SOME GEOMETRY IN HIGHDIMENSIONAL SPACES


 Juniper McCarthy
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1 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES MATH 57A. Itroductio Our geometric ituitio is derived from threedimesioal space. Three coordiates suffice. May objects of iterest i aalysis, however, require far more coordiates for a complete descriptio. For example, a fuctio f with domai [, ] is defied by ifiitely may coordiates f(t), oe for each t [, ]. Or, we could cosider f as beig determied by its Taylor series = a t (whe such a represetatio exists). I that case, the umbers a, a, a,... could be thought of as coordiates. Perhaps the associatio of Fourier coefficiets (there are coutably may of them) to a periodic fuctio is familiar; those are agai coordiates of a sort. Strage Thigs ca happe i ifiite dimesios. Oe usually meets these, gradually (reluctatly?), i a course o Real Aalysis or Fuctioal Aalysis. But ifiite dimesioal spaces eed ot always be completely mysterious; sometimes oe lucks out ad ca watch a couterituitive pheomeo developig i R for large. This might be of use i oe of several ways: perhaps the behavior for large but fiite is already useful, or oe ca deduce a iterestig statemet about lim of somethig, or a peculiarity of ifiitedimesioal spaces is illumiated. I will describe some curious features of cubes ad balls i R, as. These illustrate a pheomeo called cocetratio of measure. It will tur out that the importat law of large umbers from probability theory is just oe maifestatio of highdimesioal geometry. Alog the way, we will meet some stadard aalysis techiques. These may be familiar to varyig degrees. I thik it could be useful for you to see how multidimesioal itegrals, liear algebra, estimates, ad asymptotics appear i the aalysis of a cocrete problem. A umber of these matters are relegated to appedices. I the mai part of the expositio, I try to focus (as much as possible) o geometric pheomea; it would be reasoable to read about those first, ad oly c by Herma Flaschka.
2 MATH 57A refer to a appedix for quick clarificatio as eeded. Ultimately, you should also uderstad the appedices.. The Cube.. Volume of the cube. C (s) is the cube cetered at the origi i R with sidelegth s. I.e., C (s) = {(x,..., x ) s x j s for all j}. Its dimesioal volume is (by defiitio) Vol(C (s)) = (s) (s) (s) = (s). } {{ } times We have a obvious cosequece: Propositio.. As, the volume of C (s) teds to zero if s <, to if s >, ad it is always = for s =. From ow o, C ( ) will be my referece cube, ad I will simply write C. So Vol(C ) = for all. Notice, however, that the poit (,..., ) is a vertex of the cube, ad it has distace / from the origi. So Propositio.. The cube C has diameter, but volume. The mathematics is completely elemetary, but I hope you agree that visualizig such behavior is rather more tricky. It gets worse... Cocetratio of volume. I wat to compare the volume of C to the volume of a subcube, C ( ɛ ), where ɛ (, ) is give. We already kow from Propositio. that the latter volume teds to zero as. Hece the shell betwee the two cubes cotais most of the volume: Propositio.3. For every ɛ (, ), teds to as. Volume of shell = Vol(C C ( ɛ )) I other words, as, there is o such thig as a thi shell of small volume. All the volume of C escapes towards its surface, out of ay prescribed subcube. I order to uderstad, i some measure, how the volume cocetrates at the surface of C, look agai at (.) Vol(C C ( ɛ )) = ( ɛ).
3 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 3 Of course, ( ɛ) because < ɛ <. Now ivoke Lemma A.: lim ( + a ) = e a. This suggests that we should let the ɛ i (.) vary with. Istead of takig a subcube C ( ɛ ) whose sides are i fixed ratio to the side (legth = ) of C, we expad the subcube, ad shrik the shell, as icreases. I this way, the volume trapped i the shrikig shell does have a ozero limit: Propositio.4. For every t >, lim Vol(C C ( t )) = lim ( t ) = e t. [It is uderstood that must be large eough to esure t < ]. How to thik about this? Say you wat to kow what shell cotais half the volume of C (for large, of course). Sice e t =.5 for t = , you kow that the cube of sidelegth has volume about, with the remaiig volume cotaied i the very thi shell (of width.69...) betwee the subcube ad our referece cube, C. Later o, we will look at the cube a bit differetly, ad see that its volume also cocetrates alog the diagoal plae x + + x =. This will be related to probability theory. First, however, we will study the ball; it is more symmetrical tha the cube, ad easier to deal with..3. Surface area of the cube. I dimesios ad, cubes are ot cubes i the everyday sese of the word. The cube C (s) is a lie segmet of legth s. The cube C (s) is a square, sidelegth s. It has four dimesioal sides, each of which is a copy of C (s). The surface area of C (s) is the total dimesioal volume, i this case just ordiary legth, of the four sides, to wit 4 (s). C 3 (s) is the usual cube. Its six dimesioal sides are copies of C (s), ad the surface area of C 3 (s) is the sum of the dimesioal volumes, or the usual areas, of its sides: 6 (s). The cube C 4 (s) i four dimesios has a umber of 3dimesioal sides. Let s call them 3faces. The surface area of C 4 (s) is really the sum of the 3dimesioal volumes of the 3faces: 8 (s) 3. To see why this formula is correct, we eed to describe the 3faces. A 3face is determied by settig oe of the coordiates equal to its extreme value, ±s, ad allowig the other coordiates to vary i [ s, s].
