Part 1: Linear Algebra

1 Chapter 1

Vector in Rn. Addition and scalar multiplication. Inner product, norm • (length or magnitude) of a vector, angle between two vectors u v cos θ = · u v k kk k orthogonality and Pythagorean theorem.

Cauchy-Schwarz inequality: • u v u v . | · | ≤k kk k The textbook provides one proof in page 17 (1.14). Here we have another proof.

1. Step 1: If v = 0, the inequality is true (equal of the two sides). 2. Step 2: Introducing u v w = u · v, − v 2 k k it can be proved that w is orthogonal to v, u v u = w + · v. v 2 k k The RHS are two orthogonal vectors, therefore

u v 2 u v 2 u 2= w 2 +| · | | · | . k k k k v 2 ≥ v 2 k k k k 3. Step 3: Multiply the inequality by v 2 and take square root, we have the Cauchy-Schwarz inequalityk k

u v u v . k kk k≥ | · | Cross (vector product) u v, parallel vectors (n = 2, 3). • ×

1 2 Chapter 1

Matrix, row and column vectors, transpose, addition, scalar multipli- • cation.

l m m n Matrix multiplication, A R × ,B R × , the resulting matrix • l n ∈ ∈ is C = AB R × . Arithmetic rules of matrix multiplication cij = m ∈ k aikbkj.

n n 1 SquareP matrix A R × , identity matrix, inverse of a matrix AA− = • I. ∈

Symmetric matrix aij = aji, anti-symmetric (skew-symmetric) matrix • T T T aij = aji, orthogonal matrix AA = I, normal matrix AA = A A. Symmetric,− skew-symmetric, and orthogonal matrices are normal, but other matrices could also be normal.

H With complex matrix: conjugate transpose A , aij aji∗ . Hermitian • H H H → 1 H A = A, skew-hermitian A = A, unitary A = A− or A A = I and normal matrix AAH = AH A.−

Other special matrices: diagonal, upper and lower triangular matrices. • n n 1 Vector as a special matrix v R v R × . Matrix and vector • multiplication A Rm n and∈ v R→n, Av∈ = w Rm. ∈ × ∈ ∈ Example: A + AH is Hermitian, A AH is skew-Hermitian, A can • be made as a summation of B + C,− where B is Hermitian and C is skew-Hermitian.

3 Chapter 3

Homogeneous and non-homogeneous equations. • Degenerate equation, consistency and uniqueness. • Echelon form, pivot variables and free variables. • Elementary operations: • 1. E1–Interchange two equations.

2 2. E2–Multiply an equation by a constant. 3. E3–Add (or subtract) another equation.

Gauss elimination. Example: • x 3x 2x = 6 1 − 2 − 3 2x 4x 3x = 8 1 − 2 − 3 3x + 6x + 8x = 5 − 1 2 3 − (1)

Echelon matrix satisfies the following properties • (1). All zero rows are at the bottom of the matrix; (2). Each leading nonzero entry in a row is to the right of the leading nonzero entry in the preceding row;

Row canonical form of the matrix has two additional properties

(3). Each pivot is equal to 1; (4). Each pivot is the only nonzero entry in its column.

Row equivalence, two matrices A and B are row equivalent if the two • matrices can become each other by a set of elementary operations.

Rank is the number of pivots in the echelon form. • Gauss elimination in matrix form. Example • 1 1 2 4 5 1 1 0 10 9 − − − [A, b] = 2 2 3 1 3 0 0 1 7 7     3 3 −4 2 1 ∼ 0 0 0− 0− 0 − −     Existence: for M = [A, b] and A, the solution exist if and only if • rank(M) is the same as rank(A).

uniqueness: the solution is unique if and only if the rank of A is n, • where n is the number of unknowns.

Square system of linear equations. •

3 Inverse of coefficient matrix A. A square system of equations Ax = b • has a unique solution iff the coefficient matrix A is invertible. In this 1 case, A− b is the unique solution to the equation Ax = b.

