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on as a sum of partial fractions. ()()()()()()()()()()()()521121121121152112112115121xABxxxxxxxxABxxxxxxxxAxBx++−=++−+−+−+=+−++−+−+=−++()()5252xAxABxBxABxAB+=−+++=++−+()()()()125ABABxx+=+−++215ABAB+=−+=36B=2B=3A=−253221211xxxxx+=−+−−+− CA6-84 Systems of Equations and Matrices Repeated Linear Factors We may... |
n example. 2. How can you check that you decomposed a partial fraction correctly? In Exercises 3 – 16, find the partial fraction decomposition for the rational expressions with denominators that contain non-repeated linear factors. 3. 6. 9. 12. 15. 4. 7. 10. 13. 16. 5. 8. 11. 14. In Exercises 17 – 27, find the partial ... |
t two terms based only on the way we see patterns. The actual pattern may be different. For example, −3, −1, 1, may continue as −3, −1, 1, 3, 5, 7, or as −3, −1, 1, −30, −10, 10, , or as any of an infinite number of possibilities. In general, there are two broad ways sequence formulas are written, recursively and expli... |
the sequence. 1. The sequence , , generates the odd numbers 3, 5, 7, 9, . Computing the first few differences, we have , , and . This suggests the sequence is arithmetic. To verify this, we find This establishes that the sequence is arithmetic with common difference . 2. We write out the first several terms: . We find ... |
We express a series, or the sum of a sequence, using the following notation. Summation Notation Definition 7.3. Summation Notation: Consider the sequence . • The sum of the terms , from to , can be written as . • The sum is written as . The symbol Σ is the Greek capital letter sigma and is read as ‘sigma’ or ‘summation... |
of summing infinitely many arithmetic terms. Suppose Gauss had been asked to add up all of the natural numbers. This would mean adding progressively larger and larger numbers or, in other words, bigger and bigger terms. The sum just keeps getting larger. This sum will not be a finite number and is said to ‘go to infini... |
is 17 years. 11242++++111141664−−−−−11134kk−=132nn=3927141664−+−+11142nn−=−11485nn−=()11347jj−=250k−kx=7k=xka60189kka==n()135100nkk=−1357−−−−−1231000++++0.70.132.34.1710.1595.867− 7.2 Series CA7-35 57. payments are $50, interest rate is 1.0%, term is 30 years. 58. payments are $100,... |
()5ab+0nnn()6ab+()6660661622633644655665423324566666661234561520156kkkababkaabababababbaabababababb−=−−−−−+==++++++=++++++()423x−()()()()442323xx−=+−()()()()()()()()()()()()()()()()()()()()()()()444404411422433443243223234234442232323312316483649422781169621621681kkkxxxk... |
4, 4-29, 4-45, 446, 4-68 logarithmic scales, 4-19 logistic model, 4-66 lower triangular matrix, 6-54 M main diagonal of a matrix, 6-49, 6-54, 6-56 major axis of an ellipse, 5-30 matrix, 6-1, 6-29, 6-30 addition, 6-43 definition of, 6-43 properties of, 6-44 additive identity, 6-44 additive inverse, 6-44 augmented, 6-30,... |
for r>0 the one-to-one curveparametrizes the branch x>0 ofthe hyperbola.Exercises1.Compute the velocity vector ofthe curve in Example 4.2(3) for arbitrarytand for t=0,t=p/2,t=p,visualizing those on Fig.1.8.2.Find the unique curve such that a(0)=(1,0,5) and a¢(t)=(t2,t,et).3.Find the coordinate functions ofthe curve b=... |
to dx,dy,dz,we findA 0-form is just a differentiable function f.A 1-form is an expression fdx+g dy+h dz,just as in the preceding section.A 2-form is an expression fdx dy+g dx dz+h dy dz.A 3-form is an expression fdx dy dz.We already know how to add 1-forms:simply add corresponding coeffi-cient functions.Thus,in index no... |
222cossin,,coscossinsinsincos22222tttttt=-=,,batFtFrtrtrtrtrtt()=()()=()=-()cossincossincossin,,222222uuu123100010001=()=()=(),,,,,,,,.361.Calculus on Euclidean SpaceR2R2r2vFuurabvFIG. 1.14 Since Rnis filled by the radial lines a(t)=p+tvstarting at p,Rmis filledby their image curves b(t)=F(p+tv) starting at F(p) (Fig.1.1... |
hat each point ofOhas an eneighborhood that is entirely contained in O.In short,all points near enoughto a point ofan open set are also in the set.This definition is valid with R3replaced by Rn—or indeed any set furnished with a reasonable distance function.dpqpqpqpq,()=-()+-()+-()()11222233212.pq-=---()pqpqpq112233,,,d... |
