Tessellations II. Irena Swanson Reed College, Portland, Oregon. MathPath, Lewis & Clark College, Portland, Oregon, 9 July 2015

Transcription

1 Tessellations II Irena Swanson Reed College, Portland, Oregon MathPath, Lewis & Clark College, Portland, Oregon, 9 July 205

2 Theorem. There are semi-regular tessellations up to translations, rotations and reflections: () (2) (3) (4) (5) (6) (7) (8) (9) (0) () 6.6.6

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4 A finite part of semiregular tessellation Goal: given a specific rectangle, fill it with the semiregular tessellation: all horizontal cuts should be along edges, and the bottom row has an integer number of triangles at the base.

5 A finite part of semiregular tessellation Goal: given a specific rectangle, fill it with the semiregular tessellation: all horizontal cuts should be along edges, and the bottom row has an integer number of triangles at the base. Question: is the rectangle on the left a square?

6 Determine the number q of rows and the number p of columns for a semiregular tessellation of a rectangle whose base is b units and height is (very close to) h units.

7 base = 5, height = 7

8 base = 5, height = 7 number of rows q = 7 number of columns p = 5

9 base = 5, height = 7 number of rows q = 8 number of columns p = 5

10 base = 5, height = 7 number of rows q = 8 number of columns p = 5 Can we do better?

11 base = 5, height = 7 red: number of rows q = 8 number of columns p = 5 blue: number of rows q = 3 number of columns p = 8 Can we do better?

12 Determine the number q of rows and the number p of columns for a semiregular tessellation of a rectangle whose base is b units and height is (very close to) h units. We have b = 5 and h = 7. We want p, q with i.e., we want 7 p = 5 q p q = ,

13 Continued fractions

14 Continued fractions Start with any non-negative number r, say π.

15 Continued fractions Start with any non-negative number r, say π. Approximate it with its floor a 0 = r, i.e., the largest integer smaller than or equal to the number. For r = π, we know that a 0 = 3.

16 Continued fractions Start with any non-negative number r, say π. Approximate it with its floor a 0 = r, i.e., the largest integer smaller than or equal to the number. For r = π, we know that a 0 = 3. Now, r a 0 is a non-negative number strictly smaller than. If r a 0 equals 0, stop the process.

17 Continued fractions Start with any non-negative number r, say π. Approximate it with its floor a 0 = r, i.e., the largest integer smaller than or equal to the number. For r = π, we know that a 0 = 3. Now, r a 0 is a non-negative number strictly smaller than. If r a 0 equals 0, stop the process. In all/most of my examples we ll never get 0, so I will stop asking this question.

18 Continued fractions Start with any non-negative number r, say π. Approximate it with its floor a 0 = r, i.e., the largest integer smaller than or equal to the number. For r = π, we know that a 0 = 3. Now, r a 0 is a non-negative number strictly smaller than. If r a 0 equals 0, stop the process. In all/most of my examples we ll never get 0, so I will stop asking this question. Otherwise, write r a 0 = r for some positive number r >.

19 Continued fractions Start with any non-negative number r, say π. Approximate it with its floor a 0 = r, i.e., the largest integer smaller than or equal to the number. For r = π, we know that a 0 = 3. Now, r a 0 is a non-negative number strictly smaller than. If r a 0 equals 0, stop the process. In all/most of my examples we ll never get 0, so I will stop asking this question. Otherwise, write r a 0 = r for some positive number r >. So r = a 0 +. r

20 Repeat the process from the previous page with r in place of r. for some positive number r 2 >. a = r, and r = a + r 2

21 Repeat the process from the previous page with r in place of r. for some positive number r 2 >. a = r, and r = a + r 2 So r = a 0 + r = a 0 + a + r 2

22 Repeat the process from the previous page with r in place of r. a = r, and r = a + r 2 for some positive number r 2 >. So r = a 0 + r = a 0 + a + r 2 Continue: r = a 0 + a + a 2 + r 3 = a 0 + a + a 2 + a 3 + r 4 = = a 0 + a + a 2 + a 3 + a

23 Repeat the process from the previous page with r in place of r. a = r, and r = a + r 2 for some positive number r 2 >. So r = a 0 + r = a 0 + a + r 2 Continue: r = a 0 + a + a 2 + r 3 = a 0 + a + a 2 + a 3 + r 4 = = a 0 + These are continued fraction expansions of r. a + a 2 + a 3 + a

24 Notation for continued fractions: r = a 0 + a + a 2 + a 3 + a = [a 0 ; a, a 2, a 3, a 4,...].

25 Notation for continued fractions: r = a 0 + a + a 2 + a 3 + a = [a 0 ; a, a 2, a 3, a 4,...]. Example: π = [3; 7, 5,, 292,,,,...].

