运用taylor展开
令f(x)为3次根号下2x+1
f'(0)=2/3;当n>=2时,求得f(x)的n次导数在0处的值为(-1)^(n-1)*2^n*(2*5*...*(3n-4))/(3^n)
f(x)=f(0)+f'(0)x+f''(0)x^2+...
运用taylor展开
令f(x)为3次根号下2x+1
f'(0)=2/3;当n>=2时,求得f(x)的n次导数在0处的值为(-1)^(n-1)*2^n*(2*5*...*(3n-4))/(3^n)
f(x)=f(0)+f'(0)x+f''(0)x^2+...