为什么当n为大于等于1的自然数时，sin(1/(2^n))<1/(2^n)？

显示全部楼层 · 2010-1-5 08:10:19

验证当n=1时,满足条件然后求1/(2^n)-sin(1/(2^n))的导数就很容易证明了

千问 · 2010-1-5 08:10:19

1楼正确，具体过程如下：(sin(1/(2^n))'=cos(1/(2^n))*(-n*2^(-n-1))=2*n*(n+1)*cos(1/(2^n))(1/(2^n))'=-n*2^(-n-1)=2*n*(n+1)因为当n为大于等于1时，0<1/(2^n)<1,0<cos(1/(2^n))<1所以(sin(1/(2^n))'<(1/(2^n))'，也就是说sin(1/(2^n))增大的趋势比1/(2^n)慢，当n=1时，sin(1/(2^n))=sin0.5约等于0，小于0.5=1/(2^n)，即sin(1/(2^n))<1/(2^n)所以当n为大于等于1的自然数时，sin(1/(2^n))<1/(2^n)