Math Analysis Homework - Week 2
Class 4
T1
设 x 1 > 1 , x n + 1 = 3 x n + 1 x n + 3 , n = 1 , 2 , … x_1 > 1, x_{n+1} = \frac{3x_n + 1}{x_n + 3}, n = 1, 2, \dots x 1 > 1 , x n + 1 = x n + 3 3 x n + 1 , n = 1 , 2 , …
x n > 1 ⟹ 1 < x n + 1 = 3 x n + 1 x n + 3 < 3 Inductively , x n ∈ ( 1 , 3 ) x n + 1 − x n = 3 x n + 1 − x n 2 − 3 x n x n + 3 = 1 − x n 2 x n + 3 < 0 ∴ { x n } is bounded and decreasing lim n → ∞ x n exists . x n + 1 = 3 x n + 1 x n + 3 ⟹ lim n → ∞ x n = 3 ( lim n → ∞ x n ) + 1 ( lim n → ∞ x n ) + 3 ⟹ lim n → ∞ x n = 1 \begin{gathered}
x_n>1 \implies 1<x_{n+1}=\dfrac{3x_n+1}{x_n+3}<3 \\
\text{Inductively},x_n\in (1,3)
x_{n+1}-x_n \\
=\dfrac{3x_n+1-x_n^2-3x_n}{x_n+3} \\
=\dfrac{1-x_n^2}{x_n+3}<0 \\
\therefore \{ x_n \} \text{is bounded and decreasing} \\\lim_{n \to \infty} x_n \text{exists}. \\
x_{n+1}=\dfrac{3x_n+1}{x_n+3} \\
\implies \lim_{n \to \infty} x_{n}=\dfrac{3(\lim_{n \to \infty} x_n)+1}{(\lim_{n \to \infty} x_n)+3} \\
\implies \lim_{n \to \infty} x_n=1
\end{gathered} x n > 1 ⟹ 1 < x n + 1 = x n + 3 3 x n + 1 < 3 Inductively , x n ∈ ( 1 , 3 ) x n + 1 − x n = x n + 3 3 x n + 1 − x n 2 − 3 x n = x n + 3 1 − x n 2 < 0 ∴ { x n } is bounded and decreasing n → ∞ lim x n exists . x n + 1 = x n + 3 3 x n + 1 ⟹ n → ∞ lim x n = ( lim n → ∞ x n ) + 3 3 ( lim n → ∞ x n ) + 1 ⟹ n → ∞ lim x n = 1
T2
设 0 < a 1 < b 1 0 < a_1 < b_1 0 < a 1 < b 1 a n + 1 = a n b n , b n + 1 = a n + b n 2 , n ∈ N + a_{n+1} = \sqrt{a_n b_n}, b_{n+1} = \frac{a_n + b_n}{2}, n \in \mathbb{N}_+ a n + 1 = a n b n , b n + 1 = 2 a n + b n , n ∈ N + { a n } , { b n } \{a_n\}, \{b_n\} { a n } , { b n }
∀ n , a n < b n ⟹ { b n > a n + 1 = a n b n > a n a n < b n + 1 = a n + b n 2 < b n ⟹ { a n < b n < b n − 1 < … < b 1 , b n > a n > a n − 1 > … > a 1 ⟹ { a n } , { b n } is bounded and monotonic ⟹ A = lim n → ∞ a n , B = lim n → ∞ b n exists ⟹ { lim n → ∞ b n + 1 = lim n → ∞ a n + b n 2 lim n → ∞ a n + 1 = a n b n ⟹ { B = A + B 2 A = A B ⟹ A = B \begin{gathered}
\forall n,a_n<b_n \\
\implies \begin{cases}
b_n>a_{n+1}=\sqrt{a_n b_n}>a_n \\
a_n<b_{n+1}=\frac{a_n+b_n}{2}<b_n
\end{cases}
\\
\implies
\begin{cases}
a_n<b_n<b_{n-1}<\ldots<b_1, \\
b_n>a_n>a_{n-1}>\ldots>a_1
\end{cases} \\
\implies
\{ a_n \} ,\{ b_n \} \text{ is bounded and monotonic} \\
\implies A=\lim_{n \to \infty} a_n,B=\lim_{n \to \infty} b_n \text{ exists} \\
\implies
\begin{cases}
\lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} \frac{a_n + b_n}{2} \\
\lim_{n \to \infty} a_{n+1} = \sqrt{a_n b_n}
\end{cases} \\
\implies
\begin{cases}
B=\dfrac{A+B}{2} \\
A=\sqrt{ A B } \\
\end{cases}
\implies A=B