4 4 MATH 57A For example, {(s, x, x 3, x 4 ) x j s for j =, 3, 4} is a 3face. It is a copy of the cube C 3 (s) = {(x, x, x 3 ) x j s for j =,, 3}; except the idices have chaged (which is irrelevat). This object has the 3dimesioal volume (s) 3. There are eight 3faces, sice ay oe of the four coordiates could be fixed at +s or s, with the other three coordiates allowed to vary i [ s, s]. The surface area of C (s) is the sum of the ( )dimesioal volumes of its ( )faces. Area is ot really a good term, but we will use it to distiguish betwee the dimesioal volume of the solid dimesioal object, ad the ( )dimesioal volume of the boudary of the object. Later o, the same covetio will apply to the ball. Exercise.. How may ( )faces does C (s) have? How may ( k)dimesioal faces, for k? (A face is a vertex, ad the face, by covetio, is the whole solid cube). Exercise.. For each s (, ), compare the behavior of the volume of C (s) to the behavior of its surface area, as. The case s = should strike you as beig couterituitive. Exercise.3. Fix ɛ (, s). Let S ɛ be the slice S ɛ = C (s) {(x,..., x, z) z < ɛ}, i.e. a sort of equatorial slice of the cube. How does the ratio behave as? Vol(S ɛ ) Vol(C (s)) 3. Volume of the Ball The material about cocetratio of volume comes (at times verbatim) from P. Lévy, Problèmes cocrets d aalyse foctioelle, Gauthier Villars (95), p.9 ff. See also pp. 66 of the Math 57 otes for the computatio of the volume of a ball i R. somethig like ( )dimesioal Lebesgue measure might be preferred by cogosceti
5 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 5 x d(si ) cos si IR Figure. Volume of ball by slices 3.. Volume of the ball, part. We write B (R) for the solid ball, radius R, cetered at the origi i R. The term ball meas surface plus iterior. The surface itself is called the sphere, deoted by S (R). The superscript ( ) sigifies that the sphere i R has dimesio ( ). Thus B (R) = {(x,..., x ) R x + x R}, S (R) = {(x,..., x ) R x + x = R}. Example 3.. To be clear about dimesios, cosider B 4 () = {(x, x, x 3, x 4 ) x + x + x 3 + x 4 } R 4. This is 4dimesioal, i the sese that as log as x +x +x 3 +x 4 <, all four coordiates ca be varied idepedetly. However, S 3 () = {(x, x, x 3, x 4 ) x + x + x 3 + x 4 = } R 4. This is 3dimesioal, i the sese that oce three of the x j are fixed, the remaiig oe is largely determied (up to ± sig, i this example). There are better defiitios, but this should covey the idea. It is rather tricky to picture S 3, sice we ca t go ito four dimesios, but there are ways. I ay case, this sectio deals with the ball oly.
6 6 MATH 57A Now let K be the volume of the uit ball B () i R. The the volume of B (R) is K R. Let us calculate K (refer to Figure ). The circle represets the ball of radius. The horizotal axis is R, the set of ( )tuples (x,..., x ). The vertical axis is the last directio, x. If we were computig the volume of the ball i R 3 (i.e., = 3), we would add up the volumes of circular slices of radius cos θ ad thickess d(si θ): K 3 = π π π cos θ d(si θ). Now this itegrad π cos θ is the twodimesioal volume of the twodimesioal ball (=disk) of radius cos θ. Note that it has the form volume of twodimesioal uit ball radius = π cos θ. Usig symmetry i θ, we may thus write π K 3 = K cos 3 θ dθ. The patter cotiues for > 3. I claim that x = si θ, represeted by a horizotal lie segmet i the figure, itersects B () i a ( ) dimesioal ball of radius cos θ. Ideed, the itersectio is defied by (3.) x + + x + si θ = x + + x si θ = (cos θ), ad this is B (cos θ). So, istead of a very thi slice with a twodimesioal disk as crosssectio, we ow have a very thi (d(si θ)) slice whose crosssectio is the ball of radius cos θ i dimesio ( ). Its volume is K cos θ d(si θ). Thus, π (3.) K = K cos θ dθ = I K, where (3.3) I def = π cos θ dθ. To make further progress towards a useful formula for K, we eed to uderstad the itegral I.
