显然有
σ 2 = ∑ k = 0 25 C 25 k ( k − 25 2 ) 2 ∑ k = 0 25 C 25 k = 2 − 25 ∑ k = 0 25 C 25 k ( k − 25 2 ) 2 \sigma^2 = \dfrac{\displaystyle\sum_{k=0}^{25} \mathrm{C}_{25}^k \left( k - \dfrac{25}{2} \right)^2}{\displaystyle\sum_{k=0}^{25} \mathrm{C}_{25}^k} = 2^{-25} \sum_{k=0}^{25} \mathrm{C}_{25}^k \left( k - \dfrac{25}{2} \right)^2 σ 2 = k = 0 ∑ 25 C 25 k k = 0 ∑ 25 C 25 k ( k − 2 25 ) 2 = 2 − 25 k = 0 ∑ 25 C 25 k ( k − 2 25 ) 2
我们知道
k C n k = k n ! k ! ( n − k ) ! = n ! ( k − 1 ) ! ( n − k ) ! = n ( n − 1 ) ! ( k − 1 ) ! ( n − k ) ! = n C n − 1 k − 1 k \mathrm{C}_n^k = k \dfrac{n!}{k! (n-k)!} = \dfrac{n!}{(k-1)!(n-k)!} = n \dfrac{(n-1)!}{(k-1)!(n-k)!} = n \mathrm{C}_{n-1}^{k-1} k C n k = k k ! ( n − k )! n ! = ( k − 1 )! ( n − k )! n ! = n ( k − 1 )! ( n − k )! ( n − 1 )! = n C n − 1 k − 1
更进一步地，
k ( k − 1 ) C n k = n ( k − 1 ) C n − 1 k − 1 = n ( n − 1 ) C n − 2 k − 2 k(k-1) \mathrm{C}_n^k = n(k-1) \mathrm{C}_{n-1}^{k-1} = n(n-1) \mathrm{C}_{n-2}^{k-2} k ( k − 1 ) C n k = n ( k − 1 ) C n − 1 k − 1 = n ( n − 1 ) C n − 2 k − 2
于是
σ 2 = 2 − 25 ∑ k = 0 25 C 25 k ( k − 25 2 ) 2 = 2 − 25 ∑ k = 0 25 C 25 k ( k ( k − 1 ) − 24 k + 625 4 ) = 2 − 25 [ ( ∑ k = 0 25 k ( k − 1 ) C 25 k ) − 24 ( ∑ k = 0 25 k C 25 k ) + 625 4 ( ∑ k = 0 25 C 25 k ) ] = 2 − 25 [ ( ∑ k = 0 23 600 C 23 k ) − 24 ( ∑ k = 0 24 25 C 24 k ) + 625 4 ( ∑ k = 0 25 C 25 k ) ] = 2 − 25 ( 600 × 2 23 − 600 × 2 24 + 625 4 × 2 25 ) = 25 4 \begin{align*} \sigma^2 &= 2^{-25} \sum_{k=0}^{25} \mathrm{C}_{25}^k \left( k - \dfrac{25}{2} \right)^2 \\ &= 2^{-25} \sum_{k=0}^{25} \mathrm{C}_{25}^k \left( k(k-1) - 24k + \dfrac{625}{4} \right) \\ &= 2^{-25} \left[ \left( \sum_{k=0}^{25} k(k-1) \mathrm{C}_{25}^k \right) - 24 \left( \sum_{k=0}^{25} k \mathrm{C}_{25}^k \right) + \dfrac{625}{4} \left( \sum_{k=0}^{25} \mathrm{C}_{25}^k \right) \right] \\ &= 2^{-25} \left[ \left( \sum_{k=0}^{23} 600 \mathrm{C}_{23}^k \right) - 24 \left( \sum_{k=0}^{24} 25 \mathrm{C}_{24}^k \right) + \dfrac{625}{4} \left( \sum_{k=0}^{25} \mathrm{C}_{25}^k \right) \right] \\ &= 2^{-25} \left( 600 \times 2^{23} - 600 \times 2^{24} + \dfrac{625}{4} \times 2^{25} \right) = \dfrac{25}{4} \end{align*} σ 2 = 2 − 25 k = 0 ∑ 25 C 25 k ( k − 2 25 ) 2 = 2 − 25 k = 0 ∑ 25 C 25 k ( k ( k − 1 ) − 24 k + 4 625 ) = 2 − 25 [ ( k = 0 ∑ 25 k ( k − 1 ) C 25 k ) − 24 ( k = 0 ∑ 25 k C 25 k ) + 4 625 ( k = 0 ∑ 25 C 25 k ) ] = 2 − 25 [ ( k = 0 ∑ 23 600 C 23 k ) − 24 ( k = 0 ∑ 24 25 C 24 k ) + 4 625 ( k = 0 ∑ 25 C 25 k ) ] = 2 − 25 ( 600 × 2 23 − 600 × 2 24 + 4 625 × 2 25 ) = 4 25
于是 σ = 2.5 \sigma = 2.5 σ = 2.5
直接归纳发现不等式方向错了，我们证更强的 a n ≤ 2 2 n − 1 − 1 a_n \le 2^{2^{n-1}}-1 a n ≤ 2 2 n − 1 − 1
归纳，奠基 a 1 ≤ 2 2 0 − 1 a_1 \le 2^{2^0}-1 a 1 ≤ 2 2 0 − 1
a n + 1 = a n 2 + 1 ≤ 2 2 n − 2 2 n − 1 + 1 + 2 ≤ 2 2 n − 1 a_{n+1} = a_n^2+1 \le 2^{2^{n}} - 2^{2^{n-1}+1}+2 \le 2^{2^{n}} - 1 a n + 1 = a n 2 + 1 ≤ 2 2 n − 2 2 n − 1 + 1 + 2 ≤ 2 2 n − 1
设 b n = ( 3 / 2 ) 2 n b_n = (3/2)^{2^n} b n = ( 3/2 ) 2 n a n > b n a_n > b_n a n > b n a 4 > b 4 a_4>b_4 a 4 > b 4
下面严谨证明： 设 r n = a n b n r_n=\dfrac{a_n}{b_n} r n = b n a n r n + 1 = a n 2 + 1 b n 2 > ( a n b n ) 2 = r n 2 r_{n+1} = \dfrac{a_n^2+1}{b_n^2} > \left( \dfrac{a_n}{b_n} \right)^2 = r_n^2 r n + 1 = b n 2 a n 2 + 1 > ( b n a n ) 2 = r n 2
由 r 4 > 1 r_4 > 1 r 4 > 1 n ≥ 4 n \ge 4 n ≥ 4 r n > 1 r_n > 1 r n > 1 r n + 1 > r n 2 > r n ≥ r 4 r_{n+1} > r_n^2 > r_n \ge r_4 r n + 1 > r n 2 > r n ≥ r 4
于是
∣ a n − b n ∣ = b n ( r n − 1 ) > ( 3 / 2 ) 2 n r 4 |a_n - b_n| = b_n(r_n-1) > (3/2)^{2^n} r_4 ∣ a n − b n ∣ = b n ( r n − 1 ) > ( 3/2 ) 2 n r 4
只需讨论 d > r 4 d>r_4 d > r 4 ( 3 / 2 ) 2 n r 4 < d ⟺ n < log 3 / 2 log 2 d r 4 (3/2)^{2^n} r_4 < d \iff n < \log_{3/2} \log_{2} \dfrac{d}{r_4} ( 3/2 ) 2 n r 4 < d ⟺ n < log 3/2 log 2 r 4 d n n n
