Math Analysis Homework - Week 5
Class 1
T1
∑ i = 0 n a i n + 1 − i = 0 ⟹ ∃ x 0 ∈ ( 0 , 1 ) , ∑ i = 0 n a i x 0 n − i = 0 \begin{gathered}
\sum _{i = 0} ^{n} \dfrac{a_i}{n+1-i}=0 \\
\implies \exists x_0\in(0,1),\sum _{i = 0} ^{n} a_ix_0^{n-i}=0
\end{gathered} i = 0 ∑ n n + 1 − i a i = 0 ⟹ ∃ x 0 ∈ ( 0 , 1 ) , i = 0 ∑ n a i x 0 n − i = 0
let f ( x ) = ∑ i = 0 n a i x n − i let F ( x ) = ∑ i = 0 n a i x n − i + 1 n − i + 1 F ( 0 ) = 0 , F ( 1 ) = ∑ i = 0 n a i n + 1 − i = 0 ⟹ Rolle’s Theorem ∃ ξ ∈ ( 0 , 1 ) , f ( ξ ) = F ′ ( ξ ) = 0 ⟹ x 0 = ξ \begin{gathered}
\text{let } f(x)=\sum _{i = 0} ^{n} a_ix^{n-i} \\
\text{let } F(x)=\sum _{i = 0} ^{n} \dfrac{a_ix^{n-i+1}}{n-i+1} \\
F(0)=0,F(1)=\sum _{i = 0} ^{n} \dfrac{a_i}{n+1-i} =0 \\
\stackrel{\text{ Rolle's Theorem }}{\Longrightarrow} \exists \xi \in (0,1),f(\xi)=F'(\xi)=0 \\
\implies x_0=\xi
\end{gathered} let f ( x ) = i = 0 ∑ n a i x n − i let F ( x ) = i = 0 ∑ n n − i + 1 a i x n − i + 1 F ( 0 ) = 0 , F ( 1 ) = i = 0 ∑ n n + 1 − i a i = 0 ⟹ Rolle’s Theorem ∃ ξ ∈ ( 0 , 1 ) , f ( ξ ) = F ′ ( ξ ) = 0 ⟹ x 0 = ξ
T2
{ f ( x ) ∈ C [ a , b ] ∀ x ∈ ( a , b ) , ∃ f ′ ( x ) f ( a ) = f ( b ) = 0 ⟹ { a > 0 ⟹ ∃ ξ ∈ ( a , b ) , f ′ ( ξ ) = f ( ξ ) ξ ∀ λ , ∃ ξ , f ′ ( ξ ) = λ f ( ξ ) \begin{gathered}
\begin{cases}
f(x)\in C[a,b] \\
\forall x\in (a,b),\exists f'(x) \\
f(a)=f(b)=0
\end{cases} \\
\implies \begin{cases}
a>0 \implies \exists \xi\in(a,b),f'(\xi)=\dfrac{f(\xi)}{\xi} \\
\forall \lambda,\exists \xi,f'(\xi)=\lambda f(\xi)
\end{cases}
\end{gathered} ⎩ ⎨ ⎧ f ( x ) ∈ C [ a , b ] ∀ x ∈ ( a , b ) , ∃ f ′ ( x ) f ( a ) = f ( b ) = 0 ⟹ ⎩ ⎨ ⎧ a > 0 ⟹ ∃ ξ ∈ ( a , b ) , f ′ ( ξ ) = ξ f ( ξ ) ∀ λ , ∃ ξ , f ′ ( ξ ) = λ f ( ξ )
(1)
let F ( x ) = f ( x ) x F ( a ) = F ( b ) = 0 ⟹ Rolle’s Theorem ∃ ξ ∈ ( a , b ) , F ′ ( ξ ) = 0 ⟹ F ′ ( ξ ) = f ′ ( ξ ) ξ − f ( ξ ) ξ 2 = 0 ⟹ f ′ ( ξ ) = f ( ξ ) ξ \begin{gathered}
\text{let } F(x)=\dfrac{f(x)}{x} \\
F(a)=F(b)=0 \\
\stackrel{\text{ Rolle's Theorem }}{\Longrightarrow}\exists \xi\in (a,b),F'(\xi)=0 \\
\implies F'(\xi)=\dfrac{f'(\xi)\xi-f(\xi)}{\xi^2} =0 \\
\implies f'(\xi)=\dfrac{f(\xi)}{\xi}
\end{gathered} let F ( x ) = x f ( x ) F ( a ) = F ( b ) = 0 ⟹ Rolle’s Theorem ∃ ξ ∈ ( a , b ) , F ′ ( ξ ) = 0 ⟹ F ′ ( ξ ) = ξ 2 f ′ ( ξ ) ξ − f ( ξ ) = 0 ⟹ f ′ ( ξ ) = ξ f ( ξ ) (2)
let F ( x ) = e − λ x f ( x ) F ( a ) = F ( b ) = 0 ⟹ ∃ ξ ∈ ( a , b ) , F ′ ( ξ ) = 0 ⟹ x = ξ , F ′ ( x ) = e − λ x ( f ′ ( x ) − λ f ( x ) ) = 0 ⟹ f ′ ( x ) = λ f ( x ) \begin{gathered}
\text{let } F(x)=e^{-\lambda x}f(x) \\
F(a)=F(b)=0
\implies
\exists \xi \in (a,b),F'(\xi)=0 \\
\implies x=\xi,
F'(x)=e^{-\lambda x}(f'(x)-\lambda f(x))=0 \\
\implies f'(x)=\lambda f(x)
\end{gathered} let F ( x ) = e − λ x f ( x ) F ( a ) = F ( b ) = 0 ⟹ ∃ ξ ∈ ( a , b ) , F ′ ( ξ ) = 0 ⟹ x = ξ , F ′ ( x ) = e − λ x ( f ′ ( x ) − λ f ( x )) = 0 ⟹ f ′ ( x ) = λ f ( x )
T3
{ ∀ x ∈ [ 0 , 1 ] , ∃ f ′ ′ ′ ( x ) f ( 0 ) = f ( 1 ) = 0 F ( x ) = x 2 f ( x ) ⟹ ∃ ξ ∈ ( 0 , 1 ) , F ′ ′ ′ ( ξ ) = 0 \begin{gathered}
\begin{cases}
\forall x\in [0,1],\exists f'''(x) \\
f(0)=f(1)=0 \\
F(x)=x^2f(x)
\end{cases}\implies \exists \xi\in (0,1),F'''(\xi)=0
\end{gathered} ⎩ ⎨ ⎧ ∀ x ∈ [ 0 , 1 ] , ∃ f ′′′ ( x ) f ( 0 ) = f ( 1 ) = 0 F ( x ) = x 2 f ( x ) ⟹ ∃ ξ ∈ ( 0 , 1 ) , F ′′′ ( ξ ) = 0
