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Cantor function: Difference between revisions

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== Alternative definition ==

Define a sequence of functions f_n on the interval that converge to the cantor function.

Let f_0(x) = x.

Then f_(n+1)(x) will be defined in terms of f_n(x).

Let f_(n+1)(x) = 0.5 * f_n(3x) when 0<=x<=1/3.

Let f_(n+1)(x) = 0.5 when 1/3<=x<=2/3.

Let f_(n+1)(x) = 0.5 + 0.5 * f_n(3 (x-2/3)) when 2/3<=x<=1.

Observe that f_n converges to the cantor fuction. Also notice that the choice of starting function does not really matter, provided f_0(0) = 0 and f_0(1)=1 and f_0 is bounded.

Revision as of 17:21, 19 September 2003

The Cantor function is a function c : [0,1] → [0,1] defined as follows:

Write x in base 3. If the digit 1 occurs in x, replace it with 2 and all following digits with 0. Divide all digits by 2. Interpret the resulting string of digits as a base 2 number. The result is c(x). (It may be much easier to understand this definition by looking at the graph below than by grasping the algorithm.)

This function is the most frequently cited example of a real function that is continuous but not absolutely continuous. It has no derivative at any member of the Cantor set; its derivative is 0 elsewhere. Extended to the left with value 0 and to the right with value 1, it is the cumulative probability distribution function of a random variable that is uniformly distributed on the Cantor set. This probability distribution has no discrete part, i.e., it does not concentrate positive probability at any point. It also has no part that can be represented by a density function; integrating any putative probability density function that is not almost everywhere zero over any interval will give positive probability to some interval to which this distribution assigns probability zero.

Alternative definition

Define a sequence of functions f_n on the interval that converge to the cantor function.

Let f_0(x) = x.

Then f_(n+1)(x) will be defined in terms of f_n(x).

Let f_(n+1)(x) = 0.5 * f_n(3x) when 0<=x<=1/3.

Let f_(n+1)(x) = 0.5 when 1/3<=x<=2/3.

Let f_(n+1)(x) = 0.5 + 0.5 * f_n(3 (x-2/3)) when 2/3<=x<=1.

Observe that f_n converges to the cantor fuction. Also notice that the choice of starting function does not really matter, provided f_0(0) = 0 and f_0(1)=1 and f_0 is bounded.