\end{gathered} ∀ n , a n < b n ⟹ { b n > a n + 1 = a n b n > a n a n < b n + 1 = 2 a n + b n < b n ⟹ { a n < b n < b n − 1 < … < b 1 , b n > a n > a n − 1 > … > a 1 ⟹ { a n } , { b n } is bounded and monotonic ⟹ A = n → ∞ lim a n , B = n → ∞ lim b n exists ⟹ { lim n → ∞ b n + 1 = lim n → ∞ 2 a n + b n lim n → ∞ a n + 1 = a n b n ⟹ ⎩ ⎨ ⎧ B = 2 A + B A = A B ⟹ A = B
T3
设数列 { x n } \{x_n\} { x n } 0 < x n < 1 0 < x_n < 1 0 < x n < 1 ( 1 − x n ) x n + 1 > 1 4 , n = 1 , 2 , 3 , … (1 - x_n)x_{n+1} > \frac{1}{4}, n = 1, 2, 3, \dots ( 1 − x n ) x n + 1 > 4 1 , n = 1 , 2 , 3 , … lim n → ∞ x n = 1 2 \lim_{n \to \infty} x_n = \frac{1}{2} lim n → ∞ x n = 2 1
1 4 ( 1 − x n ) < x n + 1 < 1 ⟹ x n < 3 4 1 4 ( 1 − x n − 1 ) < x n < 3 4 ⟹ x n − 1 < 2 3 同理 ⟹ x n − 2 < 1 2 ∴ ∀ n , x n < 1 2 又 x n + 1 = 1 4 ( 1 − x n ) > x n ⟺ ( 2 x n − 1 ) 2 > 0 ⟹ True X = lim n → ∞ x n exists x n + 1 > 1 4 ( 1 − x n ) ⟹ X ≥ 1 4 ( 1 − X ) ⟹ X = 1 2 \begin{gathered}
\dfrac{1}{4(1-x_n)}<x_{n+1}<1 \\
\implies x_n<\dfrac{3}{4} \\
\dfrac{1}{4(1-x_{n-1})}<x_n<\dfrac{3}{4} \\
\implies x_{n-1}<\dfrac{2}{3} \\
\text{同理}
\implies x_{n-2}<\dfrac{1}{2} \\
\therefore \forall n,x_n<\dfrac{1}{2} \\
\text{又} x_{n+1}=\dfrac{1}{4(1-x_n)} > x_n \\
\iff (2x_n-1)^2>0
\implies \text{True} \\
X=\lim_{n \to \infty} x_n \text{ exists} \\
x_{n+1}>\dfrac{1}{4(1-x_n)} \\
\implies X\ge \dfrac{1}{4(1-X)} \\
\implies X=\dfrac{1}{2}
\end{gathered} 4 ( 1 − x n ) 1 < x n + 1 < 1 ⟹ x n < 4 3 4 ( 1 − x n − 1 ) 1 < x n < 4 3 ⟹ x n − 1 < 3 2 同理 ⟹ x n − 2 < 2 1 ∴ ∀ n , x n < 2 1 又 x n + 1 = 4 ( 1 − x n ) 1 > x n ⟺ ( 2 x n − 1 ) 2 > 0 ⟹ True X = n → ∞ lim x n exists x n + 1 > 4 ( 1 − x n ) 1 ⟹ X ≥ 4 ( 1 − X ) 1 ⟹ X = 2 1
T4
设 x 1 ∈ ( 0 , 1 ) , x n + 1 = x n ( 1 − x n ) , n ∈ N + x_1 \in (0, 1), x_{n+1} = x_n(1 - x_n), n \in \mathbb{N}_+ x 1 ∈ ( 0 , 1 ) , x n + 1 = x n ( 1 − x n ) , n ∈ N + { n x n } \{nx_n\} { n x n }
lim n → ∞ 1 x n n ⟸ Stolz Theorem lim n → ∞ 1 x n − 1 x n − 1 = lim n → ∞ 1 x n − 1 ( 1 − x n − 1 ) − 1 x n − 1 = lim n → ∞ 1 1 − x n − 1 = 1 1 − lim n → ∞ x n − 1 \begin{gathered}
\lim_{n \to \infty} \dfrac{\dfrac{1}{x_n} }{n} \\
\stackrel{\text{ Stolz Theorem }}{\Longleftarrow}
\lim_{n \to \infty} \dfrac{1}{x_n} -\dfrac{1}{x_{n-1}} \\
=\lim_{n \to \infty} \dfrac{1}{x_{n-1}(1-x_{n-1})} -\dfrac{1}{x_{n-1}} \\
=\lim_{n \to \infty} \dfrac{1}{1-x_{n-1}} \\
=\dfrac{1}{1-\lim_{n \to \infty} x_{n-1}}
\end{gathered} \\ n → ∞ lim n x n 1 ⟸ Stolz Theorem n → ∞ lim x n 1 − x n − 1 1 = n → ∞ lim x n − 1 ( 1 − x n − 1 ) 1 − x n − 1 1 = n → ∞ lim 1 − x n − 1 1 = 1 − lim n → ∞ x n − 1 1 对于x x x
{ x 1 ≤ 1 1 , x 2 < min ( x 1 , 1 − x 1 ) ≤ 1 2 x n ≤ 1 n , n ≥ 2 ⟹ x n + 1 < 1 n ( 1 − 1 n ) = n − 1 n 2 < 1 n + 1 ⟹ x n ≤ 1 n 0 < x n ≤ 1 n ⟹ Squeeze Theorem lim n → ∞ x n = 0 ⟹ lim n → ∞ 1 x n n = 1 1 − lim n → ∞ x n = 1 ⟹ lim n → ∞ n x n = 1 \begin{gathered}
