中科大瀚海星云BBS论坛
中国科大,中科大bbs,瀚海星云,瀚海星云bbs,中科大瀚海星云
[回到开始]
[上一篇][下一篇]
发信人: Sidao_Ni (Sidao Ni)，原信区: ustcbbs
标 题: Re: right problem
发信站: 中国科大BBS站 (Fri, 29 Mar 1996 08:50:22)，站内信件
上回 Canary Cat 说道...
>
> f(x)=1+x+x^2/2!+x^3/3!+...+x^(2n+1)/(2n+1)!=0
> 有且只有一个实根.
f(x) >0 for all x>=0; (1)
x-> - infinity: f(x)-> -infinity (because the highest order is 2n+1);
hence, at least one real root exists, and the root is negative, let a0 be the
least nagative;
we have :
f(x) <0; for all x<a0;
and
f'(x) = 1 + x + x^2/2!+...+x^2n/2n!=f(x) - x^2n+1/(2n+1)!
thus:
f'(x) - f(x) = -x^2n+1/(2n+1)!
(f(x) exp(-x))'= (f'(x)-f(x))exp(-x) = -x^(2n+1)/(2n+1)!exp(-x);
f(x)-f(a0) =exp(x) * [ integral( from a0 to x): -x^(2n+1)/(2n+1)!* exp(-x)];
but f(a0) = 0;
thus f(x) =exp(x) * [ integral( from a0 to x): -x^(2n+1)/(2n+1)!* exp(-x)];
for all x <0:
-x^(2n+1)/(2n+1)! >0;
exp(x)>0, exp(-x)>0;
thus the integral must be greater than zero.
f(x) >0 for all 0>x>a0; (3)
from (1) & (3) we get
f(x) >0 for all x >a0;
and (2)
f(x) <0 for all x <a0;
then
a0 is the real root and the only one real root.
石头 --
* Origin: 中国科大BBS站 <bbs@ustc.edu.cn>
[回到开始]
[上一篇][下一篇]