David M. Bressoud Macalester College, St. Paul, Minnesota Given at Allegheny College, Oct. 23, 2003

Transcription

1 David M. Bressoud Macalester College, St. Paul, Minnesota Given at Allegheny College, Oct. 23, 2003

2 The Fundamental Theorem of Calculus:. If F' ( x)= f ( x), then " f ( x) dx = F( b)! F( a). b a 2. d dx x! f t dt f x. a () = ( ) Are these statements always true? What assumptions do we have to make about f and F in order for these statements to be true?

3 S. F. LaCroix (802): Integral calculus is the inverse of differential calculus. Its goal is to restore the functions from their differential coefficients.

4 S. F. LaCroix (802): Integral calculus is the inverse of differential calculus. Its goal is to restore the functions from their differential coefficients. What if a function is not the derivative of some identifiable function? 2 e (! x )

5 S. F. LaCroix (802): Integral calculus is the inverse of differential calculus. Its goal is to restore the functions from their differential coefficients. Joseph Fourier: Put the emphasis on definite integrals b (he invented the notation! ) and defined them in a terms of area between graph and x-axis.

6 S. F. LaCroix (802): Integral calculus is the inverse of differential calculus. Its goal is to restore the functions from their differential coefficients. Joseph Fourier: Put the emphasis on definite integrals b (he invented the notation! ) and defined them in a terms of area between graph and x-axis. How do you define area?

7 S. F. LaCroix (802): Integral calculus is the inverse of differential calculus. Its goal is to restore the functions from their differential coefficients. Joseph Fourier: Put the emphasis on definite integrals b (he invented the notation! ) and defined them in a terms of area between graph and x-axis. A.-L. Cauchy: First to define the integral as the limit of the summation ( )(! ) f x x x " i! i i!

8 Bernhard Riemann (852, 867) On the representation of a function as a trigonometric series ( ) " ( ) #! i i i" Defined f x dx as limit of f x x x! b a ( )

9 Bernhard Riemann (852, 867) On the representation of a function as a trigonometric series ( ) " ( ) #! i i i" Defined f x dx as limit of f x x x! b a Key to convergence: on each interval, look at the variation of the function V sup f x inf f x i ( ) = ( )! ( ) x" [ x, x ] x" [ xi!, xi] i! i

10 Bernhard Riemann (852, 867) On the representation of a function as a trigonometric series b ( ) " ( ) #! i i i" Defined f x dx as limit of f x x x! a ( ) Key to convergence: on each interval, look at the variation of the function V sup f x inf f x i = ( )! ( ) x" [ x, x ] x" [ xi!, xi] i! i! " ( ) Integral exists if and only if Vi xi xi" can be made as small as we wish by taking sufficiently small intervals.

11 Any continuous function is integrable: Can make V i as small as we want by taking sufficiently small intervals: # #! Vi ( xi " xi" )<!( xi " xi" )= ( b a b a b " a )= #. " "

12 Any piecewise continuous function is integrable: If ( )" ( )= lim f x lim f x J, " + x! c x! c $ then take an interval of length < 2 J + where the variation is less than J + : ( ) $ V ( x " x )<( J + ) V x x ( ) + " 2 # # ( ) i i i" i i i" J + other intervals $ $ < + = $. 2 2

13 Bernhard Riemann (852, 867) On the representation of a function as a trigonometric series Riemann gave an example of a function that has a jump discontinuity in every subinterval of [0,], but which can be integrated over the interval [0,].

14 Riemann s function: f( x)= { } nx n! " 2 n= " $ x! ( nearest integer ), when this is < {}= x 2, # %$ 0, when distance to nearest integer is { nx} n 2 has n jumps of size between 0 and n 2 2

15 At x = a 2 b, gcd ( a, 2b)=, the function jumps by! 2 2 8b

16 Riemann s function: f( x)= { } nx n! " 2 n= At x = a 2 b, gcd ( a, 2b)=, the function jumps by! 2 2 8b The key to the integrability is that given any positive number, no matter how small, there are only a finite number of places where is jump is larger than that number.

17 If lim f' ( x) and lim f' ( x) exist, " + x! c x! c then they must be equal and they must equal f' () c. Mean Value Theorem: ( )! () f x f c f' ()= c lim = lim!! x" c x! c x" c f' ( k), x < k < c f ( x)! f () c f' ()= c lim = lim + + x" c x! c x" c f' ( k), c < k < x

18 If lim f' ( x) and lim f' ( x) exist, x! c x! c then they must be equal and they must equal " + f' () c. Mean Value Theorem: ( )! () f x f c f' ()= c lim = lim!! x" c x! c x" c f' ( k), x < k < c f ( x)! f () c f' ()= c lim = lim + + x" c x! c x" c f' ( k), c < k < x The derivative of a function cannot have any jump discontinuities!

