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Suppose we have a function $f:A\to\mathbb{R}$ where $A\subseteq[a,b]$, $a,b\in\mathbb{R}$ and $S\subseteq A$.

Now suppose $f$ is

$$f(x)=\begin{cases} x^2+1 & x\in S_1\\ x & x\in S_2\\ \end{cases}$$

Where $S_1$ and $S_2$ are pairwise disjoint subsets of $A=S_1\cup S_2$ where $S_1=\left\{\frac{1}{c}+\frac{\sqrt{2}}{2}:c\in\mathbb{Z}\right\}\cap[0,1]$, $S_2=\left\{\frac{1}{2^{(d_1)}}+\frac{1}{2^{(d_2)}}:d_1,d_2\in\mathbb{Z}\right\}\cap[0,1]$ and $[a,b]=[0,1]$ where $a=0$ and $b=1$.

Suppose we divide $[0,1]$ into $r$ sub-intervals $\left\{[x_{i-1},x_i]\right\}_{i=1}^{r}$, where $a=x_0\le x_1 \le ...\le x_i \le ...\le x_{r-1}\le x_r=b$; and, if $f$ is defined in that sub-interval we take a point from that sub-interval; (however, if $f$ is not defined, we set $f=0$) and multiply it by measures $\mu(S_1\cap[x_{i-1},x_i],A)$, for $f(S_1)$, and $\mu(S_2\cap[x_{i-1},x_i],A)$, for $f(S_2)$,

We define $\mu(S_1,A)$ and $\mu(S_2,A)$ as the following:

In each sub-interval $[x_{i-1},x_i]$, we cover $S_1\cap [x_{i-1},x_i]$ by $m_1$ sub-intervals, of each of the sub-intervals in $\left(I_{i}\right)_{i=1}^{r}=\left\{[x_{i-1},x_i]\right\}_{i=1}^{r}$. We define this as $\left(I_{i_1,k}\right)_{k=1}^{m_1}$, where $1 \le m_1 \le \max\left\{|S_1|,1\right\}$. Then cover $S_2\cap[x_{i-1},x_i]$ by $m_2$ sub-intervals of each of the sub-intervals in $\left(I_{i}\right)_{i=1}^{r}=\left\{[x_{i-1},x_i]\right\}_{i=1}^{r}$, which we define as $\left(I_{i_2,k}\right)_{k=1}^{m_2}$, where $1 \le m_2 \le \max\left\{|S_2|,A\right\}$. Last, cover $A=S_1\cup S_2$ by $n$ sub-intervals of $J=[a,b]$, which we define as $\left(J_{k}\right)_{k=1}^{n}$ where $1 \le n \le \max\left\{|A|,1\right\}$. All these sub-intervals have the same length which we define as $c\in\mathbb{R}^{+}$.

We wish to minimize (or find the infimum) of the total sum of the lengths of the $m_1$ sub-intervals covering $S_1\cap[x_{i-1},x_i]$, where the infimum depends on $c$. This means we must set $c$ to specific values before taking the infimum. This would give us $m_1$ divided by $n$. The same is applied with $m_2$ sub-intervals covering $S_2\cap [x_{i-1},x_i]$ and the $n$ sub-intervals covering $A\cap [x_{i-1},x_i]$. This would give us $m_2$ divided by $n$.

Next we want want to find the infimum of $m_1$ divided by $n$ over $c$, (both $m_1$ and $n$ depends on $c$). We want the same with the infimum of $m_2$ divided by $n$ over $c$. This gives us $\mu(S_1\cap[x_{i-1},x_i],A)$ and $\mu(S_2\cap[x_{i-1},x_i],A)$.

To get our average one must take

$$\sum_{i=1}^{r}(x_i^2+1)\mu(S_1\cap[x_{i-1},x_i],A)+\sum_{i=1}^{r}(x_i)\mu(S_2\cap[x_{i-1},x_i],A)$$

Questions

  1. How would we do this on Mathematica?
  2. How would we generalize this to any piece-wise function defined on a countable set?