\begin{cases}
x_1\le \dfrac{1}{1},x_2<\min(x_1,1-x_1)\le \dfrac{1}{2} \\
x_n\le \dfrac{1}{n},n\ge 2
\implies
x_{n+1}<\dfrac{1}{n}(1-\dfrac{1}{n} )=\dfrac{n-1}{n^2} <\dfrac{1}{n+1}
\end{cases} \\
\implies x_n\le \dfrac{1}{n} \\
0<x_n\le \dfrac{1}{n} \\
\stackrel{\text{Squeeze Theorem}}{\Longrightarrow} \\
\lim_{n \to \infty} x_n=0 \\
\implies
\lim_{n \to \infty} \dfrac{\dfrac{1}{x_n} }{n} =\dfrac{1}{1-\lim_{n \to \infty} x_{n}}=1 \\
\implies \lim_{n \to \infty} nx_n=1
\end{gathered} ⎩ ⎨ ⎧ x 1 ≤ 1 1 , x 2 < min ( x 1 , 1 − x 1 ) ≤ 2 1 x n ≤ n 1 , n ≥ 2 ⟹ x n + 1 < n 1 ( 1 − n 1 ) = n 2 n − 1 < n + 1 1 ⟹ x n ≤ n 1 0 < x n ≤ n 1 ⟹ Squeeze Theorem n → ∞ lim x n = 0 ⟹ n → ∞ lim n x n 1 = 1 − lim n → ∞ x n 1 = 1 ⟹ n → ∞ lim n x n = 1
T5
求极限 lim n → ∞ ( n ! e − [ n ! e ] ) \lim_{n \to \infty} (n!e - [n!e]) lim n → ∞ ( n ! e − [ n ! e ])
a n = n ! e − [ n ! e ] = lim n → ∞ { n ! e } = lim n → ∞ { ∑ i = 0 ∞ n ! i ! } = lim n → ∞ { ∑ i = n + 1 ∞ n ! i ! } = lim n → ∞ { 1 n + 1 + 1 ( n + 1 ) ( n + 2 ) + 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) + … } < lim n → ∞ { ∑ i = 1 ∞ ( n + 1 ) − i } = lim n → ∞ { 1 n } = 0 \begin{gathered}
a_n=n!e-[n!e] \\
=\lim_{n \to \infty} \{ n!e \} \\
=\lim_{n \to \infty} {\left\{ \sum _{i = 0} ^{\infty} \dfrac{n!}{i!} \right\}} \\
=\lim_{n \to \infty} {\left\{ \sum _{i = n+1} ^{\infty} \dfrac{n!}{i!} \right\}} \\
=\lim_{n \to \infty} {\left\{ \dfrac{1}{n+1} +\dfrac{1}{(n+1)(n+2)} +\dfrac{1}{(n+1)(n+2)(n+3)} +\ldots \right\}} \\
<\lim_{n \to \infty} {\left\{ \sum _{i = 1} ^{\infty} (n+1)^{-i} \right\}} \\
=\lim_{n \to \infty} {\left\{ \dfrac{1}{n} \right\}} \\
=0
\end{gathered} a n = n ! e − [ n ! e ] = n → ∞ lim { n ! e } = n → ∞ lim { i = 0 ∑ ∞ i ! n ! } = n → ∞ lim { i = n + 1 ∑ ∞ i ! n ! } = n → ∞ lim { n + 1 1 + ( n + 1 ) ( n + 2 ) 1 + ( n + 1 ) ( n + 2 ) ( n + 3 ) 1 + … } < n → ∞ lim { i = 1 ∑ ∞ ( n + 1 ) − i } = n → ∞ lim { n 1 } = 0
T6
求极限 lim n → ∞ a n ( 1 n + 1 + 1 n + 2 + ⋯ + 1 n + n ) \lim_{n \to \infty} a_n(\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + n}) lim n → ∞ a n ( n + 1 1 + n + 2 1 + ⋯ + n + n 1 )
x > ln ( x ) + 1 ⟹ 1 x ∈ ( ln ( x x − 1 ) , ln ( x + 1 x ) ) ⟹ a n ∈ ( ∑ i = n + 1 2 n ln ( i i − 1 ) , ∑ i = n + 1 2 n ln ( i + 1 i ) ) = ( ln ( 2 n n ) , ln ( 2 n + 1 n + 1 ) ) ⟹ Squeeze Theorem lim n → ∞ a n = ln ( 2 ) \begin{gathered}
x>\ln(x)+1 \\
\implies \dfrac{1}{x}\in(\ln(\dfrac{x}{x-1}),\ln(\dfrac{x+1}{x} ) ) \\
\implies a_n\in (\sum _{i = n+1} ^{2n} \ln(\dfrac{i}{i-1}),\sum _{i = n+1} ^{2n} \ln(\dfrac{i+1}{i} ) ) \\
=(\ln(\dfrac{2n}{n}),\ln(\dfrac{2n+1}{n+1} ) ) \\
\stackrel{\text{Squeeze Theorem}}{\Longrightarrow}