7 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES The itegral I. Here we collect iformatio about I. We retur to the volume K i the ext subsectio. Itegratio by parts i (3.3) gives (for > ) Hece I = [si θ cos θ] π + ( ) = ( ) π = ( )(I I ). π (cos θ cos θ) dθ cos θ si θ dθ (3.4) I = I. Sice I = π ad I =, we ca solve (3.4) recursively. Startig with I, we fid I, I 4,..., ad from I we get I 3, I 5,.... The patter is easy to ascertai: I p = π 3 5 (p ) (3.5), 4 6 (p) (3.6) I p+ = 4 6 (p) 3 5 (p + ). Remark 3.. Stadard otatio: (p + ) = (p + )!!. We will eed to kow how I behaves as. Sice the itegrad cos θ decreases with for every θ, oe expects that I. This is true, but we wat to kow how fast the decrease is. The ext lemma provides the aswer. We use the otatio a b to sigify that lim a b = ; this would be a good time to glace at Appedix B. Lemma 3.. I π. Proof. From (3.5) ad (3.6), oe sees that for all positive, (3.7) I I = π. (Aother proof: I = ( )I implies (3.8) I I = ( )I I ; the writig (3.8) for with replaced by,,..., we get the strig of equalities I I = ( )I I = ( )I I 3 = = I I = π.) Here is the idea of the rest of the proof. First we ote that by (3.4), I I. Next we show that I is trapped betwee I ad I.
8 8 MATH 57A It will follow that I I, so that (3.7) gives I π. Takig the square root will establish the Lemma. Let s do it more carefully. Because cos θ, we have for θ (, π ), ad thus I Divide by I ad use (3.7): cos θ < cos θ < cos θ < I < I. < I I <. Sice the right side has limit, it follows that Now multiply (3.7) by I I I lim =. I ad rearrage: ( ) I = π/() I I. The right side has limit =, hece so does the left side. Now take the square root. Remark 3.. Lemma 3. has a remarkable cosequece. = p +, it says that or i loghad, lim p (p + ) I p+ = π, 4 (p) (3.9) lim p 3 5 (p ) p + = π. Applied to There is a dotdotdot otatio that is used for ifiite products, aalogous to the usual otatio for ifiite series (here... stads for a certai limit of partial products, aalogous to the familiar partial sums); oe would write π = This is called Wallis product. J. Wallis, Arithmetica ifiitorum, Oxford 656 (!). The product ca be recast as Π (4 /(4 )). By the way, Wallis was a codebreaker for Oliver Cromwell, ad later chaplai to Charles II after the restoratio
9 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 9 Let us ow look at π cos θ dθ more closely. The itegrad is decreasig i θ, ad for each θ (where its value is = ), it decreases to zero with. So the mai cotributio to the itegral should come from a eighborhood of θ =. Ideed, if α > ad θ (α, π ], the cos θ < cos α ad so π π cos θ dθ < cos α dθ = ( π α) cos α. α α Now cos α approaches zero geometrically, but we saw i Lemma 3. that I approaches zero oly like. We are i the situatio of the example i Appedix B ( geometric decrease beats power growth ), which gives (cos α) : We obtai: I π α cos θ dθ. Propositio 3.. For every α (, π ], I α cos θ dθ. If, however, the limit of itegratio α is allowed to decrease as icreases, oe ca trap a proper fractio of the the total value of I i the shrikig iterval of itegratio. Cosider β From (A6) i Appedix A, ad we just saw that Thus β (3.) lim I cos θ dθ = β cos ( t ) dt β I π. cos θ dθ = cos ( t ) dt. β e t dt, β e t π Remark 3.3. By the time we get to (3.), we o loger eed β (, π/). That restrictio was used i the earlier observatio that the itegral over ay fiite subiterval [, α] [, π/] was asymptotically dt.
10 MATH 57A equivalet to I. Oce we go to the shrikig upper limit α = β/, the this upper limit will fall ito (, π/] whe is large eough, o matter the choice of β Volume of the ball, part. We had obtaied the relatio (3.), K = π K cos θ dθ = I K. Replace by ( ) to get K = I K, ad thus K = 4I I K. Now recallig (3.7) from the precedig subsectio, we obtai a recursio relatio from which K ca be determied: (3.) K = π K. Kowig that K = π ad K 3 = 4π, we fid K 3 4, K 6,... ad K 5, K 7,..., ad recogize the patter: (3.) (3.3) π p K p = p! (π) p K p+ = 3 5 (p + ). This formula has a amazig cosequece: Propositio 3.. For every R >, the volume of the ball of radius R i R teds to zero as. Proof. If = p is eve, for example, the volume will be K p R p = (πr ) p. p! Thus K p+ R p+ (πr ) = (p + )(p + ) K pr p, showig that the volume decreases rapidly, oce p becomes sufficietly large. (If you otice that the K p R p happe to be the coefficiets i the Taylor series represetatio of e x whe x = πr, this argumet should rig a bell). The proof for odd dimesios is similar. As was see i Propositio., the depedece of the volume of the cube C (s) o s ad is quite differet. There is aother curious relatio betwee cubes ad balls that should be oticed. The distace betwee the origi ad the vertex (s,..., s) of C (s) is s. Therefore, the radius (=s ) of the smallest ball
