2017 - Page 9 of 43 - Shawn Zhong

$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)$

Math 375 – Homework 11

Math 375 – 12/4

Math 375 – 11/30

$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)$

Shawn Zhong

Shawn Zhong

Math 375 – Matrix Algebra Review

Math 375 – 12/5

Math 375 – Homework 11

Math 375 – 12/4

Math 375 – 11/30

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