Riemann Integration Theory

Transcription

1 Riemann Integration Theory James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 3, 2017

2 Outline 1 Uniform Partition Riemann Sums 2 Refinements of Partitions 3 Defining the Riemann Integral 4 Fundamental Integral Estimates

3 Uniform Partition Riemann Sums To save typing, let s learn to use a Matlab function. In Matlab s file menu, choose create a new Matlab function which gives f u n c t i o n [ v a l u e 1, v a l u e 2,... ] = MyFunction ( arg1, arg2,... ) % s t u f f i n h e r e end [value1, value2,...] are returned values the function calculates that we want to save. (arg1, arg2,...) are things the function needs to do the calculations. They are called the arguments to the function. MyFunction is the name of the function. This function must be stored in the file MyFunction.m.

4 Uniform Partition Riemann Sums Our function returns the Riemann sum, RS, and use the arguments: our function f, the partition P and the Evaluation set E. Since only one value returned [RS] can be RS. f u n c t i o n RS = RiemannSum ( f, P, E ) % comments alway b e g i n w i t h a % matlab l i n e s h e r e end The name for the function RiemannSum must be used as the file name: i.e. we must use RiemannSum.m as the file name.

5 Uniform Partition Riemann Sums The Riemann sum function: 1 f u n c t i o n RS = RiemannSum ( f, P, E ) % f i n d Riemann sum dx = d i f f (P) ; RS = sum ( f ( E). dx ) ; [ s i z e P,m] = s i z e (P) ; %g e t s i z e o f P a r t i t i o n 6 c l f ; % c l e a r t h e o l d graph h o l d on % s e t h o l d to on f o r i = 1 : s i z e (P) 1 % graph r e c t a n g l e s % p l o t r e c t a n g l e code... end 11 % p l o t f u n c t i o n code... y = l i n s p a c e (P( 1 ),P( s i z e P ), 101) ; h o l d o f f ; end

6 Uniform Partition Riemann Sums Now to see graphically how the Riemann sums converge to a finite number, let s write a new function: Riemann sums using uniform partitions and midpoint evaluation sets. 1 f u n c t i o n RS = RiemannUniformSum ( f, a, b, n ) % s e t up a u n i f o r m p a r t i t i o n w i t h n+1 p o i n t s d e l t a x = ( b a ) /n ; P = [ a : d e l t a x : b ] ; % makes a row v e c t o r f o r i =1:n 6 s t a r t = a+( i 1) d e l t a x ; s t o p = a+i d e l t a x ; E ( i ) = 0. 5 ( s t a r t+s t o p ) ; end % send i n t r a n s p o s e o f P and E so we use column v e c t o r s 11 % b e c a u s e o r i g i n a l RiemannSum f u n c t i o n u s e s columns RS = RiemannSum ( f, P, E ) ; end

7 Uniform Partition Riemann Sums We can then generate a sequence of Riemann sums for different values of n. We generate a sequence of figures which converge to a fixed value. >> f x ) s i n ( 3 x ) ; >> RS = RiemannUniformSum ( f, 1,4, 10) ; >> RS= RiemannUniformSum ( f, 1,4, 20) ; >> RS = RiemannUniformSum ( f, 1,4, 30) ; >> RS= RiemannUniformSum ( f, 1,4, 40) ;

8 Uniform Partition Riemann Sums Figure: The Riemann sum with a uniform partition P 10 of [ 1, 4] for n = 10. The function is sin(3x) and the Riemann sum is

9 Uniform Partition Riemann Sums Figure: Riemann sum with a uniform partition P 20 of [ 1, 4] for n = 20. The function is sin(3x) and the Riemann sum is

10 Uniform Partition Riemann Sums Figure: Riemann sum with a uniform partition P 40 of [ 1, 4] for n = 40. The function is sin(3x) and the Riemann sum is

11 Uniform Partition Riemann Sums Figure: Riemann sum with a uniform partition P 80 of [ 1, 4] for n = 80. The function is sin(3x) and the Riemann sum is

12 Refinements of Partitions For each j = 1,..., n 1, we let x j = x j+1 x j. The collection of all finite partitions of [a, b] is denoted Π[a, b]. Definition The Refinement of a Partition: The partition π 1 = {y 0,..., y m } is said to be a refinement of the partition π 2 = {x 0,..., x n } if every partition point x j π 2 is also in π 1. If this is the case, then we write π 2 π 1, and we say that π 1 is finer than π 2 or π 2 is coarser than π 1. Definition Common Refinement: Given π 1, π 2 Π[a, b], there is a partition π 3 Π[a, b] which is formed by taking the union of π 1 and π 2 and using common points only once. We call this partition the common refinement of π 1 and π 2 and denote it by π 3 = π 1 π 2.