F ′ ( x ) = 2 x f ( x ) + x 2 f ′ ( x ) F ′ ′ ( x ) = 2 f ( x ) + 4 x f ′ ( x ) + x 2 f ′ ′ ( x ) F ( 0 ) = 0 , F ( 1 ) = 0 ⟹ Rolle’s Theorem ∃ x 1 ∈ ( 0 , 1 ) , F ′ ( x 1 ) = 0 F ′ ( 0 ) = 0 , F ′ ( x 1 ) = 0 ⟹ Rolle’s Theorem ∃ x 2 ∈ ( 0 , x 1 ) , F ′ ′ ( x 2 ) = 0 F ′ ′ ( 0 ) = 0 , F ′ ′ ( x 2 ) = 0 ⟹ Rolle’s Theorem ∃ ξ ∈ ( 0 , x 2 ) ⊂ ( 0 , 1 ) , F ′ ′ ′ ( ξ ) = 0 \begin{gathered}
F'(x)=2xf(x)+x^2f'(x) \\
F''(x)=2f(x)+4xf'(x)+x^2f''(x) \\
F(0)=0,F(1)=0 \\
\stackrel{\text{ Rolle's Theorem }}{\Longrightarrow}\exists x_1\in (0,1),F'(x_1)=0 \\
F'(0)=0,F'(x_1)=0 \\
\stackrel{\text{ Rolle's Theorem }}{\Longrightarrow}\exists x_2\in (0,x_1),F''(x_2)=0 \\
F''(0)=0,F''(x_2)=0 \\
\stackrel{\text{ Rolle's Theorem }}{\Longrightarrow}\exists \xi \in (0,x_2) \subset (0,1),F'''(\xi)=0
\end{gathered} F ′ ( x ) = 2 x f ( x ) + x 2 f ′ ( x ) F ′′ ( x ) = 2 f ( x ) + 4 x f ′ ( x ) + x 2 f ′′ ( x ) F ( 0 ) = 0 , F ( 1 ) = 0 ⟹ Rolle’s Theorem ∃ x 1 ∈ ( 0 , 1 ) , F ′ ( x 1 ) = 0 F ′ ( 0 ) = 0 , F ′ ( x 1 ) = 0 ⟹ Rolle’s Theorem ∃ x 2 ∈ ( 0 , x 1 ) , F ′′ ( x 2 ) = 0 F ′′ ( 0 ) = 0 , F ′′ ( x 2 ) = 0 ⟹ Rolle’s Theorem ∃ ξ ∈ ( 0 , x 2 ) ⊂ ( 0 , 1 ) , F ′′′ ( ξ ) = 0
T4
{ ∀ x ∈ ( 0 , a ) , ∃ f ′ ( x ) f ( 0 + ) = + ∞ ⟹ f ′ ( x ) has no lower bound on the right-hand neibourhood of 0 \begin{gathered}
\begin{cases}
\forall x\in (0,a),\exists f'(x) \\
f(0^+)=+\infty
\end{cases} \\
\implies f'(x) \text{ has no lower bound on the right-hand neibourhood of } 0
\end{gathered} { ∀ x ∈ ( 0 , a ) , ∃ f ′ ( x ) f ( 0 + ) = + ∞ ⟹ f ′ ( x ) has no lower bound on the right-hand neibourhood of 0
lim x → 0 + f ( x ) = + ∞ ∃ { a n } , a n ∈ ( 0 , 1 ) , a n < a n − 1 , lim n → ∞ a n = 0 , lim n → ∞ f ( a n ) = + ∞ ∀ M > 0 , i , ∃ k > i s . t . f ( a k ) > f ( a i ) + M ⟹ ∃ ξ ∈ ( a k , a i ) , f ′ ( ξ ) = f ( a k ) − f ( a i ) a k − a i < − M Q.E.D \begin{gathered}
\lim_{x \to 0^+} f(x)=+\infty \\
\exists \{ a_n \},a_n\in (0,1),a_n<a_{n-1},\lim_{n \to \infty} a_n=0,\lim_{n \to \infty} f(a_n)=+\infty \\
\forall M>0,i,\exists k>i \ s.t.\
f(a_k)>f(a_i)+M \\
\implies \exists \xi \in (a_k,a_i),f'(\xi)=\dfrac{f(a_k)-f(a_i)}{a_k-a_i} <-M \\
\\
\text{Q.E.D}
\end{gathered} x → 0 + lim f ( x ) = + ∞ ∃ { a n } , a n ∈ ( 0 , 1 ) , a n < a n − 1 , n → ∞ lim a n = 0 , n → ∞ lim f ( a n ) = + ∞ ∀ M > 0 , i , ∃ k > i s . t . f ( a k ) > f ( a i ) + M ⟹ ∃ ξ ∈ ( a k , a i ) , f ′ ( ξ ) = a k − a i f ( a k ) − f ( a i ) < − M Q.E.D
T5
{ f ∈ C [ a , b ] ∀ x ∈ ( a , b ) , ∃ f ′ ( x ) f is not constant or linear function ⟹ ∃ ξ ∈ ( a , b ) , ∣ f ′ ( ξ ) ∣ > ∣ f ( b ) − f ( a ) b − a ∣ \begin{gathered}
\begin{cases}
f \in C[a,b] \\
\forall x\in(a,b),\exists f'(x) \\
f \text{ is not constant or linear function} \\
\end{cases} \\
\implies \exists \xi \in (a,b),\vert f'(\xi) \vert >{\left \vert \dfrac{f(b)-f(a)}{b-a} \right \vert}
\end{gathered} ⎩ ⎨ ⎧ f ∈ C [ a , b ] ∀ x ∈ ( a , b ) , ∃ f ′ ( x ) f is not constant or linear function ⟹ ∃ ξ ∈ ( a , b ) , ∣ f ′ ( ξ ) ∣ > b − a f ( b ) − f ( a )
不妨设f ( b ) > f ( a ) f(b)>f(a) f ( b ) > f ( a )
let k = f ( b ) − f ( a ) b − a = ∃ x 0 , f ( x 0 ) ≠ ( x 0 − a ) k + f ( a ) f ( b ) − f ( a ) b − a = f ( b ) − f ( x 0 ) + f ( x 0 ) − f ( a ) b − x 0 + x 0 − a is bewteen f ( x 0 ) − f ( a ) x 0 − a , f ( b ) − f ( x 0 ) b − x 0 f ( x ) ⟹ Lagrange Mean Value Theorem ∃ x 1 , f ′ ( x 1 ) = f ( x 0 ) − a x 0 − a ∃ x 2 , f ′ ( x 2 ) = f ( b ) − f ( x 0 ) b − x 0 \begin{gathered} \\
