x⼀(4-x^2)^2 幂级数展开 并求收敛域半径

f(x)=x/(4-x^2)^2 f(0)=0
∫f(x)dx=(1/2)∫1/(4-x^2)^2dx^2=(-1/2)/(4-x^2)
=(-1/8)/(1-x^2/4)=∑(0,+∞)(-1/8)(x/2)^n |x/2|<1
所以：f(x)=[∑(0,+∞)(-1/8)(x/2)^n]'=(-1/16)∑(1,+∞)n(x/2)^(n-1) |x|<2

f= ln(1-x)-ln(1+x)
f'=-1/(1-x)-1/(1+x)= -∑(0,+∞)x^n-∑(0,+∞)(-x)^n=-∑(0,+∞)[x^n+(-x)^n] (注意奇数项抵消,|x|<1）
=-2∑(0,+∞)x^(2n)
f(x)=-2∑(0,+∞)x^(2n+1)/(2n+1) |x|<1)

f=arctan(x^2)
f'=2x/(1+x^4)=2x∑(0,+∞)(-x^4)^n=2∑(0,+∞)(-1)^nx^(4n+1) |x|<1)
f=2∑(0,+∞)(-1)^nx^(4n+2)/(4n+2) |x|<1)
当|x|=1级数收敛
所以：arctan(x^2)=∑(0,+∞)(-1)^nx^(4n+2)/(2n+1) |x|<=1