26 Notation for continued fractions: r = a 0 + a + a 2 + a 3 + a = [a 0 ; a, a 2, a 3, a 4,...]. Example: π = [3; 7, 5,, 292,,,,...]. Example: work it out for r = 2: 2 =

27 Notation for continued fractions: r = a 0 + a + a 2 + a 3 + a = [a 0 ; a, a 2, a 3, a 4,...]. Example: π = [3; 7, 5,, 292,,,,...]. Example: work it out for r = 2: 2 = [;

28 Notation for continued fractions: r = a 0 + a + a 2 + a 3 + a = [a 0 ; a, a 2, a 3, a 4,...]. Example: π = [3; 7, 5,, 292,,,,...]. Example: work it out for r = 2: 2 = [; 2 ( 2 = , = )

29 Notation for continued fractions: r = a 0 + a + a 2 + a 3 + a = [a 0 ; a, a 2, a 3, a 4,...]. Example: π = [3; 7, 5,, 292,,,,...]. Example: work it out for r = 2: 2 = [; 2, 2 ( 2 = , = , = )

30 Notation for continued fractions: r = a 0 + a + a 2 + a 3 + a = [a 0 ; a, a 2, a 3, a 4,...]. Example: π = [3; 7, 5,, 292,,,,...]. Example: work it out for r = 2: 2 = [; 2, 2, 2,...] (2 forever).

31 Example: 2 = [; 2, 2, 2,...] =

32 Example: 2 = [; 2, 2, 2,...] = This means that 2 = [0; 2, 2, 2,...] =

33 Example: This means that so 2 = [; 2, 2, 2,...] = = [0; 2, 2, 2,...] = = ,

34 Example: 2 = [; 2, 2, 2,...] = This means that 2 = [0; 2, 2, 2,...] = so = = = = + 2.

35 Example: = = [0;,,,,,,, 4, 3, 60,...].

36 As for decimal truncation, we can also truncate continued fractions.

37 As for decimal truncation, we can also truncate continued fractions. Example: 2 = [; 2, 2, 2,...]( = ). [; ] =, [; 2] = + 2 = 3 2 =.5, [; 2, 2] = [; 2, 2, 2] = + [; 2, 2, 2, 2] = = =.4, = = = 7 2 =.46, = = 4 29 ( = ),

38 As for decimal truncation, we can also truncate continued fractions. Example: 2 = [; 2, 2, 2,...]( = ). [; ] =, [; 2] = + 2 = 3 2 =.5, [; 2, 2] = [; 2, 2, 2] = = =.4, = = = 7 2 =.46, [; 2, 2, 2, 2] = = = 4 29 ( = ), These rational numbers are called convergents of the continued fraction.

39 Convergents of π: [3; ] = 3, [3; 7] = 22 7, [3; 7, 5] = , [3; 7, 5, ] = 355 3, [3; 7, 5,, 292] =

40 Convergents of : [0; ] = 0, [0;,,,,, ] = 8 3, [0; ] =, [0;,,,,,, ] = 3 2, [0;, ] = 2 [0;,, ] = 2 3 [0;,,, ] = 3 5 [0;,,,, ] = 5 8,, [0;,,,,,,, 4] = 60 97,, [0;,,,,,,, 4, 3] = 93 32,, [0;,,,,,,, 4, 3, 60] =

41 base = 6, height = number of rows q = number of columns p = 6

42 base = 6, height = number of rows q = 3 number of columns p = 6

43 base = 6, height = number of rows q = 3 number of columns p = 6 The continued fraction expansion of is [0, 2, 8,,, 4, 2, 3,, 3,,...], and its convergents are: 0, 2, 8 7, 9 9, 7 36, 77 63,...

44 base = 6, height = Convergents: 0, 2, 8 7, 9 9, 7 36, 77 63,... red: number of rows q = 9 number of columns p = 9 blue: number of rows q = 7 number of columns p = 8