1230, notes 16. Karl Theodor Wilhelm Weierstrass, November 18, / 18

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1 1230, notes 16 Karl Theodor Wilhelm Weierstrass, November 18, / 18

2 1230, notes 16 Karl Theodor Wilhelm Weierstrass, Left university without a degree (ignored what he was supposed to be studying law, economics, finance in favor of you can guess what November 18, / 18

3 1230, notes 16 Karl Theodor Wilhelm Weierstrass, Left university without a degree (ignored what he was supposed to be studying law, economics, finance in favor of you can guess what Became a high school teacher (taught math, physics, botany, history, German, and gymnastics) November 18, / 18

4 1230, notes 16 Karl Theodor Wilhelm Weierstrass, Left university without a degree (ignored what he was supposed to be studying law, economics, finance in favor of you can guess what Became a high school teacher (taught math, physics, botany, history, German, and gymnastics) Ill much of the time after 1850, yet published papers which led to a Professorship in Berlin November 18, / 18

5 1230, notes 16 Karl Theodor Wilhelm Weierstrass, Left university without a degree (ignored what he was supposed to be studying law, economics, finance in favor of you can guess what Became a high school teacher (taught math, physics, botany, history, German, and gymnastics) Ill much of the time after 1850, yet published papers which led to a Professorship in Berlin Students included Cantor, Schwarz (of Cauchy-Schwarz), Sofia Kovalevskaya, many others of note (41 students, mathematical descendants ) November 18, / 18

6 1230, notes 16 Karl Theodor Wilhelm Weierstrass, Left university without a degree (ignored what he was supposed to be studying law, economics, finance in favor of you can guess what Became a high school teacher (taught math, physics, botany, history, German, and gymnastics) Ill much of the time after 1850, yet published papers which led to a Professorship in Berlin Students included Cantor, Schwarz (of Cauchy-Schwarz), Sofia Kovalevskaya, many others of note (41 students, mathematical descendants ) Weierstrass s mathematical lineage goes back to Gauss, and before that Jacob Bernoulli, and before that Ulrich Zasius, , who according to the Mathematical Genealogy Project had two students and mathematical descendents. November 18, / 18

7 Main mathematical accomplishments: Made calculus rigorous Bolzano Weierstrass theorem, intermediate value theorem (though prior proof by Bolzano), ε δ definition of continuity, November 18, / 18

8 Main mathematical accomplishments: Made calculus rigorous Bolzano Weierstrass theorem, intermediate value theorem (though prior proof by Bolzano), ε δ definition of continuity, Weierstrass M - test for convergence of a series of functions November 18, / 18

9 Main mathematical accomplishments: Made calculus rigorous Bolzano Weierstrass theorem, intermediate value theorem (though prior proof by Bolzano), ε δ definition of continuity, Weierstrass M - test for convergence of a series of functions Weierstrass approximation theorem (continuous functions can be approximated by polynomials) November 18, / 18

10 Main mathematical accomplishments: Made calculus rigorous Bolzano Weierstrass theorem, intermediate value theorem (though prior proof by Bolzano), ε δ definition of continuity, Weierstrass M - test for convergence of a series of functions Weierstrass approximation theorem (continuous functions can be approximated by polynomials) Weierstrass function see below November 18, / 18

11 Main mathematical accomplishments: Made calculus rigorous Bolzano Weierstrass theorem, intermediate value theorem (though prior proof by Bolzano), ε δ definition of continuity, Weierstrass M - test for convergence of a series of functions Weierstrass approximation theorem (continuous functions can be approximated by polynomials) Weierstrass function see below at least 7 other important theorems November 18, / 18

12 A continuous but nowhere differentiable function. November 18, / 18

13 A continuous but nowhere differentiable function. Most believed this was not possible. (Easy if continuity is dropped.?) November 18, / 18

14 A continuous but nowhere differentiable function. Most believed this was not possible. (Easy if continuity is dropped.?) Bolzano 1830 Constructs a sequence of piecewise linear functions which converge to a nowhere differentiable function not published until 1922 November 18, / 18

15 A continuous but nowhere differentiable function. Most believed this was not possible. (Easy if continuity is dropped.?) Bolzano 1830 Constructs a sequence of piecewise linear functions which converge to a nowhere differentiable function not published until 1922 Cellérier, not published until 1890 C (x) = n=1 1 ( ) a k sin a k x, a > 1000 November 18, / 18

16 A continuous but nowhere differentiable function. Most believed this was not possible. (Easy if continuity is dropped.?) Bolzano 1830 Constructs a sequence of piecewise linear functions which converge to a nowhere differentiable function not published until 1922 Cellérier, not published until 1890 C (x) = n=1 1 ( ) a k sin a k x, a > 1000 Riemann R (x) = n=1 1 k 2 sin ( k 2 x ) But it can be shown that this is differentiable at π 2p+1 2q+1, p, q integers, and only at these points. November 18, / 18

