Question

幂级数求和函数

帮忙写一下解题步骤

Answer 1

(3) 原式 = ∑<n=1,∞>(2n-1)(x/√2)^(2n)
= ∑<n=1,∞>(2n+1)(x/√2)^(2n) - ∑<n=1,∞>2(x/√2)^(2n)
= [√2∑<n=1,∞>(x/√2)^(2n+1)]' - ∑<n=1,∞>2(x/√2)^(2n)
= [√2(x/√2)^3/(1-x/√2)]' - 2(x/√2)^2/(1-x/√2)
= [x^3/(2-√2x)]' - √2x^2/(√2-x)
= [3x^2(2-√2x)+√2x^3]/(2-√2x)^2 - √2x^2/(√2-x)
= (3x^2-√2x^3)/(√2-x)^2 - √2x^2/(√2-x)
= (3x^2-√2x^3)/(√2-x)^2 - √2x^2/(√2-x) = x^2/(√2-x)^2
收敛域： -1 < x/√2 < 1, -√2 < x < √2