Math 8 Istructor: Padraic Bartlett Lecture 6: Itegratio ad the Mea Value Theorem Week 6 Caltech - Fall, 2011 1 Radom Questios Questio 1.1. Show that ay positive ratioal umber ca be writte as the sum of reciprocals of fiitely may distict positive itegers. I other words, we ca express 29 24 = 1 1 + 1 3 + 1 8, ad i geeral fid distict 1, k for ay positive ratioal umber p q such that p q = 1 1 +... + 1 k. Questio 1.2. Hercules ad the Hydra. You are Hercules! Appropriately, the, you re doig battle with a Hydra. Mathematically, we ca describe a Hydra as a collectio of vertices i R 2 ad edges coectig these lies up, of the followig form: two head vertices A example of a hydra with 4 vertices: a eck vertex the body vertex I order for a collectio of vertices ad edges to be a hydra, we require that our hydra is coected 1 ad loopless 2 I ay hydra, we pick a special vertex to call the body of the hydra: give this defiitio, we ca say that a give vertex is distace k from the body if the shortest path from that vertex to the body ode uses k edges. All of the o-body vertices with exactly oe eighbor are called the heads of the hydra. A vertex coected to a head ode is called a eck. Battle is a rather civilized affair, ad goes as follows: Hercules, o his tur, ca cut off ay oe head of the hydra: i.e. he ca pick ay head vertex of the hydra, ad remove it from the hydra, alog with the sigle edge that coected that vertex to the hydra. 1 Give ay two vertices, there is a sequece of distict edges coectig these two vertices: i.e. your hydra does ot start already chopped ito two pieces. 2 Give ay two vertices, there is exactly oe sequece of distict edges coectig these two vertices: i.e. if the ecks of your hydra split somewhere, they do t joi back together. 1
The hydra tha respods by lookig at the eck vertex that was cut, ad movig from this vertex oe step closer to the body of the hydra. From there, the hydra creates k copies of the vertices ad edges all attached above this vertex, ad attaches them all to this vertex, where k is the umber of heads Hercules has cut off so far i the battle. To illustrate what s goig o, here s a sample battle: Hercules, tur 1 Hydra respods, tur 1 Hercules, tur 2 Hydra respods, tur 2... Hercules, tur 3 Hydra respods, tur 3 Show that Hercules will always wi, o matter how may heads the Hydra starts with, ad o matter how dumb his strategy is. I believe this holds eve for hydras with ifiitely may heads: see this referece for some thoughts? Questio 1.3. Suppose you have a Z Z grid of squares. Cosider the followig game we ca play o this board: Startig positio: place oe coi o every sigle square below the x-axis. Moves: If there are two cois i a row horizotally or vertically with a empty space ahead of them, you ca jump oe of the cois over the other i.e. you ca remove those two cois ad put a ew coi o the space directly ahead of them. I this game, how high o the y-axis ca you get a coi? Ca you get oe to height 3? Higher? Why or why ot? 2
2 The Mea Value Theorem The Mea Value Theorem abbreviated MVT is the followig result: Theorem 2.1. Suppose that f is a cotiuous fuctio o the iterval [a, b] that s differetiable o a, b. The there is some value c such that f c = fb fa. b a I other words, there is some poit c betwee fa ad fb such that the derivative at that poit is equal to the slope of the lie segmet coectig a, fa ad b, fb. The followig picture illustrates this claim: fb fa slope = fb-fa b-a = f c a c b A commo/practical example of the mea value theorem s use i day-to-day life is o toll roads: suppose that you get o a road at 2pm, travel 150 miles, ad get off the road at 4pm. The mea value theorem says that at some poit i time, you must have traveled 75mph, because i order to travel 150 miles i 2 hours your average speed must have bee 75 mph 3. The mea value theorem, much like the itermediate value theorem, is usually ot a tough theorem to uderstad: the tricky thig is realizig whe you should try to use it. Roughly speakig, you wat to use the mea value theorem wheever you wat to tur iformatio about a fuctio ito iformatio about its derivative, or vice-versa. We work two examples of how the mea value theorem is used below: 2.1 Applicatios of the Mea Value Theorem Example. Cosider the equatio x + y = x + y. 3 Toll booths i some states will actually use this iformatio to automatically prit out speedig tickets for people! Because the ticket you purchace whe you eter the toll road states whe ad where you etered the highway, ad you must show your ticket to leave the highway, they kow whe ad where you etered ad left. This therefore determies your average speed, which by the mea value theorem you must have attaied at some poit i your jourey! 3