由 B 得 { x P n } \{x_{P_n}\} { x P n } { x Q n } \{x_{Q_n}\} { x Q n } m 1 m_1 m 1 x P m 1 > 0 x_{P_{m_1}} > 0 x P m 1 > 0 ∣ y P n + m 1 ∣ \left|y_{P_{n+m_1}}\right| y P n + m 1 m 2 m_2 m 2 x Q m 2 < 0 x_{Q_{m_2}} < 0 x Q m 2 < 0 ∣ y Q n + m 2 ∣ \left|y_{Q_{n+m_2}}\right| y Q n + m 2
于是对于 m > m 1 , m 2 m > m_1, m_2 m > m 1 , m 2 ∣ k P n + m Q n + m ∣ = ∣ y P n + m ∣ + ∣ y Q n + m ∣ x P n + m − x Q n + m |k_{P_{n+m} Q_{n+m}}| = \dfrac{|y_{P_{n+m}}| + |y_{Q_{n+m}}|}{x_{P_{n+m}} - x_{Q_{n+m}}} ∣ k P n + m Q n + m ∣ = x P n + m − x Q n + m ∣ y P n + m ∣ + ∣ y Q n + m ∣
为了好看，我们记 a ( n ) = a n a(n) = a_n a ( n ) = a n
a ( a ( ⋯ a ( n + T m ) ⋯ ) ) ⏟ m 个 = a ( a ( ⋯ a ( a ( n ) + d T m − 1 ) ⋯ ) ) ⏟ m 个 = ⋯ = a ( a ( ⋯ a ( n ) ⋯ ) ) ⏟ m 个 + d m \underbrace{a\Bigl(a\bigl(\cdots a(n + T^m) \cdots\bigr)\Bigr)}_{m \text{ 个}} = \underbrace{a\biggl(a\Bigl(\cdots a\bigl(a(n) + dT^{m-1}\bigr) \cdots\Bigr)\biggr)}_{m \text{ 个}} = \cdots = \underbrace{a\Bigl(a\bigl(\cdots a(n) \cdots\bigr)\Bigr)}_{m \text{ 个}} + d^m m 个 a ( a ( ⋯ a ( n + T m ) ⋯ ) ) = m 个 a ( a ( ⋯ a ( a ( n ) + d T m − 1 ) ⋯ ) ) = ⋯ = m 个 a ( a ( ⋯ a ( n ) ⋯ ) ) + d m
先证明题目条件等价于“C C C M M M 1 1 1
必要性：Γ \Gamma Γ C C C C C C P , Q P, Q P , Q M , P , Q M,P,Q M , P , Q M P → ⋅ M Q → = 1 \overrightarrow{MP} \cdot \overrightarrow{MQ} = 1 MP ⋅ MQ = 1 ∣ M P ∣ , ∣ M Q ∣ |MP|, |MQ| ∣ MP ∣ , ∣ MQ ∣ 1 1 1 C C C M M M 1 1 1 C C C P P P ∣ M P ∣ = 1 |MP| = 1 ∣ MP ∣ = 1 P P P C C C Γ \Gamma Γ
设直线 l : y = k x l: y=kx l : y = k x M ∈ l M \in l M ∈ l C C C M M M 1 1 1 C C C l l l 1 1 1
因为 ( t + 1 t ) 2 − ( t − 1 t ) 2 = 4 \left(t+\dfrac{1}{t}\right)^2 - \left(t-\dfrac{1}{t}\right)^2 = 4 ( t + t 1 ) 2 − ( t − t 1 ) 2 = 4 P ( t + 1 t , t − 1 t ) ( t ≠ 0 ) P\left(t+\dfrac{1}{t}, t-\dfrac{1}{t}\right) \ \ (t \ne 0) P ( t + t 1 , t − t 1 ) ( t = 0 ) y = t − 1 t y=t - \dfrac{1}{t} y = t − t 1 P P P C C C