ttddttddtt123,,.fg==cossin.JJ,JJtfggfdut()=+¢-¢()Ú00,—()()—()()ffpp.—()()=∂∂()()[]Âffxipp212522.Frame Fields Substituting the formula for a¢given above,we get the usual formulafor arc length.This length involves only the restriction ofa(defined on some open interval) to the closedinterval [a,b]:atb.Such a restrictions:[... |
t frame fieldon b.Proof.By definition T=1.Since k=T¢>0,We saw above that Tand Nare orthogonal—that is,T• N=0.Then byapplying Lemma 1.8 at each point,we conclude that B=1,and Bisorthogonal to both Tand N.◆In summary,we have T=b¢,N=T¢/k,and B=T¥N,satisfying T• T=N• N=B• B=1,with all other dot products zero.The key to the s... |
d on I:-1 <s<1.Show that bhas unit speed,and compute its Frenetapparatus.bssss()=+()-()ÊËÁˆ¯˜131323232,,kkb==11•.Nabb•NNa=,kb•.N=-1ÂÂÂ662.Frame Fields 3.For the helix in Example 3.3,check the Frenet formulas by direct sub-stitution ofthe computed values ofk,t,T,N,B.4.Prove that(A formal proofuses properties ofthe cross... |
rame field.This appara-tus satisfies the extended Frenet formulas with speed factor vand can be com-puted by Theorem 4.3.Ifv=1,that is,ifais a unit-speed curve,the resultsofSection 3 are recovered.Let us consider some applications ofthe Frenet formulas.There are anumber ofnatural ways in which a given curve bgives rise t... |
ization,replace derivatives f¢(s) by f¢(t)v(t),where v(t) =ds/dt.)(b)In each case,decide whether the curve lies on a sphere,and ifso,findits radius and center:(i)a(t)=(2sint,sin2t,2sin2t);(ii)b(t)=(cos2t,sin2t,2sint);(iii)g(t)=(cost,1 +sint,2sin).21.Prove that the cubic curve g(t)=(at,bt2,ct3),abcπ0,is a cylindricalheli... |
i) in terms ofthe cylindricalframe field (with coefficients in terms ofr,J,z) and (ii) in terms ofthe spher-ical frame field (with coefficients in terms ofr,J,j):(a)U1.(b)cosJU1+sinJU2+U3.(c)xU1+yU2+zU3.3.Find a frame field E1,E2,E3such that2.7Connection FormsOnce more we state the essential point:The power ofthe Frenet for... |
ual forms ofE1,E2,E3by evaluating iton E1,E2,E3.This useful fact is the generalization to arbitrary frame fieldsofLemma 5.4 ofChapter 1.We compared a frame field E1,E2,E3to the natural frame field by meansofits attitude matrix A=(aij),for whichThe dual formulation is justwith the same coefficients.In fact,by the preceding ... |
that Cpreserves norms.By definition,||p||2=p• p;henceThus || C(p) ||=|| p|| for all points p.Since Cis linear,it follows easily thatCis an isometry:CCCpppppp()=()()==22••.CCpqpqpq()()=••.forall,FFIFFI--==11,, 3.1Isometries of R3103◆Our goal now is Theorem 1.7,which asserts that every isometry can beexpressed as an ortho... |
;hence C=tBA.As notedabove,Tis then necessarily translated by q-C(p).=ÊËÁÁˆ¯˜˜=ÊËÁÁˆ¯˜˜ÊËÁÁˆ¯˜˜=ÊËÁÁˆ¯˜˜tBbbbbbbbbbbbb100100112131122232132333111213,ttBAaaaBaaaaaaaaaaaa111213111213212223313233111213ÊËÁÁˆ¯˜˜=ÊËÁÁˆ¯˜˜ÊËÁÁˆ¯˜˜FCiiFpiFpiqi*ˆˆˆeefff()=()=()=()=()()FTCCCppqppq()=()()=-()+()=. 1103.Euclidean GeometryExercise... |
uch vector fields.4.1CorollaryLet Ybe a vector field on a curve ain R3,and let Fbe anisometry ofR3.Then =F*(Y) is a vector field on =F(a),andProof.To differentiate a vector field ,one simply differenti-ates its Euclidean coordinate functions,so¢=ÂYdydtUjj.YyUjj=¢=¢()YFY*.aYaYaYFIG. 3.6 1183.Euclidean GeometryThus by the c... |
follows.5.5CorollaryLet abe a unit speed curve in R3.Then ais a helix ifandonly ifboth its curvature and torsion are nonzero constants.Proof.For any numbers a>0 and bπ0,let ba,bbe the special helix givenin Example 3.3 ofChapter 2.Ifais congruent to ba,b,then (changing the sign ofbifnecessary) we can assume the isometr... |
nt is 2.Thus,by the criterion followingDefinition 7.9 ofChapter 1,xis regular and hence is a patch.Furthermore,xis proper,since its inverse function x-1:x(D) ÆDis given by the formulafuv=--122.∂∂∂∂∂∂∂∂∂∂∂∂ÊËÁÁÁˆ¯˜˜˜=∂∂∂∂ÊËÁÁÁˆ¯˜˜˜uuvufuuvvvfvfufv1001.xuvuvuv,,,()=--()122.p=++()=ppp122232121.FIG. 4.3 4.1Surfaces in R3133... |
