Math 375

Notes on Math 375: Topics in Multi-Variable Calculus and Linear Algebra Your comments and criticism are greatly welcomed.

Course Website

Syllabus

Lecture and Homework Schedule

Textbook

Calculus, Volume II, 2nd Edition, by Tom M. Apostol

Calendar

September

Mon (Dis)	Tue (Lec)	Wed (Dis)	Thu (Lec)	HW
-	-	-	9/7	HW1
9/11	9/12	9/13	9/14	HW2
9/18	9/19	9/20	9/21	HW3
9/25	9/26	9/27	9/28	HW4

All notes from September in PDF format

October

Mon (Dis)	Tue (Lec)	Wed (Dis)	Thu (Lec)	HW
10/2	Exam 1	10/4	10/5	HW5
10/9	10/10	10/11	10/12	HW6
10/16	10/17	10/18	10/19	HW7
10/23	10/24	10/25	10/26	-
10/30	10/31	11/1	11/2	HW8

Midterm 1 Practice

All notes from October in PDF format

November

Mon (Dis)	Tue (Lec)	Wed (Dis)	Thu (Lec)	HW
10/30	10/31	11/1	11/2	HW8
11/6	11/7	11/8	11/9	HW9
11/13	11/14	11/15	Exam 2	HW10
11/20	11/21	11/22	Holiday	-
11/27	11/28	11/29	11/30	HW11

Midterm 2 Practice 1

Midterm 2 Practice 2

All notes from November in PDF format

December

Mon (Dis)	Tue (Lec)	Wed (Dis)	Thu (Lec)	HW
12/4	12/5	12/6	12/7	HW12
12/11	12/12	12/13

Matrix Algebra Review

Lecture Outlines

Chapter 1: Linear Spaces

Week 01 - 9/7: Vector Space

Week 02 - 9/12: Proof Writing

Week 02 - 9/14: Subspace, Span of Vector Spaces, Linear Independence

Week 03 - 9/19: Linear Independence, Basis, Coordinates

Week 03 - 9/21: Theorems about Linear Dependence

Week 04 - 9/26: Inner Product, Length, Angle

Week 04 - 9/28: Distance, Triangle Inequality, Orthogonal, Gramm-Schmidt Process

Week 05 - 10/3: Midterm 1

Week 05 - 10/5: Best Approximation, Fourier Series

Chapter 2: Linear Transformations and Matrices

Week 06 - 10/10: Linear Transformations, Solving Systems of Equations

Week 06 - 10/12: Injective, Null Space, Range, Rank-Nullity Theorem

Week 07 - 10/17: Algebraic Operations on Linear Transformations, Injective, Inverse

Week 07 - 10/19: Matrix Representation of Linear Transformations, Matrix Multiplication

Week 08 - 10/24: Isomorphism Between Linear Transformations and Matrices, Solving Linear Equations using Matrix

Week 08 - 10/26: Solving Linear Equations

Chapter 3: Determinants

Week 09 - 10/31: Determinants

Week 09 - 11/2: Uniqueness Theorem, Properties of Determinants

Week 10 - 11/7: Determinant and Area, Inverse of Matrix, Minors and Cofactors

Week 10 - 11/9: Cofactor Expansion, Cramer's Rule, Linear Independence and Determinant

Chapter 4: Eigenvalues and Eigenvectors

Week 11 - 11/14: Eigenvalues, Eigenvectors

Week 11 - 11/16: Midterm 2

Week 12 - 11/21: Characteristic Polynomial, Trace, Diagonalization

Week 12 - 11/23: Thanksgiving Break

Chapter 8: Differential Calculus of Scalar and Vector Fields

Week 13 - 11/28: Open Balls, Limits, Continuity, Directional Derivative

Week 13 - 11/30: Partial Derivative, Total Derivative, Linear Approximation Formula

Week 14 - 12/5: Differentiable, Total Derivative, Continuity, Multivariable Chain Rule

