赞 本人无名 春芽
共回答了18个问题采纳率：83.3% 举报
设g(x)=∫(0,x)1/√(1+t^3)dt
g'(x)=1/√(1+x^3)=(1+x^3)^(-1/2) 用(1+x)^a的那个公式
=1+(-1/2)x^3+(-1/2)(-1/2-1)x^6/2!+(-1/2)(-1/2-1)(-1/2-2)x^9/3!+...
=1+Σ(-1/2)(-1/2-1)(-1/2-2)...(-1/2-1-2-...-(n-1))x^(3n)/n!n=1到+∞
则g(x)=x+(1/4)(-1/2)x^4+(1/7)(-1/2)(-1/2-1)x^7/2!+(1/10)(-1/2)(-1/2-1)(-1/2-2)x^10/3!+...
=x+Σ[1/(3n+1)](-1/2)(-1/2-1)(-1/2-2)...(-1/2-1-2-...-(n-1))x^(3n+1)/n!n=1到+∞
g'(x)=1/√(1+x^3)=(1+x^3)^(-1/2) 用(1+x)^a的那个公式
=1+(-1/2)x^3+(-1/2)(-1/2-1)x^6/2!+(-1/2)(-1/2-1)(-1/2-2)x^9/3!+...
=1+Σ(-1/2)(-1/2-1)(-1/2-2)...(-1/2-1-2-...-(n-1))x^(3n)/n!n=1到+∞
则g(x)=x+(1/4)(-1/2)x^4+(1/7)(-1/2)(-1/2-1)x^7/2!+(1/10)(-1/2)(-1/2-1)(-1/2-2)x^10/3!+...
=x+Σ[1/(3n+1)](-1/2)(-1/2-1)(-1/2-2)...(-1/2-1-2-...-(n-1))x^(3n+1)/n!n=1到+∞
1年前
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