Proof: • 1 1 1. Sufficiency: if A− exists, let x = A− b, we have

1 1 A(A− b) = (AA− )b = Ib = b

satisfies the equation. 2. Neccessity: If v is a solution and Av = b, then

1 1 1 v = Iv = (A− A)v = A− (Av) = A− b.

The concept of linear combination. Interpretation of Ax = b: b is a • linear combination of column vectors of the matrix A.

Homogeneous system of equations. Homogeneous equation always has • the trivial solution. When will homogeneous equation has non-trivial solution?

If u and v are solutions of the homogeneous equation, the linear com- • bination of them is also the solution.

Dimension and basis of solution to a homogeneous equation. • The non-zero solution of homogeneous equation. Example: • x + 2x 3x + 2x 4x = 0 1 2 − 3 4 − 5 x 3x + 2x = 0 3 − 4 5 2x 6x + 4x = 0 3 − 4 5 to become

x + 2x 3x + 2x 4x = 0 1 2 − 3 4 − 5 x 3x + 2x = 0 3 − 4 5

Free variables are x2, x4, x5. Let one of them equal to one and others zero, find the three basis.

4 Corresponding non-homogeneous equation, if w is a particular solution • to the non-homogeneous equation, u + w is also a solution, where u is the solution of the corresponding homogeneous equation.

Elementary matrix, elementary operations can be represented by mul- • tiplication of an elementary matrix. Non-singular matrix is a product of elementary matrices.

Example of 3 3 elementary matrices. • × 1. Exchange two rows (second and third):

1 0 0 E = 0 0 1   0 1 0   2. Multiplying a (second) row by a constant k:

1 0 0 E = 0 k 0   0 0 1   3. Add a row to another (add second to third):

1 0 0 E = 0 1 0   0 1 1   Finding inverse of a square matrix A. • E E E A = I,E E E IA = I, 1 2 ··· n 1 2 ··· n therefore E E E I is the inverse of A. Example: 1 2 ··· n 1 0 2 A = 2 1 3   4− 1 8   Counting # of operations in Gauss elimination method. •

5 Lower and upper triangular matrices. The inverse of a lower triangular • matrix is another lower triangular matrix. The inverse of an upper triangular matrix is another upper triangular matrix.

Gauss elimination can be expressed by the multiplication of a series of • ”atomic” lower triangular matrix. Its inverse gives the LU decomposi- tion.

Doolittle algorithm. • (n) (n) (n 1) 1. Define A = L A − 2. Let 1 0 ...   (n 1) − (n) 1 ai,n L =  .  , li,n = , i = n+1, ,N.  ..  − (n 1) ···  ln+1,n  an,n−  . .   . ..     0 l 1   N,n    3. (N 1) (N 1) (1) U = A − = L − L A ··· 4 Chapter 4

Vector space, definitions: V is a vector space if it satisfies two conditions • 1. If u, v V , then u + v V . ∈ ∈ 2. If u V , then ku V , where k is a scalar. ∈ ∈ Axioms of vector space:

1. (u + v) + w = u + (v + w). 2. There exists 0, where u+0 = u; there exists u, where u+( u) = 0. − − 3. 1u = u,

Examples of vector spaces: vector, matrix, polynomial, real functions. • 6 Subspace of a space: a subset and a space. • 1. An m-th degree polynomial is a subspace of an n-th degree poly- nomial if m < n. 2. Polynomial functions form a subspace of the real function space. 3. Vector subspace.

Linear combinations, spanning sets. If V is a vector space, a set of • vectors u1, u2, , un forms a spanning set of V if ALL vectors in V can be made by··· a linear combination of the set.

Linear dependence and independence. Given a set of vectors u1, u2, , um, • if a vector v can be made as a linear combination of the set, v is linearly··· dependent on the set. Otherwise v is linearly independent of the set.

Basis and dimension. A minimum set of vectors which spans a space • V is called the basis of the space. The minimum number of vectors to span a space V is called the dimension of the space.