11 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES cotaiig C (s) teds to with. Furthermore, if s >, the Vol(B (s )) (sice the volume of the iscribed cube teds to ifiity). O the other had, the largest ball cotaied i C (s) will have radius s, which is idepedet of. Eve if Vol(C (s)), the volume of this iscribed ball B (s) teds to zero. It is temptig to extrapolate to ifiite dimesios. I R (whatever that might mea), a cube is ot cotaied i ay ball of fiite radius. The cube with sides of legth = ad volume = cotais the uit ball, which has volume =. Well, this last statemet is mostly osese, but somewhere i it there is a little truth Asymptotic behavior of the volume. The formula for K ivolves factorials, which are approximated by Stirlig s formula (C), Propositio 3.3. k! πk k k e k. Vol(B (R )) (πe) π R. Proof. Apply Stirlig s formula, first i eve dimesios. From (3.), K p π p πp pp e p ; if we set p = / ad do some rearragig (exercise), we get (3.4) K (πe) π( ). Multiplyig this by (radius) = ( R), we get the desired formula. The case of odd dimesios is somethig of a mess, ad there is a subtle poit. Begi by multiplyig top ad bottom of the expressio (3.3) for K p+ by 4 6 (p) = p p!, ad also multiply from the begiig by the factor ( ) that will come from ( R), to get (recall that = p + ) (π) p p p! p+ (p + ). (p + )! Next, substitute for the two factorials from Stirlig s formula. terms are ordered so that certai combiatios are easier to see. πp (π) p p e p p p p+ (p + ) π(p + ) e p. (p + ) p+ The
12 MATH 57A First some obvious algebraic simplificatios i the last expressio: p (π) p p e p+ p p (p + ) p + (p + ) p. Moreover, p p (p + ) = ( ) p p p p p + = p ( +, p )p ad isertig this leads to p (3.5) K p+ (π) p e p+ p + ( + (p + ). )p p Now we do asymptotic simplificatios as p. The subtle poit arises here. Let RHS stad for the right side of (3.5). So far we kow that (3.6) lim p K p+ RHS =. If we replace ay part of RHS by a asymptotically equivalet expressio, obtaiig a modified expressio RHS, say, we will still have For example, We ca the modify (3.6): K p+ lim p RHS =. lim p K p+ lim p RHS p p+ =. p p+ =. This gives a ew deomiator, ad we get lim p K p+ (π) p e p+ (+ p )p (p + ) which we would write more briefly as =, (3.7) K p+ (π) p e p+ ( + (p + ). )p p A compariso of (3.5) ad (3.7) suggests that we have allowed p to ted to i some places, but ot i others. Is t this uethical? No it is ot, as log as we are replacig oe expressio whose ratio with K p+ teds to by aother expressio with the same property.
13 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 3 This same reasoig further allows us to substitute e for ( + p )p. By ow, (3.5) has bee simplified to (π) p e p+ (3.8). (p + ) ad it is a easy maipulatio to tur this ito the desired formula i terms of = p +. Corollary 3.. lim Vol(B (R )) = {, if R πe, if R > πe. Proof. Write the asymptotic expressio for the volume as ( πer ) π, ad recall that whe πer >, the umerator grows much faster tha the deomiator. The case is trivial. This is a little more like the behavior of the volume of the cube, especially if you thik ot about the side legth s of C (s), which remais fixed, but about the legth s of its diagoal: Cube: diameter = costat, Ball: radius = costat. The behavior of the volume is determied by the value of the proportioality costat. Still, there is o way to get a ozero, oifiite limitig volume for the ball Volume cocetratio ear the surface. Cosider the two cocetric balls B () ad B ( ɛ). The ratio of volumes is Vol(B ( ɛ)) Vol(B ()) = K ( ɛ) K = ( ɛ). For every ɛ, this ratio teds to zero as, which meas that every spherical shell, o matter how thi, will cotai essetially the whole volume of B (). Of course, it should be remembered that lim Vol(B ()) =, so the shell also has a small volume, it is just a sigificat fractio of the whole.
14 4 MATH 57A To see how the volume accumulates at the surface, we agai let ɛ deped o, as we did for the cube. Choosig ɛ = t, we fid that Vol(B ( t )) Vol(B ()) = K ( t ) K = ( t ) e t. The iterpretatio is exactly as for the cube (subsectio.) Volume cocetratio ear the equator for B (R). Recall that Vol(B (R)) for every R. We ask: what fractio of this vaishig (as ) volume accumulates ear the equator of the ball? Let θ < < θ, ad cosider (3.9) R θ θ cos θ dθ R π π = ( θ ) cos θ dθ + cos θ dθ. I θ The umerator is the volume of the slice bouded by the hyperplaes 3 x = R si θ, x = R si θ (see Figure ). Sice the factor R has cacelled, we will temporarily set R = ad work with the uit ball; later, R will be reiserted by had. Accordig to Propositio 3., the two itegrals o the right side of (3.9) are each asymptotically equivalet to I. Hece: Propositio 3.4. For every θ [ π, ) ad θ (, π ], the fractio of the volume of B (R) cotaied i the equatorial slice betwee the hyperplaes x = si θ ad x = si θ teds to as. If, however, θ > (θ < θ ), the the fractio of volume i the slice teds to zero. As before, to trap a proper fractio of the volume, we must allow the limits of itegratio to approach as. The results of Subsectio 3. suggest that we take θ = β ad θ = β, β < β. The sigs of the β s will ot matter ow, oly the fact that β j at a cleverly chose rate. Icidetally, sice si u u for u, the boudig hyperplaes x = si( β j ) are essetially x = β j as gets large. From equatio (3.), we ow obtai 3 a hyperplae i R is a dimesioal set defied by a sigle liear equatio i x,..., x. If this set does ot cotai the origi, it is sometimes called a affie hyperplae, but I wo t make that distictio. Hyper meas oe dimesio dow, i this cotext. The set defied by a sigle, geerally oliear, equatio f(x,..., x ) = is a hypersurface. The sphere S (R) is a hypersurface i R.