Week 14 - 12/7: Differentiable, Chain Rule, Gradient, Level Sets

Week 15 - 12/12: Multivariable Chain Rule, Jacobian Matrix

$Question 1 • Find a basis in which the matrix (■8(3&0@3&−2)) becomes diagonalized • Let A=(■8(3&0@3&−2)) • det⁡(A−λI)=|■8(3−λ&0@3&−2−λ)|=λ^2−λ−6=0 • ⇒λ_1=3, λ_2=−2 • ⇒Λ=(■8(3&0@0&−2)) • When λ_1=3 ○ A−λI=(■8(0&0@3&−5)) ○ ⇒v_1=k(5,3), k∈R • When λ_2=−2 ○ A−λI=(■8(5&0@3&0)) ○ ⇒v_2=k(0,1), k∈R • The basis is (5,3), (0,1) Exercise 8.17 Question 8 • Find a Cartesian equation for the tangent plane • to the surface xyz=a^3 at a general point (x_0,y_0,z_0 ). ○ ∇f=(█(yz@xz@xy)) ○ H={(x,y,z)∈R3│∇f(x_0,y_0,z_0 )⋅(█(x−x_0@y−y_0@z−z_0 ))=0} ○ ={(x,y,z)∈R3│y_0 z_0 (x−x_0 )+x_0 z_0 (y−y_0 )+x_0 y_0 (z−z_0 )=0} ○ ={(x,y,z)∈R3│xy_0 z_0+x_0 yz_0+x_0 y_0 z=3x_0 y_0 z_0=3a^3 } Question 2 • Let f:R2→R smooth • Let g(x,y)=f(u,v)=f(sin⁡〖(x)y,x^y 〗 ) • Find g_x,g_y in terms of f_u,f_v ○ Let h(x,y)=[█(u(x,y)@v(x,y) )]=[█(sin⁡(x)y@x^y )], Then T_g=T_f∘T_ℎ ○ T_f=[■8(∂f/∂u&∂f/∂v)]=[■8(f_u&f_v )] ○ T_ℎ=[■8(∂u/∂x&∂u/∂y@∂v/∂x&∂v/∂y)]=[■8(cos⁡(x)y&sin⁡(x)@y⋅x^(y−1)&x^y (y ln⁡(x)+1) )] ○ T_g=T_f∘T_ℎ=[■8(f_u&f_v )][■8(cos⁡(x)y&sin⁡(x)@y⋅x^(y−1)&x^y (y ln⁡(x)+1) )]=[█(f_u cos⁡(x)y+f_v y⋅x^(y−1)@f_u sin⁡(x)+f_v x^y (y ln⁡(x)+1) )]^T ○ ⇒{█(g_x=f_u cos⁡(x)y+f_v y⋅x^(y−1)@g_y=f_u sin⁡(x)+f_v x^y (y ln⁡(x)+1) )┤$