\lim_{n \to \infty} a_n=\ln(2)
\end{gathered} x > ln ( x ) + 1 ⟹ x 1 ∈ ( ln ( x − 1 x ) , ln ( x x + 1 )) ⟹ a n ∈ ( i = n + 1 ∑ 2 n ln ( i − 1 i ) , i = n + 1 ∑ 2 n ln ( i i + 1 )) = ( ln ( n 2 n ) , ln ( n + 1 2 n + 1 )) ⟹ Squeeze Theorem n → ∞ lim a n = ln ( 2 )
T7
设数列 x n = ( 1 + 1 2 ) ( 1 + 1 2 2 ) … ( 1 + 1 2 n ) x_n = (1 + \frac{1}{2})(1 + \frac{1}{2^2})\dots(1 + \frac{1}{2^n}) x n = ( 1 + 2 1 ) ( 1 + 2 2 1 ) … ( 1 + 2 n 1 ) lim n → ∞ x n \lim_{n \to \infty} x_n lim n → ∞ x n
x n is obviously incresing ln ( x n ) = ∑ i = 1 n ln ( 1 + 1 2 i ) < ∑ i = 1 n 1 2 i < 1 ⟹ x n < e ⟹ lim n → ∞ x n exists \begin{gathered}
x_n \text{ is obviously incresing} \\
\ln(x_n)=\sum_{i=1}^n \ln(1+\dfrac{1}{2^i} ) \\
<\sum _{i = 1} ^{n} \dfrac{1}{2^i} \\
<1 \\
\implies x_n<e
\implies \lim_{n \to \infty} x_n \text{exists}
\end{gathered} x n is obviously incresing ln ( x n ) = i = 1 ∑ n ln ( 1 + 2 i 1 ) < i = 1 ∑ n 2 i 1 < 1 ⟹ x n < e ⟹ n → ∞ lim x n exists
Class 5
T1
用柯西收敛准则证明数列收敛.
a n = ∑ i = 2 n sin ( i x ) i ( i + sin ( i x ) ) , x ∈ R \begin{gathered}
a_n=\sum _{i = 2} ^{n} \dfrac{\sin(ix)}{i(i+\sin(ix))} ,x\in R
\end{gathered} a n = i = 2 ∑ n i ( i + sin ( i x )) sin ( i x ) , x ∈ R
∣ a n + m − a n ∣ = ∣ ∑ i = n + 1 n + m sin ( i x ) i ( i + sin ( i x ) ) ∣ < ∑ i = n + 1 n + m ∣ sin ( i x ) i ( i + sin ( i x ) ) ∣ < ∑ i = n + 1 n + m 1 i ( i − 1 ) = ∑ i = n + 1 n + m 1 i − 1 − 1 i = 1 n − 1 n + m < 1 n ⟹ ∀ ϵ , N : = 1 ϵ + 100 ⟹ ∀ i , j > N , ∣ a i − a j ∣ < ϵ ⟹ Cauchy Convergence Theorem Q.E.D \begin{gathered}
\vert a_{n+m}-a_n \vert \\
={\left \vert \sum _{i = n+1} ^{n+m} \dfrac{\sin(ix)}{i(i+\sin(ix))} \right \vert} \\
< \sum _{i = n+1} ^{n+m} {\left \vert \dfrac{\sin(ix)}{i(i+\sin(ix))} \right \vert} \\
< \sum _{i = n+1} ^{n+m} \dfrac{1}{i(i-1)} \\
= \sum _{i = n+1} ^{n+m} \dfrac{1}{i-1} -\dfrac{1}{i} \\
=\dfrac{1}{n} -\dfrac{1}{n+m} \\
<\dfrac{1}{n} \\
\implies \forall \epsilon,N:=\dfrac{1}{\epsilon} +100 \\
\implies \forall i,j>N,\vert a_i-a_j \vert < \epsilon
\\
\stackrel{\text{Cauchy Convergence Theorem}}{\Longrightarrow} \\
\\
\text{Q.E.D}
\end{gathered} ∣ a n + m − a n ∣ = i = n + 1 ∑ n + m i ( i + sin ( i x )) sin ( i x ) < i = n + 1 ∑ n + m i ( i + sin ( i x )) sin ( i x ) < i = n + 1 ∑ n + m i ( i − 1 ) 1 = i = n + 1 ∑ n + m i − 1 1 − i 1 = n 1 − n + m 1 < n 1 ⟹ ∀ ϵ , N := ϵ 1 + 100 ⟹ ∀ i , j > N , ∣ a i − a j ∣ < ϵ ⟹ Cauchy Convergence Theorem Q.E.D
T2
b n = ∑ i = 1 n − 1 ∣ a i + 1 − a i ∣ is bounded ⟹ a n is convergent \begin{gathered}
b_n=\sum _{i = 1} ^{n-1} \vert a_{i+1}-a_i \vert \text{ is bounded}
\implies a_n \text{ is convergent}
\end{gathered} b n = i = 1 ∑ n − 1 ∣ a i + 1 − a i ∣ is bounded ⟹ a n is convergent