is given in terms ofitsEuclidean coordinate functions by a formulathen it follows from the definition above that the partial velocity functionsare given byThe subscript x(frequently omitted) is a reminder that xu(u,v) and xv(u,v)have point ofapplication x(u,v).2.2ExampleThe geographical patchin the sphere.Let Sbe the s... |
uvuU,()=()+¢()+()bb3.FIG. 4.19FIG. 4.20 1484.Calculus on a Surface8.Let Mbe the surface ofrevolution gotten by revolving the curve tÆ(g(t),h(t),0) about the xaxis (h>0).Show that:(a)Ifg¢is never zero,then Mhas a parametrization ofthe form(b)Ifh¢is never zero,then Mhas a parametrization ofthe formA quadric surfaceis a s... |
y ifthe dot product vp• pp=v• pis zero.Similarly,a vector fieldVon Sis a tangentvector fieldifand only ifV• X=0.For example,V(p)=(-p2,p1,0) defines a tangentvector field on Sthat points “due east”and vanishes at the north and southpoles (0,0,±r).We must emphsasize that only the tangentvector fields on Mbelong to thecalculus... |
urfaces the only new defini-tion we need is that ofthe exterior derivative dfofa 1-form f.4.4DefinitionLet fbe a 1-form on a surface M.Then the exterior deriv-ative dfoffis the 2-form such that for any patch xin M,As it stands,this is not yet a valid definition;there is a problem ofconsis-tency.What we have actually define... |
R2is a mapping.Pppppppp12313232222,,,()=--ÊËÁˆ¯˜.PpppRprRpr12312,,,,()=Êˈ¯yx-()()=()1Fuvuv,,,sinFuvuvxy,,()()=()sin.FIG. 4.28FIG. 4.29 1684.Calculus on a SurfaceJust as for mappings ofEuclidean space,each mapping ofsurfaces has atangent map.5.3DefinitionLet F:MÆNbe a mapping ofsurfaces.The tangent mapF*ofFassigns to ea... |
by first pulling it backto Euclidean space.Consider the one-dimensional case.Let a:[a,b] ÆMbe a curve segmenton a surface M.The pullback a*fofa 1-form fon Mto the interval [a,b]has the expression f(t)dt,where by the remarks following Example 4.7,Thus the scheme mentioned above yields the following result:ftUtUtt()=()()... |
tion ofthe parallels,and gives a similar count for the meridi- ans.(This suggests the informal notation x=dJ,h=dj,but see Ex.7 ofSec.7.)10.Let x:RÆMbe a 2-segment defined on the unit square R:0 u,v1.Iffis the 1-form on Msuch thatffxxuvuvuv()=+()=and,hpg2()Úxpg2()ÚhaÚ,xaÚ,aptmtntt()=()()x,02.fffaÚÚ=()+()ÊËÁˆ¯˜xxuvabdadtd... |
urves are con-stant at p=a(a)=a(b).A curve such as afor which a(a)=a(b) holds but not necessarily a¢(a)=a¢(b) is often called a loopatp.Since the sides band dofxare constant atp,for every 0 v1 the u-parameter curve av(u)=x(u,v) is also a loop atp.As vvaries from 0 to 1,the loop avvaries continuously from a0=ato thecurv... |
mooth overlap axiom:For any patches x,yin P,the compositefunctions y-1xand x-1yare Euclidean differentiable—and defined on opensets ofR2.This definition generalizes Definition 1.2:A surface in R3isa surface inthis sense.However,there is a technical gap in this definition that requiresattention.First,for any patch x:DÆMin a... |
rned into a surface in R3by mapping each Upin Mˆtothe point p+eUp(Euclidean coordinates) in R3for some small e>0.(Fore=1,the point would be the tip ofthe “arrow”Up.)10.(Continuation.)(a)Using the scheme above,plot the orientation covering surface BˆoftheMöbius band B.(Take e=1/4.)(b)By inspection,is Bˆconnected? orient... |
the “upward”unit normal U,which at pis (0,0,1).Along the xaxis,Ustays orthogonal to the xaxis,and as it pro-ceeds in the u1direction,Uswings from left to right (Fig.5.5).FIG. 5.5In fact,a routine computation (Exercise 3) shows that u1U=-u2.Similarly,we find u2U=-u1.Thus the shape operator ofMat pis given by the formulaT... |
es on its maximum value k1at e1,soLet e2be merely a unit tangent vector orthogonal to e1(presently we shallshow that it is also a principal vector).Ifuis any unit tangent vector at p,we writewhere c=cosJ,s=sinJ(Fig.5.13).Thus normal curvature kat pbecomesa function on the real line:k(J)=k(u(J)).For 1 i,j2,let Sijbe the... |
seems odd at first fora surface so obviously curved,but it will be amply justified in later work.Note that minimal surfaces have Gaussian curvature K 0,becauseifthen k1=-k2,so K=k1k20.Another notable class ofsurfaces consists ofthose with constantGauss-ian curvature.As mentioned earlier,Example 1.3 shows that a sphere o... |
