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展开全部
x^4+x^2+x
=x^2.(x^2-1) +2x^2 +x
=x^2.(x^2-1) +2(x^2-1) +x+2
f(x)
=(x^4+x^2+x)/(x^2-1)
=x^2 +2 + (x+2)/(x^2-1)
=x^2 +2 + 1/(x-1) + 1/(x^2-1)
=x^2 +2 + 1/(x-1) +(1/2)[ 1/(x-1) -1/(x+1) ]
=x^2 +2 + (3/2)[1/(x-1)] - (1/2)[1/(x+1) ]
f'(x) =2x -(3/2)[1/(x-1)^2] + (1/2)[1/(x+1)^2 ]
f''(x) =2 +3[1/(x-1)^3] - [1/(x+1)^3 ]
f'''(x) =-9/(x-1)^4 + 3/(x+1)^4
f'''(0) =-9 + 3 =-6
=x^2.(x^2-1) +2x^2 +x
=x^2.(x^2-1) +2(x^2-1) +x+2
f(x)
=(x^4+x^2+x)/(x^2-1)
=x^2 +2 + (x+2)/(x^2-1)
=x^2 +2 + 1/(x-1) + 1/(x^2-1)
=x^2 +2 + 1/(x-1) +(1/2)[ 1/(x-1) -1/(x+1) ]
=x^2 +2 + (3/2)[1/(x-1)] - (1/2)[1/(x+1) ]
f'(x) =2x -(3/2)[1/(x-1)^2] + (1/2)[1/(x+1)^2 ]
f''(x) =2 +3[1/(x-1)^3] - [1/(x+1)^3 ]
f'''(x) =-9/(x-1)^4 + 3/(x+1)^4
f'''(0) =-9 + 3 =-6
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用泰勒展开怎么做阿
追答
求f'''(0) : 不用利用泰勒展开,
泰勒展开, 也是要求找 f^(n)(x)
2018-07-10
展开全部
哪不同了?一题一题看,一题一题理解掌握,消化能力不强就一口一口慢慢吃慢慢消化。
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