Row space and rank of a matrix. Elementary operations will not change • the row space. Row equivalent matrices have the same row space. In echelon and row canonical forms, row vectors are linearly independent. They form the basis of the row space. The number of row in the two forms is called the rank of the matrix.

An n n square matrix with rank n. • × Example: find the basis for W = span(u , u , , u ). • 1 2 ··· r 1. Form matrix using vectors as rows. 2. Reduce to the echelon form. 3. Output nonzero rows as the basis.

The column rank is the same as the row rank. Column vectors form a • column space. Interpretation of Ax = b: x1a1 + x2a2 + xnan = b is a linear combination of column vectors of A. b must be··· in the column space, or there is no solution.

7 Example: Determine if a given vector w is a linear combination of n • vectors U = (u1, u2, un), and find the coefficient of linear combina- tion. This is to solve···

x u + x u + x u = w or Ux = w, 1 1 2 2 ··· n n where U is a matrix using u , u , , u as column vectors. 1 2 ··· n Polynomial as a vector. A polynomial is a vector whose components • are its coefficients.

Homogeneous equation Ax = 0. The solutions of the equation form • a vector space, called the null space of A. The dimension of the null space is the number of free variables in the echelon form, that is n r, where r is the rank of A. −

To determine what homegeneous equation whose null space is spanned • by a given set of vectors U = (u1, , un), test if for any vector w, the solution exists. The existence of Ux···= w requires consistency, therefore all remaining RHS in echelon form must equal to zero. This gives the set of homogeneous equations.

To find the basis of the null space of A, using the free variables in • the echelon form to find linearly independent solutions (choose one free variable as 1, others as zero, for each of them).

Column space and null space are complements. • To determine if two given sets of vectors span the same space, form • two matrices using the two sets of vectors as rows. Reduce them to the row-canonical form and compare the basis of the two rwo spaces.

Find homogeneous equations whose null space is spanned by a given • set of vectors ui: (1) Assume v is spanned by this space W , v is a linear combination of the given vectors; (2) The solution of Ux = v must sat- isfy the consistency condition, these condition gives the homogeneous equations W must satisfy, where U is formed by the column vectors ui. Find the sum of two spaces spanned by two set of vectors: using the • two set of vectors as row vectors for a matrix, reduce it to echelon form and find the basis.

8 Find the intersection of two spaces spanned by two set of vectors: (1) • Using these each set as column vectors, find the homogeneous equa- tions which the space must satisfy; (2) Combine the equations since for intersections, all equations must satisfy.

Typical problems: • 1. Given a set of vectors, tell if they are linearly independent, find dimension and basis.

2. Given a set of vectors u1, um, determine if another vector v is in the space spanned by the··· given set of vectors. 3. Given a matrix, find its row space and null space. 4. Given two set of vectors, they span spaces of U and V , find the sum and intersection of the two spaces.

5 Chapter 5

Mapping f : A B, for each a A, there is a unique b B. b is called • the image of a.→A is the domain∈ and B is the target set.∈ Function is a mapping.

Linear mapping. F (au + bv) = aF (u) + bF (v), where u, v A, and a, b • are two constants. ∈

Examples: projection mapping, differentiation mapping, integral map- • ping, zero mapping, identity mapping, matrix multiplication mapping.

Kernel of linear mapping: KerF = v V : F (v) = 0 . • { ∈ } Image of linear mapping: ImF = u U : v V,F (v) = u . • { ∈ ∃ ∈ } Kernel is a subspaces of V and image is a subspace of U. •

If v1, v2, , vm span the space of V , then F (v1),F (v2), ,F (vm) span • the subspace··· ImF . Proof. ···

Kernel of matrix mapping. For any vector v in the kernel of A, Av = 0. • KerA = nullsp(A). Av = 0 has unique solution, the kernel of A is zero.

9 c c c Image of matrix mapping. Since Ax = x1a1 + x2a2 + xmam is a • c ··· linear combination of column vectors ai , therefore image is the column space (spanned by column vectors).

