牛顿迭代法

设 f(x) f ( x ) 在 ℝ R 有二阶连续导数，且 f′(x)≠0 f ′ ( x ) ≠ 0 则
∀x0,x∈ℝ, ∀ x 0 , x ∈ R , 若 f(x0)=0 f ( x 0 ) = 0 则
f(x0)=f(x)+f′(x)Δx+12f″(ε)Δx2=0 f ( x 0 ) = f ( x ) + f ′ ( x ) Δ x + 1 2 f ″ ( ε ) Δ x 2 = 0
⇒ ⇒
x0−x=Δx=−1f′(x)[f(x)+12f″(ε)Δx2] x 0 − x = Δ x = − 1 f ′ ( x ) [ f ( x ) + 1 2 f ″ ( ε ) Δ x 2 ]
=−f(x)f′(x)−12f″(ε)f′(x)Δx2 = − f ( x ) f ′ ( x ) − 1 2 f ″ ( ε ) f ′ ( x ) Δ x 2
⇒ ⇒
x0=x−f(x)f′(x)−12f″(ε)f′(x)Δx2 x 0 = x − f ( x ) f ′ ( x ) − 1 2 f ″ ( ε ) f ′ ( x ) Δ x 2
由于 limx→x0[−12f″(ε)f′(x)Δx2]=0 lim x → x 0 [ − 1 2 f ″ ( ε ) f ′ ( x ) Δ x 2 ] = 0
因此 limx→x0[x−f(x)f′(x)]=x0 lim x → x 0 [ x − f ( x ) f ′ ( x ) ] = x 0
令 F(x)=x−f(x)f′(x) F ( x ) = x − f ( x ) f ′ ( x ) ，则 limx→x0F(x)=x0 lim x → x 0 F ( x ) = x 0
由此得到迭代公式：
xk+1=F(xk)=xk−f(xk)f′(xk) x k + 1 = F ( x k ) = x k − f ( x k ) f ′ ( x k )