作业帮高数微分证明题感觉不难,就是那里卡住了,/>
来源：学生作业帮 编辑：作业帮 分类：综合作业 时间：2024/05/17 22:01:22
高数微分证明题
感觉不难,就是那里卡住了,/>
感觉不难,就是那里卡住了,/>
F(x) = (x-1)^2f(x)
F(1)=0
F(2) = f(2)=0
F'(x) = (x-1)^2 f'(x) + 2(x-1)f(x)
F'(1) = 0
expands F(x) about 1
F(x) = F(1)+ F'(1)(x-1) +F''(a)(x-1)^2/2!
= F''(a)(x-1)^2/2!
put x=2
F(2) = F''(a)/2 =0
F''(a)=0
再问： F(x) = F(1)+ F'(1)(x-1) +F''(a)(x-1)^2/2! = F''(a)(x-1)^2/2! 没看懂，哪儿出来的，f（x）去哪儿了
再答： 这是Taylor expansion(泰勒展开) of F(x) about x=1 F(x) 在【0，1】上有二阶导数 =>存在ξ∈(1,2) F(x) = F(1)+ F'(1)(x-1)/1! +F''(ξ)(x-1)^2/2! 代入 x=2 F(2) = F(1)+ F'(1) +F''(ξ)/2 =0 =>F''(ξ)=0
F(1)=0
F(2) = f(2)=0
F'(x) = (x-1)^2 f'(x) + 2(x-1)f(x)
F'(1) = 0
expands F(x) about 1
F(x) = F(1)+ F'(1)(x-1) +F''(a)(x-1)^2/2!
= F''(a)(x-1)^2/2!
put x=2
F(2) = F''(a)/2 =0
F''(a)=0
再问： F(x) = F(1)+ F'(1)(x-1) +F''(a)(x-1)^2/2! = F''(a)(x-1)^2/2! 没看懂，哪儿出来的，f（x）去哪儿了
再答： 这是Taylor expansion(泰勒展开) of F(x) about x=1 F(x) 在【0，1】上有二阶导数 =>存在ξ∈(1,2) F(x) = F(1)+ F'(1)(x-1)/1! +F''(ξ)(x-1)^2/2! 代入 x=2 F(2) = F(1)+ F'(1) +F''(ξ)/2 =0 =>F''(ξ)=0