Rank and nullity of a linear mapping. rank(A) = Dim(ImA). nullity(A) = • Dim(KerA).

Let the dimension of A is n, then n = nullity(A)+rank(A) = dim(KerA)+ • dim(ImA).

Example: Let F : R4 R3 be a linear mapping • → F (x, y, z, t) = (x y + z + t, 2x 2y + 3z + 4t, 3x 3y + 4z + 5t) − − − – Find the basis and dimension of ImF . Use 4 simple basis vectors (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1) to find vectors which span the ImF , then reduce to linearly independent sets by matrix elim- ination. – Find the basis and dimension of kernel of the F . Solve F (x, y, z, t) = (0, 0, 0). Reduce to echelon form, express the solution in terms of free variables. The number of free variables gives the dimension of the kernel space.

The Rn space is n dimensional and needs n linearly independent vectors • as basis. For a linear mapping, if the domain is n vector, the dimension of kernel plus the dimension of image space equals n.

Example: let u1 = (1, 2, 5, 3), u2 = (2, 3, 1, 4), and u3 = (3, 8, 3, 5), • (a). Find the dimension− and basis of the set− (or space spanned− by the− three vectors). (b). extend the basis to form a basis for R4.

Find homogeneous system whose solution space is spanned by a given • set of vectors. Example: solution space spanned by a1 = (1, 1, 2, 3), a = 3, 4, 3, 8), a = (1, 3, 4, 1). To solve: 2 3 − T – Use a1, a2, a3 as column vectors for form a matrix A, let x = (x1, x2, x3, x4) be a vector. Then if equation Av = x has solution, that is, if x is a linear combination of a1, a2 and a3, Av = x must have solution.

10 – Use M = (a1, a2, a2, x) as augmented matrix, reduce M to echelon form. A and M must have the same rank. – The last few rows which will give M larger rank if not zero, gives the homogeneous equations for x.

6 Review 1

– Vector and matrix, arithmetics, different matrices (transpose, her- • mitian, diagonal, triangular, identity, symmetric, inverse). – System of equations, elementary operations, Gauss elimination. – Echelon form, Row canonical form, pivot, free variables, row equiv- alence, rank. – Consistency of equation, matrix reduction method (to solve equa- tion and to find inverse). – Homogeneous system of equations, express solution in terms of free variables. – Space, vector space, linear independence, linear combination, span- ning a space, subspace. – Dimension and basis, find dimension and basis for space spanned by a set of vectors. – Matrix row and column spaces. The consistency of the equation Ax = b. – Given solution space, find the homogeneous system. – Given two vector spaces, find the dimension of sum and intersec- tion of the two spaces. – Linear mapping, domain and range. – Kernel and Image, rank and nullity.

11 Part 2: Calculus

7 Chapter 1

Number sets, natural, integer, rational, real, complex, algebraic and • transcendental, equation and inequality.

Arithmetic and geometric means, triangle inequality. • a + b √ab, a, b 0 2 ≥ ≥ a + b a + b . | | ≤ | | | | Proof, direct proof (derivational), proof by contradiction, and proof by • induction. Using direct and inductive proofs for identity 1 1 + 2 + 3 + + n = n(n + 1). ··· 2 Use inductive proof to prove 1 12 + 22 + 32 + + n2 = n(n + 1)(2n + 1). ··· 6 8 Chapter 2

Sequence, Fibonacci sequence, golden ratio. • Convergent sequence, limit of a sequence, -N description. Example: • n 1 an = − , lim an = 1 n n + 1 →∞ To have 2  a 1 < , choose N > − | n − |  Infinity, bounded sequence, monotonically bounded sequence • Least upper bound and greatest lower bound, suprenum. •

12 Bolzano-Weierstrass theorem, if a sequence is bounded and infinite, • there is at least one limit point. Example: n 1 a = ( 1)n − n − n + 1

Cauchy convergence theorem: for given sequence an, for any  > 0, if • we can find N, such that for all n > N and m > N

a a < , | n − m| then a is convergent. | n| Series, arithmetic series, • n

Sn = an, an = an 1 + d − i=1 X geometric series, n n Sn = q i=0 X Zeno’s paradox (Achilles and turtle).

