4.1 SIGMA NOTATION AND RIEMANN SUMS

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1 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 1.1 SIGMA NOTATION AND RIEMANN SUMS Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas together to get the area of the whole regio. We will use that approach, but it is useful to have a otatio for addig a lot of values together: the sigma (Σ ) otatio. Summatio Sigma otatio A way to read the sigma otatio k the sum of k squared for k equals 1 to k equals the sum of 1 divided by k for k 7 k k= equals to k equals j the sum of to the j th power for j equals 0 to j equals j=0 a + a + a + a + a 6 +a 7 7 ai i= the sum of a sub i from i equals to i equals 7 The variable (typically i, j, or k) used i the summatio is called the couter or idex variable. The fuctio to the right of the sigma is called the summad, ad the umbers below ad above the sigma are called the lower ad upper limits of the summatio. (Fig. 1) Practice 1: Write the summatio deoted by each of the followig: (a) k, 7 (b) ( 1) j 1 j, (c) (m+1). j= m=0 I practice, the sigma otatio is frequetly used with the stadard fuctio otatio: f(k+) = f(1+) + f(+) + f(+) = f() + f() + f() ad f(xi ) = f(x 1 ) + f(x ) + f(x ) + f(x ) i=1

2 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus x 1 f(x) 1 0 g(x) 1 h(x) Example 1: Use the values i Table 1 to evaluate. f(k) ad ( + f(j )). k= j= Solutio:. f(k) =. f() +. f() +. f() +. f() =. () +. (1) +. (0) +. () = 1 k= Table 1 ( + f(j )) = (+f( )) + (+f( )) + (+f( )) = (+f(1)) + (+f()) + (+f()) j= = (+) + (+) + (+1) = 1. Practice : Use the values of f, g ad h i Table 1 to evaluate the followig: (a) g(k) (b) h(j) (c) ( g(k) + f(k 1) ) k= j=1 k= Example : Solutio: For f(x) = x + 1, evaluate f(k). f(k) = f(0) + f(1) + f() + f() = (0 +1) + (1 +1) + ( +1) + ( +1) = = 18. Practice : For g(x) = 1/x, evaluate g(k) ad g(k+1). k= The summad does ot have to cotai the idex variable explicitly: a sum from k= to k= of the costat fuctio f(k) = ca be writte as 7 f(k) or = + + =. = 1. Similarly, = =. =10. k= k= k= Sice the sigma otatio is simply a otatio for additio, it has all of the familiar properties of additio. Summatio Properties Sum of Costats: C = C + C + C C ( terms) =. C Additio: (ak + b k ) = ak + bk Subtractio: (ak b k ) = ak bk Costat Multiple: C. ak = C. ak Problems 16 ad 17 illustrate that similar patters for sums of products ad quotiets are ot valid.

3 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus Sums of Areas of Rectagles I Sectio. we will approximate the areas uder curves by buildig rectagles as high as the curve, calculatig the area of each rectagle, ad the addig the rectagular areas together. Example : Evaluate the sum of the rectagular areas i Fig., ad write Solutio: the sum usig the sigma otatio. { sum of the rectagular areas } = { sum of (base). (height) for each rectagle } Usig the sigma otatio, = (1). (1/) + (1). (1/) + (1). (1/) = 7/60. (1). (1/) + (1). (1/) + (1). (1/) = k= 1 k. Practice : Evaluate the sum of the rectagular areas i Fig., ad write the sum usig the sigma otatio. The bases of the rectagles do ot have to be equal. For the rectagular areas i Fig., rectagle base height area 1 1= f()=. = 8 =1 f()= = 16 6 = f()=. = 0 so the sum of the rectagular areas is = 7. Example : Write the sum of the areas of the rectagles i Fig. usig the sigma otatio. Solutio: The area of each rectagle is (base). (height). rectagle base height area 1 x 1 x 0 f(x 1 ) (x 1 x 0 ). f(x 1 ) x x 1 f(x ) (x x 1 ). f(x ) x x f(x ) (x x ). f(x ) The area of the k th rectagle is (x k x k 1 ). f(x k ), ad the total area of the rectagles is the sum (xk x k 1 ). f(x k ).

