求幂级数的和函数Σ(n+1)(x-1)^n
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记 S(x) = Σ (n+1)(x-1)^n, 则
∫ S(t)dt = ∫ (n+1)(t-1)^ndt
= [(t-1)^(n+1)] = (x-1)^(n+1) = (x-1)^2+(x-1)^3+(x-1)^4+......
= (x-1)^2/[1-(x-1)] = (x-1)^2/(2-x).
收敛域 -1
于是 S(x)=[(x-1)^2/(2-x)]' = (x-1)(3-x)/(2-x)^2. 收敛域 0
咨询记录 · 回答于2021-12-29
求幂级数的和函数Σ(n+1)(x-1)^n
记枯基 S(x) = Σ (n+1)(x-1)^n, 则∫ S(t)dt = ∫ (n+1)(t-1)^ndt= [(t-1)^(n+1)] = (x-1)^(n+1) = (x-1)^2+(x-1)^3+(x-1)^4+......= (x-1)^2/[1-(x-1)] = (x-1)^2/(2-x).收敛域 -1
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