15 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 5 Propositio 3.5. Let β < β. The fractio of the volume of B () cotaied i the slice bouded by the hyperplaes x = si( β j ), j =,, teds to (3.) π β β e t Remark 3.4. Propositio 3.5 remais true if oe asks about the volume betwee the hyperplaes x = β j, j =,. That volume differs from the volume i the propositio by si ( β ) cos θ dθ β ad aother similar itegral, but the limits of itegratio approach each other as ad the itegral teds to zero. The details are omitted. Remark 3.5. If you kow some probability theory, you will have oticed that (3.) represets a area uder the stadard ormal curve. It is a ofte used fact that the value of (3.) for β =.96, β = +.96 is approximately.95. I other words, for large all but about 5% of the volume of B (R) will be foud i the very thi slice betwee x = ±.96R. Remark 3.6. By rotatioal symmetry, Propositio 3.5 will hold for every equator, ot just the oe perpedicular to the x directio. Thik about that for a miute. Next, icorporate this i your thoughts: the volume also cocetrates ear the surface of the ball Volume cocetratio ear the equator for B (R ). The results of the last subsectio ca be viewed from a slightly differet agle, which is importat eough to be emphasized i its ow little subsectio. Istead of fixig R, ad lookig at the fractio of volume cotaied i the slice betwee 4 x = β ad x = β, oe ca let the radius of the ball become ifiite at the rate R ad look at the slice betwee x = β ad x = β. Keep i mid that Vol(B ( R )) or, so we may be talkig about a fractio of a tiy volume or of a huge volume. dt. 4 it is simpler to use β/ istead of si(β/ ); the two are asymptotically equivalet
16 6 MATH 57A Propositio 3.6. Let β < β. The fractio of the volume of B (R ) cotaied i the slice bouded by the hyperplaes x = β j, j =,, teds to (3.) π β Exercise 3.. Prove this propositio. β e t dt. 4. Surface area of the ball 4.. A formula for surface area. Recall that the surface of a ball is a sphere. We speak of the area of a sphere simply to distiguish that quatity from the volume of the whole ball. But keep i mid (Example 3.) that the sphere S 3 (R) is itself 3dimesioal (sittig i the 4dimesioal R 4 ). What we refer to as its area is actually its 3dimesioal volume. Likewise, the area of S (R) R is uderstood to mea its ( )dimesioal volume. The surface area of B (R) is easily expressed i terms of Vol(B (R)). Cosider the dimesioal volume of the spherical shell betwee the balls of radii R + δr ad R: Vol(B (R + δr)) Vol(B (R)) = K [(R + δr) R ] = K [R δr + ], with deotig terms i (δr) ad higher. Thik of this volume as beig approximately (4.) volume of shell = area of S (R) thickess δr of shell, i.e. volume of shell = K R δr +. (4.) becomes more accurate as δr. Hece the area of S is d (4.) dr Vol(B (R)) = K R. This type of argumet would ideed recover a formula which you kow from calculus, for the area of a regio D o a surface (i R 3 ) give by z = f(x, y): + fx + fy dx dy. D But we will eed to get at the area of the sphere by a differet, seemigly more rigorous approach ayway.