{ b n is increasing b n is bounded ⟹ lim n → ∞ b n = B a n + m − a n = ∑ i = n + 1 n + m a i − a i − 1 ≤ ∑ i = n + 1 n + m ∣ a i − a i − 1 ∣ = b n + m − b n ⟹ ∀ ϵ 1 , ∃ N s . t . n > N ⟹ ∣ b n − B ∣ < ϵ 1 ⟹ b n + m − b n < ∣ b n − B ∣ + ∣ B − b n + m ∣ = 2 ϵ 1 ϵ 1 : = ϵ 2 ⟹ ∀ x , y > N , ∣ a x − a y ∣ < ϵ ⟹ Cauchy Convergence Theorem Q.E.D \begin{gathered}
\begin{cases}
b_n \text{ is increasing} \\
b_n \text{ is bounded}
\end{cases}\implies \lim_{n \to \infty} b_n =B \\
a_{n+m}-a_n=\sum _{i = n+1} ^{n+m} a_{i}-a_{i-1} \\
\le \sum _{i = n+1} ^{n+m} \vert a_i-a_{i-1} \vert \\
=b_{n+m} - b_n
\\
\implies
\forall \epsilon_1, \exists N \\ s.t.\\
n>N \implies \vert b_n-B \vert < \epsilon_1 \\
\implies b_{n+m}-b_n< \vert b_n-B \vert + \vert B-b_{n+m} \vert =2\epsilon_1 \\
\epsilon_1:=\dfrac{\epsilon}{2} \implies \forall x,y>N,\vert a_x-a_y \vert < \epsilon \\
\stackrel{\text{Cauchy Convergence Theorem}}{\Longrightarrow} \\
\\
\text{Q.E.D}
\end{gathered} { b n is increasing b n is bounded ⟹ n → ∞ lim b n = B a n + m − a n = i = n + 1 ∑ n + m a i − a i − 1 ≤ i = n + 1 ∑ n + m ∣ a i − a i − 1 ∣ = b n + m − b n ⟹ ∀ ϵ 1 , ∃ N s . t . n > N ⟹ ∣ b n − B ∣ < ϵ 1 ⟹ b n + m − b n < ∣ b n − B ∣ + ∣ B − b n + m ∣ = 2 ϵ 1 ϵ 1 := 2 ϵ ⟹ ∀ x , y > N , ∣ a x − a y ∣ < ϵ ⟹ Cauchy Convergence Theorem Q.E.D
T3
∀ ϵ , ∃ N 1 = N ( ϵ ) s . t . i , j > N 1 ⟹ ∣ x i − x j ∣ < ϵ ⟺ x n is convergent \begin{gathered}
\forall \epsilon , \exists N_1=N(\epsilon) \\ s.t.\\
i,j>N_1 \implies \vert x_i-x_j \vert < \epsilon \\
\iff \\
x_n \text{ is convergent}
\end{gathered} ∀ ϵ , ∃ N 1 = N ( ϵ ) s . t . i , j > N 1 ⟹ ∣ x i − x j ∣ < ϵ ⟺ x n is convergent
逆向三角不等式显然.考虑正向
ϵ 1 : = 1 ∴ i > N ⟹ x i ∈ [ x N − ϵ 1 , x N + ϵ 1 ] a 1 : = x N − ϵ 1 , b 1 : = x N + ϵ 1 \begin{gathered}
\epsilon_1:=1 \therefore i>N \implies x_i \in [x_N-\epsilon_1,x_N+\epsilon_1] \\
a_1:=x_N-\epsilon_1,b_1:=x_N+\epsilon_1 \\
\end{gathered} ϵ 1 := 1 ∴ i > N ⟹ x i ∈ [ x N − ϵ 1 , x N + ϵ 1 ] a 1 := x N − ϵ 1 , b 1 := x N + ϵ 1 对 [ a i , b i ] [a_i,b_i] [ a i , b i ] ϵ i = ϵ i − 1 2 \epsilon_i=\dfrac{\epsilon_{i-1}}{2} ϵ i = 2 ϵ i − 1
N i = max ( N i − 1 , N ( ϵ i ) ) s . t . ∀ j > N i , x j ∈ [ x N i − ϵ i , x N i + ϵ i ] N_i=\max(N_{i-1},N(\epsilon_i)) \ s.t.\ \forall j>N_i,x_j\in [x_{N_i}-\epsilon_i,x_{N_i}+\epsilon_i] N i = max ( N i − 1 , N ( ϵ i )) s . t . ∀ j > N i , x j ∈ [ x N i − ϵ i , x N i + ϵ i ] 于是令a i = max ( a i − 1 , x N i − ϵ ) , b i = min ( b i − 1 , x N i + ϵ i ) a_i=\max(a_{i-1},x_{N_i}-\epsilon),b_i=\min(b_{i-1},x_{N_i}+\epsilon_i) a i = max ( a i − 1 , x N i − ϵ ) , b i = min ( b i − 1 , x N i + ϵ i ) ∣ b i − a i ∣ < 1 2 i − 1 \vert b_i-a_i \vert < \dfrac{1}{2^{i-1}} ∣ b i − a i ∣ < 2 i − 1 1