hatandhence on M,KrHzr=-+()=-+()11122232,.zuvuvx,()()=;ruvuvx,()()=+22rxy=+22KuvHuvuv=-++()=-++()1112222232,.xxxxxuvuuuvvvvEvuFuvGuUvuWWuvW=()=+=()==+=--()=++===()===10101111000011002222,,,,,,,,,,,,,,,,,,LNM,,.Kb=-12-£<102bK.KEGFbWWbWbbuHGEFEGF=--=-()=-=-+()=+--()=LNMLNM,22222422222220.5.4Computational Techniques229 23... |
Khabc=-=42224222/ /,andKr=12/.hZ=1Khabc=4222.hXUXZZZ===••.1TMp()KabcZZxaybzc==++ÊËÁˆ¯˜1222442424242, .whereFIG. 5.27 5.5The Implicit Case239(b)When K(p)π0,pis a critical point ofhifand only ifp(consideredas a vector) is orthogonal to Mat p.3.(a)Use the preceding exercises to find the critical points ofthe Gauss-ian curv... |
e,constant curves are also geodesics,but to avoid clutterthis case is often neglected.6.9ExampleGeodesics ofsome surfaces inR3.(1)Planes.Ifais a geodesic in a plane Porthogonal to u,then a¢•u=0,hence a≤•u=0.But a≤is by definition normal to P,hence collinear withu,so a≤=0.Thus ais a straight line.Since as noted above,eve... |
g the axis ofrevolution ofM,as shown in Fig.4.14.This geometric significance for gandhmeans that our results do not depend on the particular position ofMrel-ative to the coordinate axes ofR3.Because gand hare functions ofualone,we can writeand henceHere Eis the square ofthe speed ofthe profile curve and hence ofall theme... |
troid(Fig.5.41).Using the differential equation above,we deduce from the earlier formulas (3) that the principal curvatures ofBareThus the bugle surface has constant negative curvatureThis surface cannot be extended across its rim—not part ofB—to form alarger surface in R3since km(u) Æ•as uÆ0.◆When this surface was firs... |
ordinate functions,Example 8.4 ofChapter 2 givesBecause all forms (including functions) are now restricted to the surface S,the spherical coordinate function rhas become a constant:the radius rofthe sphere.In general,the forms associated with an adapted frame field obey the fol-lowing remarkable set ofequations.1.7Theor... |
e assumed to be connected.)3.1TheoremIfits shape operator is identically zero,then Mis part ofa plane in R3.Proof.By the definition ofshape operator,S=0 means that any unitnormal vector field E3on Mis Euclidean parallel,and hence can be iden-tified with a point ofR3(see Fig.6.4).Choose any point pin M.We will show that Ml... |
d Local Isometries281FIG. 6.8 4.2DefinitionAn isometry F:MÆofsurfaces in R3is a one-to-onemapping ofMonto that preserves dot products oftangent vectors.Explic-itly,ifF*is the derivative map ofF,thenfor any pair oftangent vectors v,wto M.IfF*preserves dot products,then it also preserves lengths oftangentvectors.It follow... |
in Example 4.6) carries rulings to meridians andhelices to parallels.Show that Mmust be a catenoid.(Hint:Use Ex.10.)12.Let Mbe the image ofa patch xwith E=1,F=0,and Ga functionofuonly (Gv=0).Ifthe derivative is bounded,show that there isa local isometry ofMinto a surface ofrevolution.Thus any small enough region in Mi... |
fields E1and E2whose valuesat each point x(u,v) ofx(D) arexxuvEG==anddKwqq1212=-Ÿ.ddqwqqwq11222211=Ÿ=Ÿ,;t<p/2Mp2Fp2t<p/22946.Geometry of Surfaces in R3 In Exercise 7 ofSection 4.4,we associated with each patch xthe coordi-nate functions u˜and v˜,which assign to each point x(u,v) the numbers uandv,respectively.For examp... |
atchlike 2-segment in a surface oriented byarea form dM.By definition,Now there are two cases:(1)IfdM(xu,xv) >0,we say that xis positively oriented.Then by the def-inition ofarea form,hence is the area ofx(R).(2)IfdM(xu,xv) <0,we say that xis negatively oriented.Thenhence is minusthe area ofx(R).Thus,to find the area ofa... |
ifferent points arenever parallel),and(2)either K≥0 or K£0 (Kdoes not change sign).Then the total curvature ofRis ±area ofG(R),where the sign is that ofK.(Evidently,this area does not exceed 4p.)For example,consider the torus Tas in Fig.5.21.Its Gauss map Gsendsthe outer halfOofT(where K0) in one-to-one fashion onto th... |