If either x or y are zero, this equatio holds; as well, if x = y ad is odd, this equatio also holds. Are there ay other values of x, y R ad N that are solutios of this equatio? Proof. First, otice the followig lemma we ca prove usig the mea value theorem: Lemma 1. Suppose you have a differetiable fuctio f with k distict roots a 1 < a 2 <... a k. The f has at least k 1 distict roots b 1 < b 2 <... b k 1, such that a 1 < b 1 < a 2 < b 2 <... b k 1 < a k. Proof. We kow f is differetiable ad cotiuous o [a 1, a 2 ]: therefore, by the mea value theorem, we ca fid some value b 1 such that f b 1 = fa 2 fa 1 a 2 a 1. But we kow that fa 2 = fa 1 = 0, because a 1 ad a 2 are roots of f: therefore, we actually have that f b 1 = fa 2 fa 1 a 2 a 1 = 0 0 a 2 a 1 = 0, ad that b 1 is a root of f. Repeatig this for all of the other pairs a j, a j+1 will create the k 1 roots b 1,... b k 1 that we claimed exist. What does this lemma tell us about our problem? Well: pick ay fixed ozero value of y i R, ad look at the equatio fx = x + y x y. We are curretly tryig to fid out which values of x give us a root of this equatio. Curretly, we kow the followig roots: If is eve, the oly root we kow is x = 0. If is odd, we kow that x = 0, x = y are two roots. Ca there be ay more distict roots? Well: by our earlier lemma, we kow that if our fuctio were to have k roots, its derivative would have to have at least k 1 roots. So: how may roots ca f have? Calculatig tells us that which is equal to 0 wheever f x = x + y 1 x 1, 0 = x + y 1 x 1 x 1 = x + y 1 x 1 = x + y 1. 4
Here, we have two cases. If is eve, we kow that 1 is odd, ad therefore the equatio x 1 = x + y 1 is equivalet to the claim x = x + y 0 = y, which cotradicts our ozero choice of y. So whe is eve, we caot have that f has a root: by our lemma, this meas that whe is eve, f caot have more tha 1 distict root. So the oly root of f whe is eve is x = 0, because we ve just show that there ca be o others. If is odd, we kow that 1 is eve, i which case our equatio x 1 = x + y 1 is equivalet to the claim x = x + y ±x = x + y y = 0 or y = 2x. So, i the case where is odd ad y is ozero, f has exactly oe root at x = y 2. By our lemma, this meas that whe is odd, f caot have more tha 2 distict roots. So the oly roots of f whe is odd are x = 0 ad x = y, because we ve just show that there ca be o others. The example above showed how we could tur iformatio about our fuctio specifically, kowledge of where its roots are ito iformatio about the derivative. The mea value theorem ca also be used to tur iformatio about the derivative ito iformatio about the fuctio, as we illustrate here: Example. Let pt deote the curret locatio of a particle movig i a oe-dimesioal space. Suppose that p0 = 0, p1 = 1, ad p 0 = p 1 = 0. Show that there must be some poit i time i [0, 1] where p t 4. Proof. We proceed by cotradictio: i.e. suppose istead that for every t [0, 1] we have p t < 4. What ca we do from here? Well: we have some boudary coditios p0 = 0, p1 = 1, p 0 = 0, p 1 = 0 ad oe global piece of iformatio p t < 4. How ca we tur this kowledge of the secod derivative ito iformatio about rest of the fuctio? Well: if we apply the mea value theorem to the fuctio p t, what does it say? It tells us that o ay iterval [a, b], there is some c a, b such that p b p a b a = p x = p c. I other words, it relates the first ad secod derivatives to each other! So, if we apply our kow boud p t < 4, t [0, 1], we ve just show that p b p a b a = p c < 4, 5
