Iterated maps and fractal structures

Start with two position rules

Take a position coordinate q on a line, with positive values to the right and metre as its unit. Start from the full closed interval −1 m ≤ q ≤ 2 m. We supply two rules: L(q) = q/3 - (2/3) m and R(q) = q/3 + (4/3) m. For example, a point at q = 2 m is sent to 0 m by L and to 2 m by R.

Both rules divide a difference between coordinates by 3. They also shift the result, so a negative coordinate can remain negative. These are exact mathematical maps supplied for this lesson. Applying a rule is an iteration; it is not a journey taking a stated number of seconds.

Prepare with Algebra, functions, graphs, and logarithms for fractions, function substitution and natural logarithms, and Units, standards, and physical constants for quantity-valued coordinates. The first reading can follow the intervals and the tables. The scaling section then introduces the exponent it needs locally; derivatives and a chaos theorem are not required.

Map a whole closed interval

A closed interval includes its two endpoints and every point between them. Because each supplied map has the positive slope 1/3, it preserves endpoint order. Applying L to the entire starting interval therefore gives [−1 m, 0 m]; applying R gives [1 m, 2 m]. Each image is 1 m long.

Keep both images. Their union consists of two closed intervals, with the open gap 0 m < q < 1 m removed. The points at 0 m and 1 m remain. Repeating the same two rules on every retained interval creates smaller pieces with more gaps.

Call the initial interval C₀ and the next retained set C₁. The integer n counts completed iterations, starting at zero. In compact notation, C_(n+1) = L(C_n) union R(C_n): apply both maps to the whole current set, then take every point belonging to either image.

Compare finite set stages and point states

The upper panel shows C₀, C₁, C₂ and C₃ on one position scale. Blue bars represent the full retained intervals; light circles emphasize their included endpoints. Vertical row spacing only separates iteration labels. It is not another spatial coordinate or elapsed time.

At n = 2 there are four intervals, each 1/3 m long. At n = 3 there are eight, each 1/9 m long. The gaps still have their original coordinate locations; the picture does not stretch each row to make its retained length fill the axis.

For comparison, the lower panel starts from just q₀ = 2 m and applies only L each time. Its four unjoined shapes mark 2 m, 0 m, −2/3 m and −8/9 m. A collection of intervals and one sequence of points use different update rules, even though the same map L appears in both.

Neither panel shows motion between iterations. The exact interval table below supplies the finite set values; the following section supplies the point-state table. No displayed row is the infinite fractal limit.

Follow one repeated map

For the lower comparison the rule is q_(n+1) = L(q_n), q_0 = 2 m. It produces one new coordinate per step. It does not apply R and does not replace a point by a whole interval.

Measure the offset from the left fixed point by qₙ + 1 m. Each left-map step divides this offset by 3. Starting from an offset of 3 m gives q_n = -1 m + 3^(1-n) m, so the sequence approaches −1 m. No velocity follows from this formula because n is a count, not a time.

An iterated function system uses several maps together to construct a set. A point sequence can instead specify one map at each step. Its choice rule must be stated; repeatedly using only L cannot be relabelled as the two-map set construction.

Define the infinite limit

The retained sets are nested: every next stage lies inside the preceding stage. Define the limit C as the points that remain in every stage: C = intersection_(n>=0) C_n. This is an infinite requirement, rather than the union of the finite rows in the picture.

There are N_n = 2^n distinct intervals at stage n. Each has length ell_n = (3/3^n) m, so their total length is 3*(2/3)^n m. This total tends to zero as n increases. It does not make C empty: −1 m and 2 m remain at every stage, as do many other points.

More precisely, for any positive length tolerance we can choose a stage whose intervals cover C with a smaller total length. This is the zero-length property of the ideal Cantor set. A finite stage still contains whole intervals and has a positive total length; a finite drawing cannot display all the infinitely small gaps of C.

Read the similarity scaling

Normalize an interval length by the original length 3 m. The dimensionless ratio ellₙ/(3 m) is 3⁻ⁿ, while the number of distinct pieces is 2ⁿ. Ask for an exponent d that balances these two changes: N_n*(ell_n/(3 m))^d = 1.

Substituting the count and length ratio reduces the balance to 2*(1/3)^d = 1. Taking natural logarithms gives d = ln(2)/ln(3), about 0.631. This is the similarity dimension of these two equal-ratio maps. The logarithm arguments and d are dimensionless; the metre remains the unit of q and of interval length.