17 Weierstrass First published example. W (x) = a k cos b k πx k=0 0 < a < 1 b an odd integer > 1 ab > 1 + 3π 2 November 18, / 18

18 Weierstrass First published example. W (x) = a k cos b k πx k=0 0 < a < 1 b an odd integer > 1 ab > 1 + 3π 2 Peano a space filling curve which is also nowhere differentiable. Motivated by Cantor s results that the cardinality of [0, 1] is the same as the cardinality of [0, 1] [0, 1] (unit square). November 18, / 18

19 (This thesis has 18 examples, from up to 2002, including several well known fractals.) November 18, / 18

20 Tagaka T (x) = = k=0 k=0 1 ( ) 2 k dist 2 k x, Z 1 2 k inf m 0 ( ) 2 k x, m = ( inf m 0 x, m ) 2 k=0 k November 18, / 18

21 Koch snowflake (famous fractal) November 18, / 18

22 van der Waerden (1930). V (x) = Very similar to Tagaka. k=0 1 ( ) 10 k dist 10 k x, Z November 18, / 18

23 van der Waerden (1930). V (x) = Very similar to Tagaka. k=0 1 ( ) 10 k dist 10 k x, Z But this one is easier to understand. It depends on the decimal expansion of x. November 18, / 18

24 Let φ k (x) = 1 ( ) 10 k dist 10 k x, Z = inf x m. m 0 10 k November 18, / 18

25 Let φ k (x) = 1 ( ) 10 k dist 10 k x, Z = inf x m. m 0 10 k We will pick a particular k and x [0, 1] and evaluate this function. November 18, / 18

26 Let φ k (x) = 1 ( ) 10 k dist 10 k x, Z = inf x m. m 0 10 k We will pick a particular k and x [0, 1] and evaluate this function. Let x =.326, k = 2. November 18, / 18

27 Let φ k (x) = 1 ( ) 10 k dist 10 k x, Z = inf x m. m 0 10 k We will pick a particular k and x [0, 1] and evaluate this function. Let x =.326, k = 2. Then φ k (x) is the minimum of the numbers x m 100 where m = 0, 1, 2, 3, 4, 5,..., 100. (It is obvious that the nearest of these numbers to x cannot be bigger than ) November 18, / 18

28 Let φ k (x) = 1 ( ) 10 k dist 10 k x, Z = inf x m. m 0 10 k We will pick a particular k and x [0, 1] and evaluate this function. Let x =.326, k = 2. Then φ k (x) is the minimum of the numbers x m 100 where m = 0, 1, 2, 3, 4, 5,..., 100. (It is obvious that the nearest of these numbers to x cannot be bigger than ) Then m =? November 18, / 18

29 Let φ k (x) = 1 ( ) 10 k dist 10 k x, Z = inf x m. m 0 10 k We will pick a particular k and x [0, 1] and evaluate this function. Let x =.326, k = 2. Then φ k (x) is the minimum of the numbers x m 100 where m = 0, 1, 2, 3, 4, 5,..., 100. (It is obvious that the nearest of these numbers to x cannot be bigger than ) Then m =? φ 2 (.326) = =.004. November 18, / 18

30 Next case: x =.326, k = 1. November 18, / 18

31 Next case: x =.326, k = 1. Find the minimum of the numbers x m 10. November 18, / 18

32 Next case: x =.326, k = 1. Find the minimum of the numbers x m 10. We choose between 1 10, 2 10, 3 10, 4 10,..., 1. Then m =? November 18, / 18

33 Next case: x =.326, k = 1. Find the minimum of the numbers x m 10. We choose between 1 10, 2 10, 3 10, 4 10,..., 1. Then m =? φ 1 (.326) = =.026. November 18, / 18

34 Case x =.326, k = 3. Minimize.326 m m =? November 18, / 18

35 Case x =.326, k = 3. Minimize.326 m m =? φ 3 (.326) = = 0. November 18, / 18

36 Case x =.326, k = 3. Minimize.326 m m =? φ 3 (.326) = = 0. To examine V we consider difference quotients: V (x + h) V (x). h November 18, / 18

37 Let h n = 1 10 n and consider V (x + h n ) V (x) lim. n h n November 18, / 18

38 Let h n = 1 10 n and consider V (x + h n ) V (x) lim. n h n We will show that this limit does not exist if x =.326. November 18, / 18

39 Let h n = 1 10 n and consider V (x + h n ) V (x) lim. n h n We will show that this limit does not exist if x =.326. Suppose n = 2 and evaluate φ k(x ) φ k (x) for k = 1, 2, 3. November 18, / 18

40 We have already calculated φ 1 (x) =.026 φ 2 (x) =.004 φ 3 (x) = 0. November 18, / 18

41 We have already calculated φ 1 (x) =.026 φ 2 (x) =.004 φ 3 (x) = 0. ( We next calculate φ 1 x ). ( φ 1 x + 1 ) = inf.336 m = 100 m =.036. November 18, / 18

