如何求一元n次方程的收敛域?
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发布时间：2024-05-09 08:10
共1个回答
热心网友
时间：2024-06-12 06:30
收敛半径 R = lim<n→∞>a<n>/a<n+1> = lim<n→∞>n(n+2)/(n+1)^2 = 1
x = ±1 时， 级数发散， 收敛域 -1 < x < 1.
S(x) = ∑<n=1, ∞> [n/(n+1)]x^n = ∑<n=1, ∞> [(n+1-1)/(n+1)]x^n
= ∑<n=1, ∞>x^n - ∑<n=1, ∞> [1/(n+1)]x^n
x ≠ 0 时
S(x) = x/(1-x) - (1/x)∑<n=1, ∞> [1/(n+1)]x^(n+1)
记 G(x) = ∑<n=1, ∞> [1/(n+1)]x^(n+1),
则 G'(x) = ∑<n=1, ∞>x^n = x/(1-x) = (x-1+1)/(1-x)= -1 + 1/(1-x)
G(x) = ∫<0, x> G'(t)dt + G(0) = ∫<0, x> [-1+1/(1-t)]dt + 0
= [-t-ln(1-t)]<0, x> = - x - ln(1-x)
S(x) = x/(1-x) - (1/x)[-x - ln(1-x)] = -1 + 1/(1-x) + 1 + ln(1-x)/x
= 1/(1-x) + ln(1-x)/x， -1 < x < 1.
x = 0 时 , S(x) = 0。
