From Pascals Triangle to Sierpinskis Triangle Nicoleta Babutiu - PowerPoint PPT Presentation

From Pascal’s Triangle to Sierpinski’s Triangle Nicoleta Babutiu

Q@A Today’s journey Conclusions Intro Sierpi ń ski’s Triangle Pascal’s Triangle Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 2

Do you remember “The 12 Days of Christmas” song? A partridge in a Two turtle doves Three French hens Four calling birds Five gold rings 6 geese a-laying pear tree 7 swans a-swimming 8 maids a-milking 9 ladies dancing 10 lords a-leaping 11 pipers piping 12 drummers drumming Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 3

How many gifts were given in total over the 12 days? •On the first day: 1 •On the second day: 1 + 2 •On the 3rd day: 1 + 2 + 3 …………………………. •On the 12th day: 1 + 2 + 3 +…+12 1+(1+2)+(1+2+3)+…+(1+2+3+…+12) = ? Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 4

Pascal’s Triangle Number of gifts received each day 1 Blaise Pascal - “Treatise on Arithmetical Triangle”, 1655 Running total number of gifts received each day 1 1 2 1 1 Yang Hui’s Triangle - the 13th 3 century 1 3 1 4 4 6 1 1 Tartaglia’s Triangle - in 1556 1 1 5 10 10 5 20 15 1 6 15 6 1 35 21 35 7 21 7 1 1 70 56 28 8 56 28 1 1 8 36 9 1 9 36 84 126 126 84 1 210 120 45 10 1 252 45 120 210 1 10 330 165 55 11 462 462 1 1 55 165 330 11 220 66 12 1 1 495 495 792 924 792 66 220 12 1 Answer: 364 Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 5

Pascal’s Method In how many different paths can you spell SIERPINSKI if you start at the top and proceed to the next row by moving diagonally left or right? S 1 I I 1 1 E E E 2 1 1 R R R R 3 1 3 1 P P P 4 6 4 I I 10 10 N N N 20 10 10 S S S S 30 30 10 10 K K K 60 50 50 I 110 110 I Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 6

Binomial Coefficients Binomial Theorem 1 C 0 n = 0 0 1 1 C 0 C 1 n = 1 1 1 2 1 C 2 n = 2 1 C 0 C 1 2 2 2 3 1 3 1 C 1 C 2 C 3 n = 3 C 0 3 3 3 3 4 6 C 0 C 1 C 2 C 3 n = 4 4 1 C 4 1 4 4 4 4 4 1 5 10 10 5 n = 5 C 0 1 C 1 C 2 C 3 C 4 C 5 5 5 5 5 5 5 n = 6 20 15 1 6 15 6 1 C 0 C 1 C 2 C 3 C 4 C 6 C 5 6 6 6 6 6 6 6 5 C a 3 b 2 + 5 C a 2 b 3 a 5 + C C a 1 b 4 b 1 b 5 4 C 0 C a (a+b) = + 5 + + 5 n = 5 2 5 5 5 1 3 4 5 . . . . 5 . . a 3 b 2 + a 2 b 3 + 5 a 1 b 4 b 1 a 5 + 4 1 5 a 10 10 5 1 (a+b) = + + b Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 7

Particular cases of Binomial Theorem If a = b = 1 1 0 n = 0 = 1 2 1 1 = 1 2 n = 1 1 + 1 2 1 n = 2 1 2 = 2 1 + 2 +1 1 3 1 3 n = 3 = 2 3 1 + 3 + 3 + 1 4 6 n = 4 4 1 1 4 1 + 4 + 6 + 4 + 1 = 2 If a =10, b = 1 0 . 10 0 (10+1) 1 = 1 = 1 n = 0 1 . 10 1 + 1 . 10 0 1 = 11 1 1 = (10+1) n = 1 1 . 10 2 + 2 . 10 1 +1 . 10 0 2 2 = 121 1 = 1 n = 2 (10+1) 3 1 . 10 3 3 . 10 2 + . 10 1 . 1 0 = 1 3 1 (10+1) + 3 = + 3 10 1331 n = 3 Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 8

