Math 375 – Homework 11 Dec 05, 2017 Shawn Math 375 No comments yet Math 375 – 12/4 Dec 04, 2017 Shawn Math 375 No comments yet Math 375 – 11/30 Nov 30, 2017 Shawn Math 375 No comments yet Math 375 – 11/29 Nov 29, 2017 Shawn Math 375 No comments yet Math 375 – 11/28 Nov 28, 2017 Shawn Math 375 No comments yet 1 … 48 49 50 51 52 … 84
−(−y)[∂/∂y (x^2+y^2 )])/(x^2+y^2 )^2 § =(−(x^2+y^2 )+2y^2)/(x^2+y^2 )^2 § =(y^2−x^2)/(x^2+y^2 )^2 ○ f_yx § (∂^2 f)/∂x∂y=∂/∂x (∂f/∂y)=∂/∂x (x/(x^2+y^2 )) § =([∂/∂x (x)](x^2+y^2 )−(x)[∂/∂x (x^2+y^2 )])/(x^2+y^2 )^2 § =(x^2+y^2−2x^2)/(x^2+y^2 )^2 § =(y^2−x^2)/(x^2+y^2 )^2 ○ Note: f_xy=f_yx Question 2 • Question ○ Consider 〖det:〗〖R(n×n)→R ○ Find all partial derivatives of det ○ Describe [∂det/(∂x_ij )] • Answer ○ ∂det/(∂x_11 ) § =lim_(h0)〖1/h(|■8(x_11+hx_12&⋯&x_1n@x_21&x_22&⋯&x_2n@⋮&⋯&⋱&⋮@x_n1&x_n2&…&x_nn )|−|■8(x_11&x_12&⋯&x_1n@x_21&x_22&⋯&x_2n@⋮&⋯&⋱&⋮@x_n1&x_n2&…&x_nn )|)〗 § =lim_(h0)〖1/hh■8(x_22&⋯&x_2n@⋮&⋱&⋮@x_n2&…&x_nn )|〗 § =|■8(x_22&⋯&x_2n@⋮&⋱&⋮@x_n2&…&x_nn )| § =C_11 • Theorem ○ Let X=(x_ij ) ○ [∂det/(∂x_ij )]=[■8(∂det/(∂x_11 ) (A)&⋯&∂det/(∂x_1n ) (A)@⋮&⋱&⋮@∂det/(∂x_n1 ) (A)&⋯&∂det/(∂x_nn ) (A) )]=[■8(C_11&⋯&C_1n@⋮&⋱&⋮@C_n1&⋯&C_nn )]=cof(A) • Application: use gradients to approximate ○ |■8(sin(π/2+0.1)&ln(1.1)@3&√4)|≈|■8(sin(π/2)&ln(1)@3&√4)|+0.1D_((■8(1&1@1&1)) ) |■8(1&0@3&2)|](https://shawnzhong.com/wp-content/uploads/2017/12/img_5a2b716026f7e.png)
![Partial Derivative • Infinitesimal Interpretation of Derivative ○ • Definition ○ ∂f/(∂x_k ) (x_1,…,x_n )=lim_(h0)〖(⏞(f(x_1,…,x_k+h…,x_n ) )┴(only x_k changes)−f(x_1,…,x_n ))/h ○ = The derivative of f(x_1,…x_n ) with respect to x_k, with all other variables fixed • Other Notations ○ ∂f/(∂x_k ) (x_1,…,x_n )=f_(x_k )=f^′ (x;e_k ) • Example ○ f(x,y,z)=x^2+xy^3 ○ ∂f/∂x=∂/∂x (x^2+xy^3 )=2x+y^3 ○ ∂f/∂y=∂/∂y (x^2+xy^3 )=3xy^2 ○ ∂f/∂z=∂/∂z (x^2+xy^3 )=0 (Because x^2+xy^3 does not depend on z) • Second Derivative ○ f_xx=(∂^2 f)/(∂x^2 )=∂/∂x (∂f/∂x)=∂/∂x (2x+y^3 )=2 ○ f_xy=(∂^2 f)/∂y∂x=∂/∂y (∂f/∂x)=∂/∂y (2x+y^3 )=3y^2 ○ f_yx=(∂^2 f)/∂x∂y=∂/∂x (∂f/∂y)=∂/∂x (3xy^2 )=3y^2 ○ f_yy=(∂^2 f)/(∂y^2 )=∂/∂y (∂f/∂y)=∂/∂y (3xy^2 )=6xy • Clairaut's Theorem ○ If ∂f/∂x,∂f/∂y,(∂^2 f)/∂x∂y exists and (∂^2 f)/∂x∂y is continuous at (a,b)∈R2 ○ Then (∂^2 f)/∂y∂x also exists and (∂^2 f)/∂x∂y=(∂^2 f)/∂y∂x • Example of f_xy≠f_yx ○ f(x,y)={■8(1&x 0@0&x≤0)┤ ○ ∂f/∂x={■8(0&x≠0@Does Not Exist&x=0)┤ ○ see the graph below (horizontal axis: x, vertical axis: f(x,y)) ○ ∂f/∂y=0 for all (x,y) ○ (∂^2 f)/∂x∂y=∂/∂x (∂f/∂y)=∂/∂x (0)=0 ○ (∂^2 f)/∂y∂x=∂/∂y (∂f/∂x)={■8(0&x≠0@Does Not Exist&x=0)┤ ○ Therefore (∂^2 