My Attempt

Subscript[S, 1][c_Integer, r_] := 
 Select[Union@
   Flatten[Table[
     1/2^(Subscript[d, 1]) + 1/2^(Subscript[d, 2]), {Subscript[d, 1], 
      1, c}, {Subscript[d, 2], Subscript[d, 1], c}]], 
  Between[#, 
    r] &] 

This lists all elements in $S_1$ as c$\to\infty$ and groups them into $r$ sub-intervals of $[0,1]$

Subscript[S, 2][c_Integer, q_] := 
 Select[Union@
   Flatten[Table[
     1/Subscript[u, 1] + Sqrt[2]/2, {Subscript[u, 1], 1, c}]], 
  Between[#, 
    q] &] 

This lists all elements of Set $S_2$ as and groups them into $r$ sub-intervals of $[0,1]$

Subscript[f, 1][x_] := 
 x^2 

This is the function $f=f_1$ on $x\in S_1$

Subscript[f, 1, 1][c_Integer, r_] := 
 Subscript[f, 
  1] /@ (Subscript[S, 1][c, #] & /@ Partition[Subdivide[r], 2, 1]) 

Groups $S_1$ into $r$ sub-intervals

Subscript[f, 1, 1, 1][c_Integer, r_] := 
 Subscript[f, 1, 1][c, r] /. {} -> {0}

Converts null sets to zero

Subscript[f, 1, 1, 1][30, 100];

Subscript[f, 2][
  x_] := x 

This is the function $f=f_2$ on $x\in S_2$

Subscript[f, 2, 2][c_Integer, r_] := 
 Subscript[f, 
  2] /@ (Subscript[S, 2][c, #] & /@ 
    Partition[Subdivide[r], 2, 
     1]) (* Groups Subscript[S, 2] into r sub-intervals *)
Subscript[f, 2, 2, 2][c_Integer, r_] := 
 Subscript[f, 2, 2][c, r] /. {} -> {0} 

Converts null sets to zero

In[205]:= 
Subscript[I, 1][r_] := 
 Block[{m = 10^16}, 
   With[{k = Ceiling@Log2@N@m + 2}, 
    Total[#] + 1 - Last[#] &@
     Unitize@BinCounts[
       Total[Tuples[2^Range[-k, -1], 2], {2}], {Min[#1] &, Max[#1] &, 
        1/m}]]] /@ Partition[Subdivide[r], 2, 1] 

This is supposed to give $m_1$ from where the total sum of the length of all $m_1$ sub-intervals, of each of the $r$ sub-intervals, covering the intersection of $S_1$ and the $r$ sub-intervals are as small as possible.

In[206]:= Subscript[I, 1][10]

During evaluation of In[206]:= BinCounts::bins: The bin specification {Min[#1]&,Max[#1]&,1/10000000000000000} is not a list of 2 or 3 real values.

Out[206]= If[29155653097335068495 === $SessionID, 
Out[206], Message[
MessageName[Syntax, "noinfoker"]]; Missing["NotAvailable"]; Null]

Unfortunately, there is something wrong with this function.

Subscript[I, 2][r_] := 
 Block[{m = 10^16}, 
   With[{k = Ceiling@N@m + 2}, 
    Total[#] + 1 - Last[#] &@
     Unitize@BinCounts[
       Total[Tuples[Range[-k, -1], 1], {2}], {Min[#1] &, Max[#1] &, 
        1/m}]]] /@ 
  Partition[Subdivide[r], 2, 
   1] 

This is supposed to give $m_2$ from where the total sum of the length of all $m_2$ sub-intervals, of each of the r sub-intervals, covering the intersection of $S_2$ and the $r$ sub-intervals are as small as possible

Subscript[I, 2][10]
J[r_] := Sum[
  Subscript[I, 1][i] + Subscript[I, 2][i], {i, 1, 
   r}] 

This is supposed to give $n$ from where the total sum of the length of all $n$ sub-intervals, of each of the $r$ sub-intervals, covering the intersection of $A$ and the $r$ sub-intervals are as small as possible

Subscript[Avg, 1][r_] := 
 Sum[Subscript[f, 1, 1, 1][30, i][[i, 1]]*
   Subscript[I, 1][r][[i]]/J[r], {i, 1, 
   r}] (* Subscript[I, 1][r][[i]]/A[r] is mu(Subscript[S, 1],A) as m\
\[Rule] Infinity, the whole sum is supposed to give the average over \
Subscript[f, 1] *)
Subscript[Avg, 2][r_] := 
 Sum[Subscript[f, 2, 2, 2][30, i][[i, 1]]*
   Subscript[I, 2][r][[i]]/J[r], {i, 1, 
   r}] (* Subscript[I, 1][r][[i]]/A[r] is mu(Subscript[S, 2],A) as m\
\[Rule]Infinity, the whole sum is supposed to give the average over \
Subscript[f, 2] *)
Subscript[Avg, 1][10000] + 
 Subscript[Avg, 1][10000] (* This is the combined total sum *)

During evaluation of In[217]:= General::nomem: The current computation was aborted because there was insufficient memory available to complete the computation.

