【第三章 习题18】
把习题16中的递推公式换成
x n + 1 = p − 1 p x n + α p x n − p + 1 x_{n + 1} = \frac{p - 1}{p}x_{n} + \frac{\alpha}{p}x_{n}^{- p + 1} xn+1=pp−1xn+pαxn−p+1
这里 p p p是固定的正整数，描述该序列的性质
【解】
若 x 1 > x p x_{1} > \sqrt[p]{x} x1>px ， p ≥ 2 p \geq 2 p≥2
根据柯西不等式
x n + 1 = p − 1 p x n + α p x n − p + 1 = x n p + … + x n p + α p x − p + 1 ≥ α p x_{n + 1} = \frac{p - 1}{p}x_{n} + \frac{\alpha}{p}x_{n}^{- p + 1} = \frac{x_{n}}{p} + \ldots + \frac{x_{n}}{p} + \frac{\alpha}{p}x^{- p + 1} \geq \sqrt[p]{\alpha} xn+1=pp−1xn+pαxn−p+1=pxn+…+pxn+pαx−p+1≥pα
这里的不等号是取不到等号的，因为 x 1 > α p x_{1} > \sqrt[p]{\alpha} x1>pα ，数学归纳可得
x n > α p x_{n} > \sqrt[p]{\alpha} xn>pα
又因为
x n + 1 − x n = 1 p ( α x n p − 1 − x n ) < 1 p ( α p − x n ) < 0 x_{n + 1} - x_{n} = \frac{1}{p}\left( \frac{\alpha}{x_{n}^{p - 1}} - x_{n} \right) < \frac{1}{p}\left( \sqrt[p]{\alpha} - x_{n} \right) < 0 xn+1−xn=p1(xnp−1α−xn)<p1(pα −xn)<0
所以该序列严格单调递减
x n + 1 − α 1 p = ( p − 1 ) ( x n − α 1 p ) x n p − 1 + α 1 p ( α p − 1 p − x n p − 1 ) p x n p − 1 x_{n + 1} - \alpha^{\frac{1}{p}} = \frac{(p - 1)\left( x_{n} - \alpha^{\frac{1}{p}} \right)x_{n}^{p - 1} + \alpha^{\frac{1}{p}}\left( \alpha^{\frac{p - 1}{p}} - x_{n}^{p - 1} \right)}{px_{n}^{p - 1}} xn+1−αp1=pxnp−1(p−1)(xn−αp1)xnp−1+αp1(αpp−1−xnp−1)
根据不等式
n a n − 1 ( b − a ) ≤ b n − a n ≤ n b n − 1 ( b − a ) （其中 n ≥ 1 , b > a > 0 , 等号当 n = 1 时成立） na^{n - 1}(b - a) \leq b^{n} - a^{n} \leq nb^{n - 1}(b - a)\ \ \ \ （其中n \geq 1,b > a > 0,等号当n = 1时成立） nan−1(b−a)≤bn−an≤nbn−1(b−a) （其中n≥1,b>a>0,等号当n=1时成立）
可得
x n + 1 − α 1 p ≤ ( p − 1 ) ( x n − α 1 p ) x n p − 1 − α 1 p ( p − 1 ) α p − 2 p ( x n − α 1 p ) p x n p − 1 x_{n + 1} - \alpha^{\frac{1}{p}} \leq \frac{(p - 1)\left( x_{n} - \alpha^{\frac{1}{p}} \right)x_{n}^{p - 1} - \alpha^{\frac{1}{p}}(p - 1)\alpha^{\frac{p - 2}{p}}\left( x_{n} - \alpha^{\frac{1}{p}} \right)}{px_{n}^{p - 1}} xn+1−αp1≤pxnp−1(p−1)(xn−αp1)xnp−1−αp1(p−1)αpp−2(xn−αp1)
= ( p − 1 ) ( x n − α 1 p ) ( x n p − 1 − α p − 1 p ) p = p − 1 p ( 1 − α p − 1 p x n p − 1 ) ( x n − α 1 p ) < p − 1 p ( x n − α 1 p ) = \frac{(p - 1)\left( x_{n} - \alpha^{\frac{1}{p}} \right)\left( x_{n}^{p - 1} - \alpha^{\frac{p - 1}{p}} \right)}{p} = \frac{p - 1}{p}\left( 1 - \frac{\alpha^{\frac{p - 1}{p}}}{x_{n}^{p - 1}} \right)\left( x_{n} - \alpha^{\frac{1}{p}} \right) < \frac{p - 1}{p}\left( x_{n} - \alpha^{\frac{1}{p}} \right) =p(p−1)(xn−αp1)(xnp−1−αpp−1)=pp−1(1−xnp−1αpp−1)(xn−αp1)<pp−1(xn−αp1)
所以
lim x n = α p \lim x_{n} = \sqrt[p]{\alpha} limxn=pα
由于
x n + 1 − α 1 p ≤ ( p − 1 ) 2 ( x n − α 1 p ) 2 x n p − 2 p x n p − 1 = ( p − 1 ) 2 ( x n − α 1 p ) 2 p x n < ( p − 1 ) 2 p α 1 p ( x n − α 1 p ) 2 x_{n + 1} - \alpha^{\frac{1}{p}} \leq \frac{(p - 1)^{2}\left( x_{n} - \alpha^{\frac{1}{p}} \right)^{2}x_{n}^{p - 2}}{px_{n}^{p - 1}} = \frac{(p - 1)^{2}\left( x_{n} - \alpha^{\frac{1}{p}} \right)^{2}}{px_{n}} < \frac{(p - 1)^{2}}{p\alpha^{\frac{1}{p}}}\left( x_{n} - \alpha^{\frac{1}{p}} \right)^{2} xn+1−αp1≤pxnp−1(p−1)2(xn−αp1)2xnp−2=pxn(p−1)2(xn−αp1)2<pαp1(p−1)2(xn−αp1)2
这与16题类似（ p = 2 p = 2 p=2），至少是平方级别的收敛。