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1 Math 348 Fall 217 Lecture 2: Review of Prerequisites Disclaimer. As we have a textbook, this lecture note is for guidance and supplement only. It should not be relied on when preparing for exams. In this lecture we review the prerequisites: Multivariable calculus, linear algebra, and basics of dierential equations. We will not be able to cover everything that will be needed through the semester. We will only cover the basics, and leave some more sophisticated topics such as inverse/implicit function theorem for later lectures to explain. Warning. In this lecture note we omit many technical assumptions such as dierentiability/integrability of functions when stating formulas and theorems, because in 348 they are almost always satised. However please do not quote the statements here for your other courses. Table of contents Lecture 2: Review Multivariable calculus The Euclidean space R n Operations on R n Functions from R m to R n Taking derivatives of functions from R m to R n Linear algebra Operations on matrices Linear dependence and independence Dierential equations The simplest and second simplest dierential equations Existence and uniqueness of a general rst order ODE

2 Dierential Geometry of Curves & Surfaces 1. Multivariable calculus Multivariable calculus studies functions from R m to R n with either m > 1 or n >1. In 348 we focus on functions from R to R 2 /R 3 (curves) and from R 2 to R 3 (surfaces). In the following we review concepts, theorems, and techniques that are important to us The Euclidean space R n R n is the mathematical representation of a n-dimensional at space through setting up n orthogonal axes. Along each axes, we dene a unit vector: e 1 ; :::; e n. Then each point in the space is identied by an n-tuple (x 1 ; :::; x n ). It is important to realize that R n can be interpreted in two ways, in the context of describing a moving particle (along a curve, on a surface, etc.). R n as possible locations. At each moment, the location of the particle is identied with a point in R n : At time t, the particle is at (x 1 ; :::; x n ). R n as possible velocities. To completely describe the motion of this particle, we also need to know its velocity. The velocity is also described by an n-tuple (v 1 ; :::; v n ). Such n-tuples are called vectors. Thus at each point x in the location space R n, we overlay a velocity space whose mathematical representation is also R n. This overlay will later be called the tangent space of the location space at x. Such distinction seems a bit silly now, but will be very useful later when we curve the location space. The key idea of dierential geometry is that when the location space is curved, the velocity space stays at. To visualize, think of a 2-dimensional surface as the curved locations space, then the velocity space at a location is the tangent plane of the surface there Operations on R n When R n is interpreted as the overlaying velocity space, the most important concepts are Norm. When v = (v 1 ; :::; v n ) describes a velocity vector, its speed is given by its norm: p kvk := v v n : (1) Inner product. The inner product of two vectors u = (u 1 ; :::; u n ) and v = (v 1 ; :::; v n ) is dened as u v := u 1 v u n v n : (2) The importance of inner product comes from the following property: Proposition 1. Let be the angle between two vectors u = (u 1 ; :::; u n ) and v = (v 1 ; :::; v n ), then cos = u v kuk kvk : (3) 2