eserves shape operators,we arrange to have corre-sponding unit normals on Mand .Let Ube a unit normal to M.SinceF*preserves dot products and Fcarries Mto ,it follows that F*(U) isa unit vector field everywhere normal to .Thus F*(U) is one ofthe twounit normals on ,say,(See Fig.6.20.)IfSand are the shape operators ofMand... |
Here ·V,WÒhas itsusual pointwise meaning as the function assigning to each point pthe number·V(p),W(p)Ò.The geometric structure provided by this collection ofinner products canbe described as a metric tensorg on M,that is,a function on all ordered pairsoftangent vectors v,wat points pofMsuch thatgppvwvw,,()=.cos.J=vwvw... |
d from R2,show that the mapping Fin (2) ofExample 7.3 ofChapter 1 is conformal and orientation-preserving.What is the complex function in this case?9.Let Mbe a region UÃR2furnished with a conformal metric given byruler function h.IfFis a Euclidean isometry UÆUthat preserves h(thatis,h(F) =h),show that Fis an isometry M... |
al isometry SÆPshows that like the sphere,Phas constantpositive curvature K=1 and all its geodesics are closed.The same scheme,applied when Shas radius r,gives the projective planeofradius r,with curvature◆Like the flat torus,Pcannot be found in R3.For the flat torus this failurecould be blamed on its geometry,since tori... |
for each tin I.) The covariant derivative Y¢ofYought to be —a¢Y,but neither a¢nor Yis defined on an open set ofMas required by thedefinition of—.The simplest solution is to define Y¢by a frame field formulamodeled on the covariant derivative formula in Lemma 3.1.So for a frame field E1,E2,write Y=f1E1+f2E2,and then define—=... |
)For a surface Min R3,a curve in Mwas defined to be a geodesic ifitsEuclidean acceleration is always normal to M.This is consistent with thegeneral definition above,for by Exercise 3 ofSection 3 the intrinsic acceler-ation ofa curve in MÃR3is the component tangent to Mofits Euclideanacceleration.Thus the former is zero i... |
ifMis an oriented surface in R3,thenwhere J=–(B,U).Here kis the curvature and B the binormal vector ofaas a curve in R3.7.5Clairaut ParametrizationsWe consider a special situation where extensive information about geodesicscan be obtained with a minimum ofcomputation.5.1DefinitionA Clairaut parametrizationx:DÆMis an ort... |
he parallel postulate,but thewhole idea that the Euclidean plane is,in some philosophical sense,anAbsolute,whose properties are self-evident.It had become just one geomet-ric surface among the infinitely many discovered by Riemann.3607.Riemannian GeometryFIG. 7.11 Exercises1.In the Poincaré half-plane,show that the rout... |
J<pfrom xuto xvaswhere Jjis the coordinate angle at pj.For example,in Fig.7.15 consider thesituation at p3.By the definition ofedge curves,b¢is xv,but (-g)¢is -xu,since-gis an orientation-reversing reparametrization ofg.Thus e3+J3=p.epJeJepJeJ11223344=-==-=,,,,pxpxpxpx1234=()=()=()=()acbcbdad,,,,,,,,3667.Riemannian Geom... |
la-tion—and ifthe various tori in Fig.7.20 were realized formally,they toowould have total curvature zero.The Gauss-Bonnet theorem provides a way to attack some seemingly for-midable problems.For instance,Example 2.4(1) shows that ifa single pointis removed from a sphere S,there exists a metric on the punctured spherew... |
mentally in Fig.7.25.There,in each case,the general characterofthe vector field Vnear the singularity is described by drawing some ofits integral curves (Exercise 13 ofSection 4.8) since Vsupplies all their veloc-ity vectors.ind,Vbap()=()-()jjp2.3807.Riemannian Geometry Picture Xas a unit vector field in the positive x-d... |
chapter we investigate the global structure ofgeometric surfaces,thatis,2-dimensional Riemannian manifolds.We want to know what the possi-ble surfaces are and what they are like.In maximum generality this goal isunrealistic,but under reasonable hypotheses,good results can be obtained.The central theme ofthis chapter is... |
arametrization ofNe.Ifq=x(u0,v0),then—admitting the value 0 for u—the radial segment out to qisNow let abe an arbitrary curve segment from pto qin M.We param-etrize aon the same interval as gwithout changing its arc length.For gtobe a shortest curve from pto q,we must proveConsider first the case where astays in the nei... |