\text{let } k=\dfrac{f(b)-f(a)}{b-a}=
\exist x_0,f(x_0)\ne (x_0-a)k+f(a) \\
\dfrac{f(b)-f(a)}{b-a}=\dfrac{f(b)-f(x_0)+f(x_0)-f(a)}{b-x_0+x_0-a} \text{ is bewteen } \\
\dfrac{f(x_0)-f(a)}{x_0-a} ,\dfrac{f(b)-f(x_0)}{b-x_0} \\
f(x)\stackrel{\text{ Lagrange Mean Value Theorem }}{\Longrightarrow} \\
\exists x_1,f'(x_1)=\dfrac{f(x_0)-a}{x_0-a} \\
\exists x_2,f'(x_2)=\dfrac{f(b)-f(x_0)}{b-x_0} \\
\end{gathered} let k = b − a f ( b ) − f ( a ) = ∃ x 0 , f ( x 0 ) = ( x 0 − a ) k + f ( a ) b − a f ( b ) − f ( a ) = b − x 0 + x 0 − a f ( b ) − f ( x 0 ) + f ( x 0 ) − f ( a ) is bewteen x 0 − a f ( x 0 ) − f ( a ) , b − x 0 f ( b ) − f ( x 0 ) f ( x ) ⟹ Lagrange Mean Value Theorem ∃ x 1 , f ′ ( x 1 ) = x 0 − a f ( x 0 ) − a ∃ x 2 , f ′ ( x 2 ) = b − x 0 f ( b ) − f ( x 0 ) 则 ∣ k ∣ \vert k \vert ∣ k ∣ f ′ ( x 1 ) , f ′ ( x 2 ) f'(x_1),f'(x_2) f ′ ( x 1 ) , f ′ ( x 2 )
T6
{ ∀ x ∈ [ a , + ∞ ) , ∃ f ′ ( x ) f ( a ) = 0 x ≥ a ⟹ ∣ f ′ ( x ) ∣ ≤ ∣ f ( x ) ∣ ⟹ ∀ x ∈ [ a , + ∞ ) , f ( x ) = 0 \begin{gathered}
\begin{cases}
\forall x\in [a,+\infty),\exists f'(x) \\
f(a)=0 \\
x \ge a \implies \vert f'(x) \vert \le \vert f(x) \vert
\end{cases} \\
\implies \forall x\in [a,+\infty),f(x)=0
\end{gathered} ⎩ ⎨ ⎧ ∀ x ∈ [ a , + ∞ ) , ∃ f ′ ( x ) f ( a ) = 0 x ≥ a ⟹ ∣ f ′ ( x ) ∣ ≤ ∣ f ( x ) ∣ ⟹ ∀ x ∈ [ a , + ∞ ) , f ( x ) = 0
不妨设a = 0 a=0 a = 0
let F ( x ) = e − 2 x f 2 ( x ) ≥ 0 F ′ ( x ) = 2 e − 2 x f ( x ) ( f ′ ( x ) − f ( x ) ) ≤ 0 ∀ x > a = 0 , F ( x ) ≤ F ( a ) ∵ F ( x ) ≥ 0 ⟹ F ( x ) = 0 , f ( x ) = 0 \begin{gathered}
\text{let } F(x)=e^{-2x}f^2(x)\ge 0 \\
F'(x)=2e^{-2x}f(x)(f'(x)-f(x))\le 0 \\
\forall x>a=0,F(x)\le F(a) \\
\because F(x)\ge 0 \\
\implies F(x)=0,f(x)=0
\end{gathered} let F ( x ) = e − 2 x f 2 ( x ) ≥ 0 F ′ ( x ) = 2 e − 2 x f ( x ) ( f ′ ( x ) − f ( x )) ≤ 0 ∀ x > a = 0 , F ( x ) ≤ F ( a ) ∵ F ( x ) ≥ 0 ⟹ F ( x ) = 0 , f ( x ) = 0 Solution2:
假设f ( x ) f(x) f ( x ) 0 0 0 ( c , d ) (c,d) ( c , d ) f ( c ) = 0 f(c)=0 f ( c ) = 0 f ( x ) > 0 , ∀ x ∈ ( c , d ) f(x)>0,\forall x\in (c,d) f ( x ) > 0 , ∀ x ∈ ( c , d ) g ( x ) = ln ∣ f ( x ) ∣ g(x)=\ln \vert f(x) \vert g ( x ) = ln ∣ f ( x ) ∣ ∣ g ′ ( x ) ∣ ≤ 1 \vert g'(x)\vert\le 1 ∣ g ′ ( x ) ∣ ≤ 1 g g g g ( c + ) g(c^+) g ( c + ) − ∞ -\infty − ∞
T7
{ f ∈ C [ a , b ] ∀ x ∈ ( a , b ) , ∃ f ′ ′ ( x ) f ( a ) = f ( b ) = 0 f + ′ ( a ) > 0 ⟹ ∃ ξ ∈ ( a , b ) , f ′ ′ ( ξ ) < 0 \begin{gathered}
\begin{cases}
f\in C[a,b] \\
\forall x\in(a,b),\exists f''(x) \\
f(a)=f(b)=0 \\
f'_+(a)>0
\end{cases} \\
\implies \exists \xi \in (a,b),f''(\xi)<0
\end{gathered} ⎩ ⎨ ⎧ f ∈ C [ a , b ] ∀ x ∈ ( a , b ) , ∃ f ′′ ( x ) f ( a ) = f ( b ) = 0 f + ′ ( a ) > 0 ⟹ ∃ ξ ∈ ( a , b ) , f ′′ ( ξ ) < 0
f ( a ) = f ( b ) = 0 ⟹ Rolle’s Theorem ∃ x 1 ∈ ( a , b ) , f ′ ( x 1 ) = 0 ∃ f ′ ′ ( x ) ⟹ f ′ ( x ) ∈ C ( a , b ) f + ′ ( a ) > 0 ⟹ ∃ x 2 ∈ ( a , a + δ ) , f ′ ( x 2 ) > 0 ∃ ξ ∈ ( x 2 , x 1 ) ⊂ ( a , b ) , f ′ ′ ( ξ ) = f ′ ( x 1 ) − f ′ ( x 2 ) x 1 − x 2 < 0 \begin{gathered}
f(a)=f(b)=0 \\
\stackrel{\text{ Rolle's Theorem }}{\Longrightarrow}\exists x_1\in(a,b),f'(x_1)=0 \\
\exists f''(x) \implies f'(x) \in C(a,b) \\