{ a i ≤ a i + 1 b i ≥ b i + 1 lim n → ∞ b n − a n = 0 ⟹ ∃ ! ξ ∈ [ a n , b n ] s . t . ∀ ϵ , N = min { i ∣ b i − a i < ϵ } ⟹ n > N ⟹ ∣ ξ − x n ∣ < ϵ Q.E.D \begin{gathered}
\begin{cases}
a_i\le a_{i+1} \\
b_i\ge b_{i+1} \\
\lim_{n \to \infty} b_n-a_n = 0
\end{cases}
\\
\implies \exists! \xi \in [a_n,b_n] \\ s.t.\\
\forall \epsilon,N=\min \{ i \vert b_i-a_i<\epsilon \}
\implies
n>N \implies \vert \xi-x_n\vert<\epsilon
\\
\text{Q.E.D}
\end{gathered} ⎩ ⎨ ⎧ a i ≤ a i + 1 b i ≥ b i + 1 lim n → ∞ b n − a n = 0 ⟹ ∃ ! ξ ∈ [ a n , b n ] s . t . ∀ ϵ , N = min { i ∣ b i − a i < ϵ } ⟹ n > N ⟹ ∣ ξ − x n ∣ < ϵ Q.E.D
T4
a 0 = 3 , a n = a n − 1 2 − 2 ⟹ { lim n → ∞ a n = + ∞ A n : = a n ∏ i = 0 n − 1 a i ⟹ lim n → ∞ A n = 5 \begin{gathered}
a_0=3,a_n=a_{n-1}^2-2 \\
\implies \begin{cases}
\lim_{n \to \infty} a_n=+\infty \\
A_n:=\dfrac{a_n}{\prod_{i=0}^{n-1}a_i} \implies \lim_{n \to \infty} A_n = \sqrt{5}
\end{cases}
\end{gathered} a 0 = 3 , a n = a n − 1 2 − 2 ⟹ ⎩ ⎨ ⎧ lim n → ∞ a n = + ∞ A n := ∏ i = 0 n − 1 a i a n ⟹ lim n → ∞ A n = 5
(1)
{ a 0 = 3 , a 1 = 7 , a 2 = 47 n > 2 , a n − 1 > 2 n ⟹ a n = a n − 1 2 − 2 > 2 2 n − 2 > 2 n + 1 ⟹ induction a n > 2 n + 1 ∴ lim n → ∞ a n ≥ lim n → ∞ 2 n + 1 = ∞ \begin{gathered}
\begin{cases}
a_0=3,a_1=7,a_2=47 \\
n>2,a_{n-1}>2^{n} \implies a_n = a_{n-1}^2-2>2^{2n}-2>2^{n+1}
\end{cases} \\
\stackrel{\text{induction}}{\Longrightarrow} a_n>2^{n+1} \\
\therefore
\lim_{n \to \infty} a_n\ge \lim_{n \to \infty} 2^{n+1} = \infty
\end{gathered} { a 0 = 3 , a 1 = 7 , a 2 = 47 n > 2 , a n − 1 > 2 n ⟹ a n = a n − 1 2 − 2 > 2 2 n − 2 > 2 n + 1 ⟹ induction a n > 2 n + 1 ∴ n → ∞ lim a n ≥ n → ∞ lim 2 n + 1 = ∞ (2)
a n = a n − 1 2 − 2 ⟹ ( a n − 2 ) = ( a n − 1 − 2 ) ( a n − 1 + 2 ) ⟹ ∏ i = 1 n ( a i + 2 ) = a n + 1 − 2 a 1 − 2 = a n + 1 − 2 5 ⟹ A n 2 = a n 2 ∏ i = 0 n − 1 a i 2 = a n + 1 + 2 ∏ i = 1 n ( a i + 2 ) = 5 a n + 1 + 2 a n + 1 − 2 ∴ lim n → ∞ A n = 5 a n + 1 + 2 a n + 1 − 2 = 5 \begin{gathered}
a_n=a_{n-1}^2-2 \\
\implies (a_n-2)=(a_{n-1}-2)(a_{n-1}+2) \\
\implies \prod _{i = 1} ^{n} (a_i+2 ) = \dfrac{a_{n+1}-2}{a_1-2} =\dfrac{a_{n+1}-2}{5} \\
\implies \\
A_n^2=\dfrac{a_{n}^2}{\prod _{i = 0} ^{n-1} a_i^2} \\
=\dfrac{a_{n+1}+2}{\prod _{i = 1} ^{n} (a_i+2)} \\
=5\dfrac{a_{n+1}+2}{a_{n+1}-2} \\
\therefore \lim_{n \to \infty} A_n=\sqrt{5\dfrac{a_{n+1}+2}{a_{n+1}-2} }=\sqrt 5
\end{gathered} a n = a n − 1 2 − 2 ⟹ ( a n − 2 ) = ( a n − 1 − 2 ) ( a n − 1 + 2 ) ⟹ i = 1 ∏ n ( a i + 2 ) = a 1 − 2 a n + 1 − 2 = 5 a n + 1 − 2 ⟹ A n 2 = ∏ i = 0 n − 1 a i 2 a n 2 = ∏ i = 1 n ( a i + 2 ) a n + 1 + 2 = 5 a n + 1 − 2 a n + 1 + 2 ∴ n → ∞ lim A n = 5 a n + 1 − 2 a n + 1 + 2 = 5
Class 6
T1
A , B is upper bounder , S ⊂ { x + y ∣ x ∈ A , y ∈ B } ⟹ sup S ≤ sup A + sup B \begin{gathered}