(1) impliesso gminimizes arc length as required.To prove (1) we will show that(2)rguruaur()()=-,forallq.Lrgr()==()pq,,grrr()==()qpq,where,.guuv()=()x,,0bpvavv()=()()x,024008.Global Structure of Surfaces 8.2Complete Surfaces401This is a kind ofefficiency condition on g:Each inch that gadvances bringsit one inch closer to... |
ally takes place—at the south pole ofS(Fig.8.3).For the hyperbolic plane,we know that radial geodesics from the originfollow Euclidean straight lines.From the formula for x(u,v) in Example1.7(2),we can compute†Thus the radial geodesics,after beginning in Euclidean fashion,spread apartever more rapidly than in R2,as mig... |
.In a normal neighborhood ofpŒM,call the region Deon and withinthe polar circle Cea polar diskofradius e.(a)Show that the area ofa polar disk isand hence(b)Use this formula to find the Gaussian curvature ofa sphere ofradius r.3.(a) At the pole 0in the hyperbolic plane H,find the length ofthe polarcircle Ceand the area of... |
il is completed.For details,see the Covering Homo-topy Theorem in [ST].4.7RemarkCoverings with finite multiplicity.Suppose F:ÆMis acovering with multiplicity k.(1)IfMis compact,then is compact.Proof.Mis covered by a finitenumber of2-segments.Each ofthese can be lifted to k2-segments in .Clearly these finitely many lifts c... |
n Mtheir differential maps agree,that is,ifthenF=G.Proof.IfMis complete,the proofis easy.Let mbe an arbitrary pointofM.By the Hopf-Rinow theorem (2.1) there is a vector vat the specialpoint psuch that gv(r)=mfor some number r.Then the preceding lemmagivesHence,in particular,In the general case,there is at least a broke... |
homogeneous.(c)Every isometry ofthe sphere is the restriction ofa Euclidean isome-try.(Hint:For (b) use Ex.5(c).)7.Prove:(a)IfC:SÆSis an isometry as in Exercise 6(a),there is a unique isom-etry Cp:PÆPsuch that FC=CpF,where Fis the projection SÆP.(b)The projective plane Pis frame-homogeneous.8.IfMis a connected surface ... |
Riemannian covering maps.For the horizontal arrows,the multiplicity ofthe covering is infinite.The ver-tical arrows are the two double coverings defined above.Only the flat torus and Klein bottle are compact;only the Möbius bandand Klein bottle are nonorientable.In fact,both vertical arrows represent ori-entation coverin... |
t k.Ifa geodesic segment sstarting at phas length ,there is a conjugatepoint ofpalong s.Proof.The sphere Sofradius has curvature k=1/r2,so Mhas curvature at least as large as that ofS.We know that in S,conjugatepoints arrive at arc length ,so the lemma asserts that they arriveno later on M.To prove this,let sbe a unit-... |
eorem 4.12 applies to complete the proof.◆7.7Theorem(Hadamard)IfMis a complete,simply connected surfacewith K0,then(1)Mis diffeomorphic to the Euclidean plane R2.(2)All nonconstant geodesics are one-to-one (so there are no geodesicloops or closed geodesics).(3)Through any points pπqin Mthere is a unique (up to parametr... |
new file frenet.mthat contains only the commands.Then these can beread into later work by <<frenet.m3.SurfacesA coordinate patch,say x,is given by listing its components as expressionsin two variables.For example,x[u_,v_]:={u*Cos[v],u*Sin[v],2v}454Appendix: Computer Formulas >Parameters can be handled as above for curv... |
e essentials,includingthe exterior derivative operator d.The command defformis used to specifythe degree ofthe forms involved.For example,defforms(x=0,y=0)tellsMaplethat x and y are 0-forms,that is,real-valued functions.Then thecommand d(x^2*sin(y))yields 2xsin(y)d(x)+x2cos(y)d(y).2.CurvesA curve in R3is described by g... |
38 ªL(b).Section 2.31.k=1,t=0,B=-(3,0,4)/5,center (0,1,0),radius 1.7.(a)1=||a(h)¢||=||a¢(h)h¢||=|h¢|,hence h¢=±1.(b)Let e=±1.Then =a(h) implies T=a¢(h)h¢=eT(h).Hence =k(h)N(h),and so on.NkadtdtLbaabaaa-¢=¢=()ÚÚ.Lhhdshhdshdscdcdcdaaaa()()=()=¢()-¢()=-¢¢=ÚÚÚ¢bssss()=+()(-122221//,sinh/,-215/211121610310,,,,-()-()/,/.470A... |
