Extending Pythagoras
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1 Extending Pythagoras Steven J. Miller, Williams College public_html/ Williams College, June 30, 2015
2 Goals of the Talk Often multiple proofs: Say a proof rather than the proof. Different proofs highlight different aspects. Too often rote algebra: Explore! Generalize! Conjecture! General: How to find / check proofs: special cases, smell test. Specific: Pythagorean Theorem.
3 Pythagorean Theorem
4 Geometry Gem: Pythagorean Theorem Theorem (Pythagorean Theorem) Right triangle with sides a, b and hypotenuse c, then a 2 + b 2 = c 2. Most students know the statement, but the proof? Why are proofs important? Can help see big picture.
5 Geometric Proofs of Pythagoras 5 Figure: Euclid s Proposition 47, Book I. Why these auxiliary lines? Why are there equalities?
6 Geometric Proofs of Pythagoras Figure: Euclid s Proposition 47, Book I. Why these auxiliary lines? Why are there equalities?
7 7 Pythagorean Theorem Dimensional Analysis Guessing Pythagoras Extending Pythagoras Conclusion Feeling Equations Oth Geometric Proofs of Pythagoras Figure: A nice matching proof, but how to find these slicings!
8 Geometric Proofs of Pythagoras Figure: Four triangles proof: I
9 Geometric Proofs of Pythagoras Figure: Four triangles proof: II
10 Geometric Proofs of Pythagoras Figure: President James Garfield s (Williams 1856) Proof.
11 Geometric Proofs of Pythagoras Lots of different proofs. Difficulty: how to find these combinations? At the end of the day, do you know why it s true?
12 Dimensional Analysis
13 Possible Pythagorean Theorems... c 2 = a 3 + b 3. c 2 = a 2 + 2b 2. c 2 = a 2 b 2. c 2 = a 2 + ab+b 2. c 2 = a ab+b 2.
14 Possible Pythagorean Theorems... c 2 = a 3 + b 3. No: wrong dimensions. c 2 = a 2 + 2b 2. No: asymmetric in a, b. c 2 = a 2 b 2. No: can be negative. c 2 = a 2 + ab+b 2. Maybe: passes all tests. c 2 = a ab+b 2. No: violates a+b > c.
15 Dimensional Analysis Proof of the Pythagorean Theorem Area is a function of hypotenuse c and angle x.
16 Dimensional Analysis Proof of the Pythagorean Theorem Area is a function of hypotenuse c and angle x. Area(c, x) = f(x)c 2 for some function f (similar triangles).
17 Dimensional Analysis Proof of the Pythagorean Theorem Area is a function of hypotenuse c and angle x. Area(c, x) = f(x)c 2 for some function f (similar triangles). Must draw an auxiliary line, but where? Need right angles!
18 Dimensional Analysis Proof of the Pythagorean Theorem Area is a function of hypotenuse c and angle x. Area(c, x) = f(x)c 2 for some function f (CPCTC). Must draw an auxiliary line, but where? Need right angles!
19 Dimensional Analysis Proof of the Pythagorean Theorem Area is a function of hypotenuse c and angle x. Area(c, x) = f(x)c 2 for some function f (CPCTC). Must draw an auxiliary line, but where? Need right angles! f(x)a 2 + f(x)b 2 = f(x)c 2
20 Dimensional Analysis Proof of the Pythagorean Theorem Area is a function of hypotenuse c and angle x. Area(c, x) = f(x)c 2 for some function f (CPCTC). Must draw an auxiliary line, but where? Need right angles! f(x)a 2 + f(x)b 2 = f(x)c 2 a 2 + b 2 = c 2.
21 Dimensional Analysis and the Pendulum
22 Dimensional Analysis and the Pendulum Period: Need combination of quantities to get seconds.
23 Dimensional Analysis and the Pendulum Period: Need combination of quantities to get seconds. T = f(x) L/g.
24 Guessing Pythagoras:
25 Finding the Functional Form Idea is to try and guess the correct functional form for Pythagoras. Guess will have some free parameters, determine by special cases. Natural guesses: linear, quadratic,...