Math 375 – 12/12

$Derivative of Vector Fields • Map ○ f:Rn→Rm ○ f(x_1,…,x_n )=[█(f_1 (x_1,…,x_n )@⋮@f_m (x_1,…,x_n ) )] • Example 1 ○ n=1, m=2 ○ f:R→Rn ○ f:t↦[█(f_1 (t)@f_2 (t) )]=[█(x(t)@y(t) )] ○ Also called parametric curve • Example 2 ○ f:Rn→R, where n≥2 ○ f is a function of n variables f(x_1,…,x_n ) • Example 3 ○ Rn→Rm, where n≥2, m≥2 ○ f:R2→R2 ○ f:(u,v)⟼(x,y) ○ Polar Coordinates § f(r,θ)=[█(x(r,θ)@y(r,θ) )]=[█(r cos⁡θ@r sin⁡θ )] • Example 4 ○ f(u,v)=[■8(2&3@1&−1@1&1)][█(u@v)]=[█(2u+3v@u−v@u+v)] • Definition ○ A map f:Rn→Rm is differentiable at a∈Rn if ○ f(a+y)=f(a)+T_a (y)+E(a,y)‖y‖ ○ Where T_a:Rn→Rm is a linear transformation, and lim_(y→0)⁡E(a,y)=0 • Theorem ○ f differentiable at a⇒f continuous at a • Theorem ○ If f(x)=[█(f_1 (x_1,…,x_n )@⋮@f_m (x_1,…,x_n ) )], and f_1,…,f_m have continuous partial derivatives ○ Then f is differentiable and T_a (y)=[■8((∂f_1)/(∂x_1 )&⋯&(∂f_1)/(∂x_n )@⋮&⋱&⋮@(∂f_m)/(∂x_1 )&⋯&(∂f_m)/(∂x_n ))][█(y_1@⋮@y_n )] • Example ○ f(r,θ)=[█(r cos⁡θ@r sin⁡θ )] ○ The linear transformation T_a has the matrix § mat(T_a )=[■8((∂f_1)/∂r&(∂f_1)/∂θ@(∂f_2)/∂r&(∂f_2)/∂θ)]=[■8(∂x/∂r&∂x/∂θ@∂y/∂r&∂y/∂θ)]=[■8(cos⁡θ&−r sin⁡θ@sin⁡θ&r cos⁡θ )] ○ At a=(√2,π/4) § mat(T_a )=[■8(√2∕2&−1@√2∕2&1)] § T_a [█(1@0)]=[█(√2∕2@√2∕2)] § T_a [█(0@1)]=[█(−1@1)] ○ In general § mat(T_((r,θ) ) )=[■8(cos⁡θ&−r sin⁡θ@sin⁡θ&r cos⁡θ )] § T_a [█(ε@0)]=ε[█(cos⁡θ@sin⁡θ )] § T_a [█(0@ε)]=εr[█(−sin⁡θ@cos⁡θ )] • Common notations ○ T_a (v)=df_a⋅v=Df(a)⋅v=f^′ (a)⋅v • Jacobian Matrix ○ [■8((∂f_1)/(∂x_1 )&⋯&(∂f_1)/(∂x_n )@⋮&⋱&⋮@(∂f_m)/(∂x_1 )&⋯&(∂f_m)/(∂x_n ))] is called Jacobian Matrix • Chain Rule ○ Given f:Rn→Rm, g:Rm→Rk ○ Consider h(x)=g(f(x))=g∘f(x) ○ {█(f is differentiable at x@g is differentiable at f(x) )┤⇒h=g∘f is differentiable at x ○ And d(g∘f)_x=dg_f(x) ⋅df_x ○ Another notation: (g∘f)^′ (x)=g^′ (f(x))⋅f^′ (x) ○ In components § x=[█(x_1@⋮@x_n )]∈Rn § u=[█(u_1@⋮@u_m )]=[█(f_1 (x)@⋮@f_m (x) )]=[█(f_1 (x_1,…,x_n )@⋮@f_m (x_1,…,x_n ) )]∈Rm § v=[█(v_1@⋮@v_k )]=[█(g_1 (u)@⋮@g_k (u) )]=[█(g_1 (u_1,…,u_m )@⋮@g_k (u_1,…,u_m ) )]=[█(g_1 (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))@⋮@g_k (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n )) )]∈Rn § mat[(g∘f)^′ ] § =[■8((∂v_1)/x_1 &⋯&(∂v_1)/x_n @⋮&⋱&⋮@(∂v_k)/x_1 &⋯&(∂v_k)/x_n )] § =[■8(∂/(∂x_1 ) g_1 (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))&⋯&∂/(∂x_n ) g_1 (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))@⋮&⋱&⋮@∂/(∂x_1 ) g_k (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))&⋯&∂/(∂x_n ) g_k (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n )) )] § =[■8((∂g_1)/(∂u_1 )⋅(∂f_1)/(∂x_1 )+…+(∂g_1)/(∂u_m )⋅(∂f_m)/(∂x_1 )&⋯&(∂g_1)/(∂u_1 )⋅(∂f_1)/(∂x_n )+…+(∂g_1)/(∂u_m )⋅(∂f_m)/(∂x_n )@⋮&⋱&⋮@(∂g_k)/(∂u_1 )⋅(∂f_1)/(∂x_1 )+…+(∂g_k)/(∂u_m )⋅(∂f_m)/(∂x_1 )&⋯&(∂g_k)/(∂u_1 )⋅(∂f_1)/(∂x_n )+…+(∂g_k)/(∂u_m )⋅(∂f_m)/(∂x_n ))] § =[■8((∂g_1)/(∂u_1 )&⋯&(∂g_1)/(∂u_m )@⋮&⋱&⋮@(∂g_k)/(∂u_1 )&⋯&(∂g_k)/(∂u_m ))][■8((∂f_1)/(∂x_1 )&⋯&(∂f_1)/(∂x_n )@⋮&⋱&⋮@(∂f_m)/(∂x_1 )&⋯&(∂f_m)/(∂x_n ))] § =mat(g^′ (f(x)))mat(f^′ (x))$