f'_+(a)>0 \implies \exists x_2 \in (a,a+\delta),f'(x_2)>0 \\
\exists \xi \in (x_2,x_1) \subset (a,b),f''(\xi)=\dfrac{f'(x_1)-f'(x_2)}{x_1-x_2} <0
\end{gathered} f ( a ) = f ( b ) = 0 ⟹ Rolle’s Theorem ∃ x 1 ∈ ( a , b ) , f ′ ( x 1 ) = 0 ∃ f ′′ ( x ) ⟹ f ′ ( x ) ∈ C ( a , b ) f + ′ ( a ) > 0 ⟹ ∃ x 2 ∈ ( a , a + δ ) , f ′ ( x 2 ) > 0 ∃ ξ ∈ ( x 2 , x 1 ) ⊂ ( a , b ) , f ′′ ( ξ ) = x 1 − x 2 f ′ ( x 1 ) − f ′ ( x 2 ) < 0
T8
∀ x ∈ ( 0 , + ∞ ) , ∃ f ′ ( x ) lim x → + ∞ f ′ ( x ) = + ∞ ⟹ f ( x ) is not uniformly continuous in ( 0 , + ∞ ) \begin{gathered}
\forall x \in (0,+\infty),\exists f'(x) \\
\lim_{x \to +\infty} f'(x)=+\infty
\end{gathered} \\
\implies f(x) \text{ is not uniformly continuous in } (0,+\infty) ∀ x ∈ ( 0 , + ∞ ) , ∃ f ′ ( x ) x → + ∞ lim f ′ ( x ) = + ∞ ⟹ f ( x ) is not uniformly continuous in ( 0 , + ∞ )
lim x → + ∞ f ′ ( x ) = + ∞ ϵ = 1 ∀ δ let M = 2 δ ⟹ ∃ X , x > X ⟹ f ′ ( x ) > M ∀ x > M , ∣ f ( x + δ ) − f ( x ) ∣ = ∣ δ f ′ ( ξ ) ∣ > 2 > ϵ Q.E.D \begin{gathered}
\lim_{x \to +\infty} f'(x)=+\infty \\
\epsilon=1 \\
\forall \delta \\
\text{let } M=\dfrac{2}{\delta} \implies \exists X,x>X \implies f'(x)>M \\
\forall x>M,\vert f(x+\delta)-f(x) \vert = \vert \delta f'(\xi) \vert >2>\epsilon \\
\\
\text{Q.E.D}
\end{gathered} x → + ∞ lim f ′ ( x ) = + ∞ ϵ = 1 ∀ δ let M = δ 2 ⟹ ∃ X , x > X ⟹ f ′ ( x ) > M ∀ x > M , ∣ f ( x + δ ) − f ( x ) ∣ = ∣ δ f ′ ( ξ ) ∣ > 2 > ϵ Q.E.D
T9
{ lim x → x 0 + f ( x ) = f ( x 0 ) ∀ x ∈ ( x 0 , x 0 + δ 0 ) , ∃ f ′ ( x ) ∃ f ′ ( x 0 + ) ⟹ f + ′ ( x 0 ) = f ′ ( x 0 + ) \begin{gathered}
\begin{cases}
\lim_{x \to x_0^+} f(x)=f(x_0) \\
\forall x \in (x_0,x_0+\delta_0),\exists f'(x) \\
\exists f'(x_0^+)
\end{cases} \\
\implies f'_+(x_0)=f'(x_0^+)
\end{gathered} ⎩ ⎨ ⎧ lim x → x 0 + f ( x ) = f ( x 0 ) ∀ x ∈ ( x 0 , x 0 + δ 0 ) , ∃ f ′ ( x ) ∃ f ′ ( x 0 + ) ⟹ f + ′ ( x 0 ) = f ′ ( x 0 + )
⟹ Lagrange Mean Value Theorem f ( x 0 + h ) − f ( x 0 ) h = f ′ ( ξ ) , ξ ∈ ( x 0 , x 0 + h ) f + ′ ( x 0 ) = lim h → 0 + f ( x 0 + h ) − f ( x 0 ) h = lim ξ → x 0 + f ′ ( ξ ) = f ′ ( x 0 + ) \begin{gathered}
\stackrel{\text{ Lagrange Mean Value Theorem }}{\Longrightarrow} \\
\dfrac{f(x_0+h)-f(x_0)}{h}=f'(\xi),\xi \in (x_0,x_0+h) \\ \\
f'_+(x_0)=
\lim_{h \to 0^+} \dfrac{f(x_0+h)-f(x_0)}{h}=\lim_{\xi \to x_0^+} f'(\xi) =f'(x_0^+)\\
\end{gathered} ⟹ Lagrange Mean Value Theorem h f ( x 0 + h ) − f ( x 0 ) = f ′ ( ξ ) , ξ ∈ ( x 0 , x 0 + h ) f + ′ ( x 0 ) = h → 0 + lim h f ( x 0 + h ) − f ( x 0 ) = ξ → x 0 + lim f ′ ( ξ ) = f ′ ( x 0 + )
Class 2
T1
x ∈ [ 0 , 1 ] , p ≥ 2 ⟹ ( 1 + x 2 ) p + ( 1 − x 2 ) p ≤ 1 2 ( 1 + x p ) \begin{gathered}
x\in [0,1],p\ge 2 \\
\implies
(\dfrac{1+x}{2} )^p+(\dfrac{1-x}{2} )^p\le \dfrac{1}{2} (1+x^p)
\end{gathered} x ∈ [ 0 , 1 ] , p ≥ 2 ⟹ ( 2 1 + x ) p + ( 2 1 − x ) p ≤ 2 1 ( 1 + x p )
f ( x ) = 1 2 ( 1 + x p ) − ( 1 + x 2 ) p − ( 1 − x 2 ) p f ′ ( x ) = p 2 ( x p − 1 − ( 1 + x 2 ) p − 1 + ( 1 − x 2 ) p − 1 ) p − 1 > 1 ⟹ a p − 1 + b p − 1 < ( a + b ) p − 1 ∴ ( 1 − x 2 ) p − 1 + x p − 1 < ( 1 + x 2 ) p − 1 ⟹ f ′ ( x ) < 0 ⟹ f ( x ) is decreasing f ( x ) > f ( 1 ) = 0 Q.E.D \begin{gathered}
f(x)=\dfrac{1}{2} (1+x^p)-(\dfrac{1+x}{2} )^p-(\dfrac{1-x}{2} )^p \\
f'(x)=\dfrac{p}{2}(x^{p-1} -(\dfrac{1+x}{2} )^{p-1}+ (\dfrac{1-x}{2} )^{p-1}) \\
p-1>1 \implies a^{p-1}+b^{p-1}<(a+b)^{p-1} \\