3 Math 348 Fall 217 Exercise 1. Obviously is not uniquely dened. What does this not matter? In particular, uv =()u?v, that is u is perpendicular(orthogonal) to v. Cross product/vector product. Let u=(u 1 ;u 2 ;u 3 );v =(v 1 ;v 2 ;v 3 ) be two vectors in R 3. Their cross product (also called vector product) is dened as u v := (u 2 v 3 u 3 v 2 ; u 3 v 1 u 1 v 3 ; u 1 v 2 u 2 v 1 ): (4) Exercise 2. Prove the following. i. u v = v u; ii. Let a; b 2 R be arbitrary. Then (a u + b w) v = a u v + b w v: (5) iii. u v = if and only if u and v are parallel. iv. (u v) u = ; (u v) v =. v. (u v) w = (v w) u = (w u) v. Also note that where is the angle between u; v. ku vk = kuk kvk sin (6) Remark 2. It turns out that reasonable cross-products can only be dened in R 3 and R 7. See W. S. Massey, Cross products of vectors in higher dimensional Euclidean spaces, The American Mathematical Monthly, 9(1): , When R n is interpreted as the underlying location space, the most important concepts are Distance. Let x = (x 1 ; :::; x n ) and y = (y 1 ; :::; y n ). For now, the only distance we know is the distance along straight lines, given by the Pythagorean theorem: p d(x; y) = (x 1 y 1 ) (x n y n ) 2 : (7) Remark 3. We see that d(x; y) = kx yk. However this will cease to be true when the location space is curved. One important theme in 348 is to generalize (7) for curved spaces. Area. The foundation of the denition of area is the following agreement: i. The unit square has area 1. ii. The area of a union of disjoint regions is the sum of the areas of these regions. 1. The idea of the proof is that if a cross product can be dened on R n such that i) v w is bilinear in v and w, ii) v w is perpendicular to both v; w, iii) jv wj 2 + (v w) 2 = kvk 2 kwk 2, then R n+1 can be made into a normed division ring with identity. By Hurwitz's Theorem n + 1 = 1; 2; 4; 8 and the conclusion follows. 3

4 Dierential Geometry of Curves & Surfaces iii. Moving a region around rigidly (that is not changing the distance of any pair of two points in the region) does not change its area. From these one could develop the whole theory of integration in two-variable calculus. Example 4. Let x = (x 1 ; x 2 ; x 3 ); y = (y 1 ; y 2 ; y 3 ) in R 3. Then the area of the triangle (; ; ) (x 1 ; x 2 ; x 3 ) (y 1 ; y 2 ; y 3 ) (; ; ) is given by 1 kx yk. 2 Volume. The foundation of the denition of volume is the following agreement: i. The unit cube has volume 1. ii. The volume of a union of disjoint regions is the sum of the volumes of these regions. iii. Moving a region around rigidly (that is not changing the distance of any pair of two points in the region) does not change its volume. With these agreement one could develop the whole theory of integration in three-variable calculus. Example 5. Let x = (x 1 ; x 2 ; x 3 ); y = (y 1 ; y 2 ; y 3 ); z = (z 1 ; z 2 ; z 3 ) in R 3. Then the volume of the parallelopiped spanned by these three vectors is given by V = j(x y) zj: (8) Exercise 3. Check your old textbook and explain what the sign of (x y) z means Functions from R m to R n Let R m be a region in R m. A function f: 7! R n is a correspondence between points in and points in R n satisfying the following: Each x 2 corresponds to exactly one y 2 R n. One classies functions into two categories: linear and nonlinear. In 348, we will see that linear functions are often considered in the context of velocity spaces. The most important functions are linear functions (linear transformations) and bilinear forms. Linear functions. A function f: 7! R n is linear if and only if for every a; b 2 R and every x; y 2, one has f(a x + b y) = a f(x) + b f(y): (9) Exercise 4. Prove that if f is linear then f() =. Exercise 5. Let f: R 7! R be linear. Prove that there is a constant c 2 R such that f(x) = c x. Notation. We often denote a linear function by a capital letter such as T First letter of transformation. 4