26 Linear Attempt Guess linear relation: c = αa+βb: what are α,β?
27 Linear Attempt Guess linear relation: c = αa+βb: what are α,β? Consider special cases: a 0 have very thin triangle so b c and thus β = 1. b 0 have very thin triangle so a c and thus α = 1.
28 Linear Attempt Guess linear relation: c = αa+βb: what are α,β? Consider special cases: a 0 have very thin triangle so b c and thus β = 1. b 0 have very thin triangle so a c and thus α = 1. Question: Does c = a+b make sense?
29 Linear Attempt: Analyzing c = a+b (so a = b = 1 implies c = 2) So, if linear, must be c = a+b. Using: Area rectangle x by y is xy. Area right triangle of sides x by y is 1 2 xy.
30 Linear Attempt: Analyzing c = a+b (so a = b = 1 implies c = 2) So, if linear, must be c = a+b. Using: Area rectangle x by y is xy. Area right triangle of sides x by y is 1 2 xy. Figure: Four triangles and a square, assuming c = a+b and a = b = 1.
31 Linear Attempt: Analyzing c = a+b (so a = b = 1 implies c = 2) So, if linear, must be c = a+b. Using: Area rectangle x by y is xy. Area right triangle of sides x by y is 1 2 xy. Figure: Four triangles and a square, assuming c = a+b and a = b = 1. Calculate area of big square two ways: Four triangles, each area : total is 2. Square of sides 2: area is 2 2 = 4. Contradiction! Cannot be linear!
32 Quadratic Attempt: Guess quadratic: c 2 = αa 2 +γab+βb 2 : what are α,β,γ?
33 Quadratic Attempt: Guess quadratic: c 2 = αa 2 +γab+βb 2 : what are α,β,γ? Consider special cases: as before get α = β = 1; difficulty γ. Figure: Four triangles and a square: a = b = 1.
34 Quadratic Attempt: Guess quadratic: c 2 = αa 2 +γab+βb 2 : what are α,β,γ? Consider special cases: as before get α = β = 1; difficulty γ. Figure: Four triangles and a square: a = b = 1. Equating areas: c 2 = 4 ( ), so c 2 = 2 or c = 2.
35 Quadratic Attempt: Guess quadratic: c 2 = αa 2 +γab+βb 2 : what are α,β,γ? Consider special cases: as before get α = β = 1; difficulty γ. Figure: Four triangles and a square: a = b = 1. Equating areas: c 2 = 4 ( ), so c 2 = 2 or c = 2. Thus 2 = 1+γ1 1+1, so γ = 0 and c 2 = a 2 + b 2.
36 Warnings: Not a proof: just shows that if quadratic, must be c 2 = a 2 + b 2. In lowest terms: = 1 4, = 1 5, = 1 2,
37 Warnings: Not a proof: just shows that if quadratic, must be c 2 = a 2 + b 2. In lowest terms: = 1 4, = 1 5, = 1 2, =
38 Warnings: Not a proof: just shows that if quadratic, must be c 2 = a 2 + b 2. In lowest terms: = 1 4, = 1 5, = 1 2, = 1 4.
39 Warnings: Not a proof: just shows that if quadratic, must be c 2 = a 2 + b 2. In lowest terms: = 1 4, = 1 5, = 1 2, = = 1 4, = 1 5, = 1 2,
40 Warnings: Not a proof: just shows that if quadratic, must be c 2 = a 2 + b 2. In lowest terms: = 1 4, = 1 5, = 1 2, = = 1 4, = 1 5, = 1 2, = 1 4.
41 Extending Pythagoras: The Sphere
42 Pythagoras on a Sphere What should the Pythagorean Theorem be on a sphere?
43 Spherical Coordinates Spherical Coordinates: ρ [0, R], θ [0,π], φ [0, 2π). x = ρ sin(θ) cos(φ). y = ρ sin(θ) sin(φ). z = ρ cos(θ). Note z = ρ cos(θ), then (x, y) from circle of radius r = ρ sin(θ) and angle φ.
44 Special Cases What could the Pythagorean Formula be on a sphere?
45 Special Cases What could the Pythagorean Formula be on a sphere? If a, b, c small relative to radius R should reduce to planar Pythagoras. Can have equilateral right triangle with a = b = c. Only depends on ratios a/r, b/r, c/r.