for ay a < b [0, 1]. I particular, if we set a = 0, b = t ad remember our boudary coditio p 0 = 0, we ve prove that p t p 0 t 0 = p t 0 t = p t < 4 t p t < 4t. Similarly, if we let a = 1 t ad b = 1, we get p 1 p 1 t 1 1 t = 0 p 1 t t = p 1 t t p 1 t < 4t. < 4 Excellet! We ve tured iformatio about the secod derivative ito iformatio about the first derivative. Preted, for the momet, that you re back i your high school calculus courses, ad you kow how to fid atiderivatives. I this situatio, you ve got a fuctio pt with the followig properties: p0 = 0, p1 = 1, p t < 4t, ad p 1 t < 4t. What do you kow about pt? Well: if p t < 4t, you ca itegrate to get that pt < 2t 2 +C, for some costat C: usig our boudary coditio p0 = 0 tells us that i specific we ca pick C = 0, ad we have pt < 2t 2, t 0, 1. Similarly, if we use our observatio that p1 t > 4t, we ca itegrate to get that p1 t > 2t 2 + C: usig our boudary coditio p1 = 1 tells us that i specific we ca pick C = 1, which gives us p1 t > 2t 2 + 1, t 0, 1. But what happes if we plug i t = 1 2? I our first boud, we have p 1 2 < 2 1 2 2 = 1 2. Coversely, i our secod boud we have p 1 1 2 > 2 1 2 2 + 1 = 1 2 : i other words, at 1 2 our fuctio must both be greater tha ad less tha 1 2! This is clearly impossible, so we ve reached a cotradictio... Assumig, of course, that we kow how to perform atidifferetiatio. Which we do t at least, ot officially! How ca we solve this problem usig just the mea value theorem? 6
Earlier, we tured iformatio about the secod derivative ito iformatio about the first derivative. Ca we do that trick agai to get iformatio about the origial fuctio? Well: let s try applyig the mea value theorem to the fuctio f, o the iterval [0, t]. This tells us that there is some value of c 0, t such that ft f0 t 0 = f c. If we use our boudary coditio f0 = 0 ad our boud f c < 4c < 4t, c 0, t, this becomes ft 0 t = f c < 4t ft < 4t 2. This is... ot the boud of 2t 2 that we got by atidifferetiatig. What ca we do here? Well: what happes if we take this boud, ad look at the iterval [t, 2t]? Applyig the mea value theorem there tells us that there is some c t, 2t such that f2t ft = f c 2t t f2t ft = t f c < t 4c < t 4 2t = 8t 2 f2t < 8t 2 + f t < 4t 2 + 8t 2. Similarly, if we look at [2t, 3t] ad apply the mea value theorem, we get that there is some c 2t, 3t such that f3t f2t = f c 3t 2t f3t f2t = t f c < t 4c < t 4 3t = 12t 2 f3t < 12t 2 + f 2t < 4t 2 + 8t 2 + 12t 2, ad by a simple iductive argumet that if we look at [ 1t, t], we ll get the boud ft < 4t 2 + 8t 2 + 12t 2 +... + 4t 2 = 41 + 2 +... + t 2 But we kow that the sum of the first positive itegers is just +1 2 : so we ve show that that + 1 ft < 4 t 2 = 2t 2 + 1. 2 So: take ay y [0, 1], ad write y = y. The, applyig the above boud tells us fy = f y y 2 < 2 + 1 = 2y2 + 1 2 = 2y 2 + 1. Lettig go to ifiity tells us that fy < 2y 2. Idetical calculatios show that p1 y > 2y 2 + 1, ad therefore that i particular for y = 1 2 we have p 1 2 both greater tha ad less tha 1 2, a cotradictio. 7
3 Itegratio Defiitio 3.1. A fuctio f is itegrable o the iterval [a, b] if ad oly if the followig holds: For ay ɛ > 0, there is a partitio a = t 1 < t 2 <... < t 1 < t = b of the iterval [a, b] such that sup fx t i+1 t i if fx t i+1 t i < ɛ x t i,t i+1 x t i,t i+1 Oe way to iterpret the sums above is through the followig picture: a=t 1 t 2 t 3 t 4 t 5 t 6 t 7 b=t 8 Specifically, thik of the if-sum as the area of the blue rectagles i the picture below, ad thik of the sup-sum as the area of the red rectagles i the picture below. The, the differece of these two sums ca be thought of as the area of the gray-shaded rectagles i the picture above. Thus, we re sayig that a fuctio fx is itegrable iff we ca fid collectios of red rectagles a upper limit o the area uder the curve of fx ad collectios of blue rectagles a lower limit o the area uder the curve of fx such that the area of these upper ad lower approximatios are arbitrarily close to each other. Note that the above coditio is equivalet to the followig claim: if fx is itegrable, we ca fid a sequece of partitios {P } such that the area of the gray rectagles with respect to the P partitios goes to 0 i.e. a sequece of partitios {P } such that lim sup fx t i+1 t i if fx t i+1 t i = 0. x t i,t i+1 x t i,t i+1 8
