Question

这个幂级数展开怎么求

展开是个等比数列吧，想知道首项可以含x吗

Answer 1

f(x) = x/(x^2-2x+2) = [(x-1)+1]/[1+(x-1)^2]
= (x-1)/[1+(x-1)^2] + 1/[1+(x-1)^2]
记 u = x-1, f(u) = u/(1+u^2) + 1/(1+u^2)
S(u) = ∫<0, u>f(t)dt = ∫<0, u>[t/(1+t^2)+1/(1+t^2)]dt
= [(1/2)ln(1+t^2)+arctant]<0, u> = (1/2)ln(1+u^2)+arctanu
= (1/2)∑<n=1,∞>(-1)^(n-1)u^(2n)/n + ∑<n=1,∞>(-1)^(n-1)u^(2n-1)/(2n-1)
= ∑<n=1,∞>(-1)^(n-1)[u^(2n)/(2n)+u^(2n-1)/(2n-1)]
f(u) = S'(u) = ∑<n=1,∞>(-1)^(n-1)[u^(2n-1)+u^(2n-2)]
收敛域 ： -1 < u^2 ≤ 1, -1 ≤ u ≤ 1.
f(x) = ∑<n=1,∞>(-1)^(n-1)[(x-1)^(2n-1)+(x-1)^(2n-2)],
收敛域 ： -1 ≤ x-1 ≤ 1, 即 0 ≤ x ≤ 2.