DSP

牛顿迭代法

2019-07-13 16:59发布

站内文章 / DSP
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设 f(x)" role="presentation" style="position: relative;">f(x) 在 R" role="presentation" style="position: relative;">R 有二阶连续导数，且 f′(x)≠0" role="presentation" style="position: relative;">f′(x)≠0 则
∀x0,x∈R," role="presentation" style="position: relative;">∀x0,x∈R, 若 f(x0)=0" role="presentation" style="position: relative;">f(x0)=0 则
f(x0)=f(x)+f′(x)Δx+12f″(ε)Δx2=0" role="presentation" style="position: relative;">f(x0)=f(x)+f′(x)Δx+12f″(ε)Δx2=0
⇒" role="presentation" style="position: relative;">⇒
x0−x=Δx=−1f′(x)[f(x)+12f″(ε)Δx2]" role="presentation" style="position: relative;">x0−x=Δx=−1f′(x)[f(x)+12f″(ε)Δx2]
=−f(x)f′(x)−12f″(ε)f′(x)Δx2" role="presentation" style="position: relative;">=−f(x)f′(x)−12f″(ε)f′(x)Δx2
⇒" role="presentation" style="position: relative;">⇒
x0=x−f(x)f′(x)−12f″(ε)f′(x)Δx2" role="presentation" style="position: relative;">x0=x−f(x)f′(x)−12f″(ε)f′(x)Δx2
由于 limx→x0[−12f″(ε)f′(x)Δx2]=0" role="presentation" style="position: relative;">