f)/∂x∂y≠(∂^2 f)/∂y∂x Total Derivative & Linear Approximation Formula • Illumination ○ f(x+Δx,y+Δy)−f(x,y) ○ =f(x+Δx,y+Δy)−f(x+Δx,y)+f(x+Δx,y)−f(x,y) ○ =[f(x+Δx,y)−f(x,y)]+[f(x+Δx,y+Δy)−f(x+Δx,y)] ○ =(f(x+Δx,y)−f(x,y))/Δx×Δx+(f(x+Δx,y+Δy)−f(x+Δx,y))/Δy×Δy ○ ≈∂f/∂x×Δx+∂f/∂y×Δy • Theorem ○ If f_x and f_y are continuous, then there exist functions ε_x and ε_y ○ f(x+Δx,y+Δy)=f(x,y)+∂f/∂x (x,y)Δx+∂f/∂y (x,y)Δy+ε_x Δx+ε_y Δy ○ Where ε_x,ε_y→0 as Δx,Δy→0 ○ Note § (f(x+Δx,y+Δy)−f(x+Δx,y))/Δy=∂f/∂y (x,y)+ε_y § (f(x+Δx,y)−f(x,y))/Δx=∂f/∂x (x,y)+ε_x • Linear Approximation ○ f(x_1+Δx_1,…,x_n+Δx_n ) ○ =f(x_1,…,x_n )+f_(x_1 ) (x_1,…,x_n )Δx_1+…+f_(x_n ) (x_1,…,x_n )Δx_n+ε_1 Δx_1+…+ε_n Δx_n ○ Where ε_k→0 as Δx_1,…,Δx_n→0 • Linear Approximation (Vector Notation) ○ x=(x_1,…,x_n )∈Rn ○ Δx=(Δx_1,…,Δx_n )∈Rn ○ ε=(ε_1,…,ε_n )∈Rn ○ f(x+Δx)=f(x)+∇ ⃗f(x)⋅Δx+ε⋅Δx ○ Where § ∇ ⃗f(x)=(∂f/(∂x_1 ) (x),…,∂f/(∂x_n ) (x)) is called the gradient of f § ∇ ⃗f(x)⋅Δx=f_(x_1 ) (x_1,…,x_n )Δx_1+…+f_(x_n ) (x_1,…,x_n )Δx_n § ε⋅Δx=ε_1 Δx_1+…+ε_n Δx_n • Example ○ f(x,y)=x^2+xy^3 ○ Find the linear approximation at (x,y)=(1,2) ○ Calculate f(1,2),f_x (1,2),f_y (1,2) § f(1,2)=1^2+1⋅2^3=9 § f_x (1,2)=[2x+y^3 ]_█(x=1@y=2)=2+2^3=10 § f_y (1,2)=[3xy^2 ]_█(x=1@y=2)=3⋅1⋅2^2=12 § ∇ ⃗f(1,2)=[█(10@12)] ○ f(1+Δx,2+Δy) § =f(1,2)+f_x (1,2)Δx+f_y (1,2)Δy+ε_x Δx+ε_y Δy § =⏟(9+10Δx+12Δy)┬approximation+⏟(ε_x Δx+ε_y Δy)┬error ○ f(1.01,1.99)=f(1+0.01,2−0.01)≈9+10⋅0.01−12⋅0.01=8.89 ○ Tangent plane at (1,2)](https://shawnzhong.com/wp-content/uploads/2017/12/img_5a2b71bf9db27.png)
![Question 1 (from Monday) • Given ○ Let V be a vector space and let T:V→V be a linear map ○ Suppose x,y∈V are eigenvectors of T with eigenvalues λ and μ. • Prove ○ If ax+by (a≠0, b≠0) is an eigenvector of T, then λ=μ • Proof ○ Tx=λx, Ty=μy ○ ⇒T(ax+by)=aλx+bμy ○ Denote the eigenvalue for ax+by to be k ○ ⇒T(ax+by)=k(ax+by) ○ ⇒aλx+bμy=akx+bky ○ ⇒a(λ−k)x−b(μ−k)y=0 ○ If x,y are linearly independet § a(λ−k)=b(μ−k)=0 § Because a≠0, b≠0 § ⇒λ=μ=k ○ If x,y are linearly dependet § x=cy for some c § Tx=cTy=cμy=μ(cy)=μx § ⇒λ=μ Question 2 • Given ○ Let A be a real n×n matrix such that A^2=−I • Note ○ [■8(0&a@−1/a&0)]^2=−I, (a≠0) • Proof: A is invertibe ○ A(−A)=−A^2=−(−I)=I ○ ⇒A^(−1)=−A ○ ⇒A is invertibe • Proof: n is even ○ Suppose n is odd ○ det〖A^2 〗=(detA )^2≥0 ○ det(−I)=−1<0 ○ Which makes a contradiction ○ Therefore n is even • Proof: A has no real eigenvalues ○ Suppose ∃λ∈R, x∈Rn, s.t. Ax=λx ○ A^2 x=−Ix=−x=λ^2 x ○ So λ^2=−1⇒λ=±i ○ Which makes a contradiction ○ Therefore A has no real eigenvalues • Proof: detA=1 (when n=2) ○ A=[■8(a&b@c&d)] ○ A^2=[■8(a^2+bc&ab+bd@ac+cd&d^2+bc)] ○ {█(a^2+bc=d^2+bc=−1@ab+bd=ac+cd=0)┤ ○ ⇒ad−bc=1 • Proof: detA=1 (general case) ○ (detA )^2=det〖A^2 〗=det(−I)=(−1)^n=1 ○ ⇒detA=±1 ○ Ax=λx⇒(Ax) ̅=(λx) ̅⇒Ax ̅=λ ̅x ̅ ○ Therefore the eigenvalues come in complex conjugate pairs ○ detA=(λ_1 (λ_1 ) ̅ )(λ_2 (λ_2 ) ̅ )⋯(λ_k (λ_k ) ̅ )≥0 ○ Therefore detA=1 Question 3 • Given ○ Let T:V→V be a finite-dimensional real linear transformation ○ T has no real eigenvalues • Proof: n is even ○ Suppose n is odd ○ f(λ)=−λ^n+a_(n−1) λ^(n−1)+…+a_1 λ+a_0 ○ As λ→∞, f(λ)⇒−∞ ○ As λ→−∞, f(λ)⇒∞ ○ By the Intermediate Value Theorem ○ f(λ) must have a real root ○ Which makes a contradiction ○ Therefore n is even • Proof: n=dimV](https://shawnzhong.com/wp-content/uploads/2017/12/img_5a2b7202133d6.png)
![Open Balls and Open Sets • Open Interval • Closed Interval • Interior Point ○ E⊆Rn is a subset ○ p∈E is an interior point if there is an r 0 ○ such that B_r (p)⊆E ○ where B_r (p) is the open disc of radius centered at p ○ B_r (p)={x∈Rn│‖x−p‖ r} • Koch s Snowflake • Open Sets ○ E⊆Rn is open if all x∈E are interior points in E • Example • Boundary Point ○ A point p∈Rn is a boundary point for E if for every r 0 ○ B_r (p) contains x,y with x∈E and y∉E Limits and Continuity • Limits ○ lim_(x→a)f(x)=L⟺lim_(‖x−a‖→0)‖f(x)−L‖=0 ○ If x→a, then f(x)→L • Properties ○ If f(x)→L∈Rm,g(x)→M∈Rm, when x→a, then ○ f(x)±g(x)→L±M ○ f(x)⋅g(x)→L⋅M ○ ‖f(x)‖→‖L‖ ○ f(x)/g(x) →L/M ○ (only when n=1, f(x),g(x)∈Rn) • Graph ○ Graph of f={(x,y,z)|z=f(x,y)} • Continuity ○ f:Rn→Rm is continuous at a∈Rn ○ if lim_(x→a)f(x)=f(a) • Continuous Function Example ○ f(x_1,…,x_n )=x_k ○ f:Rn→R • Properties ○ If f,g is continuous ○ Then f±g, fg, f/g (g(a)≠0) are continuous • Example ○ f:R2→R ○ f(x,y)={■8(xy/(x^2+y^2 )&(x,y)≠(0,0)@0&x=y=0)┤ ○ f is continuous at all point except (0,0) ○ Let (x,y)→(0,0) along a straight line with angle θ ○ x=rcosθ, y=rsinθ ○ f(x,y)=xy/(x^2+y^2 )=(r^2 sinθ cosθ)/(r^2 cos^2θ+r^2 sinθ )=cosθ sinθ ○ Note that f(x,y) does not depend on r ○ lim_█((x,y)→(0,0)@along line@with angle θ)f(x,y)=sinθ cosθ ○ When θ=π/2⇒f=0, when θ=π/4⇒f=1/2⋯ ○ Therefore we get the counter plot near origin ○ And the graph near 0 Derivative • Directional Derivative ○ D_hf(x)=∇_hf(x)= f^′ (x;h⃗ )=df_x⋅h ○ =lim_(t→0)〖(f(x+th⃗ )−f(x))/t〗 ○ =[d/dt f(x+th⃗ )]_(t=0) • Example ○ f:Rn→R ○ f(x)=‖x‖^2 ○ f^′ (x;h⃗ ) ○ =[d/dt f(x+th]_(t=0) ○ =[d/dt ‖x+th^2 ]_(t=0) ○ =[d/dt (h2 t^2+(2hx)t+x^2 )]_(t=0) ○ =[2h2 t+2hx]_(t=0) ○ =2x⋅h • Partial Derivative • Total Derivative - • •](https://shawnzhong.com/wp-content/uploads/2017/12/img_5a2b7266cc275.png)
