Question

微积分求解 请问这题怎么写？

Answer 1

cosx-1与-x^2/2等阶，所以(1+ax^2)^(1/3)-1=-x^2/2,
(1+ax^2)^(1/3)=1-x^2/2,
1+ax^2=(1-x^2/2)^3,
这个a好象不是一个常数，而是一个x的表达式哦，a³-3a²b+3ab²-b²
ax^2=3x^4/4-3x^2/2-x^6/8，
所以a=-x^4/8+3x^2/4-3/2

Answer 2

x->0
(1+ax^2)^(1/3) = 1+(1/3)ax^2 +o(x^2)
1-(1+ax^2)^(1/3) = -(1/3)ax^2 +o(x^2)
cosx -1 = -(1/2)x^2 +o(x^2)
=>
-(1/3)a =-1/2
a=3/2

追问

哥 请问有木有纸

最后两步 符号 看不懂

追答

1-(1+ax^2)^(1/3) 与 cosx -1 等价
=>-(1/3)ax^2 +o(x^2)与 -(1/2)x^2 +o(x^2) 等价
=>-(1/3)a = -1/2
=> a= 3/2

追问

哥 o是什么？

(1+ax^2)^(1/3) = 1+(1/3)ax^2 +o(x^2) 还有哥 (1/3)ax^2 +o(x^2) 这是怎么来的

追答

o(x^2) : 即代表所有x^n 项 （n>2)
根据泰勒公式
cosx = 1-(1/2)x^2 +(1/24)x^4 +....
有无数的项目， 用o(x^2)代表所有x^n 项 （n>2)
cosx =1-(1/2)x^2 +o(x^2)
cosx -1=-(1/2)x^2 +o(x^2)

追问

给下面的程过去也没有右边的1/3

追答

consider
f(x) =(1+x)^(1/3) => f(0) = 1
f'(x)=(1/3)(1+x)^(-2/3) => f'(0) = 1/3
(1+x)^(1/3) = 1 + (1/3)x + o(x)
x= ax^2
=>
(1+ax^2)^(1/3) = 1 + (1/3)ax^2 + o(x^2)