5 Math 348 Fall 217 Proposition 6. Let T : R m 7! R n be linear. Then there are m n numbers a 11 ; a 12 ; :::; a 1m ; a 21 ; :::; a n1 ; :::; a nm 2 R, such that for every v 2, the following holds for w = T (v): w 1 = a 11 v a 1m v m (1) w n = a n1 v a nm v m : (11) In other words, we have T (v) = A v where A is the matrix Exercise 6. Prove Proposition a 11 a 1m a n1 a nm Bilinear forms. A bilinear form is a special type of nonlinear function: B:R 2n 7!R, satisfying for every a; b 2 R and every u; v; w 2 R n, 1 A. B(a u + b w; v) = a B(u; v) + b B(w; v); (12) B(u; a w + b v) = a B(u; w) + b B(u; v): (13) Exercise 7. Prove that the inner product: f(u; v) := u v is a bilinear form. Exercise 8. Prove or disprove: The area A(u; v) := ku vk is a bilinear form. Proposition 7. Let B: R 2n 7! R be a bilinear form. Then there are n 2 numbers a 11 ; :::; a 1n ; a 21 ; :::; a n1 ; :::; a nn such that Exercise 9. Let A := inner product. n B(u; v) = X i=1 a11 a n1 a nn 1.4. Taking derivatives of functions from R m to R n Differentiability. Xn u i a ij v j : (14) j=1 1 A. Prove that B(u; v) = u A v = (A u) v where denotes Definition 8. A function f: R m 7! R n is dierentiable at x 2 if and only if the following holds: There is a linear function T : R m 7! R n such that kf(x + v) f(x) T (v)k lim = : (15) v! kvk The linear function T is called the dierential of f at x and will be denoted Df(x). Remark 9. Note that f is dened on which belongs to the location space R m, and f takes value in another location space R n ; while T is dened on R m which is the velocity overlay space on top of, centered at x, and takes value in another velocity space centered at f(x). 5

6 Dierential Geometry of Curves & Surfaces By Proposition 6 there is a matrix A such that T (v) = A v for all v 2 R m. The numbers a ij in the matrix A is given by the partial derivatives of f at x: Definition 1. Let f: R m 7! R. Its j-th partial derivative at x = (x 1 ; :::; x m ) 2 is dened (x) := df(x 1; :::; x j 1 ; t; x j+1 ; :::; x m ) j t=xj : (16) Exercise 1. Prove that a j (x). Note. This matrix (representation of the dierential of f) is called the Jacobian matrix of f at x. Exercise 11. Let f: R m 7! R n be dierentiable everywhere. Prove that if Df(x) = for all x 2 R m, then f is a constant. Chain Rule. Let f(x): R m 7! R n and g(y): R n 7! R k. Then the composite function g f: R m 7! R k is dened as (g f)(x) = g(f(x)): (17) The partial derivatives of g f can be calculated f) i = X j i k(x) : j Exercise 12. Find a calculus textbook and work on a few chain rule problems. Higher order derivatives. We consider the simplest case f: R m 7! R. At every x 2, it has a dierential which is a linear function Df(x): R m 7! R. As this linear function can be represented by a 1 m matrix, it can also be identied with an m-vector. Thus at each point x 2, we have a vector g(x) 2 R m. Dierentiating this function g we obtain the second order derivative of f at x, denoted D 2 f(x). Exercise 13. Prove or disprove: Let x 2 be arbitrary. The second order derivative of f can be seen as a bilinear form on R 2m = R m R m, where each R m is the velocity space at x. Exercise 14. Prove that the matrix representing the bilinear form is given by a ij : j As we can see here, for functions between R m and R n with m or n > 1, higher order derivatives get more and more complicated. However there is one exception. Exercise 15. What happens if we take successive derivatives of a function f: R 7! R n? Note. The matrix representation of D 2 f(x) is called the Hessian matrix of f at x. Exercise 16. What are the Jacobian and Hessian matrices for f: R 7! R? Taylor expansion. 6

7 Math 348 Fall 217 Let f: R m 7! R. Then there holds f(x + v) = f(x) + Df(x)(v) D2 f(x)(v; v) + R (2) where lim v! krk kvk 2 =. (2) is the Taylor expansion of f at x to second order. Exercise 17. Prove the following. Examples. m Df(x)(v) = X v i ; D 2 f(x)(v; v) = i i=1 Xm 2 j v i v j : (21) Example 11. Let f(x; y):= e xy. We calculate its Taylor expansion at (; ) to second order. We have Therefore the expansion reads f(; ) = e = y exy (; ) @y = x exy (; ) 2 = 2 y2 e xy ) = 2 f f = exy f (; ) = 2 = 2 x2 e xy f (; ) = f(u; v) = f(; ) + Df(; )(u; v) D2 f(; )(u; v) + R = 1 + u v + R: (22) Remark 12. (Interpreting f and Df) Functions are often called maps. For classical dierential geometry this is pertinent. Consider, e.g., Google Map with GPS. This can be seen as a function f from R 2 to the earth surface. Thus f(x ) is the real world location of a car that is at location x on the map. More importantly, if the car symbol is moving with velocity v on the map, the velocity of the car itself in the real world is given by [Df(x )](v) (remember that Df(x ) is a function!). It is woth mentioning that, although the car is located in a curved surface (earth surface), its possible velocities at a particular location (time) form a plane, the tangent plane of earth surface at location f(x ). 2. Linear algebra The fundamental idea of calculus is to study functions through their rst and second (or higher) derivatives. We have seen that such derivatives can be viewed as linear transformations between R m and R n. Therefore to understand them we need linear algebra. 7

