Class of fractal curves derived from recursion on the base-2 representation of the integers

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
1












Consider the recurrence




$displaystyle
T(n) = begincases
1 &textif n = 0 \
0 &textif n = 1 \
T(lfloor n/2rfloor) + T(n bmod 2) &textotherwise
endcases
$




If you plot the value of this recurrence at the nonnegative integers, you get a fractal curve:



enter image description here



(The red line is $lfloorlog_2 nrfloor$; if you set $T(1)=1$ instead of $T(1)=0$, the recurrence degenerates to that function.)



Does this fractal have a name? Is it a member of a more general class of curves?




recurrence-relations fractals




share|cite|improve this question























    up vote
    2
    down vote

    favorite
    1












    Consider the recurrence




    $displaystyle
    T(n) = begincases
    1 &textif n = 0 \
    0 &textif n = 1 \
    T(lfloor n/2rfloor) + T(n bmod 2) &textotherwise
    endcases
    $




    If you plot the value of this recurrence at the nonnegative integers, you get a fractal curve:



    enter image description here



    (The red line is $lfloorlog_2 nrfloor$; if you set $T(1)=1$ instead of $T(1)=0$, the recurrence degenerates to that function.)



    Does this fractal have a name? Is it a member of a more general class of curves?




    recurrence-relations fractals




    share|cite|improve this question





















      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      Consider the recurrence




      $displaystyle
      T(n) = begincases
      1 &textif n = 0 \
      0 &textif n = 1 \
      T(lfloor n/2rfloor) + T(n bmod 2) &textotherwise
      endcases
      $




      If you plot the value of this recurrence at the nonnegative integers, you get a fractal curve:



      enter image description here



      (The red line is $lfloorlog_2 nrfloor$; if you set $T(1)=1$ instead of $T(1)=0$, the recurrence degenerates to that function.)



      Does this fractal have a name? Is it a member of a more general class of curves?




      recurrence-relations fractals




      share|cite|improve this question











      Consider the recurrence




      $displaystyle
      T(n) = begincases
      1 &textif n = 0 \
      0 &textif n = 1 \
      T(lfloor n/2rfloor) + T(n bmod 2) &textotherwise
      endcases
      $




      If you plot the value of this recurrence at the nonnegative integers, you get a fractal curve:



      enter image description here



      (The red line is $lfloorlog_2 nrfloor$; if you set $T(1)=1$ instead of $T(1)=0$, the recurrence degenerates to that function.)



      Does this fractal have a name? Is it a member of a more general class of curves?





      recurrence-relations fractals





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 22 at 17:13









      zwol

      266111




      266111




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          2
          down vote













          I'm not sure that "fractal" is the correct term for this curve, in spite of the seeming self-similarity. If we plot your curve over the interval $[0,2^m]$ for several choices of $m$, we find that the lengths are $2^m+1-2$:



          enter image description here



          As the curves roughly double in length with each step, we'd expect a dimension of 1.



          In the context of fractal geometry, it might make more sense to scale these curves so that each lies over the unit interval. If we plot them in correct aspect ratio, we see a picture that looks like so:



          enter image description here



          Note that the limiting object is just the unit interval; the dimension is again 1.






          share|cite|improve this answer





















          • You're considering the entire domain of the recurrence, but isn't it more interesting if we consider intervals [2^n, 2^(n+1) - 1] for successive values of n as successive iterations of the fractal generator?
            – zwol
            Jul 24 at 13:12










          • Hmm... I. think I'm considering the limiting object when we grab pieces over the intervals $[0,2^n]$. I don't think changing the intervals to $[2^n,2^n+1-1]$ would make a huge difference. In any event, if we want to apply the ideas of fractal geometry, then it's natural to consider a limit of some sort.
            – Mark McClure
            Jul 24 at 13:48










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );








           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859580%2fclass-of-fractal-curves-derived-from-recursion-on-the-base-2-representation-of-t%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote













          I'm not sure that "fractal" is the correct term for this curve, in spite of the seeming self-similarity. If we plot your curve over the interval $[0,2^m]$ for several choices of $m$, we find that the lengths are $2^m+1-2$:



          enter image description here



          As the curves roughly double in length with each step, we'd expect a dimension of 1.



          In the context of fractal geometry, it might make more sense to scale these curves so that each lies over the unit interval. If we plot them in correct aspect ratio, we see a picture that looks like so:



          enter image description here



          Note that the limiting object is just the unit interval; the dimension is again 1.






          share|cite|improve this answer





















          • You're considering the entire domain of the recurrence, but isn't it more interesting if we consider intervals [2^n, 2^(n+1) - 1] for successive values of n as successive iterations of the fractal generator?
            – zwol
            Jul 24 at 13:12










          • Hmm... I. think I'm considering the limiting object when we grab pieces over the intervals $[0,2^n]$. I don't think changing the intervals to $[2^n,2^n+1-1]$ would make a huge difference. In any event, if we want to apply the ideas of fractal geometry, then it's natural to consider a limit of some sort.
            – Mark McClure
            Jul 24 at 13:48














          up vote
          2
          down vote













          I'm not sure that "fractal" is the correct term for this curve, in spite of the seeming self-similarity. If we plot your curve over the interval $[0,2^m]$ for several choices of $m$, we find that the lengths are $2^m+1-2$:



          enter image description here



          As the curves roughly double in length with each step, we'd expect a dimension of 1.