4 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus Practice : Write the sum of the areas of the shaded rectagles i Fig. 6 usig the sigma otatio. Area Uder A Curve Riema Sums Suppose we wat to calculate the area betwee the graph of a positive fuctio f ad the iterval [a, b] o the x axis (Fig. 7). The Riema Sum method is to build several rectagles with bases o the iterval [a, b] ad sides that reach up to the graph of f (Fig. 8). The the areas of the rectagles ca be calculated ad added together to get a umber called a Riema Sum of f o [a, b]. The area of the regio formed by the rectagles is a approximatio of the area we wat. Example : Approximate the area i Fig. 9a betwee the graph of f ad the iterval [, ] o the x axis by summig the areas of the rectagles i Fig. 9b. Solutio: The total area of rectagles is ()() + (1)() = 11 square uits. I order to effectively describe this process, some ew vocabulary is helpful: a partitio of a iterval ad the mesh of the partitio. A partitio P of a closed iterval [a,b] ito subitervals is a set of +1 poits { x 0 = a, x 1, x, x,..., x 1, x = b} i icreasig order, a= x 0 < x 1 < x < x <... < x 1 < x = b. (A partitio is a collectio of poits o the axis ad it does ot deped o the fuctio i ay way.) The poits of the partitio P divide the iterval ito subitervals (Fig. 10): [x 0, x 1 ], [x 1, x ], [x, x ],..., ad [x 1, x ] with legths x 1 = x 1 x 0, x = x x 1, x = x x,..., ad x = x x 1. The poits x k of the partitio P are the locatios of the vertical lies for the sides of the rectagles, ad the bases of the rectagles have legths x k for k = 1,,,...,.

5 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus The mesh or orm of partitio P is the legth of the logest of the subitervals [x k 1,x k ], or, equivaletly, the maximum of x k for k = 1,,,...,. For example, the set P = {,,.6,.1, 6} is a partitio of the iterval [,6] (Fig. 11) ad divides the iterval [,6] ito subitervals with legths x 1 = 1, x = 1.6, x =. ad x =.9. The mesh of this partitio is 1.6, the maximum of the legths of the subitervals. (If the mesh of a partitio is "small," the the legth of each oe of the subitervals is the same or smaller.) Practice 6: P = {,.8,.8,., 6., 7, 8} is a partitio of what iterval? How may subitervals does it create? What is the mesh of the partitio? What are the values of x ad x? A fuctio, a partitio, ad a poit i each subiterval determie a Riema sum. Suppose f is a positive fuctio o the iterval [a,b] P = { x 0 = a, x 1, x, x,..., x 1, x = b} is a partitio of [a,b] c k is a x value i the k th subiterval [x k 1, x k ] : x k 1 c k x k. The the area of the k th rectagle is f(c k ). ( x k x k 1 ) = f(c k ). x k. (Fig. 1) Defiitio: A summatio of the form f(c k ). x k is called a Riema Sum of f for the partitio P. This Riema sum is the total of the areas of the rectagular regios ad is a approximatio of the area betwee the graph of f ad the x axis. Example 6: Fid the Riema sum for f(x) = 1/x ad the partitio {1,, } usig the values c 1 = ad c =. (Fig. 1) Solutio: The subitervals are [1,] ad [,] so x 1 = ad x = 1. The the Riema sum for this partitio is f(ck ). x k = f(ck ). x k = f(c 1 ). x 1 + f(c ). x = f(). () + f(). (1) = 1 () + 1 (1) = 1.7.