根据
1 1 + k 2 [ t − 1 t − k ( t + 1 t ) ] 2 = 1 1 + k 2 [ ( 1 − k ) t − ( 1 + k ) 1 t ] 2 = 1 1 + k 2 [ ( 1 − k ) 2 t 2 + ( 1 + k ) 2 ( 1 t ) 2 − 2 ( 1 − k 2 ) ] \begin{align*} \dfrac{1}{1+k^2} \left[t-\dfrac{1}{t}-k\left(t+\dfrac{1}{t}\right)\right]^2 &= \dfrac{1}{1+k^2} \left[(1-k)t - (1+k)\dfrac{1}{t}\right]^2 \\ &= \dfrac{1}{1+k^2} \left[ (1-k)^2 t^2 + (1+k)^2 \left(\dfrac{1}{t}\right)^2 - 2(1-k^2) \right] \\ \end{align*} 1 + k 2 1 [ t − t 1 − k ( t + t 1 ) ] 2 = 1 + k 2 1 [ ( 1 − k ) t − ( 1 + k ) t 1 ] 2 = 1 + k 2 1 [ ( 1 − k ) 2 t 2 + ( 1 + k ) 2 ( t 1 ) 2 − 2 ( 1 − k 2 ) ]
分类讨论：
(i) 当 k = 1 k = 1 k = 1 LHS = 2 t 2 \text{LHS} = \dfrac{2}{t^2} LHS = t 2 2 k = − 1 k=-1 k = − 1 LHS = 2 t 2 \text{LHS} = 2t^2 LHS = 2 t 2 t t t LHS ≤ 1 \text{LHS} \le 1 LHS ≤ 1
(ii) 当 k 2 ≠ 1 k^2 \ne 1 k 2 = 1
LHS ≥ 2 1 + k 2 [ ∣ 1 − k 2 ∣ − ( 1 − k 2 ) ] \text{LHS} \ge \dfrac{2}{1+k^2} \left[ |1-k^2| - (1-k^2) \right] LHS ≥ 1 + k 2 2 [ ∣1 − k 2 ∣ − ( 1 − k 2 ) ]
且等号在 t = ± ∣ 1 + k 1 − k ∣ t = \pm\sqrt{\left|\dfrac{1+k}{1-k}\right|} t = ± 1 − k 1 + k 2 1 + k 2 [ ∣ 1 − k 2 ∣ − ( 1 − k 2 ) ] \dfrac{2}{1+k^2} \left[ |1-k^2| - (1-k^2) \right] 1 + k 2 2 [ ∣1 − k 2 ∣ − ( 1 − k 2 ) ] LHS \text{LHS} LHS
(a) k 2 < 1 k^2<1 k 2 < 1 RHS = 0 \text{RHS}=0 RHS = 0
(b) k 2 > 1 k^2>1 k 2 > 1 RHS = 4 ( k 2 − 1 ) 1 + k 2 ≤ 1 \text{RHS} = \dfrac{4(k^2-1)}{1+k^2} \le 1 RHS = 1 + k 2 4 ( k 2 − 1 ) ≤ 1 1 < k 2 ≤ 5 3 1 < k^2 \le \dfrac{5}{3} 1 < k 2 ≤ 3 5
综上所述，k k k [ − 15 3 , 15 3 ] \left[ -\dfrac{\sqrt{15}}{3}, \dfrac{\sqrt{15}}{3} \right] [ − 3 15 , 3 15 ]
记 K ( x , y ) = ( x − 1 ) 2 + y 2 K(x,y)=(x-1)^2+y^2 K ( x , y ) = ( x − 1 ) 2 + y 2 K K K
由 1 1 1
Γ : ( x − 1 + K ) 2 − y 2 − 4 K 2 = 0 \varGamma: (x-1+K)^2-y^2-4K^2=0 Γ : ( x − 1 + K ) 2 − y 2 − 4 K 2 = 0
即