Math 375 – 12/11

$Quiz Question • Given ○ A:R3→R3 ○ det⁡〖(A−2I)=0〗 ○ tr (A)=−2 ○ det⁡(A)=6 • Question: Eigenvalue of A ○ det⁡〖(A−2I)=0〗⇒2 is an eigenvalue ○ tr (A)=−2⇒∑_(i=1)^n▒λ_i =λ_1+λ_2+λ_3=−2 ○ det⁡(A)=6⇒∏_(i=1)^n▒λ_i =λ_1 λ_2 λ_3=6 ○ {█(█(λ_1=2@λ_1+λ_2+λ_3=−2)@λ_1 λ_2 λ_3=6)┤⇒{█(λ_2=−3@λ_3=−1)┤ ○ Therefore eigenvalues are 2, −1, −3 • Question: Characteristic Polynomial ○ f(λ)=(λ−λ_1 )(λ−λ_2 )(λ−λ_3 )=(λ−2)(λ+1)(λ+3) • Note: f(λ)=det⁡(λI−A) Question 1 • Given ○ f:R2\{0}→R ○ f(x,y)=xy/(x^2+y^2 ), ∀(x,y)∈R2 • Question: Find the direction of steepest decedent at (1,3) ○ ∇f(x,y)=[█(∂f/∂x@∂f/∂y)]=[█(y(y^2−x^2 )/(x^2+y^2 )^2 @x(x^2−y^2 )/(x^2+y^2 )^2 )] ○ ∇f(1,3)=[█(3(3^2−1^2 )/(1^2+3^2 )^2 @1(1^2−3^2 )/(1^2+3^2 )^2 )]=[█(6/25@−2/25)] • Question: Find the line best approximate the level set at (1,3) ○ ∇f(1,3)⋅n ⃗=0⇒n ⃗=[█(1@3)] ○ x+3y+c=0 ○ 1+3⋅3+c=0 ○ ⇒c=−10 ○ l: x+3y−10=0 ○ Alternative: ∇f(1,3)⋅[█(x−1@y−3)]=0 • Question: Estimate f(0.8,3.05) ○ f(0.8,3.05) ○ =f(1−0.2,3+0.05) ○ ≈f(1,3)+∇f(1,3)[█(−0.2@0.05)] ○ =3/(1^2+3^3 )+[█(6/25@−2/25)][█(−0.2@0.05)] ○ =0.248 Question 2 • Find a basis in which the matrix (■8(3&0@3&−2)) becomes diagonalized • Let A=(■8(3&0@3&−2)) • det⁡〖(A−λI)=λ^2−λ−6=0〗 • ⇒λ_1=3, λ_2=−2 • When λ_1=3 ○ A−λI=(■8(0&0@3&−5)) ○ ⇒v_1=(5,3) • When λ_2=−2 ○ A−λI=(■8(5&0@3&0)) ○ ⇒v_2=(0,1) • The basis is (5,3), (0,1)$

