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ln(x+(1+x^2)^(1/2))幂级数展开
RT
人气:485 ℃ 时间:2020-06-07 19:09:32
解答
f=ln(x+(1+x^2)^(1/2)),求导:
f'=1/(1+x^2)^(1/2)=(1+x^2)^(-1/2)
=1+(-1/2)x^2+(-1/2)(-1/2-1)x^4+(-1/2)(-1/2-1)(-1/2-2)x^6+...|x|
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