Question

高等数学，函数展开成x-1的幂级数？

求解

Answer 1

xlnx = (x-1)lnx + lnx = (x-1)ln[1-(1-x)] + ln[1-(1-x)]

1/(1-x) = 1+x+x^2+…+x^n + …
Integrating from 0 to x,
ln(1-x) = x+x^2/2+…+x^(n+1)/(n+1)+…
ln[1-(1-x)] = (1-x)+(1-x)^/2+...+ (1-x)^n/n + ... = -(x-1)+(x-1)^2/2+...+(-1)^n (x-1)^n/n+...; n from 1 to infinity (1)
(x-1)ln[1-(1-x)] = (x-1)[(1-x)+(1-x)^2/2+…+(1-x)^(n+1)/(n+1) + …]
= -(x-1)^2 + (x-1)^3/2 - …+ (-1)^n (x-1)^(n+1)/n+…; n from 1 to infinity (2)
(1)+(2): xlnx = Answer