Theorem of infinite series • S = a + a + + a + n 0 1 ··· n ··· a lim | n+1| < 1 n a →∞ | n| the series is convergent.

Series • ∞ 1 S = n n(n + 1) n=1 X Harmonic series diverges. • n 1 1 1 1 S = = 1 + + + + n i + 1 2 3 ··· n i=1 X 13 9 Chapter 3

Functions, bounded and unbounded, monotonic functions. • Inverse and implicit functions, multi-valued function. • Polynomial, algebraic and transcendental functions. • Power function, logarithmic function, trigonometric function. • Limit of a function, -δ description, left and right limits. • lim x2 = 4. x 2 → Given  > 0, find δ > 0, such that when x 2 < δ, x2 4 < . In this case, if we choose x 2 < δ =  , we| will− have| | − | | − | 2 x2 4 = x 2 x + 2 < 2 x 2 =  | − | | − || | | − | suppose x is sufficiently close to 2 such that x > 0.

Left limit, right limit and existence of limit. • Example: • sin x f(x) = x Sandwich theorem.

The limit of e: • n 1 1 e = lim(1 + x) x = lim 1 + x 0 n n → →∞   Discuss limits: • x lim x , lim | |. x 0 x 0 → | | → x Examples of limit: • x cos x 1 lim , lim − x 0 √x + 9 3 x 0 x2 → − →

14 • ax 1 y 1 lim − = ln a lim = ln a 1 = ln a, x 0 x 0 → x → ln(1 + y) limx 0 ln(1 + y) y → where y = ax 1. − Continuity of function, three conditions for continuity, piece-wise con- • tinuity.

Continuity of functions: • x2 3x + 1 sin x f(x) = − , f(x) = x , f(x) = x 1 , f(x) = . x 1 | | | − | x − 10 Chapter 4

Derivative as rate of change, condition for existence, piecewise differ- • entiability.

Derivation of derivative, examples: • 1. Direct derivation: y = xn, y = sin x. 2. y = ax,

x+∆x x ∆x a a x a 1 x y0 = lim − = a lim − = a ln a ∆x 0 ∆x 0 → ∆x → ∆x 3. Inverse function derivative, example y = arcsin x, y = arctan x.

Product rule, chain rule, quotient rule, high order derivatives. • Examples of function not having derivative y = x etc. • | | Monotone of function, local extreme values, maximum and minimum, • inflection. 1 y = x3 4x 3 − Example: maximum enclosed area problem. •

15 Rolle’s theorem, • f(a) = f(b) = 0, f 0(ξ) = 0. → . To prove it, assume two cases: (1) f(x) 0, and (2) f(x) = 0 at some points. In case (2), there must be maximum≡ and minimum,6 inspect derivative at either maximum or minimum. mean value theorem, • f(b) f(a) − = f 0(ξ) b a − Inspect f(b) f(a) F (x) = f(x) f(a) (x a) − − − − b a − for Rolle’s theorem. Cauchy’s generalization. • f(b) f(a) f 0(ξ) − = g(b) g(a) g (ξ) − 0 Inspect f(b) f(a) F (x) = f(x) f(a) (g(x) g(a)) − − − − g(b) g(a) − for Rolle’s theorem. L’Hospital’s rules is a result of Cauchy’s generalization. • x x Examples: y = x , limx 0 x . • → 2x 1 cos x 2 e 1 lim − , lim(cos x)1/x , lim − . x 0 2 x 0 x 0 → x → → x Derivative of inverse function and implicit function. • Curvature • dθ κ = ds