8 Dierential Geometry of Curves & Surfaces 2.1. Operations on matrices Determinant. Determinant is dened for square matrices. Determinant is uniquely dened. Let f(v 1 ; :::; v n ): R n R n 7! R be such that i. f is linear in each of the v i 's. That is, e.g., f is a linear function of v 1 when v 2 ; :::; v n are xed. ii. f(e 1 ; :::; e n ) = 1. Where e i is the vector with one 1 at the ith entry and 's elsewhere. iii. f changes sign when two columns are switched. e.g. f(v 1 ; v 2 ; v 3 ; :::; v n ) = f(v 2 ; v 1 ; v 3 ; :::; v n ). Then f(v 1 ; :::; v n ) = det V where V is the matrix with v 1 as its rst column, v 2 as its second column, and so on. Exercise 18. Prove the above claim for n = 2; Let A := a 11 a 12 a 21 a 22. Then det A := a 11 a 22 a 12 a 21 : (23) Exercise 19. Let u; v; x; y 2 R 3. Prove that u x v x (u v) (x y) = det u y v y 3 3. Let A a 1 11 a 12 a 13 a 21 a 22 a 23 A. Then a 31 a 32 a 33 : (24) det A := a 11 a 22 a 33 + a 12 a 23 a 31 + a 21 a 32 a 13 a 13 a 22 a 31 a 12 a 21 a 33 a 23 a 32 a 11 : Exercise 2. Let u; v; w 2 R 3. Let A u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 (25) 1 A. Prove that det A = (u v) w. A square matrix A is invertible, that is there is a matrix B such that A B = B A = I A if and only if det A =/. Eigenvalues and eigenvectors. In most cases the matrix A under study is the mathematical representation of a linear function T. We could try to understand an n n matrix through its n 2 entries. However a more ecient way is through its eigenvalues/eigenvectors. The eigenvalues of an n n matrix A are the solutions to the equation det( I A) = : (26) 8

9 Math 348 Fall 217 The eigenvectors corresponding to an eigenvalue are those vectors satisfying ( I A) v = () A v = v: (27) In many situations, one could choose n linearly independent eigenvectors fv 1 ; :::; v n g to form a coordinate system, that is every vector u 2 R n can be uniquely represented as u = a 1 v a n v n where a 1 ; :::; a n 2 R. Then the linear function T can be very easily understood: T (u) = 1 a 1 v n a n v n : (28) The most useful situation for us is when A is symmetric: a ij = a ji for all i; j = 1; 2; :::; n. In this case we can choose n eigenvectors v 1 ; :::; v n satisfying i. v i?v j, for all i =/ j; ii. kv i k = 1 for all i = 1; 2; :::; n Linear dependence and independence Linear dependence/independence. Let v 1 ; :::; v k 2 R n. We say they are linearly dependent if there are a 1 ; :::; a k 2 R, not all equal to, such that a 1 v a k v k = : (29) If no such a 1 ; :::; a k exist, we say the vectors are linearly independent. Exercise 21. Prove that, if v 1 ; :::; v k are linearly independent, then the following holds. If a 1 ; :::; a k 2 R satisfy a 1 v a k v k =, then a 1 = a 2 = = a k =. Exercise 22. Prove that if v 1 ; :::; v k are linearly dependent, then at least one of them is redundant in the sense that it equals a linear combination of the other vectors. n vectors v 1 = (v 11 ; :::; v 1n ); :::; v n = (v n1 ; :::; v nn ) 2 R n are linearly independent if and only if det A =/ where A v 1 11 v 1n A: (3) v n1 v nn n vectors v 1 ; :::; v n 2 R n are linearly independent if and only if they form a base of R n, that is every u 2 R n can be written uniquely as a linear combination of v 1 ; :::; v n. 3. Dierential equations A dierential equation is an equation involving the derivative (and/or higher order derivatives) of functions The simplest and second simplest dierential equations In the following x(t): R 7! R is a single variable function. The simplets ODE. 9

