Question

求幂级数∑(2n-1)(x∧2n)/4∧n的收敛域及和函数？

在线等 挺急的

Answer 1

安排数列的一般项将x^n换为(1/4)^n(4^n)*x^n结果得到：∑(2n-1)(x^2n/4^n)=∑(2n-1)(1/4)^n(4^n)*x^n
使用Weierstrass M条件来确定收敛性：(1)数列{(2n-1)(1/4)^n(4^n)}在[-4,4]上收敛；(2)f(x)存在并连续；
由Weierstrass M条件的存在性和唯一性定理可知，收敛域是[-4,4]，和函数为f(x)=x^2/4+x^4/16+x^6/64+…
希望你喜欢。

Answer 2

收敛半径的平方
R^2 = lim<n→∞>a<n>/a<n+1>
= lim<n→∞>(2n-1)4^(n+1)/[(2n+1)4^n] = 4
R = 2,
x = ±2 时， 级数变为 ∑(2n-1) 发散， 故收敛域为 (-2, 2).
S(x) = ∑<n=1, ∞>(2n-1)(x∧2n)/4∧n = ∑<n=1, ∞>(2n-1)(x/2)^(2n)
= ∑<n=1, ∞>(2n+1)(x/2)^(2n) - 2∑<n=1, ∞>(x/2)^(2n)
= S1(x) - 2(x/2)^2/[1-(x/2)^2] = S1(x) - 2x^2/(4-x^2)
其中 S1(x) = ∑<n=1, ∞>(2n+1)(x/2)^(2n)
= [∑<n=1, ∞>(x/2)^(2n+1)]' = {(x/2)^3/[1-(x/2)^2]}' = [x^3/(8-2x^2)]'
= [3x^2(8-2x^2)-x^3(-4x)]/(8-2x^2)^2 = (24x^2-2x^4)/(8-2x^2)^2
= (12x^2-x^4)/[2(4-x^2)^2]
S(x) = x^2(12-x^2)/[2(4-x^2)^2] - 2x^2/(4-x^2)
= (12x^2-x^4)/[2(4-x^2)^2] - 2x^2/(4-x^2)
= [12x^2-x^4-4x^2(4-x^2)]/[2(4-x^2)^2]
= (-4x^2+3x^4)/[2(4-x^2)^2], x∈(-2, 2)