6 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 6 Practice 7: Calculate the Riema sum for f(x) = 1/x o the partitio {1,, } usig the values c 1 =, c =. Practice 8: What is the smallest value a Riema sum for f(x) = 1/x ad the partitio {1,, } ca have? (You will eed to select values for c 1 ad c.) What is the largest value a Riema sum ca have for this fuctio ad partitio? Table shows the results of a computer program that calculated Riema sums for the fuctio f(x) = 1/x with differet umbers of subitervals ad differet ways of selectig the poits c i i each subiterval. Whe the mesh of the partitio is small (ad the umber of subitervals large), all of the ways of selectig the c i lead to approximately the same umber for the Riema sums. For this decreasig fuctio, usig the left edpoit of the subiterval always resulted i a su that was larger tha the area. Choosig the right ed poit gave a value smaller that the area. Why? Table : Riema sums for f(x) = 1/x o the iterval [1,] Values of the Riema sum for differet choices of c k umber of mesh c k = left edge c k = "radom" poit c k = right edge subitervals = x k 1 i [x k 1, x k ] = x k As the mesh gets smaller, all of the Riema Sums seem to be approachig the same value, approximately (l = ) Example 7: Fid the Riema sum for the fuctio f(x) = si(x) o the iterval [0, π] usig the partitio {0, π/, π/, π} with c 1 = π/, c = π/, c = π/. Solutio: The subitervals (Fig. 1) are [0, π/], [π/, π/], ad [π/,π] so x 1 = π/, x = π/ ad x = π/. The Riema sum for this partitio is f(ck ). x k = si(π/). (π/) + si(π/). (π/) + si(π/). (π/) =. π + 1. π +. π.18. Practice 9: Fid the Riema sum for the fuctio ad partitio i the previous example, but use c 1 = 0, c = π/, c = π/.

7 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 7 Two Special Riema Sums: Lower & Upper Sums Two particular Riema sums are of special iterest because they represet the extreme possibilities for Riema sums for a give partitio. Defiitio: Suppose f is a positive fuctio o [a,b], ad P is a partitio of [a,b]. Let m k be the x value i the kth subiterval so that f(m k ) is the miimum value of f i that iterval, ad let M k be the x value i the kth subiterval so that f(m k ) is the maximum value of f i that iterval. LS P : US P : f(mk ). x k is the lower sum of f for the partitio P. f(mk ). x k is the upper sum of f for the partitio P. Geometrically, the lower sum comes from buildig rectagles uder the graph of f (Fig. 1a), ad the lower sum (every lower sum) is less tha or equal to the exact area A: LS P A for every partitio P. The upper sum comes from buildig rectagles over the graph of f (Fig. 1b), ad the upper sum (every upper sum) is greater tha or equal to the exact area A: US P A for every partitio P. The lower ad upper sums provide bouds o the size of the exact area: LS P A US P. For ay c k value i the kth subiterval, f(m k ) f(c k ) f(m k ) ; so, for ay choice of the c k values, the Riema sum RS P = f(ck ). x k satisfies f(mk ). x k f(ck ). x k f(mk ). x k or, equivaletly, LS P RS P US P. The lower ad upper sums provide bouds o the size of all Riema sums. The exact area A ad every Riema sum RS P for partitio P both lie betwee the lower sum ad the upper sum for P (Fig. 16). Therefore, if the lower ad upper sums are close together the the area ad ay Riema sum for P must also be close together. If we kow that the upper ad lower sums for a partitio P are withi uits of each other, the we ca be sure that every Riema sum for partitio P is withi uits of the exact area.

8 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 8 Ufortuately, fidig miimums ad maximums ca be a time cosumig busiess, ad it is usually ot practical to determie lower ad upper sums for "wiggly" fuctios. If f is mootoic, however, the it is easy to fid the values for m k ad M k, ad sometimes we ca explicitly calculate the limits of the lower ad upper sums. For a mootoic bouded fuctio we ca guaratee that a Riema sum is withi a certai distace of the exact value of the area it is approximatig. Theorem: If f is a positive, motoically icreasig, bouded fuctio o [a,b], the for ay partitio P ad ay Riema sum for P, distace betwee the Riema sum ad the exact area distace betwee the upper sum ad the lower sum {f(b) f(a)}. (mesh of P ). Proof: The Riema sum ad the exact area are both betwee the upper ad lower sums so the distace betwee the Riema sum ad the exact area is less tha or equal to the distace betwee the upper ad lower sums. Sice f is mootoically icreasig, the areas represetig the differece of the upper ad lower sums ca be slid ito a rectagle (Fig. 17) whose height equals f(b) f(a) ad whose base equals the mesh of P. The the total differece of the upper ad lower sums is less tha or equal to the area of the rectagle, {f(b) f(a)}. (mesh of P). PROBLEMS I problems 1 6, rewrite the sigma otatio as a summatio ad perform the idicated additio. 1. k k=. (1 + j). (1 + ) j=1 =1. si(πk). cos(πj) 6. 1/k j=0 I problems 7 1, rewrite each summatio usig the sigma otatio. Do ot evaluate the sums