17 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 7 x d si cos IR Figure. Surface area by slices Let L deote the surface area of the uit ball B () i R. We kow from geeral priciples that the ball B (R) will have surface area L R ; this will be used shortly. Now refer to Figure, which agai represets the uit ball B () i R. (If you are at all ueasy with the use of i what follows, work through the steps usig = 3). Recall from (3.) that the lie segmet x = si θ represets the ball B (cos θ); as just remarked, it will have surface area L (cos θ). This is ( )dimesioal volume. The outside arc of the shaded area is the surface of a cylider put together as follows: the base is B (cos θ) ad the edge (surface) of the base is S (cos θ); the height is dθ. For small dθ, the side of the cylider is early perpedicular to the base, ad the the outside arc of the shaded regio represets a cotributio circumferece height = L (cos θ) dθ to the total surface area. Therefore (4.3) L = π π L cos θ dθ = I L. To obtai a formula for L, we ca proceed i three ways. First, i the formula L = K, which is (4.) at R =, use the kow expressios for K. Secod, i (4.3), start with = 3 ad L 3 = L =
18 8 MATH 57A Figure 3. Volume vs. surface area π, ad use the kow expressios for I. Third, i (4.3), replace L by I 3 L 3 (which is (4.3) after the replacemet ( ) ( )), to get L = 4I I 3 L 3 = π L 3 (last step by (3.7)). The use L = π ad L = 4π to get started o a recursive determiatio of L. Presumably, all these methods lead to the same aswer: Propositio 4.. L p = 4π (π)p (p )!! L p = π πp p!. Remark. It is worth uderstadig exactly why cos θ becomes cos θ, ad ot cos θ, which would be aalogous to the chage i the expoet of R. For the volume, we had which was the multiplied by volume of base of slice (radius) thickess of slice = d(si θ) = cos θ dθ. For the surface area, it was surface area of base of slice (radius) (here the power does just drop by ), but this was oly multiplied by the arclegth dθ. So the differece is the bit of rectagle that sticks out beyod the circular arc i Figure 3. Exercise 4.. Compare the behavior of the volume ad surface area of B (R ), for all R >. Oe value of R should be peculiar.
19 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES Surface area cocetratio ear the equator. The surface area of B (R) cocetrates ear the (or ay) equator. The calculatio has already bee doe. Look at (3.9). If R is replaced by R, ad cos θ is replaced by cos θ, (3.9) represets the fractio of surface area of B (R) cotaied betwee the hyperplaes x = si θ j, j =,. Everythig we kow about the limits remais true, word for word. I will ot restate the results, you do it: Exercise 4.. State precisely the propositio about surface area cocetratio ear the equator. 5. The weak law of large umbers I this sectio, cocetratio of volume for the cube about its equator is show to be a expressio of oe of the basic theorems of probability theory. I will use several techical probability defiitios i a ituitive way; otherwise, there would be too may digressios. 5.. The probability settig. Let C = C agai be the iterval ( ) = [, ], ad cosider the experimet of choosig a umber at radom from this iterval. All results are to be cosidered equally likely. More precisely, the probability that a radomly chose umber x falls ito a subiterval [a, b] [, ] is defied to be its legth, b a. We write P (x [a, b]) = b a. (So P (x [, ]) = ). I a setup like this, where the possible choices form a cotiuum, the probability that a radom x will be exactly equal to a give umber is zero. For example, P (x = ) =. A justificatio might be that you would ever kow x to all its decimal places. More mathematically, this is the oly possible logical cosequece of our startig defiitio of the probability: P (x [, ]) = =. Now we go to C (uit side legth, remember), ad defie the probability that a radomly chose poit lies i a regio D C to be the volume, Vol(D). Agai, the whole cube has probability. Oe ca imagie pickig x = (x,..., x ) as a whole, e. g. by throwig a dart ito the dimesioal cube. Or ad this is what I will do oe ca build up the radom x oe coordiate at a time. First choose x C at radom. The, idepedetly of this first choice, pick x C. Cotiue. The probability that a x built i such a way lies i D is agai Vol(D). This is ot easy to prove with full mathematical rigor, oe essetially eeds the theory of Lebesgue measure. I will illustrate for the case = ad D = a rectagle. The key techical otio, which caot be avoided, is idepedece.
20 MATH 57A Summary. A experimet with a umber (maybe ifiite) of possible outcomes is performed. A evet A is a subset of the set of possible outcomes. A probability measure assigs umbers betwee ad to evets. The iterpretatio of P (A) = p is that if the experimet is repeated a large umber N times, the about pn of the results will fit the criteria defiig A. (Every sigle statemet I just made requires clarificatio ad much more mathematical precisio.) Example 5.. Suppose a 6sided die is tossed 6, times. Let A =, 4, or 6 shows, ad B =, 3, 4, or 6 shows. Assumig that each of the six outcomes,, 3, 4, 5, 6 is equally likely, we have P (A) = ad P (B) = 3. This meas, ituitively, that A should happe o about half, or 3, of the tosses. Of these 3,, about oe third each will result i i, 4, 6, so B happes o /3 or, tosses. Tosses resultig i or 3 are o loger possible, sice we are restrictig attetio to outcomes where A has occurred. I other words, the probability of B was 3 at the outset, ad it was still after the results were restricted to the outcomes, 4, 6 i A. Oe says: 3 (5.) the probability of B give A = the probability of B. We caot make a better predictio about the occurrece of B amogst the outcomes satisfyig A tha we could about the occurrece of B amogst all possible outcomes. Example 5.. Suppose ow B is,3, or 5 shows. The oe of the 3, tosses fittig the descriptio of A will fit the descriptio of B. Thus: the probability of B give A =. If we kow that A has occurred, we kow with certaity that B has ot occurred. We cosider the A ad B i Example 5. to be idepedet. Example 5. illustrates a extreme case of oidepedece. Aother extreme case would be B =,, 4, or 6 shows. If we kow that A has happeed, it is certai that B has also happeed, or the probability of B give A =. The techical meaig of idepedece is a rephrasig of (5.): (5.) P (A ad B both happe ) = P (A)P (B). To uderstad this, rewrite it as (if P (A) ) (5.3) P (A ad B both happe ) P (A) = P (B).
