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Blancmange curve

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In mathematics, the blancmange curve is a fractal curve constructible by midpoint subdivision. It is also known as the Takagi curve, after Teiji Takagi who described it in 1903, or as the Takagi–Landsberg curve, a generalization of the curve. The name blancmange comes from its resemblance to a pudding of the same name. It is a special case of the more general de Rham curve.

The blancmange function is defined on the unit interval by

{\rm blanc}(x) = \sum_{n=0}^\infty {s(2^{n}x)\over 2^n},

where s(x) is defined by s(x)=\min_{n\in{\bold Z}}|x-n|, that is, s(x) is the distance from x to the nearest integer. The infinite sum defining blanc(x) converges absolutely for all x, but the resulting curve is a fractal. The blancmange function is continuous (indeed, uniformly continuous) but nowhere differentiable.

The Takagi–Landsberg curve is a slight generalization, given by

T_w(x) = \sum_{n=0}^\infty w^n s(2^{n}x)

for a parameter w; thus the blancmange curve is the case w = 1 / 2. The value H = − log2w is known as the Hurst parameter. For w = 1 / 4, one obtains the parabola: the construction of the parabola by midpoint subdivision was described by Archimedes.

The function can be extended to all of the real line: applying the definition given above shows that the function repeats on each unit interval.

Contents

[edit] Graphical construction

The blancmange curve can be visually built up out of sawtooth functions if the infinite sum is approximated by finite sums of the first few terms. In the illustration below, progressively finer sawtooth functions (shown in red) are added to the curve at each stage.

n = 0 n ≤ 1 n ≤ 2 n ≤ 3

[edit] Integrating Blancmange Curve

Given that the integral of blanc(x) from 0 to 1 is 1/2, the identity blanc(x) = blanc(2x) / 2 + s(x) allows the integral over any interval to be computed by the following relation. The computation is recursive with computing time on the order of log of the accuracy required.


\begin{align}
I(x) &= \int_0^x{\rm blanc}(x)\,dx,\\
I(x) &=\begin{cases}
1/2+I(x-1) & \text{ if }x \geq 1\\
1/2-I(1-x) & \text{ if }1/2 < x < 1 \\
I(2x)/4+x^2/2 & \text{ if } 0 \leq x \leq 1/2  \\
-I(-x) & \text{ if } x < 0
\end{cases} \\

\int_a^b{\rm blanc}(x)\,dx &= I(b) - I(a).
\end{align}

[edit] See also

[edit] References

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