13 Refinements of Partitions The relation is a partial ordering of Π[a, b]. It is not a total ordering, since not all partitions are comparable. There is a coarsest partition, also called the trivial partition. It is given by π 0 = {a, b}. We may also consider uniform partitions of order k. Let h = (b a)/k. Then π = {x 0 = a, x 0 + h, x 0 + 2h,..., x k 1 = x 0 + (k 1)h, x k = b}. Theorem Refinements and Common Refinements If π 1, π 2 Π[a, b], then π 1 π 2 if and only if π 1 π 2 = π 2. Proof If π 1 π 2, then π 1 = {x 0,..., x p } {y 0,..., y q } = π 2. Thus, π 1 π 2 = π 2, and we have π 1 π 2 = π 2. Conversely, suppose π 1 π 2 = π 2. By definition, every point of π 1 is also a point of π 1 π 2 = π 2. So, π 1 π 2.

14 Defining the Riemann Integral We abuse notation somewhat by using the notation σ π to denote that the Evaluation set σ is chosen from the subintervals determined by the partition π. Definition We say f B[a, b] is Riemann Integrable on [a, b] if there exists a real number, I, such that for every ɛ > 0 there is a partition, π 0 Π[a, b] such that S(f, π, σ) I < ɛ for any refinement, π, of π 0 and any evaluation set, σ π. We denote this value, I, by I RI (f ; a, b)

15 Defining the Riemann Integral We denote the set of Riemann integrable functions on [a, b] by RI [a, b]. Also, it is readily seen that the number RI (f ; a, b) in the definition above, when it exists, is unique. So we can speak of Riemann Integral of a function, f. We also have the following conventions. 1 RI (f ; a, b) = RI (f ; b, a) 2 RI (f ; a, a) = 0 3 f is called the integrand. Theorem RI [a, b] is a vector space over R and the mapping I R : RI [a, b] R defined by I R (f ) = RI (f ; a, b) is a linear mapping.

16 Defining the Riemann Integral Proof Let f 1, f 2 RI [a, b], and let α, β R. For any π Π[a, b] and σ π, we have S(αf 1 + βf 2, π, σ) = π (αf 1 + βf 2 )(s j ) x j = α π f 1 (s j ) x j + β π f 2 (s j ) x j = αs(f 1, π, σ) + βs(f 2, π, σ). Since f 1 is Riemann integrable, given ɛ > 0, there is a real number I 1 = RI (f 1, a, b) and a partition π 1 Π[a, b] such that S(f 1, π, σ) I 1 < ɛ 2( α +1) ( ) for all refinements π of π 1 and all σ π.

17 Defining the Riemann Integral Proof Likewise, since f 2 is Riemann integrable, there is a real number I 2 = RE(f 2 ; a, b) and a partition π 2 Π[a, b] such that S(f 2, π, σ) I 2 < ɛ 2( β +1) ( ) for all refinements π of π 2 and all σ π. Let π 0 = π 1 π 2. Then π 0 is a refinement of both π 1 and π 2. So, for any refinement, π, of π 0, and any σ π, we have Equation and Equation are valid. Hence, S(f 1, π, σ) I 1 < S(f 2, π, σ) I 2 < ɛ 2( α +1) ɛ 2( β +1).

18 Defining the Riemann Integral Proof Thus, for any refinement π of π 0 and any σ π, it follows that S(αf 1 + βf 2, π, σ) (αi 1 + βi 2 ) = αs(f 1, π, σ) + βs(f 2, π, σ) αi 1 βi 2 α S(f 1, π, σ) I 1 + β S(f 2, π, σ) I 2 ɛ < α 2( α +1) + β ɛ 2( β +1) < ɛ. This shows that αf 1 + βf 2 is Riemann integrable and that the value of the integral RI (αf 1 + βf 2 ; a, b) is given by αri (f 1 ; a, b) + βri (f 2 ; a, b). It then follows immediately that I R is a linear mapping.