          In the context of fractal geometry, it might make more sense to scale these curves so that each lies over the unit interval. If we plot them in correct aspect ratio, we see a picture that looks like so:



          enter image description here



          Note that the limiting object is just the unit interval; the dimension is again 1.






          share|cite|improve this answer





















          • You're considering the entire domain of the recurrence, but isn't it more interesting if we consider intervals [2^n, 2^(n+1) - 1] for successive values of n as successive iterations of the fractal generator?
            – zwol
            Jul 24 at 13:12










          • Hmm... I. think I'm considering the limiting object when we grab pieces over the intervals $[0,2^n]$. I don't think changing the intervals to $[2^n,2^n+1-1]$ would make a huge difference. In any event, if we want to apply the ideas of fractal geometry, then it's natural to consider a limit of some sort.
            – Mark McClure
            Jul 24 at 13:48












          up vote
          2
          down vote










          up vote
          2
          down vote









          I'm not sure that "fractal" is the correct term for this curve, in spite of the seeming self-similarity. If we plot your curve over the interval $[0,2^m]$ for several choices of $m$, we find that the lengths are $2^m+1-2$:



          enter image description here



          As the curves roughly double in length with each step, we'd expect a dimension of 1.



          In the context of fractal geometry, it might make more sense to scale these curves so that each lies over the unit interval. If we plot them in correct aspect ratio, we see a picture that looks like so:



          enter image description here



          Note that the limiting object is just the unit interval; the dimension is again 1.






          share|cite|improve this answer













          I'm not sure that "fractal" is the correct term for this curve, in spite of the seeming self-similarity. If we plot your curve over the interval $[0,2^m]$ for several choices of $m$, we find that the lengths are $2^m+1-2$:



          enter image description here



          As the curves roughly double in length with each step, we'd expect a dimension of 1.



          In the context of fractal geometry, it might make more sense to scale these curves so that each lies over the unit interval. If we plot them in correct aspect ratio, we see a picture that looks like so:



          enter image description here



          Note that the limiting object is just the unit interval; the dimension is again 1.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 23 at 16:13









          Mark McClure

          22.5k34069




          22.5k34069











          • You're considering the entire domain of the recurrence, but isn't it more interesting if we consider intervals [2^n, 2^(n+1) - 1] for successive values of n as successive iterations of the fractal generator?
            – zwol
            Jul 24 at 13:12










          • Hmm... I. think I'm considering the limiting object when we grab pieces over the intervals $[0,2^n]$. I don't think changing the intervals to $[2^n,2^n+1-1]$ would make a huge difference. In any event, if we want to apply the ideas of fractal geometry, then it's natural to consider a limit of some sort.
            – Mark McClure
            Jul 24 at 13:48
















          • You're considering the entire domain of the recurrence, but isn't it more interesting if we consider intervals [2^n, 2^(n+1) - 1] for successive values of n as successive iterations of the fractal generator?
            – zwol
            Jul 24 at 13:12










          • Hmm... I. think I'm considering the limiting object when we grab pieces over the intervals $[0,2^n]$. I don't think changing the intervals to $[2^n,2^n+1-1]$ would make a huge difference. In any event, if we want to apply the ideas of fractal geometry, then it's natural to consider a limit of some sort.
            – Mark McClure
            Jul 24 at 13:48















          You're considering the entire domain of the recurrence, but isn't it more interesting if we consider intervals [2^n, 2^(n+1) - 1] for successive values of n as successive iterations of the fractal generator?
          – zwol
          Jul 24 at 13:12




          You're considering the entire domain of the recurrence, but isn't it more interesting if we consider intervals [2^n, 2^(n+1) - 1] for successive values of n as successive iterations of the fractal generator?
          – zwol
          Jul 24 at 13:12












          Hmm... I. think I'm considering the limiting object when we grab pieces over the intervals $[0,2^n]$. I don't think changing the intervals to $[2^n,2^n+1-1]$ would make a huge difference. In any event, if we want to apply the ideas of fractal geometry, then it's natural to consider a limit of some sort.
          – Mark McClure
          Jul 24 at 13:48




          Hmm... I. think I'm considering the limiting object when we grab pieces over the intervals $[0,2^n]$. I don't think changing the intervals to $[2^n,2^n+1-1]$ would make a huge difference. In any event, if we want to apply the ideas of fractal geometry, then it's natural to consider a limit of some sort.
          – Mark McClure
          Jul 24 at 13:48












           

          draft saved


          draft discarded


























           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859580%2fclass-of-fractal-curves-derived-from-recursion-on-the-base-2-representation-of-t%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What is the equation of a 3D cone with generalised tilt?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
          Read more

          Relationship between determinant of matrix and determinant of adjoint?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
          Read more

          Color the edges and diagonals of a regular polygon

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
          Read more