9 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 9 I problems 1 1, use the values of a k ad b k i Table ad verify that the value i part (a) does equal the value i part (b). k a k b k 1 1 Table 1. (a) (ak + b k ) (b) ak + bk 1. (a) (ak b k ) (b) ak bk 1. (a) ak (b). ak For problems 16 18, use the values of a k ad b k i Table ad verify that the value i part (a) does ot equal the value i part (b). 16. (a) ak. bk (b) ak. bk 17. (a) (ak ) (b) ( ak ) 18. (a) ak /b k (b) ( ak ) / ( bk ) For problems 19 0, f(x) = x, g(x) = x, ad h(x) = /x. Evaluate each sum. 19. f(k) 0. f(k) 1. f(j). f(1+i) j=0 i=0. g(m). g( f(k) ). g (j) 6. k. g(k) m=1 j= h(k) 8. h(i) 9. f(). h() 0. g(k). h(k) k= i=1 =1 For problems 1 6, write out each summatio ad simplify the result. These are examples of "telescopig sums" ( (k) (k 1) ). ( (k) (k 1) ). ( k 1 k+1 ) 8. ( (k+1) (k) ). ( k+1 k ) 6. (xk x k 1 )

10 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 10 For problems 7, (i) list the subitervals determied by the partitio P, (ii) fid the values of x i, (iii) fid the mesh of P, ad (iv) calculate xi. i=1 7. P = {,,., 6, 7 } 8. P = {,.6,,.,,., 6 } 9. P = {, 1, 0, 1., } 0. P is give i Fig P is give i Fig P is give i Fig. 0.. For x i = x i x i 1, verify that xi = legth of the iterval [a, b]. i=1 For problems 8, (i) sketch the graph of f o the give iterval, (ii) draw vertical lies at each poit of the give partitio, (iii) evaluate each f(c i ) ad sketch the correspodig rectagle, ad (iv) calculate ad add together the areas of the rectagles.. f(x) = x + 1, P = { 1,,, } (a) c 1 = 1, c =, ad c =. (b) c 1 =, c =, ad c =.. f(x) = x, P = { 0, 1, 1., } (a) c 1 = 0, c = 1, ad c =. (b) c 1 = 1, c = 1., ad c = f(x) = x, P = { 0,,, 10 } (a) c 1 = 1, c =, ad c = 9. (b) c 1 = 0, c =, ad c = f(x) = si(x), P = {0, π/, π/, π } (a) c 1 = 0, c = π/, ad c = π/. (b) c 1 = π/, c = π/, ad c = π. 8. f(x) = x, P = { 0, 1, } (a) c 1 = 0, c =. (b) c 1 = 1, c =. For problems 9, sketch the fuctio ad fid the smallest possible value ad the largest possible value for a Riema sum of the give fuctio ad partitio. 9. f(x) = 1 + x (a) P = { 1,,, } (b) P = { 1,,,, } (c) P = { 1, 1.,,,, } 0. f(x) = 7 x (a) P = { 0,, } (b) P = { 0, 1,, } (c) P = { 0,., 1, 1.,, } 1. f(x) = si(x) (a) P = { 0, π/, π } (b) P = { 0, π/, π/, π } (c) P = { 0, π/, π/, π }. f(x) = x x + (a) P = { 0,, } (b) P = { 0, 1,, } (c) P = { 0,., 1,,., } Upper ad Lower Sum Problems. Suppose LS P = 7.6 ad US P = 7.0 for a positive fuctio f ad a partitio P of the iterval [ 1, ]. We ca be certai that every Riema sum for the partitio P is withi what distace of the exact value of the area betwee the graph of f ad the iterval [ 1,]? (b) What if LS P = 7.7 ad US P = 7.90?