\therefore (\dfrac{1-x}{2})^{p-1}+x^{p-1}<(\dfrac{1+x}{2})^{p-1} \\
\implies f'(x)<0 \\
\implies f(x) \text{ is decreasing} \\
f(x)>f(1)=0 \\
\\
\text{Q.E.D}
\end{gathered} f ( x ) = 2 1 ( 1 + x p ) − ( 2 1 + x ) p − ( 2 1 − x ) p f ′ ( x ) = 2 p ( x p − 1 − ( 2 1 + x ) p − 1 + ( 2 1 − x ) p − 1 ) p − 1 > 1 ⟹ a p − 1 + b p − 1 < ( a + b ) p − 1 ∴ ( 2 1 − x ) p − 1 + x p − 1 < ( 2 1 + x ) p − 1 ⟹ f ′ ( x ) < 0 ⟹ f ( x ) is decreasing f ( x ) > f ( 1 ) = 0 Q.E.D
T2
x ∈ ( 0 , π 2 ) ⟹ 2 x < sin x + tan x \begin{gathered}
x\in (0,\dfrac{\pi}{2} )\implies 2x<\sin x+\tan x
\end{gathered} x ∈ ( 0 , 2 π ) ⟹ 2 x < sin x + tan x
f ( x ) = sin x + tan x − 2 x f ′ ( x ) = cos x + 1 cos 2 x − 2 f ′ ′ ( x ) = sin x ( 2 cos 3 x − 1 ) > 0 ⟹ f ′ ( x ) is increasing ⟹ f ′ ( x ) > f ′ ( 0 ) = 0 ⟹ f ( x ) is increasing f ( x ) > f ( 0 ) = 0 Q.E.D \begin{gathered}
f(x)=\sin x+\tan x-2x \\
f'(x)=\cos x+\dfrac{1}{\cos^2 x}-2 \\
f''(x)=\sin x(\dfrac{2}{\cos^3 x} -1)>0 \\
\implies f'(x) \text{ is increasing} \\
\implies f'(x)>f'(0)=0 \\
\implies f(x) \text{ is increasing} \\
f(x)>f(0)=0 \\
\\
\text{Q.E.D}
\end{gathered} f ( x ) = sin x + tan x − 2 x f ′ ( x ) = cos x + cos 2 x 1 − 2 f ′′ ( x ) = sin x ( cos 3 x 2 − 1 ) > 0 ⟹ f ′ ( x ) is increasing ⟹ f ′ ( x ) > f ′ ( 0 ) = 0 ⟹ f ( x ) is increasing f ( x ) > f ( 0 ) = 0 Q.E.D
T3
{ ∀ x ∈ [ 0 , + ∞ ) , ∃ f ′ ( x ) f ( 0 ) = 0 f ′ ( x ) is strictly increasing ⟹ f ( x ) x is strictly increasing \begin{gathered}
\begin{cases}
\forall x\in[0,+\infty),\exists f'(x) \\
f(0)=0 \\
f'(x) \text{ is strictly increasing}
\end{cases} \\
\implies \dfrac{f(x)}{x} \text{ is strictly increasing}
\end{gathered} ⎩ ⎨ ⎧ ∀ x ∈ [ 0 , + ∞ ) , ∃ f ′ ( x ) f ( 0 ) = 0 f ′ ( x ) is strictly increasing ⟹ x f ( x ) is strictly increasing
f ( x ) = f ( x ) − f ( 0 ) = ( x − 0 ) f ′ ( ξ ) ξ ∈ ( 0 , x ) ⟹ f ′ ( ξ ) < f ′ ( x ) ⟹ f ( x ) < x f ′ ( x ) ⟹ ( f ( x ) x ) ′ = x f ′ ( x ) − f ( x ) x 2 > 0 Q.E.D \begin{gathered}
f(x)=f(x)-f(0)=(x-0)f'(\xi) \\
\xi\in(0,x) \implies f'(\xi)<f'(x) \\
\implies f(x)<xf'(x) \\
\implies (\dfrac{f(x)}{x} )'=\dfrac{xf'(x)-f(x)}{x^2} >0 \\
\\
\text{Q.E.D}
\end{gathered} f ( x ) = f ( x ) − f ( 0 ) = ( x − 0 ) f ′ ( ξ ) ξ ∈ ( 0 , x ) ⟹ f ′ ( ξ ) < f ′ ( x ) ⟹ f ( x ) < x f ′ ( x ) ⟹ ( x f ( x ) ) ′ = x 2 x f ′ ( x ) − f ( x ) > 0 Q.E.D
T4
calculate the extremum point for
f ( x ) = arcsin 2 x 1 + x 2 \begin{gathered}
f(x)=\arcsin \dfrac{2x}{1+x^2}
\end{gathered} f ( x ) = arcsin 1 + x 2 2 x
f ′ ( x ) = 1 1 − ( 2 x 1 + x 2 ) 2 2 ( 1 + x 2 ) − 4 x 2 ( 1 + x 2 ) 2 = 2 − 2 x 2 x 4 + 2 x 2 + 1 − 4 x 2 ( 1 + x 2 ) = 2 ( 1 − x 2 ) ∣ 1 − x 2 ∣ ( 1 + x 2 ) x < − 1 ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing x ∈ ( − 1 , 1 ) ⟹ f ′ ( x ) > 0 , f ( x ) is increasing x > 1 ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing ⟹ minimum: ( − 1 , − π 2 ) , maximum: ( 1 , π 2 ) \begin{gathered}
f'(x)=\dfrac{1}{\sqrt{1-(\dfrac{2x}{1+x^2} )^2}}\dfrac{2(1+x^2)-4x^2}{(1+x^2)^2} \\
=\dfrac{2-2x^2}{\sqrt{x^4+2x^2+1-4x^2}(1+x^2)} \\
=\dfrac{2(1-x^2)}{\vert 1-x^2 \vert (1+x^2)} \\
x<-1 \implies f'(x)<0,f(x) \text{ is decreasing} \\
x\in (-1,1) \implies f'(x)>0,f(x) \text{ is increasing} \\
x>1 \implies f'(x)<0,f(x) \text{ is decreasing} \\
