0
$\begingroup$

I am creating a code that calculates or approximates

$$D(S\cap[a,b])=\lim_{n\to\infty}\frac{\left|S\cap{F_n\cap[a,b]}\right|}{\left|F_n\cap[a,b]\right|}$$

where $D$ is the density of $S\cap[a,b]$ (in $A\cap[a,b]$), $[a,b]$ is an interval for $a,b\in\mathbb{R}$, $F_n$ is the Folner Sequence of $A$, and $S\subseteq A$. For more information, click here (replace $G,X,i,g$ with $A,S,n,a$), here and here.

I attempted to approximate $D(S\cap[0,1])$ when $S=\left\{\frac{m^2}{n^2}:m,n\in\mathbb{Z},n\neq 0\right\}$, $A=\mathbb{Q}$ and $F_n=\left\{\frac{p}{2^k(2q+1)}:p,k,q\in\mathbb{Z},2^k \le n, |2q+1|\le n, \left|\frac{p}{2^k(2q+1)}\right|\le n\right\}$.

(In my code I replaced $A$ with A[x_,y_,z_,...], $F_n$ with F[x_,y_,z_,...], $S$ with S[x_,y_,z_,...], and $D$ with d. I also set $a=0$ and $b=1$.)

Clear[A, F, f, p, Ff, S, X, Y, d, j, k];
A[p_, k_, q_] := p/((2^k)*(2*q + 1));
F[p_, n_] := 
  Table[A[p, k, q], {k, 0, Floor[Log[2, n]]}, {q, 0, 
    Floor[(n - 1)/2]}];
f[n_, a_, b_] := 
  p /. Table[
    Solve[a <= A[p, k, q] <= b, p, Integers], {k, 0, 
     Floor[Log[2, n]]}, {q, 0, Floor[(n - 1)/2]}];
Ff[n_, a_, b_] := 
  DeleteDuplicates@
   Flatten@Table[
     F[f[n, a, b][[v]][[u]], n][[v]][[u]], {v, 1, 
      Floor[Log[2, n]] + 1}, {u, 1, Floor[(n - 1)/2] + 1}];
S[j_, k_] := j^2/k^2;
X[n_, a_, b_] := 
  Count[Resolve[
      Exists[{j, k}, 
       S[j, k] == # && {j, k} а \[Element] Integers]] & /@ 
    Ff[n, a, b], True];
Y[n_, a_, b_] := Length[Ff[n, a, b]];
d[n_, a_, b_] := N[(X[n, a, b])/Y[n, a, b]];

Ff[4, 1, 2]
X[4, 1, 2]
Y[4, 1, 2]
d[4, 1, 2]
Limit[d[n, 1, 2], n -> Infinity]

I get

Out[102]= {1, 2, 4/3, 5/3, 3/2, 7/6, 11/6, 5/4, 7/4, 13/12, 17/12, \
19/12, 23/12}

Out[103]= 0

Out[104]= 13

Out[105]= 0.

During evaluation of In[93]:= Table::iterb: Iterator {v,1,1+Floor[Log[n]/Log[2]]} does not have appropriate bounds.

During evaluation of In[93]:= Table::iterb: Iterator {v,1,1+Floor[Log[n]/Log[2]]} does not have appropriate bounds.

During evaluation of In[93]:= Table::iterb: Iterator {v,1,1+Floor[Log[n]/Log[2]]} does not have appropriate bounds.

During evaluation of In[93]:= General::stop: Further output of Table::iterb will be suppressed during this calculation.

During evaluation of In[93]:= Table::nliter: Non-list iterator (Resolve[\!\(\*SubscriptBox[\(\[Exists]\), \({j, k}\)]\((S[j, k] == #1 && {j, k}\ а \[Element] \*TemplateBox[{},
"Integers"])\)\)]&)[{v,1,1+Floor[Log[n]/Log[2]]}] at position 2 does not evaluate to a real numeric value.

Out[106]= 0.

Why is X[4,1,2] not equal to 1? Why is Limit[d[n,1,2],n->Infinity] showing an error? How do we correct both?

Improve this question
$\endgroup$
2
  • $\begingroup$ In RHS of definition of X you see that symbol a is not colored in green by Mathematica. This means this is not the same symbol as used in X[n_, a_, b_], just change it to right a. Probably you typed this in different keyboard layout. Then X[4,1,2] gives 1. $\endgroup$ – Alx Oct 2 '19 at 2:28
  • $\begingroup$ @Alx How do we correctly solve Limit[d[n,a,b],n->Infinity]? If I set S[j_,k_]:=j/k^3 I still get $0$ and error messages instead of $1$. $\endgroup$ – Arbuja Oct 2 '19 at 13:35