I other words, there s a series of partitios P such that these upper ad lower sums both coverge to the same value: i.e. a collectio of partitios P such that lim sup fx t i+1 t i = lim x t i,t i+1 If this happes, the we defie b a fxdx = lim sup fx t i+1 t i = lim x t i,t i+1 if fx t i+1 t i. x t i,t i+1 if fx t i+1 t i, x t i,t i+1 ad say that this quatity is the itegral of fx o the iterval [a, b]. For coveiece s sake, deote the upper sums of fx over a partitio P as Ufx, P, ad the lower sums as Lfx, P. This discussio, hopefully, motivates why we ofte say that the itegral of some fuctio fx is just the area uder the curve of fx. Pictorially, we are sayig that a fuctio is itegrable if ad oly if we ca come up with a well-defied otio of area for this fuctio; i other words, if sufficietly fie upper bouds for the area beeath the curve the sup- sums are arbitrarily close to sufficietly fie lower bouds for the area beeath the curve the if-sums. The defiitio of the itegral, sadly, is a tricky oe to work with: the sups ad ifs ad sums over partitios amout to a to of otatio, ad it s easy to get lost i the symbols ad have o idea what you re actually maipulatig. If you ever fid yourself feelig cofused i this way, just remember the picture above! Basically, there are three thigs to iteralize about this defiitio: the area of the red rectagles correspods to the upper-boud sup-sums, the area of the blue rectagles correspods to the lower-boud if-sums, ad if these two sums ca be made to be arbitrarily close to each other i.e. the area of the gray rectagles ca be made arbitrarily small the we have a good idea of what the area uder the curve is, ad ca say that b a fx is just the limit of the area of those red rectagles uder icreasigly smaller partitios which is also the limit of the area of the blue rectagles. The itegral is a difficult thig to work with usig just the defiitio: later o, we ll develop lots of tools to help us actually do otrivial thigs with the itegral. To illustrate how workig with the defiitio goes, though, let s work two examples: 3.1 Calculatig the Itegral Example. The itegral of ay costat fuctio fx = C from a to b exists; furthermore, b a cdx = C b a. 9
Proof. Pick ay costat fuctio fx = C. To use our defiitio of the itegral, we eed to fid a sequece of partitios P such that lim Ufx, P Lfx, P goes to 0. How ca we do this? Well: what kids of partitios of [a, b] ito parts eve exist? Oe partitio that ofte comes i hady is the uiform partitio, where we break [a, b] ito pieces all of the same legth: i.e. the partitio { P = a, a + b a, a + 2b a, a + 3b a,... a + b a = b. I almost ay situatio where you eed a partitio, this will work excelletly! I particular, oe advatage of this partitio is that the legths t i+1 t i i the upper ad lower sums are all the same: they re specifically b a. Let s see what this partitio does for our itegral. If we look at Ufx, P, where P is the uiform partitio defied above, we have 1 Ufx, P = i=0 1 = i=0 1 = sup x a+i b a b a,a+i+1 x a+i b a b a C i=0 sup =C b a + C b a = C b a. b a,a+i+1 fx } a + i + 1 b a a ib a fx b a, because fx is a costat fuctio, +... + C b a Similarly, if we look at Lfx, P, we have 1 Lfx, P = if i=0 x a+i b a b a,a+i+1 fx 1 b a = if i=0 x a+i b a,a+i+1 b a fx 1 b a = C i=0 =C b a + C b a = C b a. a + i + 1 b a, because fx is a costat fuctio, +... + C b a a ib a Therefore, we have that the limit lim Ufx, P Lfx, P = lim C b a C b a = lim 0 = 0, 10
ad cosequetly our itegral exists ad is equal to lim Ufx, P = C b a. Example. The fuctio fx = x p is itegrable o [0, b] for ay p N ad b R +. Furthermore, the itegral of this fuctio is bp+1 p+1. Proof. Our uiform partitio, where we broke our iterval up ito equal parts, worked pretty well for us above! Let s see if it ca help us i this problem as well. If we let P = {0, b, 2 b,... b = b, we have that Ufx, P is just 1 sup x k b,k+1 b x p k + 1 b k b 1 = k + 1 b p b = bp+1 p+1 k + 1 p, k=1 ad that the lower-boud sum, if, is 1 if x k b,k+1 xp b k + 1 b k b = 1 k b p b = bp+1 p+1 k p. k=1 Takig the limit of their differece, we have that lim Ufx, P b p+1 1 Lfx, P = lim b p+1 = lim p+1 p+1 1 k + 1 p k + 1 p b p+1 p+1 1 k p 1 k p b p+1 = lim p+1 1p + 2 p + 3 p +... + p 0 p + 1 p + 2 p +... + 1 p = lim = lim = 0. b p+1 p+1 p b p+1 Thus, by our defiitio, the fuctio x p is itegrable o [0, b]! Furthermore, we have that the itegral of this fuctio is just b p+1 1 lim Ufx, P = lim p+1 k + 1 p = lim b p+1 p+1 k p. k=1 11
So: it suffices to uderstad what the sum k=1 kp is, for ay p. Ufortuately, doig that is rather hard: see Faulhaber s formula for some discussio about this matter. I our ext set of lectures, we ll come up with some workarouds for this issue! 12