\implies \text{minimum:} (-1,-\dfrac{\pi}{2} ),\text{maximum:} (1,\dfrac{\pi}{2} )
\end{gathered} f ′ ( x ) = 1 − ( 1 + x 2 2 x ) 2 1 ( 1 + x 2 ) 2 2 ( 1 + x 2 ) − 4 x 2 = x 4 + 2 x 2 + 1 − 4 x 2 ( 1 + x 2 ) 2 − 2 x 2 = ∣1 − x 2 ∣ ( 1 + x 2 ) 2 ( 1 − x 2 ) x < − 1 ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing x ∈ ( − 1 , 1 ) ⟹ f ′ ( x ) > 0 , f ( x ) is increasing x > 1 ⟹ f ′ ( x ) < 0 , f ( x ) is decreasing ⟹ minimum: ( − 1 , − 2 π ) , maximum: ( 1 , 2 π )
T5
calculate the extremum point for
f ( x ) = ( ∑ i = 0 n x i i ! ) e − x , n ≥ 1 \begin{gathered}
f(x)=(\sum _{i = 0} ^{n} \dfrac{x^i}{i!} )e^{-x},n\ge 1
\end{gathered} f ( x ) = ( i = 0 ∑ n i ! x i ) e − x , n ≥ 1
f ′ ( x ) = e − x ( ∑ i = 0 n − 1 x i i ! − ∑ i = 1 n x i i ! ) = − x n e − x n ! n m o d 2 = 0 : f ′ ( x ) ≤ 0 ( f ′ ( x ) = 0 ⟺ x = 0 ) ⟹ f ( x ) has no extremum n m o d 2 = 1 : sign f ′ ( x ) = − sign x ⟹ f ( x ) has maximum ( 0 , 1 ) \begin{gathered}
f'(x)=e^{-x}(\sum _{i = 0} ^{n-1} \dfrac{x^i}{i!} -\sum _{i = 1} ^{n} \dfrac{x^i}{i!} ) \\
=-\dfrac{x^ne^{-x}}{n!} \\
n \bmod 2=0:f'(x)\le 0(f'(x)=0 \iff x=0) \\
\implies f(x) \text{ has no extremum} \\
n \bmod 2=1:\operatorname{sign}f'(x)=-\operatorname{sign}x \\
\implies f(x) \text{ has maximum } (0,1)
\end{gathered} f ′ ( x ) = e − x ( i = 0 ∑ n − 1 i ! x i − i = 1 ∑ n i ! x i ) = − n ! x n e − x n mod 2 = 0 : f ′ ( x ) ≤ 0 ( f ′ ( x ) = 0 ⟺ x = 0 ) ⟹ f ( x ) has no extremum n mod 2 = 1 : sign f ′ ( x ) = − sign x ⟹ f ( x ) has maximum ( 0 , 1 )
T6
x f ′ ′ ( x ) + 3 x ( f ′ ( x ) ) 2 = 1 − e − x ⟹ { x = c ( c ≠ 0 ) is an extremum ⟹ f ( c ) is a minimum x = c ( c = 0 ) is an extremum ⟹ f ( c ) is a minimum \begin{gathered}
xf''(x)+3x(f'(x))^2=1-e^{-x} \\
\implies \begin{cases}
x=c(c\ne 0) \text{ is an extremum } \implies f(c)\text{ is a minimum} \\
x=c(c=0) \text{ is an extremum } \implies f(c) \text{ is a minimum}
\end{cases}
\end{gathered} x f ′′ ( x ) + 3 x ( f ′ ( x ) ) 2 = 1 − e − x ⟹ { x = c ( c = 0 ) is an extremum ⟹ f ( c ) is a minimum x = c ( c = 0 ) is an extremum ⟹ f ( c ) is a minimum
(1)
x = c is an extremum ⟹ f ′ ( c ) = 0 c f ′ ′ ( c ) + 3 c ( f ′ ( c ) ) 2 = 1 − e − c ⟹ sign f ′ ′ ( c ) = sign 1 − e − c c = 1 ⟹ lim x → c f ′ ( x ) − f ′ ( c ) x − c = f ′ ( x ) x − c > 0 ⟹ sign f ′ ( x ) = sign ( x − c ) ⟹ f ( x ) is a minimum \begin{gathered}
x=c \text{ is an extremum} \\
\implies f'(c)=0 \\
cf''(c)+3c(f'(c))^2=1-e^{-c} \\
\implies \operatorname{sign} f''(c)=\operatorname{sign} \dfrac{1-e^{-c}}{c} =1 \\
\implies \lim_{x \to c} \dfrac{f'(x)-f'(c)}{x-c}=\dfrac{f'(x)}{x-c} >0 \\
\implies \operatorname{sign} f'(x)=\operatorname{sign} (x-c) \\
\implies f(x) \text{ is a minimum}
\end{gathered} x = c is an extremum ⟹ f ′ ( c ) = 0 c f ′′ ( c ) + 3 c ( f ′ ( c ) ) 2 = 1 − e − c ⟹ sign f ′′ ( c ) = sign c 1 − e − c = 1 ⟹ x → c lim x − c f ′ ( x ) − f ′ ( c ) = x − c f ′ ( x ) > 0 ⟹ sign f ′ ( x ) = sign ( x − c ) ⟹ f ( x ) is a minimum (2)
f ′ ( 0 ) = 0 lim x → 0 ( f ′ ′ ( x ) + 3 f ′ 2 ( x ) ) = lim x → 0 1 − e − x x = 1 ⟹ f ′ ′ ( 0 ) = 1 ⟹ same as (1)’s proof f ( x ) is a minimum \begin{gathered}
f'(0)=0 \\
\lim_{x \to 0} (f''(x)+3f'^2(x))=\lim_{x \to 0} \dfrac{1-e^{-x}}{x} =1 \\
\implies f''(0)=1 \\
\stackrel{\text{ same as (1)'s proof }}{\Longrightarrow} f(x) \text{ is a minimum}
