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求函数ln(x+√(1+x^2))在x=0处的幂级数展式,并求展开式成立的区间
人气:453 ℃ 时间:2020-04-01 07:27:26
解答
(ln(x+√(1+x^2)))'=1/(√(1+x^2))=(1+x^2)^(-1/2)(1+x^2)^(-1/2)=1-(1/2)x^2+(-1/2)(-1/2-1)/2!(x^4)+(-1/2)(-1/2-1)(-1/2-2)/3!(x^6)+...=1-(1/2)x^2+(-1/2)(-3/2)/2!(x^4)+(-1/2)(-3/2)(-5/2)/3!(x^6)+...=1-(1/2...
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