parts with equal area and perimeter (some special cases are known only).

Comment by Per Alexandersson

2013-05-12T17:14:29Z

haha, nice there!

Comment by Per Alexandersson

2013-05-06T08:12:28Z

Correct me if I'm wrong, but for the non-skew version, $K_{\lambda,w}>0$ iff $\lambda \geq_d w$ in dominance order. And I suppose that $\lambda \geq_d w \Leftrightarrow n\lambda \geq_d nw$ is easy to show...

Comment by Per Alexandersson

2013-05-05T15:01:22Z

Oh, read it as "to $10^{-4}$ precision". Silly me...

Comment by Per Alexandersson

2013-05-05T14:25:05Z

This is nonsense. Also, the product of two non-primes is not necessary composed, so I believe you example is faulty. Vote to close.

Comment by Per Alexandersson

2013-04-30T08:18:46Z

marshall: I think it is ok, I read the question too quickly.

Comment by Per Alexandersson

2013-04-30T06:46:13Z

You probably would like to be more specific about "random". There is no canonical way to choose random elements from R.

Comment by Per Alexandersson

2013-04-05T14:47:49Z

So what is the question?

Comment by Per Alexandersson

2013-03-08T07:44:53Z

Solve numerically, with 10-12 digits, and see Plouffes inverter, or search in wolframalpha. If it is expressible in known constants, you might get a hit. Knowing the exact value may now indicate how to PROVE it is the exact value you got.

Comment by Per Alexandersson

2013-03-05T21:56:35Z

This is for research level questions. Your question is not on research level. Ask over at math.stackexchange

Comment by Per Alexandersson

2013-03-05T17:26:21Z

Terry Tao: Right, that is a possibility, but I can see why the poster believed that the current version was true; I did some quick numerics (that is how I got the example), and the counterexamples are quite rare, (with appropriate definition on "rare").

Comment by Per Alexandersson

2013-03-05T10:56:24Z

It sounds a bit like a multi-fractal, (en.wikipedia.org/wiki/Multifractal_system) which has "mixed" fractal dimensions. If your data do not have sufficient resolution, it might be an artifact that it eventually becomes zero. Now, some DLA-systems (en.wikipedia.org/wiki/…) have something similar happening in them, if I recall correctly.

Comment by Per Alexandersson

2013-02-27T06:38:18Z

I'd say this is almost better to ask over at mathematica.stackexchange, my bet is that if you ask nicely, you will even get working code for your problem.