The diagonals in Pascal’s Triangle 1, 1, 1, 1, 1 ,1, … A constant sequence The sequence of natural numbers 1, 2, 3, 4, 5, 6 , … 1 The sequence of triangular numbers 1 1 2 1 The sequence of tetrahedral numbers 1 3 1 3 1 4 6 4 1 1 1, 5, 15, 35, 70, … The sequence of 4-simplex numbers 1 5 1 10 10 5 20 15 1 6 15 6 1 21 35 35 1 7 21 7 1 70 28 8 56 56 28 1 1 8 36 9 36 84 126 126 84 1 1 9 1 Henri Poincaré, about algebraic topology Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 9

Fibonacci sequence in Pascal’s Triangle F 0 = F 1 = 1 = F + 0 F n F n-1 n-2 0, 1, 1, 2, 3, 5, 8, ….. 1 1 1 1 2 1 3 1 3 1 4 1 1 6 4 1 5 1 10 10 5 Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 10

Other properties of Pascal’s Triangle 1 Divisibility 1 1 2 Hockey-stick identity 1 1 3 1 3 1 6 4 4 1 1 1+3+6+10=20 1 1 5 10 10 5 + C 2 2 + C 2 20 15 + C 1 6 15 6 1 C = C 3 6 3 2 4 5 2 35 21 35 7 21 7 1 1 n 70 56 28 8 56 28 1 1 8 C C k = 36 9 1 9 36 84 126 126 84 1 r n+1 k+1 r=k 210 120 45 10 1 252 45 120 210 1 10 330 165 55 11 462 462 1 1 55 165 330 11 220 66 12 1 1 495 495 792 924 792 66 220 12 1 Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 11

Is there some geometry in Pascal’s Triangle? Number Number Number Number Number Number of of of of of of points segments triangles quadri- penta- hexa- laterals gons gons n=0 1 1 1 n=1 1 n=2 2 1 n=3 3 1 3 1 1 4 6 4 1 n=4 1 n=5 1 5 10 10 5 1 6 15 20 15 6 1 n=6 R M Z A Q Y S P U B N V T X Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 12

Probabilities in Pascal’s Triangle? Total number of Possible outcomes outcomes n=0 H T 1 n=1 2 TT 2 HH HT, TH 2 n=2 3 TTH,THT,HTT HHH n=3 HHT,HTH,THH 2 TTT n = number of toss * X= number of heads 1/2 1/2 C n k 2/4 1/4 1/4 P(X=k)= 2 n C 3 3 1 1/8 3/8 3/8 1/8 = P(X=1)= 2 3 8 Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 13

Sierpinski’s Triangle Wac ł aw Sierpi ń ski described the Sierpinski Triangle in 1915. It is a fractal with the overall shape of an equilateral triangle, subdivided recursively into smaller equilateral triangles. https://www.pinterest.ca/leterrae/italy-tour-cosmatesque-pavements/ However, similar patterns appear already in the 13th- century Cosmati mosaics in the cathedral of Anagni, Italy. Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 14

Constructing the Sierpinski Triangle 1. Shrinking and duplication Step 2: Shrink the Step 1: Start with an Step 3: Repeat step 2 triangle to 1/2 height equilateral triangle with each of the and 1/2 width, smaller triangles make three copies Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 15

Constructing the Sierpinski Triangle 2. Removing triangles Step 3: Repeat step Step 2: Subdivide it Step 1: Start with an 2 with each of the into four smaller equilateral triangle remaining smaller congruent equilateral triangles infinitely triangles and remove the central triangle Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 16

Constructing the Sierpinski Triangle 3. Pascal's Triangle n=0 1 1 1 1 n=1 1 1 n=2 1 2 1 1 0 1 n=3 3 1 1 3 1 1 1 1 n=4 4 1 0 1 0 1 6 4 1 0 n=5 1 5 10 10 5 1 1 1 0 0 1 1 6 15 20 15 0 1 0 1 1 6 1 1 0 1 n=6 1 1 n=7 7 21 35 35 21 7 1 1 1 1 1 1 1 1 Pascal’s Triangle Pascal’s Triangle (even/odd numbers) Nicoleta Babutiu CEMC: Bringing Teachers Together Virtually | August 20, 2020 17