Math 375 – 12/7

$Differentiable • Theorems ○ f differentiable ⇒ f continuous ○ f differentiable ⇒f has partial derivative ○ T_a (v)=∂f/(∂x_1 ) (a) v_1+…+∂f/(∂x_n ) (a) v_n ○ f has continuous partial derivative ⇒f is differentiable • Example ○ There is a function f which has partial derivative everywhere ○ but is not continuous at (0,0) Chain Rule • Formula ○ d/dt f(x_1 (t),…,x_n (t))=∂f/(∂x_1 ) (dx_1)/dt+…+∂f/(∂x_n ) (dx_n)/dt • Proof ○ Let g(t)=f(x_1 (t),…,x_n (t)) § g(t+Δt)−g(t)=f(x_1 (t+Δt),…,x_n (t+Δt))−f(x_1 (t),…,x_n (t)) ○ Define Δx_k=x_k (t+Δt)−x_k (t), (k=1,…,n) § g(t+Δt)−g(t)=∂f/(∂x_1 ) (x)Δx_1+…+∂f/(∂x_n ) (x)Δx_n+Error ○ Derivative of g(t) § g^′ (t)=lim_(Δt→0)⁡〖(g(t+Δt)−g(t))/Δt〗 § =lim_(Δt→0)⁡(∂f/(∂x_1 ) (x) (Δx_1)/Δt+…+∂f/(∂x_n ) (x) (Δx_n)/Δt+Error/Δt) § =∂f/(∂x_1 ) (x) lim_(Δt→0)⁡((Δx_1)/Δt)+…+∂f/(∂x_1 ) (x) lim_(Δt→0)⁡((Δx_n)/Δt)+lim_(Δt→0)⁡(Error/Δt) ○ Note that § lim_(Δt→0)⁡((Δx_k)/Δt)=lim_(Δt→0)⁡((x_k (t+Δt)−x_k (t))/Δt)=(dx_k)/dt § lim_(Δt→0)⁡(Error/Δt)=0 ○ Therefore § g^′ (t)=∂f/(∂x_1 ) (dx_1)/dt+…+∂f/(∂x_n ) (dx_n)/dt • Gradient ○ g^′ (t)=f_(x_1 ) (x(t)) x_1^′ (t)+…+f_(x_n ) (x(t)) x_n^′ (t) ○ g^′ (t)=(█(f_(x_1 )@⋮@f_(x_n ) ))(█(x_1^′ (t)@⋮@x_n^′ (t) ))=∇ ⃗f(x ⃗ )⋅x ⃗^′ (t)=‖∇ ⃗f(x ⃗ )‖⋅‖x ⃗^′ (t)‖ cos⁡θ ○ ∇ ⃗f is called gradient ○ x ⃗^′ (t) is called velocity vector • Interpretation ○ (x_1 (t),…,x_n (t)) cordinates of a point moving in Rn (t=time) ○ Velocity vector: v(t)=x ⃗^′ (t)=lim_(Δt→0)⁡〖(x ⃗(t+Δt)−x ⃗(t))/Δt〗=(x_1^′ (t),…,x_n^′ (t)) ○ Example: Linear motion with constant velocity § x ⃗(t)=(x_1 (t),…,x_n (t))=p ⃗+t ⃗=(p_1+tv_1,…,p_n+tv_n ) § v(t)=x ⃗′(t)=(v_1,..,v_n ) ○ Example: Circular motion § x ⃗(t)=(█(cos⁡t@sin⁡t )) § v(t)=x ⃗′(t)=(█(−sin⁡t@cos⁡t )) ○ Example § f(x,y)=x^2+y^2 § ∇ ⃗f(x,y)=(█(f_x@f_y ))=(█(2x@2y)) • Theorem ○ f(x ⃗(t)) does not depend on t, if and only if ○ ∇ ⃗f(x ⃗(t))⊥x ⃗′(t) for all t • Proof ○ f(x ⃗(t)) constant for atb ○ ⟺d/dt (f(x ⃗(t)))=0 for atb ○ ⟺∇ ⃗f(x ⃗(t))⋅x ⃗(t)=0 for atb ○ ⟺∇ ⃗f(x ⃗(t))⊥x ⃗′(t) ○ Note: By convention 0 ⃗⊥any vector • Application: Gradient Decent ○ To decrease/increase f(x ⃗(t)), how should we choose x^′ (t) ○ Maximize/Minimize Condition § Maximal if cos⁡θ=+1, i.e. θ=0 § Minimal if cos⁡θ=−1, i.e. θ=π ○ Steepest Ascent/Descent ○ Pseudocode Level Sets • Definition ○ D⊆Rn open ○ If f:D→R, then ○ The level set of f at level c is ○ {x ⃗∈Rn│f(x ⃗ )=c}=f^(−1) (c) • Example ○ f(x,y,z)=x^2+y^2+z^2 ○ f^(−1) (1)={(x,y,z)∈R3 |x^2+y^2+z^2=1}=unit sphere ○ f^(−1) (0)={(x,y,z)∈R3 |x^2+y^2+z^2=0}=origin • Tangent ○ Let S⊆Rn, p∈S ○ Then v ⃗∈Rn is tangent to S ○ If there is a path x ⃗(t) with x ⃗(t_0 )=p and x^′ (t)=v ⃗, and x ⃗(t)∈S for all t • Theorem ○ If S=f^(−1) (c)={x ⃗∈Rn│f(x ⃗ )=c} ○ And v ⃗ is tangent to S at p∈S ○ Then ∇ ⃗f(p)⊥v ⃗ • Proof ○ Given v ⃗ is tangent to S ○ So there is a path x(t)∈S for all t with {█(x ⃗(t_0 )=p@x ⃗^′ (t)=v ⃗ )┤ ○ Since f(x ⃗(t))=c for all t ○ We have 0=d/dt f(x ⃗(t))=∇ ⃗f(x ⃗(t))⋅x ⃗^′ (t) ○ So at t=t_0, 0=∇ ⃗f(p)⋅v ⃗$

Shawn Zhong

Shawn Zhong

Math 375

Math 375

Course Website

Textbook

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October

November

December

Lecture Outlines

Chapter 1: Linear Spaces

Chapter 2: Linear Transformations and Matrices

Chapter 3: Determinants

Chapter 4: Eigenvalues and Eigenvectors

Chapter 8: Differential Calculus of Scalar and Vector Fields

Math 375 – 12/13

Math 375 – 12/12

Math 375 – 12/11

Math 375 – 12/7

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