21 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES The left side is the fractio of the outcomes satisfyig A for which B also happes. I the settig of Example 5., it would be, 3, = 3 6, 6, = 3 = P (B). Example 5.3. We will pick a poit x C at radom, by choosig its two coordiates idepedetly. Let A be the evet x [a, b ], ad let A be the evet x [a, b ]. We have P (A ) = b a, P (A ) = b a. The choices of x ad x are to be made idepedetly. This meas that we require that P (x [a, b ] ad x [a, b ]) = (b a )(b a ). But this is the same as sayig that P (x = (x, x ) rectagle with base [a, b ] ad height [a, b ]) = Vol(rectagle). The geeral statemet P (x D) = Vol(D) is deduced by approximatig the regio D by a uio of small rectagles (compare the defiitio of a multiple itegral as limit of Riema sums). Laws of large umbers give precise meaig to statemets that somethig should happe o average. For example, if we choose, radomly ad idepedetly, a large umber N of poits i C = [, ], say a, a,..., a N, ay positive a j should sooer or later be early balaced by a egative oe, ad the average a + a + + a N (5.4) N should be close to zero. The larger N is, the better the law of averages should apply, ad this ratio should be closer to zero. Now, as just explaied, these idepedetly chose a j ca be thought of as the coordiates of a radomly picked poit a = (a,..., a N ) i the highdimesioal cube C N. Oe versio of the law of averages says that the volume of the regio i C N where (5.4) is ot close to zero becomes egligible as N. I other words, we have cocetratio of volume agai: ( Vol {a = (a,..., a N ) C N a + a + + a N N ) is small}. I ow tur to the geometric problem, ad revert to my usual otatio (x,..., x ) C. The iterest is i cocetratio of volume about the plae Π : x + + x =. Notice that this plae has ormal (,,..., ), ad that the lie alog the ormal is i fact the mai
22 MATH 57A diagoal of the cube, passig through the corer (,,..., ). The distace to the corer is. We wat to look at slices bouded by plaes Π ±η : x +... = ±η; possibly η will deped o we shall see. The hoest approach would be to fid a ice, more or less explicit, formula for the volume of such a equatorial slice, as there was for a ball, ad the to apply the geometric fact of cocetratio of volume to obtai a probabilistic applicatio. However, I cofused myself tryig to fid a volume formula; there are too may corers whe you slice the cube by a plae Π η, ad I gave up. The less satisfyig approach is to prove the probabilistic result as it is doe i probability texts (usig a simple iequality, see 57 Notes, p. 386), ad to reiterpret it i geometric laguage. This is what I have to do. 5.. The weak law of large umbers. For x = (x,..., x ) C, set S (x) = x + + x ; this is a fuctio whose domai is the cube. I will sometimes omit the argumet x. Further, give ɛ >, defie the equatorial slice (5.5) S ɛ = {x C S (x) < ɛ}. Theorem 5. (Weak law of large umers). For every ɛ >, Equivaletly, (5.6) lim Vol ( lim Vol(S ɛ) =. {x C S ) (x) > ɛ} =. Remark 5.. The weak law is really a much more geeral theorem. I have stated it i the cotext of our cube example. But the x j might also be chose at radom from the twoelemet set {, }, with P ( ) = P () =. These two umbers could sigify tails ad heads i a toss of a fair coi. We would the be workig with a discrete cube, {, }. Theorem 5. gives oe precise meaig to the law of averages for coi tosses: there are, o average, about as may heads as there are tails. Remark 5.. If there is a weak law, there should be a strog law, ad there is. It is set i the limitig cube, C, where ow x = (x, x,...) is a tuple. It says that those x for which x + + x N (5.7) lim = N N
23 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 3 occupy total volume i C. Evidetly, some more theory is eeded here, to give meaig to the otio of volume i C, ad to clarify the ature of possible sets of zero volume (where the limit (5.7) fails to hold). The proof of Theorem 5. is based o a useful iequality, which is stated here for the cube C, although ay regio of itegratio could be used. For typig coveiece, I use a sigle itegral sig to stad for a fold multiple itegral, ad I abbreviate dx dx = d x: dx dx d x. } {{ } C C Lemma 5. (Chebyshev s iequality). Let F : C R be a fuctio for which σ def = F (x) d x is fiite. For δ >, let The A δ = {x C F (x) > δ}. (5.8) Vol(A δ ) < σ δ. Proof. (of lemma) Sice A δ C, F (x) d x F (x) d x = σ. A δ C For x A δ we have (δ) < (F (x)), so δ d x < F (x) d x. A δ A δ The left side is just δ A δ d x = δ Vol(A δ ). Combie this strig of iequalities to get as desired. δ Vol(A δ ) < σ, Proof. (of Theorem) We apply Chebyshev s iequality to F = S. First, we ote that (S ) = x j + x i x j. j= i= j= j i
24 4 MATH 57A The itegral over C of the double sum vaishes, sice i the iterated multiple itegral, x i dx i =. A typical cotributio from the simple sum is ( ) x = x dx dx... dx = C. Thus, σ =. Chebyshev s iequality ow gives ( ) Vol {x S (x) > δ} Set δ = ɛ i this relatio, ad you get ( (5.9) Vol {x S ) (x) > ɛ} < δ. < ɛ. This teds to zero as, ad the theorem is proved Coectio with cocetratio of volume. We have see that the volume of the regio C where S < ɛ teds to as. The geometric meaig of this fact still eeds a little more clarificatio. To this ed, we switch to a coordiate system i which oe directio is perpedicular to the plaes Π η : x + +x = η. This ew directio is aalogous to the x directio i our discussio of the ball. Let e,..., e be the stadard basis, Thus, e j = (,...,, }{{},,..., ). j th place x = x e + + x e. Now let f be the uit vector (,..., ) = (e + + e ) i the directio ormal to Π. The orthogoal complemet to f is of course the ( )dimesioal vector subspace Π = {x R x + + x = }.