For this separated Cantor construction the same value is its Hausdorff dimension, a formal description of how size scales at small lengths. That identification uses a theorem for similarity copies with an appropriate separation condition. A first reading can use the count–length balance; the full general dimension theorem is an optional mathematical return, not proved here.

Check separation and endpoints

The separation condition is concrete here. The open interval −1 m < q < 2 m is mapped to the disjoint open intervals (−1 m, 0 m) and (1 m, 2 m), both inside the original open interval. The two branches therefore make genuinely separate pieces rather than duplicate labels for one image.

If we declared L twice instead, there would be 2ⁿ possible branch words but only one distinct interval at each stage. Repeated images coincide, and the limiting set is the single fixed point −1 m. Counting the duplicated words in the same scaling equation would not establish the dimension of that set.

Also check which points survive. A point at 1/2 m is removed at the first stage. A point at −1/2 m survives the first stage but falls into the second-stage open gap (−2/3 m, −1/3 m). In contrast, 0 m and 1 m are retained endpoints at every stage. Membership in one finite stage alone does not establish membership in C.

Keep the model and scale limits

The exact set C is a shifted and rescaled version of the classical Cantor set. With x = (q+1 m)/(3 m), its normalized coordinates lie between 0 and 1. They are precisely the numbers having a base-three expansion that uses only digits 0 and 2. An endpoint can have more than one expansion: 1/3 = 0.1000… in base three also equals 0.0222…, so the first form does not exclude it.

Its exact self-similarity means C is the union of L(C) and R(C), with the same pattern recurring at smaller scales. This is a statement about the infinite mathematical set. A natural or manufactured pattern can only be tested over a stated range of measured scales; similarity in a finite image does not establish an exact all-scale fractal.

Keep the model question explicit: are you following a chosen point rule, constructing a union of images, or testing a scale relation? An iteration count is not automatically time, a branch word is not automatically a distinct piece, and an exact mathematical scaling law is not an empirical measurement.

Worked examples

Try it yourself

1. For the original separated construction, find the number of intervals, each interval length and total retained length at n = 4.

   Show the answer and reasoning

   16 intervals; each 1/27 m long; total retained length 16/27 m.

   Use 2⁴ = 16 and 3/3⁴ m = 1/27 m. Multiplying the count by the individual length gives the total.

2. At which first stage are 1/2 m and −1/2 m removed? Are 0 m and 1 m removed?

   Show the answer and reasoning

   1/2 m is removed at stage 1; −1/2 m at stage 2. The endpoints 0 m and 1 m remain in every stage.

   The first open gap is (0 m, 1 m). The point −1/2 m lies in the later open gap (−2/3 m, −1/3 m). Closed endpoints remain included.

3. The single-map sequence has q₂ = −2/3 m. Apply only L once more; give q₃ and its offset from −1 m.

   Show the answer and reasoning

   q₃ = −8/9 m; its offset from −1 m is 1/9 m.

   L(−2/3 m) = −2/9 m − 2/3 m = −8/9 m. Add 1 m to obtain the offset. This is one point update, not a union of two images.

4. Reverse the alternating rule: apply R and then L. Find the two-step fixed point and its intermediate coordinate.

   Show the answer and reasoning

   The two-step fixed point is −1/4 m; the intermediate coordinate after R is 5/4 m.

   L(R(q)) = q/9 − (2/9) m. Solving q = q/9 − (2/9) m gives q = −1/4 m, and R(−1/4 m) = 5/4 m.

5. Solve 2(1/3)ᵈ = 1 for d. Does d have unit m because the coordinate uses metres?

   Show the answer and reasoning

   d = ln(2)/ln(3), about 0.631; it is dimensionless.

   Taking logarithms gives ln 2 − d ln 3 = 0. The coordinate has a length unit, but the scaling equation contains only a count and a ratio of lengths.

6. At n = 4 a declared pair of maps is L and L. There are 16 written branch words. How many distinct intervals are actually retained, and what is their total length?

   Show the answer and reasoning

   One distinct interval is retained, with total length 1/27 m.

   Every word gives the same four-fold left composite. Its image interval has length 3/3⁴ m = 1/27 m; counting it repeatedly would count overlap rather than separate pieces.

What to carry forward

• A map can update one point or act on every point of a set; the update rule must say which.
• The finite Cantor stages have 2ⁿ closed intervals and total length 3(2/3)ⁿ m; their infinite intersection remains nonempty.
• The similarity exponent is dimensionless, and identifying a set dimension requires the relevant separation conditions.
• A finite drawing and a physical scaling range do not establish an exact all-scale fractal.