A,B \text{is upper bounder},S\subset\{ x+y\vert x\in A,y\in B \} \\
\implies \sup S\le \sup A+\sup B
\end{gathered} A , B is upper bounder , S ⊂ { x + y ∣ x ∈ A , y ∈ B } ⟹ sup S ≤ sup A + sup B
显然S S S sup S \sup S sup S
若sup S > sup A + sup B \sup S>\sup A+\sup B sup S > sup A + sup B M = sup A + sup B M=\sup A+\sup B M = sup A + sup B
则∀ x ∈ A , y ∈ B , x ≤ sup A , y ≤ sup B ⟹ x + y ≤ M \forall x\in A,y\in B,x\le \sup A,y\le \sup B \implies x+y \le M ∀ x ∈ A , y ∈ B , x ≤ sup A , y ≤ sup B ⟹ x + y ≤ M
于是M M M sup S \sup S sup S
故sup S ≤ sup A + sup B \sup S\le \sup A+\sup B sup S ≤ sup A + sup B
eg. 当A , B , S A,B,S A , B , S { x + y ∣ x ∈ A , y ∈ B } \{ x+y\vert x\in A,y\in B \} { x + y ∣ x ∈ A , y ∈ B }
T2
{ A , B aren’t empty , α ≥ 0 , C = A + α B = { x ∣ x = a + α b , a ∈ A , b ∈ B } ⟹ sup ( A + α B ) = M = sup A + α sup B \begin{gathered}
\begin{cases}
A,B \text{ aren't empty},\alpha\ge 0, \\
C=A+\alpha B =\{ x\vert x=a+\alpha b,a\in A,b\in B \}
\end{cases} \\
\implies \sup(A+\alpha B)=M=\sup A+\alpha \sup B
\end{gathered} { A , B aren’t empty , α ≥ 0 , C = A + α B = { x ∣ x = a + α b , a ∈ A , b ∈ B } ⟹ sup ( A + α B ) = M = sup A + α sup B
∀ c ∈ C = a + α b , a ≤ sup A , b ≤ sup B ⟹ a + α b ≤ sup A + α sup B = M ∀ M ′ < M , ϵ = M − M ′ let X = sup A − ϵ 3 , Y = sup B + ϵ 3 α According to the definition of supremum, ∃ a > X ∈ A , y > Y ∈ B ⟹ ∃ c = a + b ∈ C , c > X + Y > M ′ ⟹ sup C = sup A + α sup B \begin{gathered}
\forall c\in C=a+\alpha b, \\
a\le \sup A,b\le \sup B \implies a+\alpha b\le \sup A+\alpha \sup B=M \\
\forall M'<M,\epsilon=M-M' \\
\text{let }X=\sup A-\dfrac{\epsilon}{3} ,Y=\sup B+\dfrac{\epsilon}{3\alpha} \\
\text{According to the definition of supremum, } \exists a>X\in A,y>Y\in B \\
\implies \exists c=a+b\in C,c>X+Y>M'
\implies \sup C=\sup A+\alpha \sup B
\end{gathered} ∀ c ∈ C = a + α b , a ≤ sup A , b ≤ sup B ⟹ a + α b ≤ sup A + α sup B = M ∀ M ′ < M , ϵ = M − M ′ let X = sup A − 3 ϵ , Y = sup B + 3 α ϵ According to the definition of supremum, ∃ a > X ∈ A , y > Y ∈ B ⟹ ∃ c = a + b ∈ C , c > X + Y > M ′ ⟹ sup C = sup A + α sup B
T3
A ⊂ B ⟹ sup A ≤ sup B , inf A ≥ inf B \begin{gathered}
A\subset B \\
\implies \sup A\le \sup B,\inf A \ge \inf B
\end{gathered} A ⊂ B ⟹ sup A ≤ sup B , inf A ≥ inf B
if sup A > sup B let M = sup B ⟹ Def of supremum ∃ a ∈ A , a > M = sup B { a ∈ A ⟹ A ⊂ B a ∈ B a > sup B ⟹ False ∴ sup A ≤ sup B \begin{gathered}
\text{if }\sup A>\sup B \\
\text{let }M=\sup B
\stackrel{\text{Def of supremum}}{\Longrightarrow} \exists a\in A,a>M=\sup B \\
\begin{cases}
a\in A \stackrel{A\subset B}{\Longrightarrow} a\in B \\
a>\sup B
\end{cases}
\implies \text{False} \\
