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Class of fractal curves derived from recursion on the base-2 representation of the integers
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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:
(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
asked Jul 22 at 17:13
zwol
266111
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up vote
2
down vote
favorite
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:
(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
asked Jul 22 at 17:13
zwol
266111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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:
(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
asked Jul 22 at 17:13
zwol
266111
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:
(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
asked Jul 22 at 17:13
zwol
266111
asked Jul 22 at 17:13
zwol
266111
asked Jul 22 at 17:13
zwol
266111
asked Jul 22 at 17:13
zwol
266111
266111
add a comment |Â
add a comment |Â
1 Answer
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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 $2^m+1-2$:
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:
Note that the limiting object is just the unit interval; the dimension is again 1.
answered Jul 23 at 16:13
Mark McClure
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
add a comment |Â
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 $2^m+1-2$:
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:
Note that the limiting object is just the unit interval; the dimension is again 1.
answered Jul 23 at 16:13
Mark McClure
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
add a comment |Â
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 $2^m+1-2$:
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:
Note that the limiting object is just the unit interval; the dimension is again 1.
answered Jul 23 at 16:13
Mark McClure
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
add a comment |Â
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 $2^m+1-2$:
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:
Note that the limiting object is just the unit interval; the dimension is again 1.
answered Jul 23 at 16:13
Mark McClure
22.5k34069
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 $2^m+1-2$:
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:
Note that the limiting object is just the unit interval; the dimension is again 1.
answered Jul 23 at 16:13
Mark McClure
22.5k34069
answered Jul 23 at 16:13
Mark McClure
22.5k34069
answered Jul 23 at 16:13
Mark McClure
22.5k34069
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
add a comment |Â
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
add a comment |Â
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