cosx,e^x怎么展开为幂级数xx
答案:2 悬赏:20
解决时间 2021-11-09 07:03
- 提问者网友:心裂
- 2021-11-08 18:21
cosx,e^x怎么展开为幂级数xx
最佳答案
- 二级知识专家网友:晚安听书人
- 2021-11-08 19:05
f(x)=e^x
f'(x)=e^x
f^(n)(x) =e^x
f^(n)(0) =1
e^x = f(0) +f'(0)x/1! + f''(0)x^2/2!+...
= 1+ x/1!+x^2/2!+.....
g(x) =cosx =>g(0) =1
g'(x) = -sinx =>g'(0) =0
g''(x) = -cosx =>g''(0) =-1
g'''(x) = sinx =>g'''(0) =0
g''''(x) = cosx =>g''''(0) =1
g^(n)(x) = g^(n-4)(x)
g(x) =g(0) +g'(0)x/1! + g''(0)x^2/2! +...
=1-x^2/2! +x^4/4! +.....
f'(x)=e^x
f^(n)(x) =e^x
f^(n)(0) =1
e^x = f(0) +f'(0)x/1! + f''(0)x^2/2!+...
= 1+ x/1!+x^2/2!+.....
g(x) =cosx =>g(0) =1
g'(x) = -sinx =>g'(0) =0
g''(x) = -cosx =>g''(0) =-1
g'''(x) = sinx =>g'''(0) =0
g''''(x) = cosx =>g''''(0) =1
g^(n)(x) = g^(n-4)(x)
g(x) =g(0) +g'(0)x/1! + g''(0)x^2/2! +...
=1-x^2/2! +x^4/4! +.....
全部回答
- 1楼网友:而你却相形见绌
- 2021-11-08 19:27
函数展开成幂级数的方法是:1)求出f(x) 的各阶导函数,并且它们在x=0处的各阶导数值,如果某一阶导数不存在,则函数无法展开成幂级数;2)写出幂级数 f(0)+f'(0)x+[f''(0)/2!]x^2+...+[f(n)(0)/n!]x^n+...(其中f(n)(0)表示在x=0处的n阶导数值),并求其收敛半径r;
3)考察x在区间(-r,r)内时余项r(n)的极限是否为零,r(n)=[f(n+1)(a)/(n+1)!]x^(n+1),a是0到x之间的某个数,若为零则上式就是展开式。
cos(x)=1-x^2/2!+x^4/4!-...+(-1)^n*x^2n/(2n)!+...,x属于r
e^x=1+x+x^2/2!+...+x^n/n!+...
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