19 Riemann s function: f( x)= { } nx n! " 2 n= At x = a 2 b, gcd ( a, 2b)=, the function jumps by! 2 2 8b The key to the integrability is that given any positive number, no matter how small, there are only a finite number of places where is jump is larger than that number. Conclusion: F x x ( )= ()! f t dt exists and is well - defined 0 for all x, but F is not differentiable at any rational number with an even denominator.

20 The Fundamental Theorem of Calculus:. If F' ( x)= f ( x), then " f ( x) dx = F( b)! F( a). b a 2.If f does not have a jump discontinuity at x, then d dx x! f t dt f x. a () = ( )

21 If F is differentiable at x = a, can F '(x) be discontinuous at x = a?

22 If F is differentiable at x = a, can F '(x) be discontinuous at x = a? Yes!

23 " 2 x sin ( ) Fx x, x! 0, ( )= # $ 0, x = 0.

24 " 2 x sin ( ) Fx x, x! 0, ( )= # $ 0, x = 0. F' ( x)= 2xsin( x)! cos ( ) x, x " 0.

25 " 2 x sin ( ) Fx x, x! 0, ( )= # $ 0, x = 0. F' ( x)= 2xsin( x)! cos ( ) x, x " 0. F( h)! F( 0) F' ( 0)= lim h" 0 h 2 h lim sin ( ) = h h" 0 h = lim hsin ( h)= 0. h" 0

26 " 2 x sin ( ) Fx x, x! 0, ( )= # $ 0, x = 0. F' ( x)= 2xsin( x)! cos ( ) x, x " 0. F( h)! F( 0) F' ( 0)= lim h" 0 h 2 h lim sin ( ) x = h F h" 0 h = lim hsin ( h)= 0. h" 0 lim! 0 ' F' ( 0) ( x) does not exist, but does exist (and equals 0).

27 # 2xsin( f x x x x ( )= )! cos ( ), " 0, $ % 0, x = 0. Is f integrable? x % 0? " 2 x sin ( ) f t dt x, x! 0, () = # $ 0, x = 0.

28 # 2xsin( f x x x x ( )= )! cos ( ), " 0, $ % 0, x = 0. Is f integrable? V sup f x inf f x i = ( )! ( ) x" [ x, x ] x" [ xi!, xi] i! i! " ( ) Integral exists if and only if Vi xi xi" can be made as small as we wish by taking sufficiently small intervals.

29 # 2xsin( f x x x x ( )= )! cos ( ), " 0, $ % 0, x = 0. Is f integrable? Yes! V sup f x inf f x i = ( )! ( ) x" [ x, x ] x" [ xi!, xi] i! i! " ( ) Integral exists if and only if Vi xi xi" can be made as small as we wish by taking sufficiently small intervals.

30 Consider: Fx " $ x # %$ 2 sin 2, x 0, x 0, x = 0. ( )= ( )! F' #% ( ) xsin cos, x, ( x)= x! ( ) " 0 $ x x &% 0, x = 0. Is F' integrable? x? " 2 $ x & F' t dt 0 # %$ sin 2, x 0, x 0, x = 0. () = ( )!

31 Consider: Fx " $ x # %$ 2 sin 2, x 0, x 0, x = 0. ( )= ( )! F' #% ( ) xsin cos, x, ( x)= x! ( ) " 0 $ x x &% 0, x = 0. Is F' integrable? x? " 2 $ x & F' t dt 0 # %$ sin 2, x 0, x 0, x = 0. () = ( )! No!

32 Consider: Fx " $ x # %$ 2 sin 2, x 0, x 0, x = 0. ( )= ( )! F' #% ( ) xsin cos, x, ( x)= x! ( ) " 0 $ x x &% 0, x = 0. Is F' integrable? x? " 2 $ x & F' t dt 0 # %$ sin 2, x 0, x 0, x = 0. () = ( )! But the right side does equal lim # x!" 0! F' () t dt

33 What if the derivative, F ', is bounded? Will the derivative always be integrable?

34 What if the derivative, F ', is bounded? Will the derivative always be integrable? No!

35 Cantor s Set First described by H.J.S. Smith, 875

36 Cantor s Set First described by H.J.S. Smith, 875 Then by Vito Volterra, 88

37 Cantor s Set First described by H.J.S. Smith, 875 Then by Vito Volterra, 88 And finally by Georg Cantor, 883

38 Cantor s Set 0 /3 2/3 Remove [/3,2/3]

39 Cantor s Set 0 /9 2/9 /3 2/3 7/9 8/9 Remove [/3,2/3] Remove [/9,2/9] and [7/9,8/9]

40 Cantor s Set 0 /9 2/9 /3 2/3 7/9 8/9 Remove [/3,2/3] Remove [/9,2/9] and [7/9,8/9] Remove [/27,2/27], [7/27,8/27], [9/27,20/27], and [25/27,26/27], and so on What s left?