\end{gathered} f ′ ( 0 ) = 0 x → 0 lim ( f ′′ ( x ) + 3 f ′2 ( x )) = x → 0 lim x 1 − e − x = 1 ⟹ f ′′ ( 0 ) = 1 ⟹ same as (1)’s proof f ( x ) is a minimum
T7
p > 1 , x ∈ [ 0 , 1 ] ⟹ 1 2 p − 1 ≤ x p + ( 1 − x ) p ≤ 1 \begin{gathered}
p>1,x\in [0,1] \implies \\
\dfrac{1}{2^{p-1}} \le x^p+(1-x)^p\le 1
\end{gathered} p > 1 , x ∈ [ 0 , 1 ] ⟹ 2 p − 1 1 ≤ x p + ( 1 − x ) p ≤ 1
x ≤ 1 ⟹ x p ≤ x ⟹ x p + ( 1 − x ) p ≤ x + 1 − x = 1 f ( x ) = x p + ( 1 − x ) p f ′ ( x ) = p x p − 1 − p ( 1 − x ) p − 1 = p ( x p − 1 − ( 1 − x ) p − 1 ) x > 1 2 ⟹ x > 1 − x ⟹ f ′ ( x ) > 0 x < 1 2 ⟹ f ′ ( x ) < 0 ⟹ ( 1 2 , f ( 1 2 ) ) is a minimum , f ( x ) ≥ f ( 1 2 ) = 2 − ( p − 1 ) \begin{gathered}
x\le 1 \implies x^p\le x \implies x^p+(1-x)^p\le x+1-x=1 \\
f(x)=x^p+(1-x)^p \\
f'(x)=px^{p-1}-p(1-x)^{p-1}=p(x^{p-1}-(1-x)^{p-1}) \\
x>\dfrac{1}{2} \implies x>1-x \implies f'(x)>0 \\
x<\dfrac{1}{2} \implies f'(x)<0 \\
\implies (\dfrac{1}{2},f(\dfrac{1}{2})) \text{ is a minimum} ,f(x)\ge f(\dfrac{1}{2})=2^{-(p-1)}
\end{gathered} x ≤ 1 ⟹ x p ≤ x ⟹ x p + ( 1 − x ) p ≤ x + 1 − x = 1 f ( x ) = x p + ( 1 − x ) p f ′ ( x ) = p x p − 1 − p ( 1 − x ) p − 1 = p ( x p − 1 − ( 1 − x ) p − 1 ) x > 2 1 ⟹ x > 1 − x ⟹ f ′ ( x ) > 0 x < 2 1 ⟹ f ′ ( x ) < 0 ⟹ ( 2 1 , f ( 2 1 )) is a minimum , f ( x ) ≥ f ( 2 1 ) = 2 − ( p − 1 )
T8
{ f , g : R → R f ′ ′ ( x ) + f ′ ( x ) g ( x ) − f ( x ) = 0 f ( a ) = f ( b ) = 0 ( a < b ) ⟹ ∀ x ∈ [ a , b ] , f ( x ) = 0 \begin{gathered}
\begin{cases}
f,g:R\to R \\
f''(x)+f'(x)g(x)-f(x)=0 \\
f(a)=f(b)=0(a<b)
\end{cases} \\
\implies \forall x\in [a,b],f(x)=0
\end{gathered} ⎩ ⎨ ⎧ f , g : R → R f ′′ ( x ) + f ′ ( x ) g ( x ) − f ( x ) = 0 f ( a ) = f ( b ) = 0 ( a < b ) ⟹ ∀ x ∈ [ a , b ] , f ( x ) = 0
此证明已弃用.但应该是正确的.
f ′ ′ ( x ) + f ′ ( x ) g ( x ) − f ( x ) = 0 ⟺ ∀ x , f ′ ( x ) ≠ 0 ∨ f ′ ′ ( x ) = f ( x ) ∀ x 0 ∈ ( a , b ) , f ′ ( x 0 ) = 0 Assume f ( x 0 ) ≠ 0 Without loss of generality,assume f ( x 0 ) > 0 ⟹ f ′ ′ ( x 0 ) = f ( x 0 ) > 0 ⟹ ∃ δ , ∃ x 0 ′ ∈ ( x 0 , δ ) , f ′ ( x 0 ′ ) > 0 ∃ ξ , f ′ ( ξ ) = f ( x 0 ) − f ( b ) x 0 − b < 0 ⟹ Darbox Theorem ∃ x 1 ∈ ( x 0 , ξ ) , f ′ ( x 1 ) = 0 Repeat this you get { x n } , x i ∈ ( x i − 1 , ξ ) , f ′ ( x i ) = 0 lim n → ∞ x n = ξ , lim n → ∞ f ′ ( x n ) = 0 ≠ f ′ ( ξ ) Ridiculous! ⟹ f ′ ( x ) = 0 ⟹ f ( x ) = f ′ ′ ( x ) = 0 Assume f ( x 0 ) ≠ 0 , a 0 = a , b 0 = b ∃ x 1 ∈ ( a i , x 0 ) , x 2 ∈ ( x 0 , b i ) , f ′ ( x 1 ) = f ( x 0 ) x 0 − a i ≠ 0 f ′ ( x 2 ) = f ( x 0 ) x 0 − b i ≠ 0 f ′ ( x 1 ) f ′ ( x 2 ) < 0 ⟹ ∃ x 3 , f ′ ( x 3 ) = 0 ⟹ f ′ ( x 3 ) = f ( x 3 ) = 0 f ′ ( x 1 ) ≠ 0 ⟹ ∃ x 1 ′ ∈ N ( x 1 ) , f ( x 1 ′ ) ≠ 0 same for ∃ x 2 ′ ∈ N ( x 2 ) , f ( x 2 ′ ) ≠ 0 ⟹ ∃ f ( a i ) = f ( x 3 ) = 0 , x 1 ′ ∈ ( a i , x 3 ) , f ( x 1 ′ ) ≠ 0 ∃ f ( x 3 ) = f ( b i ) = 0 , x 2 ′ ∈ ( x 3 , b i ) , f ( x 2 ′ ) ≠ 0 let [ a i + 1 , b i + 1 ] = the shorter one in [ a i , x 3 ] , [ x 3 , b i ] ⟹ { b i + 1 − a i + 1 ≤ 1 2 ( b i − a i ) [ a i , b i ] ⊃ [ a i + 1 , b i + 1 ] ∃ c i ∈ [ a i , b i ] , f ( c i ) ≠ 0 ⟹ Nested Intervals Principle ∃ ξ ∈ [ a i , b i ] lim n → ∞ a n = ξ lim n → ∞ c n = ξ lim n → ∞ f ( a n ) = lim n → ∞ f ( c n ) Ridiculous! ⟹ ∀ x ∈ ( a , b ) , f ( x ) = 0 Q.E.D \begin{gathered}
f''(x)+f'(x)g(x)-f(x)=0 \\
\iff \forall x,f'(x)\ne 0 \lor f''(x)=f(x) \\
