这道微积分题有人会解吗？麻烦啦
(a)
erf(x)=(2/√π)∫(0->x) e^(-t^2) dt
x=a
erf(a)=(2/√π)∫(0->a) e^(-t^2) dt (1)
x=b
erf(b)=(2/√π)∫(0->b) e^(-t^2) dt (2)
(2)-(1)
erf(b)-erf(a)
=(2/√π)∫(0->b) e^(-t^2) dt -(2/√π)∫(0->a) e^(-t^2) dt
=(2/√π)∫(a->b) e^(-t^2) dt
=(2/√π)∫(a->b) e^(-x^2) dx
(b)
erf(x)=(2/√π)∫(0->x) e^(-t^2) dt =>erf(0)=0
erf'(x)=(2/√π)e^(-x^2) =>erf'(0)/1!=2/√π
erf''(x)=(-1/√π)x.e^(-x^2) =>erf''(0)/2!=0
erf'''(x)=(-1/√π)(1-2x^2).e^(-x^2) =>erf'''(0)/3!=-(1/6)(2/√π)
n>1
consider
e^(-x^2) = 1-x^2+(1/2)x^4+....+[(-1)^n/n!]x^(2n) + o(x^(2n))
2n-1 是奇数 =>erf^(2n-1)(0)/(2n-1)!=(-1)^(n-1).(2/√π) [1/(n!.(2n-1))]
2n 是偶数 =>erf^(2n)(0)/(2n)!= 0
Maclaurin series of erf(x)
erf(x)
=erf(0) +(erf'(0)/1!)x +....
=(2/√π) [ x - (1/6)x^3 +....+(-1)^n.{ 1/[(n+1)!.(2n+1)] }.x^(2n+1) +..... ]