46 Special Cases What could the Pythagorean Formula be on a sphere? If a, b, c small relative to radius R should reduce to planar Pythagoras. Can have equilateral right triangle with a = b = c. Only depends on ratios a/r, b/r, c/r. Maybe a relation involving cosines of a/r, b/r, c/r as arc length is related to angle!
47 Cosine Guess cos(u) = 1 u 2 /2!+u 4 /4! 1 u 2 /2 (u small). Ingredients (will consider R large relative to a, b, c: cos(a/r) cos(b/r) cos(c/r) a 2. R 2 b 2. R 2 c 2. R 2 Algebra: cos(c/r) cos(a/r) cos(b/r): c 2 R 2 ( a 2 R 2 )( b 2 R 2 ) c 2 = a 2 + b 2 2a2 b 2 R 2. = 1 a2 + b 2 2R 2 + a2 b 2 4R 4
48 Spherical Proof Three points: P 0, P A, P B : P 0 : (R, 0, 0). P A : (R,θ A,φ A ): P A P 0 = θ A 2π R = a R. P B : (R,θ B,φ B ): P B P 0 = θ B 2π R = b R. Length P B P A is θ AB 2π R, where θ AB angle between P A P 0 and P B P 0. Proof follows from dot product: P Q = P Q cos(θpq ).
49 Spherical Proof: Continued Cartesian Coordinates for Dot Product: Remember right triangle: can take φ A = 0, φ B = π/2. P A P 0 : (R sinθ A, 0, R cosθ A ), length is R. P B P 0 : (0, R sinθ B, R cosθ B ), length is R. P B P A = 0+0+R 2 cosθ A cosθ B. Dot product now gives cos(θ AB ) = Substituting yields P B P A P B P 0 P A P 0 = P B P A R 2. cos(θ AB ) = R2 cosθ A cosθ B R 2 = cosθ A cosθ B, proving spherical Pythagoras!
50 Next Step: Generalize Keep going! Generalize further! What s the next natural candidate?
51 Next Step: Generalize Keep going! Generalize further! What s the next natural candidate? Hyperbolic! Guess:
52 Next Step: Generalize Keep going! Generalize further! What s the next natural candidate? Hyperbolic! Guess: cosh(c) = cosh(a) cosh(b), where cosh is the hyperbolic cosine! cos(x) = 1 2 ( e ix + e ix), cosh(x) = 1 ( e x + e x). 2
53 Conclusion
54 Conclusion Math is not complete explore and conjecture! Different proofs highlight different aspects. Get a sense of what to try / what might work.
55 55 Pythagorean Theorem Dimensional Analysis Guessing Pythagoras Extending Pythagoras Conclusion Feeling Equations Oth Feeling Equations
56 Sabermetrics Sabermetrics is the art of applying mathematics and statistics to baseball. Danger: not all students like sports (Red Sox aren t making life easier!). Lessons: not just for baseball; try to find the right statistics that others miss, competitive advantage (business, politics).
57 Estimating Winning Percentages Assume team A wins p percent of their games, and team B wins q percent of their games. Which formula do you think does a good job of predicting the probability that team A beats team B? Why? 57 p + pq p + q + 2pq, p pq p + q + 2pq, p + pq p + q 2pq p pq p + q 2pq
58 Estimating Winning Percentages p + pq p+q + 2pq, p+pq p + q 2pq, p pq p+q + 2pq, p pq p + q 2pq How can we test these candidates? Can you think of answers for special choices of p and q?
59 Estimating Winning Percentages p + pq p+q + 2pq, p+pq p + q 2pq, p pq p+q + 2pq, p pq p + q 2pq Homework: explore the following: p = 1, q < 1 (do not want the battle of the undefeated). p = 0, q > 0 (do not want the Toilet Bowl). p = q. p > q (can do q < 1/2 and q > 1/2). Anything else where you know the answer?
60 Estimating Winning Percentages p + pq p+q + 2pq, p+pq p + q 2pq, p pq p+q + 2pq, p pq p + q 2pq Homework: explore the following: p = 1, q < 1 (do not want the battle of the undefeated). p = 0, q > 0 (do not want the Toilet Bowl). p = q. p > q (can do q < 1/2 and q > 1/2). Anything else where you know the answer?
