Math 521 - 4/27 May 03, 2018 Shawn Math 521 No comments yet Math 521 - 4/25 May 03, 2018 Shawn Math 521 No comments yet Math 521 - 4/23 May 03, 2018 Shawn Math 521 No comments yet
![Theorem 5.9: Extended Mean Value Theorem • Statement ○ Given § f and g are continuous real-valued functions on [a,b] § f,g are differentiable on (a,b) ○ Then there is a point x∈(a,b) at which § [f(b)−f(a)] g^′ (x)=[g(b)−g(a)] f^′ (x) • Proof ○ Let h(t)≔[f(b)−f(a)]g(t)−[g(b)−g(a)]f(t), (a≤t≤b) ○ Then h is continuous on [a,b] and differentiable on (a,b) ○ We want to show that h′(x)=0 for some x∈(a,b) ○ By definition of h, h(a)=f(b)g(a)−f(a)g(b)=h(b) ○ If h is constant § h′ (x)=0 on all of (a,b), and we are done ○ If h is not constant § ∃t∈(a,b) s.t. h(t) h(a)=h(b) or h(t) h(a)=h(b) § By Theorem 4.16, ∃x∈(a,b) s.t. § h(x) is either a global maximum or a global minimum § By Theorem 5.8, h′ (x)=0 Theorem 5.10: Mean Value Theorem • Statement ○ If f and g are continuous real-valued functions on [a,b] ○ And f,g are differentiable on (a,b) ○ Then ∃x∈(a,b) s.t. f(b)−f(a)=(b−a) f^′ (x) • Proof ○ Let g(x)=x in Theorem 5.9 Theorem 5.11: Derivative and Monotonicity • Suppose f is differentiable on (a,b) • If f^′ (x)≥0,∀x∈(a,b), then f is monotonically increasing • If f^′ (x)=0,∀x∈(a,b), then f is constant • If f^′ (x)≤0,∀x∈(a,b), then f is monotonically decreasing Theorem 5.15: Taylor s Theorem • Statement ○ Suppose § f is a real-valued function on [a,b] § Fix a positive integer n § f^((n−1) ) is continuous on (a,b) § f^((n) ) (t) exists ∀t∈(a,b) ○ Let α,β∈[a,b], where a≠β ○ Define P(t)=∑_(k=0)^(n−1)▒〖(f^((k) ) (α))/k! (t−α)^k 〗 ○ Then ∃x between α and β s.t. ○ f(β)=P(β)+(f^((n) ) (x))/n! (β−α)^n • Note ○ When n=1, this is the Meal Value Theorem • Proof ○ Without loss of generality, suppose α β ○ Define M∈R by § f(β)=P(β)+M(β−α)^n § Then we want to show that § n!M=f^((n) ) (x) for some x∈[α,β] ○ Define difference function g by § g(t)=f(t)−P(t)−M(t−α)^n, where a≤t≤b § Then g(β)=0 by our choice of M § Taking derivative n times on both side, we get § g^((n) ) (t)=f^((n) ) (t)−n!M, where a≤t≤b § Note that P(t) disappears, since its degree is n−1 ○ Then we only need to show g^((n) ) (x)=0 for some x∈[α,β] § P^((k) ) (α)=f^((k) ) (α),for 0≤k≤n−1, by definition of P § Therefore, g(α)=g^′ (α)=…=g^((n−1) ) (α)=0 § Also, g(β)=0, by definition of M § By the Mean Value Theorem, g^′ (x_1 )=0 for some x_1∈[α,β] § g^′ (α)=0, so g^′′ (x_2 )=0 for some x_2∈[α,x_1 ] § After n steps, g^((n) ) (x_n )=0 for some x_n∈[α,x_(n−1) ] § So, x_n∈[α,β]](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aebe3879b0e7.png)
![Definition 5.1: Derivative • Let f be defined (and real-valued) on [a,b] • ∀x∈[a,b], let ϕ(t)=(f(t)−f(x))/(t−x) (a t b, t≠x) • Define f^′ (x)=(lim)_(t→x)ϕ(t), provided that this limit exists • f′ is called the derivative of f • If f′ is defined at point x, f is differentiable at x • If f′ is defined ∀x∈E⊂[a,b], then f is differentiable on E Theorem 5.2: Differentiability Implies Continuity • Statement ○ Let f be defined on [a,b] ○ If f is differentiable at x∈[a,b] then f is continuous at x • Proof ○ lim_(t→x)(f(t)−f(x))=lim_(t→x)((f(t)−f(x))/(t−x) (t−x))=lim_(t→x)(f^′ (x)(t−x))=0 ○ So lim_(t→x)f(t)=f(x) Theorem 5.5: Chain Rule • Statement ○ Given § f is continuous on [a,b], and f^′ (x) exists at x∈[a,b] § g is defined on I⊃im(f), and g is