41 Cantor s Set 0 /9 2/9 /3 2/3 7/9 8/ ! " ! $ # % " =! + " $ # 3% + " $ & # # 3% 2 3 =! ( 3 2 =! = 0.! " $ # 3% + $!' % What s left has measure 0.

42 A set has measure 0 if for we can put the set inside a union of intervals whose total lengths are as close to 0 as we wish. Examples: Any finite set has measure zero. The Cantor set has measure 0.! " # $ n n = 2,,!% 2 & ' ) 2 (, * , - ) * 4 (, total length < ) (, + + * , - " 2 ( = ,

43 0, 0.,. 99,, " 3, $ 3, "! 4, + $ 4, # % # % [ )( ]! + " # 2 3 " # ! 5, + $ 5, "! 6, + $ 6, "! 7, + $ 7, % # % # %! 8, + $ %, "! 9 + # , $ 5 0 %, "! + $ # 5 0, 5 0 %,!

44 0, 0.,. 99,, " 3, $ 3, "! 4, + $ 4, # % # % [ )( ]! + " # 2 3 " # ! 5, + $ 5, "! 6, + $ 6, "! 7, + $ 7, % # % # %! 8, + $ %, "! 9 + # , $ 5 0 %, "! + $ # 5 0, 5 0 %,! Every rational number in [0,] is in at least one of these intervals. The sum of the lengths of these intervals is less than 2 + +! ! # " $ = + % =

45 Base three numbers: 02! +! 3 + 2! 9 + 0! 27 +! 8 = 03 s 3 s 9 s 27 s 8 s

46 Base three numbers: /3 s /9 s /27 s /8 s! +! 3 + 2! 9 + 0! 27 +! 8 = 03 2! + 3 0! + 9! ! =

47 Base three numbers: /3 s = /9 s /27 s tercimal number /8 s! +! 3 + 2! 9 + 0! 27 +! 8 = 03 2! + 3 0! + 9! ! =

48 Cantor s set eliminates all tercimals from 0. to 0.2. From 0.0 to 0.02 and 0.2 to 0.22 From 0.00 to 0.002, 0.02 to 0.022, 0.20 to 0.202, and 0.22 to Cantor s set eliminates all tercimals that terminate or contain a.

49 Cantor s set eliminates all tercimals from 0. to 0.2. From 0.0 to 0.02 and 0.2 to 0.22 From 0.00 to 0.002, 0.02 to 0.022, 0.20 to 0.202, and 0.22 to Cantor s set eliminates all tercimals that terminate or contain a. What s left? = ! =! " 9 = 4

50 The size of what s left is still uncountable. Each element of the Cantor set corresponds to a unique number in base 2: 4 = ! = ! = We get all of the base 2 numbers except those that terminate (rational numbers with denominators a power of 2).

51 The Fundamental Theorem of Calculus:. If F' ( x)= f ( x), then " f ( x) dx = F( b)! F( a). b a 2.If f does not have a jump discontinuity at x, then d dx x! f t dt f x. a () = ( )

52 Vito Volterra, 88: There exists a function, F(x), whose derivative, F '(x), exists and is bounded for all x, but the derivative, F '(x), cannot be integrated.

53 Create a new set like the Cantor set except the first middle piece only has length /4 each of the next two middle pieces only have length /6 the next four pieces each have length /64, etc. " % The amount left has size! $ !' # & " =! + " % # 4& + " % $ # # 4& =! ( 4! 2 = " % # 4& + %!' &

54 We ll call this set SVC (for Smith-Volterra-Cantor). It has some surprising characteristics:. SVC contains no intervals - no matter how small a subinterval of [0,] we take, there will be points in that subinterval that are not in SVC. SVC is nowhere dense.

55 We ll call this set SVC (for Smith-Volterra-Cantor). It has some surprising characteristics:. SVC contains no intervals - no matter how small a subinterval of [0,] we take, there will be points in that subinterval that are not in SVC. SVC is nowhere dense. 2. Given any collection of disjoint subintervals of [0,], if the sum of the lengths is greater than /2, then they must contain at least one point in SVC.

56 We ll call this set SVC (for Smith-Volterra-Cantor). It has some surprising characteristics:. SVC contains no intervals - no matter how small a subinterval of [0,] we take, there will be points in that subinterval that are not in SVC. SVC is nowhere dense. 2. Given any collection of disjoint subintervals of [0,], if the sum of the lengths is greater than /2, then they must contain at least one point in SVC. 3. Given any partition of [0,] into subintervals, the sum of the lengths of the intervals that contain points in SVC will always be at least /2. SVC has measure /2.

