求In (1+x)展开的欧拉方法

求In (1+x)展开的欧拉方法

上面一篇文章已经推导了 e^w=1+w,另外对于1+w，也有这样的关系成立x=(1+w)^j-1,x为任意值.

推导

1+x=(1+w)^j=e^jw

Go

In (1+x)=jw

Go

In (1+x)=j* { (x+1)^(1/j)-1 }

Go

Newton method

In (1+x)=j* { 1+x^1*(1/j)/1!+x^2*(1/j)*(1/j-1)/2!+x^3*(1/j)*(1/j-1)*(1/j-2)/3!...-1 }

Go

In (1+x)=j* { x^1*(1/j)/1!+x^2*(1/j)*(1/j-1)/2!+x^3*(1/j)*(1/j-1)*(1/j-2)/3!... }

Go

In (1+x)= { x^1/1!+x^2*(1/j-1)/2!+x^3*(1/j-1)*(1/j-2)/3!... }

Go

1/j几乎为0

In (1+x)= { x^1/1!+x^2*(0-1)/2!+x^3*(0-1)*(0-2)/3!... }

Go

In (1+x)= { x^1/1!-x^2/2+x^3/3-x^4/4... }

与前面的用积分方法获得的结果一样