Mathematics Archives - Page 42 of 60 - Shawn Zhong

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 ⃗$

Math 375 – 12/6

Math 375 – Homework 12

Math 375 – Matrix Algebra Review

Math 375 – 12/5

$Example • f:R2→R • f(x,y)={■8(xy/(x^2+y^2 )&(x,y)≠(0,0) @0&x=y=0)┤ • Counterplot • Graph • Partial Derivative ○ ∂f/∂x,∂f/∂y exist at all points in R2 including the origin ○ When (x,y)≠(0,0), we have x^2+y^2≠0 § So xy/(x^2+y^2 ) is differentiable as function of x ○ When (x,y)=(0,0) § ∂f/∂x=lim_(h0)⁡〖(f(x+hy)−f(x,y))/h=lim_(h0)⁡〖(⏞(f(h0) )┴0−⏞(f(0,0) )┴0)/h=lim_(h0)⁡〖0/h=0 § Similarly ∂f/∂y=0 ○ Note § Both ∂f/∂x,∂f/∂y exist, but f is not differentiable everywhere § There we need a different definition for differentiable Differentiable • Definition ○ f:Rn→R is a differentiable at a∈Rn if ○ f(a+v)=⏟(f(a)+T_a (v) )┬(linear approximation)+⏟(‖v‖E(a,v) )┬(error term) ○ Where § T_a=Rn→R is a linear map § lim_(v→0)⁡E(a,v)=0 ○ Alternative formulation § lim_(v→0)⁡〖(f(a+v)−f(a)−T_a (v))/‖v‖ 〗=0 § Where E(a,v)=(f(a+v)−f(a)−T_a (v))/‖v‖ ○ Find T_a (v) § T_a (v)=T_a (v_1 e_1+…+v_n e_n )=v_1 ⏞(T_a (e_1 ) )┴(c_1 )+…+v_n ⏞(T_a (e_n ) )┴(c_n ) § For c_1=T_a (e_1 ), choose v=he_1 § lim_(h0)⁡〖(f(a+h�_1 )−f(a)−T_a (h�_1 ))/h=0 § ⇒lim_(h0)⁡((f(a_1+ha_2,…,a_n )−f(a_1,a_2,…,a_n ))/h(T_a (h�_1 ))/h=0 § ⇒∂f/(∂x_1 ) (a_1,…,a_n )−lim_(h0)⁡〖(c_1 h/h=0 § ⇒∂f/(∂x_1 ) (a_1,…,a_n )−c_1=0 § ⇒∂f/(∂x_1 ) (a_1,…,a_n )=c_1 § Similarly c_k=∂f/(∂x_k ) (a) Total Derivative • Definition ○ If f:Rn→R is differentiable at a, then ○ T_a (v)=v_1 ∂f/(∂x_1 ) (a)+…+v_n ∂f/(∂x_n ) ○ Here, the linear map T_a:Rn→R is called the total derivative of f at a • Alternative notations ○ f^′ (a) ○ df_a ○ Df_a ○ Df(a) • Theorem ○ If f is differentiable at a then f is continuous at a • Proof ○ We want to show lim_(h0)⁡f(a+v)=f(a) § lim_(v→0)⁡〖f(a+v)−f(a)〗 § =lim_(v→0)⁡(f(a+v)−f(a)−T_a (v)+T_a (v)) § =lim_(v→0)⁡(‖v‖ (f(a+v)−f(a)−T_a (v))/‖v‖ +T_a (v)) § =lim_(v→0)⁡(‖v‖)⋅lim_(v→0)⁡((f(a+v)−f(a)−T_a (v))/‖v‖ )+lim_(v→0)⁡(T_a (v)) § =0⋅0+0 § =0 ○ Therefore lim_(h0)⁡f(a+v)=f(a) ○ Note § ‖T_a (v)‖=‖c_1 v_1+…+c_n v_n ‖=‖c⋅v‖≤‖c‖⋅‖v‖ § So ‖T_a (v)‖≤‖c‖⋅‖v‖ § If v→0, then T_a (v)→0 • Properties of differentiable functions ○ Differentiable ⇒ Continuous ○ Differentiable ⇒ Partial derivative exist • Example ○ For f(x,y)={■8(xy/(x^2+y^2 )&(x,y)≠(0,0) @0&x=y=0)┤ ○ Partial derivatives exist at (0,0), but not continuous at (0,0) ○ Therefore f(x,y) is NOT DIFFERENTIABLE • Therorem ○ If ∂f/(∂x_1 ),…,∂f/(∂x_n ) exist and are continuous at a ○ Then f is differentiable at a, and the total derivative is given below ○ T_a (v)=v_1 ∂f/(∂x_1 ) (a)+…+v_n ∂f/(∂x_n ) (a) Continuity • Definition ○ f is contunuous at x=a if ○ lim_(x→0)⁡f(x)=f(a) ○ lim_(‖x−a‖→0)⁡‖f(x)−f(a)‖=0 • Example ○ For f(x,y)=xy/(x^2+y^2 ), (x,y)≠(0,0) ○ ∂f/∂x,∂f/∂y exist and continuous when (x,y)≠(0,0) ○ ⇒f Fréchet differentiable everywhere except at (0,0) Chain Rule • Definition ○ g(t)=f(x_1 (t),…,x_n (t)) ○ dg/dt=df(x_1 (t),…,x_n (t))/dt=∂f/(∂x_1 ) (dx_1)/dt+…+∂f/(∂x_n ) (dx_n)/dt • Proof ○ Let a=x(t)=(x_1 (t),…,x_n (t)) ○ Let v=x(t+h−x(t), then lim_(h0)⁡v=0 ○ lim_(h0)⁡〖(f(x(t+h)−f(x(t)))/h ○ (To be continued)$

Shawn Zhong

Shawn Zhong

Mathematics

Math 375 – 12/7

Math 375 – 12/6

Math 375 – Homework 12

Math 375 – Matrix Algebra Review

Math 375 – 12/5

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