57 Volterra s construction: " 2 x sin ( ) Start with the function Fx x, x! 0, ( )= # $ 0, x = 0. Restrict to the interval [0,/8], except find the largest value of x on this interval at which F '(x) = 0, and keep F constant from this value all the way to x = /8.

58 Volterra s construction: To the right of x = /8, take the mirror image of this function: for /8 < x < /4, and outside of [0,/4], define this function to be 0. Call this function f x. ( )

59 Volterra s construction: f x ( ) is a differentiable function for all values of x, but lim f ' x and lim f ' x x! 0 ( ) ( ) + " x! 4 do not exist

60 Now we slide this function over so that the portion that is not identically 0 is in the interval [3/8,5/8], that middle piece of length /4 taken out of the SVC set.

61 We follow the same procedure to create a new function, f2( x), that occupies the interval [0,/6] and is 0 outside this interval.

62 f2( x) We slide one copy of into each interval of length /6 that was removed from the SVC set.

63 Volterra s function, V(x), is what we obtain in the limit as we do this for every interval removed from the SVC set. It has the following properties:. V is differentiable at every value of x, and its derivative is bounded (below by.03 and above by.023).

64 Volterra s function, V(x), is what we obtain in the limit as we do this for every interval removed from the SVC set. It has the following properties:. V is differentiable at every value of x, and its derivative is bounded (below by.03 and above by.023). 2. If a is a left or right endpoint of one of the removed intervals, then the derivative of V at a exists (and equals 0), but we can find points arbitrarily close to a where the derivative is +, and points arbitrarily close to a where the derivative is.

65 We ll call this set SVC (for Smith-Volterra-Cantor). It has some surprising characteristics:. SVC contains no intervals - no matter how small a subinterval of [0,] we take, there will be points in that subinterval that are not in SVC. SVC is nowhere dense. 2. Given any collection of disjoint subintervals of [0,], if the sum of the lengths is greater than /2, then they must contain at least one point in SVC. 3. Given any partition of [0,] into subintervals, the sum of the lengths of the intervals that contain points in SVC will always be at least /2. SVC has measure /2.

66 We ll call this set SVC (for Smith-Volterra-Cantor). It has some surprising characteristics:. SVC contains no intervals - no matter how small a subinterval of [0,] we take, there will be points in that subinterval that are not in SVC. SVC is nowhere dense. 2. Given any collection of disjoint subintervals of [0,], if the sum of the lengths is greater than /2, then they must contain at least one point in SVC. 3. Given any partition of [0,] into subintervals, the sum of the lengths of the intervals that contain points in SVC will always be at least /2. SVC has measure /2.

67 No matter how we partition [0,], the pieces that contain endpoints of removed intervals must have lengths that add up to at least /2. The pieces on which the variation of V ' is at least 2 must have lengths that add up to at least /2.

68 Bernhard Riemann (852, 867) On the representation of a function as a trigonometric series ( )! # ( ) i i " i" Defined definite integral as limit of f x x x Key to convergence: on each interval, look at the variation of the function V sup f x inf f x i = ( )! ( ) x" [ x, x ] x" [ xi!, xi] i! i! " ( ) Integral exists if and only if Vi xi xi" can be made as small as we wish by taking sufficiently small intervals.

69 Conclusion: Volterra s function V can be differentiated and has a bounded derivative, but its derivative, V ', cannot be integrated: d dx V x v x x ( )= ( ), but # v ( t ) dt! V ( x )" V ( 0. 0 )

70 The Fundamental Theorem of Calculus:. If F has bounded variation and then b " f ( x) dx = F( b)! F( a). a F' ( x)= f ( x), 2.If f does not have a jump discontinuity at x, then d dx x! f t dt f x. a () = ( )

71 The Lebesgue Integral Henri Lebesgue Ph.D. thesis: On a generalization of the definite integral, 90

72 f ( x )= 2xsin ( x )! cos ( x )

73 f ( x )= 2xsin ( x )! cos ( x )

74 f ( x )= 2xsin ( x )! cos ( x )

75 f ( x )= 2xsin ( x )! cos ( x )

76 f ( x )= 2xsin ( x )! cos ( x )

77 f ( x )= 2xsin ( x )! cos ( x )

78 With Lebesgue s definition, the derivative of Volterra s function is integrable. Theorem If f ' exists and is bounded on [a,b], then f ' is integrable (in the Lebesgue sense) and " b a () = ( )! ( ) f ' t dt f b f a. Theorem If f is integrable on [a,b], then d dx! x a measure 0. f () t dt = f ( x), except, possibly, on a set of