设n次多项式方程p(x)=0的n个根均为实数 它们是 ξ1≥ξ2≥…≥ξn n≥2.

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设n次多项式方程p(x)=0的n个根均为实数,它们是

  ξ1≥ξ2≥…≥ξn, n≥2.

热搜:设等差数列an的前n项和为sn若s4>=10s5<=15

答案解析

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  • 可写p(x)为
       latex.action?latex=cCh4KT1hX25ccHJvZCBfe2k9MX1ebih4LVx4aSBfaSk%3D
       求导得
       latex.action?latex=cCcoeCk9YV9uXHN1bSBfe2k9MX1eblxwcm9kIF97XGJlZ2lue2FycmF5fXtjfWo9MVxcalxuZXEg%0D%0AaVxlbmR7YXJyYXl9fV5uKHgtXHhpIF9qKQ%3D%3D
       当x>ξ1.时,
       latex.action?latex=XGZyYWN7cCh4KX17cCcoeCl9PVxmcmFje1xwcm9kIF97aT0xfV5uKHgtXHhpIF9pKX17XHN1bSBf%0D%0Ae2k9MX1eblxwcm9kIF97XGJlZ2lue2FycmF5fXtjfWo9MVxcalxuZXEgaVxlbmR7YXJyYXl9fV57%0D%0Abn0oeC1ceGkgX2opfe%2B8njA%3D
       牛顿迭代公式为
       latex.action?latex=eF97aysxfT14X2stXGZyYWN7cCh4X2spfXtwJyh4X2spfQ%3D%3D,k=0,1,2,…  (2.33)
       因而
       latex.action?latex=eF8xPXhfMC1cZnJhY3twKHhfMCl9e3AnKHhfMCl977yceF8w  (2.34)
       另外
       latex.action?latex=eF8xLVx4aSBfMT14XzAtXGZyYWN7cCh4XzApfXtwJyh4XzApfS1ceGkgXzE9eF8wLVx4aSBfMS1c%0D%0AZnJhY3twKHhfMCktcCcoXHhpIF8xKX17cCcoeF8wKX0%3D
       latex.action?latex=PSh4XzAtXHhpIF8xKVxsZWZ0WyAxLVxmcmFje3AnKFxldGEgXzApfXtwJyh4XzApfSBccmlnaHRd%0D%0A
       其中η0∈(ξ1,x0).由latex.action?latex=MO%2B8nFxmcmFje3AnKFxldGEgXzApfXtwJyh4XzApfe%2B8nDE%3D及上式得
       x1-ξ1>0,  (2.35)
       综合(2.34)和(2.35)式有
       ξ1<x1<x0.
       类似地,如果ξ1<xk,可证有ξ1<xk+1<xk.这就说明{xk}严格单调下降以ξ1为下界,因而必收敛.设latex.action?latex=XGxpbSBfe2tcdG8gXGluZnR5IH14X2s9XGJhciB7eH0%3D,易知latex.action?latex=XGJhciB7eH1cZ2VxIFx4aSBfMQ%3D%3D.
       在(2.33)式的两边取极限得
       latex.action?latex=XGJhcnt4fT1cYmFye3h9LVxmcmFje3AoXGJhcnt4fSl9e3AnKFxiYXJ7eH0pfQ%3D%3D
       因而latex.action?latex=cChcYmFye3h9KT0w,即latex.action?latex=XGJhcnt4fQ%3D%3D为p(x)=0的根.故latex.action?latex=XGJhcnt4fT1ceGkgXzE%3D.综上,我们得到迭代序列严格单调递减收敛于ξ1.$当x<ξn时,
       latex.action?latex=XGZyYWN7cCh4KX17cCcoeCl9PVxmcmFje1xwcm9kIF97aT0xfV5uKHgtXHhpIF9pKX17XHN1bSBf%0D%0Ae2k9MX1eblxwcm9kIF97XGJlZ2lue2FycmF5fXtjfWo9MVxcalxuZXEgaVxlbmR7YXJyYXl9fV57%0D%0Abn0oeC1ceGkgX2opfe%2B8nDA%3D
       因而
       latex.action?latex=eF8xPXhfMC1cZnJhY3twKHhfMCl9e3AnKHhfMCl977yeeF8w  (2.36)
       另外
       latex.action?latex=eF8xLVx4aSBfbj14XzAtXGZyYWN7cCh4XzApfXtwJyh4XzApfS1ceGkgX24%3D
       latex.action?latex=PXhfMC1ceGkgX24teF8wLVxmcmFje3AoeF8wKS1wKFx4aSBfbil9e3AnKHhfMCl9
       latex.action?latex=PSh4XzAtXHhpIF9uKVxsZWZ0WyAxLVxmcmFje3AnKFxiYXJ7XGV0YSB9XzApfXtwJyh4XzApfSBc%0D%0AcmlnaHRd
       其中latex.action?latex=XGJhcntcZXRhIH1fMA%3D%3D∈(x0,ξn).由latex.action?latex=MO%2B8nFxmcmFje3AnKFxiYXJ7XGV0YSB9XzApfXtwJyh4XzApfe%2B8nDE%3D及上式得
       x1-ξn<0,  (2.37)
       综合(2.36)和(2.37)式有
       x0<x1<ξn.
       类似地可证明若xk<ξn,有xk<xk+1<ξn.因而{xk}严格单调递增并以ξn为上界.类似于(1)的证明有latex.action?latex=XGxpbSBfe2tcdG8gXGluZnR5IH14X2s9XHhpIF9u.