61 Estimating Winning Percentages p pq p+q 2pq = p(1 q) p(1 q)+(1 p)q Homework: explore the following: p = 1, q < 1 (do not want the battle of the undefeated). p = 0, q > 0 (do not want the Toilet Bowl). p = q. p > q (can do q < 1/2 and q > 1/2). Anything else where you know the answer?
62 Estimating Winning Percentages: Proof Figure: First see how A does, then B.
63 Estimating Winning Percentages: Proof Figure: Two possibilities: A has a good day, or A doesn t.
64 Estimating Winning Percentages: Proof Figure: B has a good day, or doesn t.
65 Estimating Winning Percentages: Proof Figure: Two paths terminate, two start again.
66 Estimating Winning Percentages: Proof Figure: Probability A beats B.
67 Lessons Special cases can give clues. Algebra can suggests answers. Better formula: Bill James Pythagorean Won-Loss formula.
68 Numerical Observation: Pythagorean Won-Loss Formula Parameters RS obs : average number of runs scored per game; RA obs : average number of runs allowed per game; γ: some parameter, constant for a sport. James Won-Loss Formula (NUMERICAL Observation) Won Loss Percentage = RS obs γ RS obs γ + RA obs γ γ originally taken as 2, numerical studies show best γ is about Used by ESPN, MLB. See for a derivation.
69 Other Gems
70 Sums of Integers S n := n = n(n+1) n2.
71 Sums of Integers S n := n = n(n+1) n2. Proof 1: Induction. Proof 2: Grouping: 2S n = (1+n)+(2+(n 1))+ +(n+1).
72 Sums of Integers S n := n = n(n+1) n2. Proof 1: Induction. Proof 2: Grouping: 2S n = (1+n)+(2+(n 1))+ +(n+1). Instead of determining sum useful to get sense of size.
73 Sums of Integers S n := n = n(n+1) n2. Proof 1: Induction. Proof 2: Grouping: 2S n = (1+n)+(2+(n 1))+ +(n+1). Instead of determining sum useful to get sense of size. Have n 2 n 2 S n n; thus S n is between n 2 /4 and n 2, have the correct order of magnitude of n.
74 Sums of Integers S n := n = n(n+1) n2. Proof 1: Induction. Proof 2: Grouping: 2S n = (1+n)+(2+(n 1))+ +(n+1). Instead of determining sum useful to get sense of size. Have n n 2 2 S n n; thus S n is between n 2 /4 and n 2, have the correct order of magnitude of n. Can improve: divide and conquer again: lather, rinse, repeat... n n n 2n n 3n 4 4 S n, so 6 16 n2 S n.
75 75 Pythagorean Theorem Dimensional Analysis Guessing Pythagoras Extending Pythagoras Conclusion Feeling Equations Oth Geometric Irrationality Proofs: a b 2b a b a b Figure: Geometric proof of the irrationality of 2. b
76 Geometric Irrationality Proofs: b a b a Figure: Geometric proof of the irrationality of 3
77 77 Pythagorean Theorem Dimensional Analysis Guessing Pythagoras Extending Pythagoras Conclusion Feeling Equations Oth Geometric Irrationality Proofs: Figure: Geometric proof of the irrationality of 5.
78 Geometric Irrationality Proofs: Figure: Geometric proof of the irrationality of 5: the kites, triangles and the small pentagons.
79 Geometric Irrationality Proofs: Figure: Geometric proof of the irrationality of 6.
80 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1.
81 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets.
82 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets. Example: 8 cookies and 5 people (C = 8, P = 5):
83 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets. Example: 8 cookies and 5 people (C = 8, P = 5):
84 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets. Example: 8 cookies and 5 people (C = 8, P = 5):
85 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Solved x1 + +x P = C, x i 0. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets. Example: 8 cookies and 5 people (C = 8, P = 5):
Extending Pythagoras
Extending Pythagoras Steven J. Miller, Williams College sjm1@williams.edu, Steven.Miller.MC.96@aya.yale.edu http://web.williams.edu/mathematics/sjmiller/public_html/ Williams College, April 8, 2015 Goals
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