differentiable at f(x) ○ If h(t)=g(f(t)) (a≤t≤b), then ○ h is differentiable at x, and h^′ (x)=g^′ (f(x))⋅f^′ (x) • Proof ○ Let y=f(x) ○ By the definition of derivative § f(t)−f(x)=(t−x)(f^′ (x)+u(t)), where t∈[a,b], lim_(t→x)u(t)=0 § g(s)−g(y)=(s−y)(g^′ (y)+v(s)), where s∈I, lim_(s→y)v(s)=0 ○ Let s=f(t), then § h(t)−h(x)=g(f(t))−g(f(x)) § =(f(t)−f(x))(g^′ (y)+v(s)) § =(t−x)(f^′ (x)+u(t))(g^′ (y)+v(s)) ○ If t≠x, then § (ht)−hx))/(t−x)=(f^′ (x)+u(t))(g^′ (y)+v(s)) ○ As t→x § u(t)→0, and v(s)→s § So s=f(t)→f(x)=y by continuity § Therefore h′ (x)=lim_(t→x)〖(ht)−hx))/(t−x)〗=f^′ (x) g^′ (y)=g^′ (f(x)) f^′ (x) Definition 5.7: Local Maximum and Local Minimum • Let X be a metric space, f:X→R • f has a local maximum at p∈X if ∃δ 0 s.t. ○ f(q)≤f(p),∀q∈X s.t. d(p,q) δ • f has a local minimum at p∈X if ∃δ 0 s.t. ○ f(q)≥f(p),∀q∈X s.t. d(p,q) δ Theorem 5.8: Local Extrema and Derivative • Statement ○ Let f be defined on [a,b] ○ If f has a local maximum (or minimum) at x∈(a,b) ○ Then f^′ (x)=0 if it exists • Proof ○ By Definition 5.7, choose δ, then § a x−δ x x+δ b ○ Suppose x−δ t x § (f(t)−f(x))/(t−x)≥0 § Let t→x (with t x), then f^′ (x)≥0 ○ Suppose x t x+δ § (f(t)−f(x))/(t−x)≤0 § Let t→x (with t x), then f^′ (x)≤0 ○ Therefore f^′ (x)=0](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aea9e3142970.png)
![Definition 2.45: Connected Set • Let X be a metric space, and A,B⊂X • A and B are separated if ○ A∪B ̅=∅ and A ̅∪B=∅ ○ i.e. No point of A lies in the closure of B and vice versa • E⊂X is connected if ○ E is not a union of two nonempty separated sets Theorem 2.47: Connected Subset of R • Statement ○ E⊂R is connected if and only if E has the following property ○ If x,y∈E and x z y, then z∈E • Proof (⟹) ○ By way of contrapositive, suppose ∃x,y∈E, and z∈(x,y) s.t. z∉E ○ Let A_z=E∩(−∞,z) and B_z=E∩(z,+∞) ○ Then A_z and B_z are separated and E=A_z∪B_z ○ Therefore E is not connected • Proof (⟸) ○ By way of contrapositive, suppose E is not connected ○ Then there are nonempty separated sets A and B s.t. E=A∪B ○ Let x∈A,y∈B. Without loss of generality, assume x y ○ Let z≔sup(A∩[x,y]). Then by Theorem 2.28, z∈A ̅ ○ By definition of E, z∉B. So, x≤z y ○ If z∉A § x∈A and z∉A § ⇒x z y § ⇒z∉E ○ If z∈A § Since A and B are separated, z∉B ̅ § So ∃z_1 s.t. z z_1 y and z_1∉B § Then x z_1 y, so z_1∉E Theorem 4.22: Continuous Mapping of Connected Set • Statement ○ Let X,Y be metric spaces ○ Let f:X→Y be a continuous mapping ○ If E⊂X is connected then f(E)⊂Y is also connected • Proof ○ Suppose, by way of contradiction, that f(E) is not connected ○ "i.e." f(E)=A∪B, where A,B⊂Y are nonempty and separated ○ Let G≔E∩f^(−1) (A) and H≔E∩f^(−1) (B) ○ Then E=G∪H, where G,H≠∅ ○ Since A⊂A ̅, we have G⊂f^(−1) (A ̅ ) ○ Since f is continuous and A ̅ is closed, f^(−1) (A ̅ ) is also closed ○ Therefore G ̅⊂f^(−1) (A ̅ ), and hence f(G ̅ )⊂A ̅ ○ Since f(H)=B and A ̅∩B=∅, we have G ̅∩H=∅ ○ Similarly, G∩H ̅=∅ ○ So, G and H are separated ○ This is a contradiction, therefore f(E) is connected Theorem 4.23: Intermediate Value Theorem • Statement ○ Let f:R→R be continuous on [a,b] ○ If f(a) f(b) and if c statifies f(a) c f(b) ○ Then ∃x∈(a,b) s.t. f(x)=c • Proof ○ By Theorem 2.47, [a,b] is connected ○ By Theorem 4.22, f([a,b]) is a connected subset of R ○ By Theorem 2.47, the result follows](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aea9aa22812b.png)