• x2 y2 + = 1 a2 b2

16 11 Chapter 5

The Newton Leibniz quarrel, who invented integral. • Definite integral as a limit of Riemann sum. • b n f(x)dx = lim f(ξk)∆xk n a →∞ Z Xk=1 Definite integral is a linear operator • b b b (C1f(x) + C2g(x))dx = C1 f(x)dx + C2 g(x)dx. Za Za Za Integral on intervals • a b c b f(x)dx = 0, f(x)dx = f(x)dx + f(x)dx. Za Za Za Zc a b a b a f(x)dx = f(x)dx+ f(x)dx = 0, f(x)dx = f(x)dx − Za Za Zb Za Zb • b m(b a) f(x)dx M(b a), m f(x) M, x [a, b]. − ≤ ≤ − ≤ ≤ ∈ Za

• N N b b a a , f(x)dx f(x) dx i ≤ | i| ≤ | | i=1 i=1 Za Za X X

The mean value theorem • 1 b f(ξ) = f(x)dx b a − Za Newton-Leibniz formula • b f(x)dx = F (b) F (a),F 0(x) = f(x) − Za 17 Part 1: x if F (x) = f(t)dt, then F 0(x) = f(x). Za Part 2: If F (x) is an antiderivative of f(x), then

x a f(x)dx = F (x) + C, since f(x)dx = 0, wehave C = F (a) − Za Za Therefore we have the Newton-Leibniz formula

b f(x)dx = F (b) F (a) − Za To calculate the definite integral, we can first find the antiderivative, • or indefinite integral.

Indefinite integrals of basic functions, power, exponential, trigonomet- • ric, inverse trigonometric.

Substitution method as an inverse of chain rule, example • dx dx tan xdx, , √1 x2 1 + x2 Z Z − Z 1 cos2 x sin x √1 + cos xdx = − dx = dx 1 cos x √1 cos x Z Z r − Z − Integration by part as an inverse to the product rule •

(uv)0 = u0v + v0u, uv0 = (uv)0 u0v, udv = uv vdu − − Z Z example: x cos x, xex, ex sin x. Z Z Z Partial fraction, integrable functions • A 1 A 2x + A ,, , Bx + C (Bx + C)2 x2 + Bx + C x2 + Bx + C Z Z Z Z

18 Example: • 1 1 2x2 , , x2 a2 x2 + a2 (x 1)2(x + 1) Z − Z Z − Example •

• 4x + 1 4x + 1 , x2 + 2x + 2 x2 + 2x 3 Z Z − Properties of definite integral, linear operator, properties of bounds, • b b f(x)dx f(x) dx, a < b ≤ | | Za Za

The mean value theorem • F (b) F (a) 1 b − = f(x)dx = f(ξ) b a b a − − Za

• d x f(t)dt = f(x) dx Za Example: • 3 d x (sin t + t)dt dt 2 Zx Numerical integral, rectangular rule, trapezoidal rule, Simpson’s rule. • Area between two curves y = f(x) and y = g(x) • b A = (f(x) g(x))dx. − Za Arc-length Example: arc-length of curve y = cosh x from x = 0 to • x = ln 2. Volume of revolution about x-axis for y = f(x). • x2 x2 V = πy2dx == π(f(x))2dx Zx1 Zx1

19 1 Volume of revolution about y-axis for y = f(x), x = f − (y) = g(y). • y2 y2 V = πx2dy == π(g(y))2dy Zy1 Zy1 Examples: cylinder, sphere. • Example of Riemann sum • 1 1 1 1 1 1 1 lim + = lim + + n n + 1 n + 2 ··· n + n n n 1 + 1/n 1 + 2/n ··· 1 + n/n →∞   →∞   1 1 = dx 1 + x Z0 Improper integral, example • 4 ∞ 2 x dx x e− dx, dx √4 x Z0 Z0 −

Differential dy = f 0(x)dx. • 12 Chapter 6

Multi-variable functions, domain and range. • Definition of limit. • Limit, iterative limit and continuity, examples • x2 y2 x4 lim − , lim . (x,y) (0,0) x2 + y2 (x,y) (0,0) y2 + x4 → →

Partial derivative, special example • xy (x, y) = (0, 0) f(x, y) = x2+y2 6 0 Otherwise  both of its partial derivatives exist, but function is not continuous.

