Question

求一道高数题答案.将函数f(x)=xarctan[(1+x)/(1-x)]展开成x的幂级数.要完整的步骤,

Answer 1

令g(x)=arctan[(1+x)/(1-x)],g(0)=π/4

∫[0->x]g'(t)dt = g(x)-g(0)=g(x)-π/4

g'(x)=[(1+x)/(1-x)]'/[1+(1+x)²/(1-x)²]=1/(1+x²)

g(x)=∫[0->x]g'(t)dt+π/4=∫[0->x] 1/(1+t²)dt+π/4

易知1/(1+t²)=1-t^2+t^4-t^6+…… |t|x] (1-t^2+t^4-t^6+……) dt

=π/4+(x-x^3/3+x^5/5-x^7/7+……)

f(x)=xg(x)=πx/4+(x^2-x^4/3+x^6/5-x^8/7+……)=πx/4+∑[(-1)^n][x^(2n+2)]/(2n+1) [n=0->+∞]