42 We have already calculated φ 1 (x) =.026 φ 2 (x) =.004 φ 3 (x) = 0. ( We next calculate φ 1 x ). ( φ 1 x + 1 ) = inf.336 m = 100 m =.036. Then ( φ 1 x ) φ1 (x) = = November 18, / 18

43 Also, ( φ ) = inf.336 m = = m 100 = φ 2 (.326) November 18, / 18

44 Also, ( φ ) = inf.336 m = = m 100 Hence, if x =.326 then = φ 2 (.326) ( φ 2 x ) φ2 (x) = November 18, / 18

45 Also, ( φ ) = inf.336 m = = m 100 Hence, if x =.326 then = φ 2 (.326) ( φ 2 x ) φ2 (x) = k = 3 : ( φ 3 x ) φ3 (x) = 0, because both terms in the numerator are zero. November 18, / 18

46 Similarly, if x =.326 and k > 3, ( φ k x ) φ3 (x) = November 18, / 18

47 Similarly, if x =.326 and k > 3, Hence., ( φ k x ) φ3 (x) = V (x + h 2 ) V (x) h 2 = φ 1 (x + h 2 ) φ 1 (x) h 2 = 1. November 18, / 18

48 Similarly, if x =.326 and k > 3, Hence., ( φ k x ) φ3 (x) = V (x + h 2 ) V (x) h 2 = φ 1 (x + h 2 ) φ 1 (x) h 2 = 1. So for a given n, = V (x + h n ) V (h n ) φ = k (x ± h n ) φ k (x) h n ±h k=1 n n 1 n 1 φ k (x ± h n ) φ k (x) = ±h k=1 n k=1 ±1 November 18, / 18

49 Similarly, if x =.326 and k > 3, Hence., ( φ k x ) φ3 (x) = V (x + h 2 ) V (x) h 2 = φ 1 (x + h 2 ) φ 1 (x) h 2 = 1. So for a given n, = V (x + h n ) V (h n ) φ = k (x ± h n ) φ k (x) h n ±h k=1 n n 1 n 1 φ k (x ± h n ) φ k (x) = ±h k=1 n k=1 and this is odd if n is even and it is even if n is odd. Hence V (x) does not exist. ±1 November 18, / 18

50 The Cantor Function November 18, / 18

51 Define f : [0.1] [0, 1] as follows: The Cantor Function November 18, / 18

52 Define f : [0.1] [0, 1] as follows: The Cantor Function If x [0, 1], express x as a tertiary decimal. November 18, / 18

53 The Cantor Function Define f : [0.1] [0, 1] as follows: If x [0, 1], express x as a tertiary decimal. If f has a 1 in its expansion, replace everything after the first 1 by 0. November 18, / 18

54 The Cantor Function Define f : [0.1] [0, 1] as follows: If x [0, 1], express x as a tertiary decimal. If f has a 1 in its expansion, replace everything after the first 1 by 0. Replace all remaining 2 s by 1 s November 18, / 18

55 The Cantor Function Define f : [0.1] [0, 1] as follows: If x [0, 1], express x as a tertiary decimal. If f has a 1 in its expansion, replace everything after the first 1 by 0. Replace all remaining 2 s by 1 s The result is the binary decimal for f (x). November 18, / 18

56 The Cantor Function Define f : [0.1] [0, 1] as follows: If x [0, 1], express x as a tertiary decimal. If f has a 1 in its expansion, replace everything after the first 1 by 0. Replace all remaining 2 s by 1 s The result is the binary decimal for f (x). November 18, / 18

57 For example, if x ( 1 3, 2 ) 3, then f (x) = 1 2. If x ( 1 9, 2 9) then f (x) = 1 4. If x ( 7 9, 8 ) 9 then f (x) = 3 4, etc. November 18, / 18

58 For example, if x ( 1 3, 2 ) 3, then f (x) = 1 2. If x ( 1 9, 2 9) then f (x) = 1 4. If x ( 7 9, 8 ) 9 then f (x) = 3 4, etc. Therefore f (x) is constant on all the removed intervals in the Cantor set. It is non-decreasing on [0, 1]. November 18, / 18

59 The derivative f (x) is zero on all the removed intervals, and so on a set of measure 1. f (0) = 0, f (1) = 1. November 18, / 18

60 The derivative f (x) is zero on all the removed intervals, and so on a set of measure 1. f (0) = 0, f (1) = 1. It turns out that f is continuous. November 18, / 18

61 The derivative f (x) is zero on all the removed intervals, and so on a set of measure 1. f (0) = 0, f (1) = 1. It turns out that f is continuous. 1 0 f (x) dx = 0 f (1) f (0) = 1. November 18, / 18

62 The derivative f (x) is zero on all the removed intervals, and so on a set of measure 1. f (0) = 0, f (1) = 1. It turns out that f is continuous. 1 0 f (x) dx = 0 f (1) f (0) = 1. The fundamental theorem of calculus does not apply to this function. November 18, / 18