\therefore \sup A\le \sup B
\end{gathered} if sup A > sup B let M = sup B ⟹ Def of supremum ∃ a ∈ A , a > M = sup B { a ∈ A ⟹ A ⊂ B a ∈ B a > sup B ⟹ False ∴ sup A ≤ sup B 取 C = − A , D = − B , C ⊂ D , sup C ≤ sup D ⟹ inf A ≥ inf B C=-A,D=-B,C\subset D,\sup C\le \sup D \implies \inf A\ge \inf B C = − A , D = − B , C ⊂ D , sup C ≤ sup D ⟹ inf A ≥ inf B
T4
∀ x ∈ A , y ∈ B , x ≤ y ⟹ sup A ≤ inf B \begin{gathered}
\forall x\in A,y\in B,x\le y \\
\implies \sup A\le \inf B
\end{gathered} ∀ x ∈ A , y ∈ B , x ≤ y ⟹ sup A ≤ inf B
假设sup A > inf B \sup A>\inf B sup A > inf B M ∈ ( sup A , inf B ) M\in (\sup A,\inf B) M ∈ ( sup A , inf B ) sup A \sup A sup A ∃ a ∈ A , a > M \exists a\in A,a>M ∃ a ∈ A , a > M inf B \inf B inf B ∃ b ∈ B , b < M \exists b\in B,b<M ∃ b ∈ B , b < M a > M > b a>M>b a > M > b ∀ x ∈ A , y ∈ B , x ≤ y \forall x\in A,y\in B,x\le y ∀ x ∈ A , y ∈ B , x ≤ y
故sup A ≤ inf B \sup A\le \inf B sup A ≤ inf B
Class 7
T1
lim x → x 0 x = x 0 \begin{gathered}
\lim_{x \to x_0} \sqrt{ x } =\sqrt x_0
\end{gathered} x → x 0 lim x = x 0
if x 0 ≠ 0 x_0\ne 0 x 0 = 0
∣ x − x 0 ∣ = ∣ x − x 0 x + x 0 ∣ ≤ δ x 0 ∴ δ : = x 0 ϵ 2 ⟹ ∀ ϵ , x ∈ N ( x 0 , δ ) , ∣ x − x 0 ∣ < ϵ \begin{gathered}
\vert \sqrt x-\sqrt {x_0} \vert \\
={\left \vert \dfrac{x-x_0}{\sqrt x+\sqrt {x_0}} \right \vert} \\
\le \dfrac{\delta}{\sqrt{x_0} } \\
\therefore \delta:=\dfrac{\sqrt{x_0}\epsilon}{2} \implies \\
\forall \epsilon,x\in N(x_0,\delta),\vert \sqrt x-\sqrt {x_0} \vert <\epsilon
\end{gathered} ∣ x − x 0 ∣ = x + x 0 x − x 0 ≤ x 0 δ ∴ δ := 2 x 0 ϵ ⟹ ∀ ϵ , x ∈ N ( x 0 , δ ) , ∣ x − x 0 ∣ < ϵ if x 0 = 0 , x 0 = 0 x_0=0,\sqrt x_0=0 x 0 = 0 , x 0 = 0
δ : = ϵ 2 4 ⟹ ∀ ϵ , x ∈ N ( x 0 , δ ) , ∣ x − x 0 ∣ = ϵ 4 < ϵ \begin{gathered}
\delta:=\dfrac{\epsilon^2}{4} \implies \forall \epsilon,x\in N(x_0,\delta), \\
\vert \sqrt x-\sqrt {x_0} \vert =\dfrac{\epsilon}{4} <\epsilon
\end{gathered} δ := 4 ϵ 2 ⟹ ∀ ϵ , x ∈ N ( x 0 , δ ) , ∣ x − x 0 ∣ = 4 ϵ < ϵ Q.E.D \begin{gathered}
\text{Q.E.D}
\end{gathered} Q.E.D
T2
lim x → + ∞ ( x + 1 − x − 1 ) = 0 \begin{gathered}
\lim_{x \to +\infty} (\sqrt{ x+1 } -\sqrt{ x-1 } ) =0
\end{gathered} x → + ∞ lim ( x + 1 − x − 1 ) = 0
x + 1 − x − 1 = 2 x + 1 + x − 1 < 1 x ∴ ∀ ϵ ∈ ( 0 , 1 ) , δ : = 4 ϵ 2 , x > δ ⟹ x + 1 − x − 1 < 1 x = ϵ 2 < ϵ Q.E.D \begin{gathered}
\sqrt{ x+1 } -\sqrt{ x-1 } \\
=\dfrac{2}{\sqrt{x+1}+\sqrt{x-1}} \\
<\dfrac{1}{\sqrt{x}} \\
\therefore \forall \epsilon \in (0,1),
\delta:=\dfrac{4}{\epsilon^2}, \\
x>\delta \implies \sqrt{x+1}-\sqrt{ x-1 } <\dfrac{1}{\sqrt{ x } } =\dfrac{\epsilon}{2} < \epsilon
\\
\text{Q.E.D}
\end{gathered} x + 1 − x − 1 = x + 1 + x − 1 2 < x 1 ∴ ∀ ϵ ∈ ( 0 , 1 ) , δ := ϵ 2 4 , x > δ ⟹ x + 1 − x − 1 < x 1 = 2 ϵ < ϵ Q.E.D