25 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 5 The process of GramSchmidt orthogoaliztio i liear algebra guaratees that we ca start with the uit vector f, ad costruct orthoormal vectors f,..., f which, together with f, provide a ew orthoormal basis of R. (These f j are far from uique; certaily oe possible set ca be writte dow, but that would be more cofusig tha helpful). The poit is that a give x i the cube ow has two coordiate expressios: x e + + x e = y f + + y f + y f. Take ier products of both sides with f, ad remember that the f j are orthoormal. You fid (5.) y = (x + + x ) = S. Now go back to (5.9). Isertig (5.), we covert it ito ( Vol {y y > ɛ ) } < ɛ. Fially, otice that the corer of the cube, where all x j =, correspods to y =. We may therefore phrase the weak law of large umbers i more geometric laguage, as follows. Propositio 5.. The diagoal of C has legth. The fractio of the (uit) volume of C cotaied i the slice S ɛ (see (5.5)) approaches as. I other words, a slice S ɛ whose width is a arbitrarily small fractio of the legth of the diagoal of C will evetually eclose almost all the volume of C. Remark 5.3. Recall that the volume of C also cocetrates ear the surface! Exercise 5.. Verify that oe possible choice for the ew basis vectors f j is, for j =,..., j f j = (,...,,,,..., ). j(j + ) j(j + ) j(j + ) } {{ } j terms
26 6 MATH 57A A. appedix: compoud iterest I the mai part of these otes, we eed two facts. First, equatio (A) below, which ofte arises i a discussio of compoud iterest i begiig calculus. It is fairly easy to prove. Secod, the rather more subtle formula (A) β lim cos ( t ) dt = β lim cos ( t β ) dt = e t which is a cosequece of (A) ad some other stuff. The first subsectio explais the ideas, ad everythig afterwards fills i the techical mathematical details. There are lots of them, because they will reappear durig the Math 57 year, ad you might as well get a brief itroductio. The details could well detract from the importat issues o first readig. Study the earlier material, ad the go through the details i the appedices later (but do go through them). A.. The ideas. The first result should be familiar. Lemma A.. For every a R, (A) lim ( + a ) = e a. I will actually require a cosequece of a refiemet of Lemma A.. This geeralizatio of Lemma A. says, i a mathematically precise way, that if b, the we still have (A3) lim ( + a + b ) = e a, ad that the approach to the limit is equally fast for all a i ay fixed fiite iterval [ A, A]. (Techical term: the covergece is uiform). The relevat applicatio is this uexpected corollary: dt, Corollary A.. (A4) lim cos ( t ) = e t. Covergece is uiform o fiite itervals, i this sese: give β > ad ɛ >, there is a iteger N > depedig o β ad ɛ but ot o t, such that > N implies (A5) cos ( t ) e t < ɛ for all t [ β, β].
27 SOME GEOMETRY IN HIGHDIMENSIONAL SPACES 7 Figure 4. Illustratio of (A4) Shelve that defiitio for the momet, ad let me reveal why (A4) should be true ad why I wat this uiform covergece. From the Taylor series for cos u we have cos u = u +. Suppose all the terms are just ot there. The by (A), (cos( t )) = ( t ) which has limit e t /! But the are there, i the guise of the correctio term i Taylor s formula. So oe ecouters the more complicated limit (A3). I will the wat to coclude that (A6) β lim cos ( t ) dt = β lim cos ( t β ) dt = e t This seems plausible, give (A4), but it is ot always true that the limit of the itegrals of a sequece of fuctios is equal to the itegral of the limit fuctio. I other words, equality = may fail for some sequeces of fuctios. It helps to look at some graphs. dt.
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