ofthe corresponding neighborhoods Up,qis a neighborhood ofqthatdoes not meet R.Section 4.81.IfMis orientable it has a nonvanishing 2-form m.Then f(t)=m(a¢(t),Y(t)) is a differentiable function on [a,b].By (ii),f(a)f(b) <0;hence fissomewhere zero on a<t<b.This contradicts (i).5.(a)The function pÆd(0,p) is continuous on ... |
ust carry asymptotic unit vectors to asymptotic unit vectors,and carry Uzto ±Uz.One such Cis .Chapter 7Section 7.11.(a)The speed squared is ·a¢,a¢Ò=a¢•a¢/h2(a).(b)·hUi,hUjÒ=Ui•Uj=dij.3.(a)The definition J(e1)=e2,J(e2)=-e1is independent ofthe choiceofpositively oriented frame field e1,e2,since for another positivelyorient... |
ector field on the sphere S(undefined at thepoles).IfAis the antipodal map,then A*(E)=E,so Etransfers to Pvia the projection SÆP.The unique singularity has index 1.3.The condition implies F(M)=N.Ifqis in N,then each point ofF-1(q)has a neighborhood mapped diffeomorphically onto a neighborhood ofq.The intersection Vofall ... |
Isometric imbedding,429Isometric immersion,429Isometric invariant,289,321Isometric surfaces,283Isometry ofEuclidean space,100decomposition theorem,105determined by frames,109–110tangent map,107–108Isometry group,See alsoEuclideansymmetry groupofEuclidean space,107 (Ex.7)ofa geometric surface,426498Index Isometry ofsurf... |
s. Example 2.1.1. Consider equation ut +tux = 0. Then equation of the integral curve is dt 1 = dx 2t2 = C and therefore u = ϕ(x − 1 t or equivalently tdt − dx = 0 which solves as x − 1 2t2) is a general solution to this equation. One can see easily that u = f (x − 1 2t2) is a solution of IVP. Example 2.1.2. (a) Conside... |
0; ut + x2ux = 0; ut + (t2 + 1)ux = 0; (x + 1)2ut + ux = 0; (x2 − 1)ut + ux = 0. ut + tux = 0; ut + t2ux = 0; ut + txux = 0; ut + (x2 + 1)ux = 0; (x + 1)ut + ux = 0; (x2 + 1)ut + ux = 0; (b) Consider IVP problem u|t=0 = f (x) as −∞ < x < ∞; does solution always exists? If not, what conditions should satisfy f (x)? Cons... |
solution is a superposition of two waves u1 = ϕ(x + ct) and u2 = ψ(x − ct) running to the left and to the right respectively with the speed c. So c is a propagation speed. Remark 2.3.2. Adding constant C to ϕ and −C to ψ we get the same solution u. However it is the only arbitrariness. Visual examples (animation) Chap... |
at t = 0 means exactly that ξ = η, but then uξ = 0 there. Really, uξ is a linear combination of ut and ux but both of them are 0 as t = 0. Therefore ϕ′(ξ) = 0 and ˜uξ = 1 4c2 ξ η ˜f (ξ, η′) dη′ where we flipped limits and changed sign. Integrating with respect to ξ we arrive to ˜u = 1 4c2 ξ ξ′ ˜f (ξ′, η′) dη′ dξ′ = η 1... |
t′) − cosx + 2(t − t′) cos(t′) dt′ sin(x) sin(2(t − t′)) cos(t′) dt′ sin(x) sin(x) sin(2(t − t′) + t′) + sin(2(t − t′) − t′) dt′ sin(2t − t′) + sin(2t − 3t′) dt′ t 0 t 0 sin(x) cos(2t − t′) + 1 3 cos(2t − 3t′) t′=t t′=0 sin(x) cos(t) − cos(2t) . = 1 3 Remark 2.5.4. Sure, separation of variables would be simpler here. ... |
sin(x) =⇒ ϕ(x) = 2 cos(x) + D and since ϕ must be continuous at 0 we get D = −4, ϕ(x) = 2 cos(x) − 4 and we get solution u = 2 cos(x + 2t) + 6 cos(x − 2t) − 4 in the zone {(x, t) : t < 0, 0 < x < −2t}. Chapter 2. 1-Dimensional Waves 68 t u = −2 cos(x + 2t) − 6 cos(x − 2t) + 4 u = −2 cos(x + 2t) + 6 cos(x − 2t) x u = 2 ... |
ed waves. In particular, consider ϕ(x) = eikx. (c) Discuss connection to 2. Problem 4. (a) Find solution utt − c2uxx = 0, u|t=0 = ϕ(x), (ux + αu)|x=0 = 0, ut|t=0 = cϕ′(x) t > 0, x > 0, x > 0, t > 0 (separately in x > ct, and 0 < x < ct). (b) Discuss reflected wave. In particular, consider ϕ(x) = eikx. Problem 5... |
AUx + BU = F where E, A, B are n × n-matrices, U is unknown n-vector (column) and F is known n-vector (column). (2.8.1) 2.8.2 Completely separable systems Assume that E and A are constant matrices, E is non-degenerate, E−1A has real eigenvalues λ1, . . . , λn and is diagonalizable: E−1A = Q−1ΛQ with Λ = diag(λ1, . . . ... |
√ π x√ 4kt −∞ e−z2 dz =: 1 2 + 1 2 erf x √ 4kt with erf(z) := 2 √ π z 0 e−z2 dz (3.1.12) (erf) and that an upper limit in integral tends to ∓∞ as t → 0+ and x ≶ 0. Then since an integrand is very fast decaying at ∓∞ we using (3.1.10) arrive to Statement (ii). Remark 3.1.4. (a) One can construct U (x, t) as a self-simi... |
