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Hausdorff and box dimension of Kiesswetter's function

variant on Sierpinski carpet: rescue the tablecloth!Hausdorff Measure and Hausdorff DimensionIs the Hausdorff dimension less than the box counting dimension?Relationship between the Hausdorff dimension and the Box-counting dimensionVariable Dimensionality ManifoldsSet with equal Hausdorff and topological dimension but larger box counting dimensionfractal dimension computingIn an IFS, given the open set condition $sum_{i = 1}^N s_i^n < 1$ holds.The relation between a fractal and its code space.Computing the lipschitz constant of an affine IFS

0

$begingroup$

Massopoust, Interpolation and Approximation with Splines and Fractals defines the Kiesswetter's fractal function from the following RB operator:

$
(Tf)(x) =
begin{cases}
-frac{f(4x)}{2} & x in [0,1/4] \
frac{-1+f(4x-1)}{2} & x in [1/4,1/2] \
frac{f(4x-2)}{2} & x in [1/2,3/4] \
frac{1+f(4x-3)}{2} & x in [3/4,1]
end{cases}
$

so that the Kiesswetter's fractal function is the unique fixed point of $T$.

Of course he also specifies the functions that make the corresponding IFS.

In my class we mentionned that the Hausdorff and the box dimension of this fractal is $3/2$? Why is this the case?

real-analysis functional-analysis measure-theory fractals

share|cite|improve this question

asked yesterday

JavierJavier

2,06621234

$endgroup$

add a comment | 

0

$begingroup$

Massopoust, Interpolation and Approximation with Splines and Fractals defines the Kiesswetter's fractal function from the following RB operator:

$
(Tf)(x) =
begin{cases}
-frac{f(4x)}{2} & x in [0,1/4] \
frac{-1+f(4x-1)}{2} & x in [1/4,1/2] \
frac{f(4x-2)}{2} & x in [1/2,3/4] \
frac{1+f(4x-3)}{2} & x in [3/4,1]
end{cases}
$

so that the Kiesswetter's fractal function is the unique fixed point of $T$.

Of course he also specifies the functions that make the corresponding IFS.

In my class we mentionned that the Hausdorff and the box dimension of this fractal is $3/2$? Why is this the case?

real-analysis functional-analysis measure-theory fractals

share|cite|improve this question

asked yesterday

JavierJavier

2,06621234

$endgroup$

add a comment | 

$begingroup$

Massopoust, Interpolation and Approximation with Splines and Fractals defines the Kiesswetter's fractal function from the following RB operator:

$
(Tf)(x) =
begin{cases}
-frac{f(4x)}{2} & x in [0,1/4] \
frac{-1+f(4x-1)}{2} & x in [1/4,1/2] \
frac{f(4x-2)}{2} & x in [1/2,3/4] \
frac{1+f(4x-3)}{2} & x in [3/4,1]
end{cases}
$

so that the Kiesswetter's fractal function is the unique fixed point of $T$.

Of course he also specifies the functions that make the corresponding IFS.

In my class we mentionned that the Hausdorff and the box dimension of this fractal is $3/2$? Why is this the case?

real-analysis functional-analysis measure-theory fractals

share|cite|improve this question

asked yesterday

JavierJavier

2,06621234

$endgroup$

share|cite|improve this question

asked yesterday

JavierJavier

2,06621234

share|cite|improve this question

asked yesterday

JavierJavier

2,06621234

asked yesterday

JavierJavier

2,06621234

asked yesterday

JavierJavier

2,06621234