( x − 1 ) 2 − y 2 + 2 K ( x − 1 ) − 3 K 2 = 0 ( 1 ) (x-1)^2 - y^2 + 2K(x-1) - 3K^2 = 0 \qquad (1) ( x − 1 ) 2 − y 2 + 2 K ( x − 1 ) − 3 K 2 = 0 ( 1 )
又设 O 1 ( 1 − m , n ) , O 2 ( 1 + m , n ) O_1(1-m,n), O_2(1+m,n) O 1 ( 1 − m , n ) , O 2 ( 1 + m , n )
⊙ O 1 : ( x − ( 1 − m ) ) 2 + ( y − n ) 2 = m 2 + n 2 \odot O_1: \bigl( x-(1-m) \bigr)^2 + (y-n)^2 = m^2 + n^2 ⊙ O 1 : ( x − ( 1 − m ) ) 2 + ( y − n ) 2 = m 2 + n 2
即
K − 2 m ( x − 1 ) − 2 n y = 0 ( 2 ) K-2m(x-1)-2ny = 0 \qquad (2) K − 2 m ( x − 1 ) − 2 n y = 0 ( 2 )
同理 ⊙ O 2 \odot O_2 ⊙ O 2
K + 2 m ( x − 1 ) − 2 n y = 0 ( 3 ) K+2m(x-1)-2ny = 0 \qquad (3) K + 2 m ( x − 1 ) − 2 n y = 0 ( 3 )
因为 Q , R s.t. ( 2 ) , S , T s.t. ( 3 ) Q,R \mathrel{\text{s.t.}} (2), S,T \mathrel{\text{s.t.}} (3) Q , R s.t. ( 2 ) , S , T s.t. ( 3 ) Q , R , S , T s.t. Q,R,S,T \mathrel{\text{s.t.}} Q , R , S , T s.t.
4 n 2 y 2 − 4 m 2 ( x − 1 ) 2 + K 2 − 4 n y K = 0 ( 4 ) 4n^2y^2 - 4m^2(x-1)^2 + K^2 - 4nyK = 0 \qquad (4) 4 n 2 y 2 − 4 m 2 ( x − 1 ) 2 + K 2 − 4 n yK = 0 ( 4 )
所以 Q , R , S , T s.t. 2 ( m 2 + n 2 ) y ( 1 ) + ( 4 ) Q,R,S,T \mathrel{\text{s.t.}} 2(m^2+n^2)y ~(1) + (4) Q , R , S , T s.t. 2 ( m 2 + n 2 ) y ( 1 ) + ( 4 )
= 2 ( n 2 − m 2 ) ( x − 1 ) 2 + 2 ( n 2 − m 2 ) y 2 + 2 ( m 2 + n 2 ) ( 2 K ( x − 1 ) − 3 K 2 ) + K 2 − 4 n y K = K [ 2 ( n 2 − m 2 ) + 4 ( m 2 + n 2 ) ( x − 1 ) + ( 1 − 6 m 2 − 6 n 2 ) ( ( x − 1 ) 2 + y 2 ) − 4 n y ] = 0 \begin{align*} &\mathrel{\phantom{=}} 2(n^2-m^2)(x-1)^2 + 2(n^2-m^2)y^2 + 2(m^2+n^2)\bigl(2K(x-1)-3K^2\bigr) + K^2 - 4nyK \\ &= K \Bigl[ 2(n^2-m^2) + 4(m^2+n^2)(x-1) + (1-6m^2-6n^2) \bigl( (x-1)^2+y^2 \bigr) - 4ny \Bigr] = 0 \end{align*} = 2 ( n 2 − m 2 ) ( x − 1 ) 2 + 2 ( n 2 − m 2 ) y 2 + 2 ( m 2 + n 2 ) ( 2 K ( x − 1 ) − 3 K 2 ) + K 2 − 4 n yK = K [ 2 ( n 2 − m 2 ) + 4 ( m 2 + n 2 ) ( x − 1 ) + ( 1 − 6 m 2 − 6 n 2 ) ( ( x − 1 ) 2 + y 2 ) − 4 n y ] = 0
于是 Q , R , S , T Q,R,S,T Q , R , S , T