19 Fundamental Integral Estimates Theorem Let f RI [a, b]. Let m = inf a x b f (x) and let M = sup a x b f (x). Then m(b a) RI (f ; a, b) M(b a). Proof If π Π[a, b], then for all σ π, we see that m x j π π f (s j ) x j M x j. π

20 Fundamental Integral Estimates Proof But π x j = b a, so m(b a) π f (s j ) x j M(b a), or m(b a) S(f, π, σ) M(b a), for any partition π and any σ π. Now, let ɛ > 0 be given. Then there exist π 0 Π[a, b] such that for any refinement, π of π 0 and any σ π, RI (f ; a, b) ɛ < S(f, π, σ) < RI (f ; a, b) + ɛ. Hence, for any such refinement, π, and any σ π, we have m(b a) S(f, π, σ) < RI (f ; a, b) + ɛ

21 Fundamental Integral Estimates Proof and M(b a) S(f, π, σ) > RI (f ; a, b) ɛ. Since ɛ > 0 is arbitrary, it follows that m(b a) RI (f ; a, b) M(b a). Theorem The Riemann integral is order preserving. That is, if f, f 1, f 2 RI [a, b], then (i) (ii) f 0 RI (f ; a, b) 0; f 1 f 2 RI (f 1 ; a, b) RI (f 2 ; a, b).

22 Fundamental Integral Estimates Proof If f 0 on [a, b], then inf x f (x) = m 0. Hence, by our fundamental estimates results for integration, we have b This proves the first assertion. a f (x)dx m(b a) 0. To prove (ii), let f = f 2 f 1. Then f 0, and the second result follows from the first.

23 Fundamental Integral Estimates Homework 10 For the given function f, interval [a, b] and choice of n, you ll calculate the corresponding uniform partition Riemann sum using the functions RiemannSum in file RiemannSum.m and RiemannUniformSum in file RiemannUniformSum.m. You can download these functions as files from the class web site. Save them in your personal class directory. Create a new word document in single space with matlab fragments in bold font. The document starts with your name, MATH 4540, Section, HW number and the date.

24 Fundamental Integral Estimates Homework 10 Continued Create a new word document for this homework. Do the document in single space. Do matlab fragments in bold font. The document starts with your name, MATH 4540, Section Number, Date and Homework number. For each value of n, do a save as and save the figure with a filename like HW#Problem#a[ ].png where [ ] is where you put the number of the graph. Something like HW10a.png, HW#Problem#b.png etc. Insert this picture into the doc resizing as needed to make it look good. Explain in the doc what the picture shows.

25 Fundamental Integral Estimates Homework 10 Continued Something like this: Jim Peterson MATH 4540, Section Number today s date and HW Number, Problem 1: Let f (t) = sin(5t) on the interval [ 1, 4] and generate the uniform Riemann sum approximations for n = 10, 20, 40 and 80. % add e x p l a n a t i o n h e r e >> f x ) s i n ( 5 x ) ; % add e x p l a n a t i o n h e r e >> RS = RiemannUniformSum ( f, 1,4,10) % add e x p l a n a t i o n h e r e >> RS = RiemannUniformSum ( f, 1,4,20) % add e x p l a n a t i o n h e r e >> RS = RiemannUniformSum ( f, 1,4,40) % add e x p l a n a t i o n h e r e >> RS = RiemannUniformSum ( f, 1,4,80)

26 Fundamental Integral Estimates Homework 10 Continued Three are MatLab uniform partition problems like discussed above and the others are theory based Let f (t) = t 2 2t + 3 on the interval [ 2, 3] with n = 8, 16, 32 and Let f (t) = sin(2t) on the interval [ 1, 5] with n = 10, 40, 60 and Let f (t) = t 2 + 8t + 5 on the interval [ 2, 3] with n = 4, 12, 30 and Prove that the relation is a partial ordering of Π[a, b].