During evaluation of In[217]:= Throw::sysexc: Uncaught SystemException returned to top level. Can be caught with Catch[\[Ellipsis], _SystemException].

Out[218]= SystemException["MemoryAllocationFailure"]

During evaluation of In[217]:= BinCounts::bins: The bin specification {Min[#1]&,Max[#1]&,1/10000000000000000} is not a list of 2 or 3 real values.

During evaluation of In[217]:= BinCounts::bins: The bin specification {Min[#1]&,Max[#1]&,1/10000000000000000} is not a list of 2 or 3 real values.

During evaluation of In[217]:= General::nomem: The current computation was aborted because there was insufficient memory available to complete the computation.

During evaluation of In[217]:= Throw::sysexc: Uncaught SystemException returned to top level. Can be caught with Catch[\[Ellipsis], _SystemException].

Out[222]= SystemException["MemoryAllocationFailure"]

Unfortunately, as you see here I am not getting any results.

How do we fix the errors in my code? Is my attempt correct?

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I can't understand what your code does but I think this might fix your BinCounts problem:

  1. Don't use the letter I - it's reserved. I replaced it with ii
  2. Use a Module because m isn't accessible in that inner With scope.
  3. Move the & just outside the BinCounts and before /@
Subscript[ii, 1][r_] := 
 Module[{m = 10^16}, 
   With[{k = Ceiling@Log2@N@m + 2}, 
    Total[#] + 1 - Last[#] &@
      Unitize[BinCounts[
        Total[Tuples[2^Range[-k, -1], 2], {2}], {Min[#1], Max[#1], 1/m}]] & /@ 
          Partition[Subdivide[r], 2, 1]]]

Subscript[ii,1][10]
(* returns: {1282, 52, 51, 3, 1, 52, 2, 2, 1, 1} *)

You should consider removing as many Subscript's from your code as this can quickly become hell, and introduce subtle bugs. It's also harder to copy-paste readable code on this site if it has subscripts, Greek, or special formatting for sums, integrals, overbars etc.

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10
  • $\begingroup$ On the last edit I fixed ]] too, make sure you pick up the latest change if you use this. $\endgroup$ – flinty Jul 9 '20 at 23:02
  • $\begingroup$ How do we fix this code Subscript[ii, 2][r_] := Module[{m = 10^16}, With[{k = Ceiling@N@m + 1}, Total[#] + 1 - Last[#] &@ Unitize[BinCounts[Total[1/k], {2}], {Min[#1], Max[#1], 1/m}]] & /@Partition[Subdivide[r], 2, 1]]. Here I'm splitting $[x_{i-1},x_{i}]$, a sub-interval of $[a,b]=[0,1]$, into $m_2$ sub-intervals and counting the number of ${m_2}^{\prime}$ sub-intervals, out of $m$ sub-intervals, that cover $S_2\cap[x_{i-1},x_i]$. $\endgroup$ – RajanArak Jul 9 '20 at 23:56
  • $\begingroup$ You're using BinCounts on a number, not a list BinCounts[Total[1/k], {2}] $\endgroup$ – flinty Jul 10 '20 at 16:42
  • $\begingroup$ @flinity I tried Subscript[ii, 2][r_] := Module[{m = 10^16}, With[{k = Ceiling@N@m + 1}, Total[#] + 1 - Last[#] &@ Unitize[BinCounts[Total[1/Range[1,k]], {2}], {Min[#1], Max[#1], 1/m}]] & /@Partition[Subdivide[r], 2, 1]] and I still don't get a proper answer. $\endgroup$ – RajanArak Jul 10 '20 at 16:44
  • $\begingroup$ Total[1/Range[1, k]] is always a number too. $\endgroup$ – flinty Jul 10 '20 at 16:46

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