11 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 11. Suppose we divide the iterval [ 1, ] ito 100 equally wide subitervals ad calculate a Riema sum for f(x) = 1 + x by radomly selectig a poit c i i each subiterval. We ca be certai that the value of the Riema sum is withi what distace of the exact value of the area betwee the graph of f ad the iterval [ 1, ]? (b) What if we take 00 equally log subitervals?. If we divide the iterval [, ] ito 0 equally wide subitervals ad calculate a Riema sum for f(x) = 1 + x by "radomly" selectig a poit c i i each subiterval (ay poit c i i the subiterval), the we ca be certai that the Riema sum is withi what distace of the exact value of the area betwee f ad the iterval [, ]? 6. If f is mootoic decreasig o [a, b] ad we divide the iterval [a, b] ito equally wide subitervals (Fig. 1), the we ca be certai that the Riema sum is withi what distace of the exact value of the area betwee f ad the iterval [a, b]. Summig Powers of Cosecutive Itegers Explicit formulas for some commoly ecoutered summatios are kow ad are useful for explicitly evaluatig some Riema sums ad their limits. The formulas below are icluded here for your referece. They will ot be used or eeded i the followig sectios. The summatio formula for the first positive itegers is relatively well kow, has several easy but clever proofs, ad eve has a iterestig story. (+1) ( 1) + = k = Proof: Let S represet the sum ( ) + ( 1) +. Rearragig the summads, we also kow S = + ( 1) + ( ) Addig these represetatios of S together, S = ( ) + ( 1) + + S = + ( 1) + ( ) S = (+1) + (+1) + (+1) (+1) + (+1) + (+1) =. (+1) so S = (+1), the desired formula.

12 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 1 It is said that Gauss discovered this formula for himself at the age of. His teacher, plaig o keepig the class busy for a while, asked the studets to add the itegers from 1 to 100. Gauss thought a few miutes, wrote his aswer o his slate, ad tured it i. Accordig to the story, he the sat smugly while his classmates struggled with the problem. 7. Fid the sum of the first 100 positive itegers i ways: (1) usig Gauss' formula, ad () usig Gauss' method. 8. Fid the sum of the first 10 odd itegers. (Each odd iteger ca be writte i the form k 1 for k = 1,,,... ) 9. Fid the sum of the itegers from 10 to 0. Formulas for other iteger powers of the first itegers have bee discovered: 1 k = + 1 = (+1) 1 k = = (+1)(+1) 6 1 k = = ( (+1) ) 1 k = = (+1)(+1)( + 1) 0 I problems 60 6, use the properties of summatio ad the formulas for powers to evaluate each sum ( + k + k ) k. (k + 1) k. (k ) Sectio.1 PRACTICE Aswers Practice 1: (a) k = (c) (m+1) = m=0 7 (b) ( 1) j 1 j = j=

13 .1 Sigma Notatio ad Riema Sums Cotemporary Calculus 1 Practice : (a) g(k) = g() + g() + g() + g() = 1 + ( ) + + = 7 k= (b) h(j) = h(1) + h() + h() + h() = = 1 j=1 (c) ( g(k) + f(k 1) ) = (g() + f()) + (g() + f()) + (g() + f()) = ( +)+(+1)+(+0) = 10 k= Practice : 1 For g(x) = 1/x, g(k) = g() + g() + g() = = 1 1 ad k= 1 g(k+1) = g() + g() + g() = 1. Practice : Rectagular areas = = 1 = 1 j j=1. Practice : f(x 0 )(x 1 x 0 ) + f(x 1 )(x x 1 ) + f(x )(x x ) = j=1 f( x j 1 ) (x j x j 1 ) or f( x k ) (x k+1 x k ). Practice 6: Iterval is [, 8]. 6 subitervals. mesh = legth of logest subiterval = 1.. x =.8 ad x = x x 1 =.8.8 = 1. Practice 7: RS = ()( 1 ) + (1)( 1 ) = 1. Practice 8: smallest RS = ()( 1 ) + (1)( 1 ) = 0.9 largest RS = ()( 1 ) + (1)( 1 ) =. Practice 9: RS = (0)( π ) + (1)( π ) + (1)( π ).6.