10 Dierential Geometry of Curves & Surfaces Consider dx = f(t): (31) Its solution, by denition, is the indenite integral Z x(t) = f(s) ds (32) which by the fundamental theorem of calculus further equals Z t f(s) ds + C (33) where C is an arbitrary constant. If we further specify x(t ) = x, then this constant is xed. Exercise 23. What is the solution to Exercise 24. Solve dx = f(t); x(t ) = x? (34) dx = f(x); x(t ) = x : (35) The next simplest ODE. Consider + c x(t) = f(t); x() = x : (36) where c 2 R. Multiplying both sides by e ct and apply Leibniz rule of dierentiation of product of functions, we reach d (ect x(t)) = e ct f(t); (37) which is the simplest ODE now. The solution of (37) is Z t e ct x(t) = x + e cs f(s) ds (38) which now gives Exercise 25. Solve In general, the equation Z t x(t) = e ct x + e c(s t) f(s) ds: (39) + c x(t) = f(t); x(t ) = x : (4) + c(t) x(t) = f(t) (41) could be solved by multiplying both sides by e C(t) where C(t) satises C (t) = c(t). Exercise 26. Solve + c(t) x(t) = f(t); x() = x : (42) 1

11 Math 348 Fall Existence and uniqueness of a general rst order ODE General situation we face in 348. In 348 we often need to study systems of dierential equations: dx 1 (t) dx n (t) = f 1 (x 1 (t); :::; x n (t)); x 1 () = x 1 (43) = f n (x 1 (t); :::; x n (t)): x n () = x n (44) Taking advantage of multivariable calculus, we can re-write the above to a more compact form: = f(x(t)); x() = x (45) where x: R 7! R n, f: R n 7! R n, and x = (x 1 ; :::; x n ). The one ODE system that we could really solve. For general f there is little hope explicitly solving (45). However there is a special, very useful, situation that we could solve. Consider the case where f i (x 1 ; :::; x n ) = a i1 x a in x n ; i = 1; 2; :::; n (46) where a ij 2 R for all i; j = 1; 2; :::; n. In other words, we have = A x(t); x() = x : (47) Now to further simplify the situation, we assume A has n linearly independent eigenvectors v 1 ; :::; v n. Writing x(t) = z 1 (t) v z n (t) v n and x = z 1 v z n v n we obtain dz i (t) = i z i (t); z i () = z i : (48) Exercise 27. Justify (48). Exercise 28. Finish the solution. Existence and uniqueness theorem. Although there is little hope explicitly solving (45), we have the following qualitative understanding. Theorem 13. Let f: R n 7! R n be dierentiable with continuous derivatives. Then there is T > and a unique function x(t) satisfying x() = x, and for all < t < T, = f(x): (49) 11

12 Dierential Geometry of Curves & Surfaces Exercise 29. Let A(t) a 1 11(t) a 1n (t) a n1(t) a nn(t) for every i; j = 1; 2; :::; n. Prove that the solution to exists and is unique. A be such that a ij (t) are continuous functions of t = A(t) x(t) (5) Remark 14. We will see later that existence of curves with certain desirable properties would be equivalent to the existence of solution to an ODE of the form (45). Then Theorem 13 would guarantee us that this curve exists. 12

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