20 Techniques of partial derivative, example • 2 y f(x, y) = x tan− 1 , f , f , f , at (1, 1) x x y xy

2 1 x High order partial direvatives. Example: verify e− 4νt satisfies • √4πνt ut = νuxx. Chain rule, total derivative, example • z = x2 + y2, x = sin ωt, y = cos ωt

Verify that u = φ(x at) satisfies u + au = 0. • − t x Differentials, for a multi-variable function f(x, y), the differential is • df = fxdx + fydy.

Condition for total differential, since f = f , The following form • xy yx P (x, y)dx + Q(x, y)dy

is a total differential if Py = Qx. Example: Determine whether the following is a total differential • (6xy y2)dx + (2xey x2)dy − − (x sin x + y2)dx + (2xy + y2)dy

Euler Theorem on homogeneous functions • F (λx , λx , , λx ) = λpF (x , x , , x ) 1 2 ··· n 1 2 ··· n Example, F (x, y) = x3 + xy2 + x2y + 3y3. ∂F ∂F ∂F x1 + x1 + + xn = pF ∂x1 ∂x2 ··· ∂xn p Consider ui = λxi and F (ui) = λ F (xi),

∂F ∂F ∂ui ∂F p 1 = = x = pλ − F (x ) ∂λ ∂u ∂λ i ∂u i i i i i X X Let λ = 1, the theorem is proved.

21 Derivative of implicit functions. • F (x, y, z) = 0, z = f(x, y) → dF = Fxdx + Fydy + Fzdz = 0, dz = fxdx + fydy Compare the two, we have

Fx Fy fx = , f= −Fz − Fz Jacobian of transformation • ∂(u, v) u u u = u(x, y), v = v(x, y), = x y ∂(x, y) vx vy

Jacobian of transformation to polar coordinates

x = ρ cos φ ∂(x, y) = ρ y = ρ sin φ ∂(ρ, φ)  Jacobian of transformation to cylindrical coordinates x = ρ cos φ ∂(x, y, z) y = ρ sin φ = ρ  ∂(ρ, φ, z)  z = z Jacobian of transformation to spherical coordinates x = r cos φ sin θ ∂(x, y, z) y = r sin φ sin θ = r2 sin θ  ∂(r, θ, φ)  z = r cos θ Transformation The necessary and sufficient condition for one-to-one • transformation x = F (u, v), y = G(u, v) the Jacobian ∂(x, y) = 0. ∂(u, v) 6 The area differential under the transformation is dxdy = Jdudv. Mean Value Theorem • f(x +h, y +k) f(x , y ) = hf (x +θh, y +θk)+kf (x +θh, y +θk). 0 0 − 0 0 x 0 0 y 0 0

22 13 Chapter 8

Maxima and Minima: necessary condition for z = f(x, y) to have • local extreme value ∂f ∂f = 0, = 0 ∂x ∂y for sufficient condition, introduce

∆ = f f f 2 xx yy − xy

1. maximum if ∆ > 0 and fxx < or fyy < 0.

2. minimum if ∆ > 0 and fxx > or fyy > 0. 3. saddle point if ∆ < 0. 4. inconclusive if ∆ = 0

Lagrange Multipliers To find the extreme value of function F (x, y, z) • under the constraint condition φ(x, y, z) = 0, consider the auxiliary function G(x, y, z) = F (x, y, z) + λφ(x, y, z) and let ∂G ∂G ∂G = 0, = 0, = 0 ∂x ∂y ∂z Example: • f(x, y) = x3 + y3 3x 12y + 20. − − Example: A rectangular box, open at the top, is to have volume of 32 • cubic feet. What must be the dimensions so that the total surface is a minimum? 64 64 xyz = 32,S = xy + 2yz + 2xz = xy + + x y x = y = 4, z = 2. Use direct method and Lagrangian multiplyer.