1 and imply integral representation (3.2.20) (or its n-dimensional variant). Visual examples (animation) Similarly to Theorem 3.1.5 and following it remark we have now: Theorem 3.2.2. Let u satisfy equation (3.2.16) in domain Ω ⊂ Rt × Rn x with f ∈ C ∞(Ω). Then u ∈ C ∞(Ω). Remark 3.2.3. (a) This “product trick” works f... |
dition ux|x=0 = 0; Problem 3. Solve IBVP problem for a heat equation t > 0, 0 < x < ∞ ut = kuxx u|t=0 = g(x) u|x=0 = h(t) (5) with g(x) = 0, h(t) = 1 t < 1, 0 t ≥ 1; g(x) = 1 x < 1, 0 x ≥ 1, h(t) = 0; g(x) = 1 − x x < 1, x ≥ 1, 0 h(t) = 0; g(x) = 1 − x2 x < 1, x ≥ 1, 0 h(t) = 0; g(x) = e−ax, h(t) = 0; g(x) = xe... |
n (4.1.14) can be rewritten as un(x, t) = Cn cos( cπnt l + ϕn) sin( πnx l ) represents a standing wave: Standing Wave which one can decompose into a sum of running waves Standing Wave Decomposition and the general discussion of standing waves could be found in Standing Wave Discussion. Points which do not move are call... |
re real and non-negative (which is also the case). Problem 1. Justify Example 4.2.6 and Example 4.2.7 Consider eignevalue problem with Robin boundary conditions X ′′ + λX = 0 X ′(0) = αX(0), X ′(l) = −βX(l), 0 < x < l, with α, β ∈ R. (a) Prove that positive eigenvalues are λn = ω2 n and the corresponding eigenfunctions... |
tions. (d)-(e) In particular (v − w) is orthogonal to w and then ∥v∥2 = ∥(v − w) + w∥2 = ∥v − w∥2 + 2 Re (v − w, w) =0 +∥w∥2. (a)-(b) Consider w′ ∈ K. Then ∥v − w′∥2 = ∥v − w∥2 + ∥w − w′∥2 because (w − w′) ∈ K and therefore it is orthogonal to (v − w). 4.3.4 Orthogonal systems: approximation. II Now let {un}n=1,2,..., ... |
, we can replace f (x) by its 2l-periodic continuation from J to R. Then we can take any interval of the length 2l and result will be the same. So we take [x − l, x + l]. Now SN (x + sin(k(N + 1 2)(x − y)) sin(k(x − y)) sin(k(N + 1 2)(x − y)) sin(k(x − y)) sin(k(N + 1 2)(x − y)) sin(k(x − y) sin(k(N + 1 2)(x − y)) sin(... |
m ∈ N. Problem 6. Decompose into Fourier series with respect to sin((n + 1 2)x) (n = 0, 1, . . .) on interval [0, π] and sketch the graph of the sum of such Fourier series: (a) 1; (b) x; (c) x(π − x); (d) sin(mx) with m ∈ N; (e) cos(mx) with m ∈ N; (f) sin((m − 1 2)x) with m ∈ N. Problem 7. Using Fourier method find al... |
ther than above eigenvalues and eigen- functions; (ii) un(x) = Hn(x)e− x2 2 where Hn(x) is a polynomial of degree n (and it is even/odd for even/odd n); (iii) All un(x) are orthogonal; ∥un∥ = √ πn!. (iv) System {un} is complete. Proof. (i) The second of equalities (4.C.5) LZ = Z(L − 1) implies that if u is an eigenfunc... |
a G(k) dk + b c+ε G(k) dk if there is a singularity at c ∈ (a, b). Often instead of pv is used original (due to Cauchy) vp (valeur principale) and some other notations. This is more general than the improper integrals studied in the end of Calculus I (which in turn generalize Riemann integrals). Those who took Complex... |
c in lower half-plane {z : Im(z) > 0}, ∞ −∞ |f (ξ − iη)|2 dξ ≤ M ∀η ≥ 0 (5.2.B.4) Chapter 5. Fourier transform 183 (ii) There exists a function u, u(x) = 0 for x < 0 and ∞ −∞ |u(x)|2 dx ≤ M (5.2.B.5) such that f (z) = ˆu(z) for z : Im(z) ≤ 0. Remark 5.2.B.1. To consider functions holomorphic in in the lower half-plane ... |
e arrive to ˆu(ξ, y) = A(ξ)e−|ξ|y + B(ξ)e|ξ|y. (5.3.17) Indeed, characteristic equation α2 − ξ2 has two roots α1,2 = ±|ξ|; we take ±|ξ| instead of just ±ξ because we need to control signs. We discard the second term in the right-hand expression of (5.3.17) because it is unbounded. However if we had Cauchy problem (i.e.... |
ld be represented in the form of the appropriate Fourier integral. max |u| < ∞. Chapter 5. Fourier transform 199 Problem 10. Solve the following equations explicitly for u(x) (that means do not leave your answer in integral form!). (a) e− x2 2 = 1 2 ∞ −∞ e−|x−y|u(y) dy; (b) e−x2 = ∞ 0 f (x − y)e−ydy; (c) e− x2 4 = 0 −∞... |
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