\forall x_0\in (a,b),f'(x_0)=0 \\
\text{Assume } f(x_0)\ne 0 \\
\text{Without loss of generality,assume } f(x_0)>0 \\
\implies f''(x_0)=f(x_0)>0 \\
\implies \exists \delta,\exists x_0'\in (x_0,\delta),f'(x_0')>0 \\
\exists \xi,f'(\xi)=\dfrac{f(x_0)-f(b)}{x_0-b} <0 \\
\stackrel{\text{ Darbox Theorem }}{\Longrightarrow}\exists x_1\in (x_0,\xi),f'(x_1)=0 \\
\text{Repeat this you get } \{ x_n \} ,x_i\in (x_{i-1},\xi),f'(x_i)=0 \\
\lim_{n \to \infty} x_n=\xi,\lim_{n \to \infty} f'(x_n)=0\ne f'(\xi) \\
\text{Ridiculous!} \\
\implies f'(x)=0 \implies f(x)=f''(x)=0 \\
\text{Assume } f(x_0)\ne 0,a_0=a,b_0=b \\
\exists x_1\in (a_i,x_0),x_2\in (x_0,b_i), \\
f'(x_1)=\dfrac{f(x_0)}{x_0-a_i} \ne 0 \\
f'(x_2)=\dfrac{f(x_0)}{x_0-b_i} \ne 0 \\
f'(x_1)f'(x_2)<0 \implies \exists x_3,f'(x_3)=0 \\
\implies f'(x_3)=f(x_3)=0 \\
f'(x_1)\ne 0 \implies \exists x_1'\in N(x_1),f(x_1')\ne 0 \\
\text{same for } \exists x_2'\in N(x_2),f(x_2')\ne 0 \\
\implies \exists f(a_i)=f(x_3)=0,x_1'\in(a_i,x_3),f(x_1')\ne 0 \\
\exists f(x_3)=f(b_i)=0,x_2'\in (x_3,b_i),f(x_2')\ne 0 \\
\text{let } [a_{i+1},b_{i+1}]=\text{the shorter one in } [a_i,x_3],[x_3,b_i] \\
\implies \begin{cases}
b_{i+1}-a_{i+1}\le \dfrac{1}{2} (b_i-a_i) \\
[a_i,b_i]\supset [a_{i+1},b_{i+1}] \\
\exists c_i\in [a_i,b_i],f(c_i)\ne 0
\end{cases}
\\
\stackrel{\text{ Nested Intervals Principle }}{\Longrightarrow} \\
\exists \xi \in [a_i,b_i] \\
\lim_{n \to \infty} a_n=\xi \\
\lim_{n \to \infty} c_n=\xi \\
\lim_{n \to \infty} f(a_n)=\lim_{n \to \infty} f(c_n) \\
\text{Ridiculous!} \\
\implies \forall x\in (a,b),f(x)=0
\\
\text{Q.E.D}
\end{gathered} f ′′ ( x ) + f ′ ( x ) g ( x ) − f ( x ) = 0 ⟺ ∀ x , f ′ ( x ) = 0 ∨ f ′′ ( x ) = f ( x ) ∀ x 0 ∈ ( a , b ) , f ′ ( x 0 ) = 0 Assume f ( x 0 ) = 0 Without loss of generality,assume f ( x 0 ) > 0 ⟹ f ′′ ( x 0 ) = f ( x 0 ) > 0 ⟹ ∃ δ , ∃ x 0 ′ ∈ ( x 0 , δ ) , f ′ ( x 0 ′ ) > 0 ∃ ξ , f ′ ( ξ ) = x 0 − b f ( x 0 ) − f ( b ) < 0 ⟹ Darbox Theorem ∃ x 1 ∈ ( x 0 , ξ ) , f ′ ( x 1 ) = 0 Repeat this you get { x n } , x i ∈ ( x i − 1 , ξ ) , f ′ ( x i ) = 0 n → ∞ lim x n = ξ , n → ∞ lim f ′ ( x n ) = 0 = f ′ ( ξ ) Ridiculous! ⟹ f ′ ( x ) = 0 ⟹ f ( x ) = f ′′ ( x ) = 0 Assume f ( x 0 ) = 0 , a 0 = a , b 0 = b ∃ x 1 ∈ ( a i , x 0 ) , x 2 ∈ ( x 0 , b i ) , f ′ ( x 1 ) = x 0 − a i f ( x 0 ) = 0 f ′ ( x 2 ) = x 0 − b i f ( x 0 ) = 0 f ′ ( x 1 ) f ′ ( x 2 ) < 0 ⟹ ∃ x 3 , f ′ ( x 3 ) = 0 ⟹ f ′ ( x 3 ) = f ( x 3 ) = 0 f ′ ( x 1 ) = 0 ⟹ ∃ x 1 ′ ∈ N ( x 1 ) , f ( x 1 ′ ) = 0 same for ∃ x 2 ′ ∈ N ( x 2 ) , f ( x 2 ′ ) = 0 ⟹ ∃ f ( a i ) = f ( x 3 ) = 0 , x 1 ′ ∈ ( a i , x 3 ) , f ( x 1 ′ ) = 0 ∃ f ( x 3 ) = f ( b i ) = 0 , x 2 ′ ∈ ( x 3 , b i ) , f ( x 2 ′ ) = 0 let [ a i + 1 , b i + 1 ] = the shorter one in [ a i , x 3 ] , [ x 3 , b i ] ⟹ ⎩ ⎨ ⎧ b i + 1 − a i + 1 ≤ 2 1 ( b i − a i ) [ a i , b i ] ⊃ [ a i + 1 , b i + 1 ] ∃ c i ∈ [ a i , b i ] , f ( c i ) = 0 ⟹ Nested Intervals Principle ∃ ξ ∈ [ a i , b i ] n → ∞ lim a n = ξ n → ∞ lim c n = ξ n → ∞ lim f ( a n ) = n → ∞ lim f ( c n ) Ridiculous! ⟹ ∀ x ∈ ( a , b ) , f ( x ) = 0 Q.E.D 上面那个太麻烦了
考虑区间中的最大值f ( x 0 ) f(x_0) f ( x 0 ) f ′ ( x 0 ) = 0 , f ( x 0 ) ≥ 0 , f ′ ′ ( x 0 ) ≤ 0 f'(x_0)=0,f(x_0)\ge 0,f''(x_0)\le 0 f ′ ( x 0 ) = 0 , f ( x 0 ) ≥ 0 , f ′′ ( x 0 ) ≤ 0 f ( x 0 ) = 0 f(x_0)=0 f ( x 0 ) = 0
最小值同理.