Example: Find the maximum of function f(x, y) = x + 2y, subject to • the constraint x2 +y2 = 1. Draw picture to help analyzing the solution.

Some special examples • 23 1. Evaluate 1 x 1 I = − ln x Z0 Consider 1 xα 1 1 φ(α) = − dx, φ0(α) = , φ(α) = ln(α+1), φ(1) = ln 2 ln x α + 1 Z0 2. Evaluate ∞ α2k2+βk I(β) = e− dk. Z−∞ dI β 2 2 = I,I(β) = Ceβ /4α dβ 2α2

∞ α2k2 C = I(0) = e− dk Z−∞ 14 Review 2

Sequences, series, limit, Cauchy theorem, least upper bound. • Functions, limit of function, continuity, derivatives. • Product rule, chain rule, derivative of inverse/implicit function. • Maximum and minimum. • Rolle’s theorem and mean value theorem. • L’Hospital’s rule. • Definite and indefinite integrals, Newton Leibniz formula. • Riemann sum and using Riemann sum to find certain limits. • Techniques of integral, substitution method, integration by part. • Partial fraction. • Area and volume of revolution. • Improper integral. • 24 Multivariable functions, limit and continuity. • Partial derivative, differential, total differential. • Transformation and Jacobian. • Lagrangian multiplier. • 15 Chapter 9

Definition of double and triple integrals •

f(x, y)dxdy = lim f(µk, ηk)∆Ak max(∆A ) 0 A k → Z Xk

f(x, y, z)dxdydz = lim f(µk, ηk, ξk)∆Vk max(∆V ) 0 V k → Z Xk Iterated integrals. Find the mass of body, with density function ρ = xy, • shape bounded by y = x2 and y = √2 x2 on the interval 0 x 1. Find center of mass. − ≤ ≤

1 √2 x2 − 7 M = yxdydx = 2 24 Z0 Zx Find the total mass of the plate in first quadrabt between y = x2 and • y = √x if the density if given as ρ = 1 + y.

1 √x 1 √y M = (1 + y)dxdy = (1 + y)dydx = (1 + y)dxdy 2 2 ZA Z0 Zx Z0 Zy Find moment of inertial of a square, a circle, with uniform density. • Evaluate the volume of the region common to tie intersecting cylinder • x2 + y2 = a2 and x2 + z2 = a2.

a √a2 x2 √a2 x2 − − V = 8 dzdydx Z0 Z0 Z0

25 Transformation to polar coordinate • α2(x2+y2) ∞ x2 e− dxdy, e− dx ZZR Z0 Moment of inertial of circular disk ρ(x2 + y2)dxdy ZZR Area of ellipse

dxdy, x = aρ cos φ, y = bρ sin φ ZZE Transformation to cylindrical coordinate, volume of cone, bounded by • paraboloid, cylinder and z = 0 plane. Transformation to spherical coordinate, volume of sphere, volume of • ellipsoid. Moment of inertial of sphere. Find the area of the region in the xy-plane bounded by the lemniscate • ρ2 = a2 cos 2φ. Find the polar moment of inertial in the xy plane bounded by x2 y2 = • 1, x2 y2 = 9, xy = 2, and xy = 4. − − ∂(x, y) 1 = ∂(u, v) √u2 + v2 Prove • x t x F (u)du dt = (x u)F (u)du 0 0 0 − Z Z  Z x Proof: I(x) = LHS and J(x) = RHS, I0(x) = J 0(x) = 0 F (u)du, therefore I(x) J(x) = c. Since I(0) = J(0) = 0, therefore c = 0. This R can also be written− as x x x F (x)dx2 = (x u)F (u)du − Z0 Z0 Z0 The result can be generalized to give x x x x n 1 n 1 F (x)dx = (x u) − F (u)du ··· (n 1)! − Z0 Z0 Z0 − Z0

26