Newto-Raphson

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1. Newton-Raphson

解决 $f(x)=0$ 问题的逼近方式

x n e w = x o l d - f ( x ) f ' ( x ) f' (x) = \partial f ( x ) \partial x

$x^{new} = x^{old}-\frac{f(x)}{f'(x)}\\ f'(x) = \frac{\partial f(x)}{\partial x}$

1级泰勒展开：

f (x + Δ x) = f (x) + Δ x f' (x)

$f(x+\Delta x) = f(x)+\Delta xf'(x)$
令

f(x+Δx)≈0 $f(x+\Delta x) \approx 0$

Δx $\Delta x$ 是x使得f(x)趋近于0的方向。那么

f (x) + Δ x f' (x) = 0 \to Δ x = - f ( x ) f ' ( x )

$f(x)+\Delta xf'(x) = 0\\ \to \Delta x = -\frac{f(x)}{f'(x)}$

x n e w = x o l d - \nabla - 2 f (x o l d) \times \nabla f (x o l d)

$x^{new} = x^{old} - \nabla^{-2}f (x^{old})\times\nabla f (x^{old})\\$
矩阵形式

x n e w = x o l d - H - 1 \nabla f (x o l d) H = \nabla 2 f

$x^{new} = x^{old} - H^{-1}\nabla f(x^{old})\\ H = \nabla^2f$
H是Hessian Matrix.

泰勒二级展开

f (X + Δ X) = f (X) + \nabla T f (X) \times Δ X + 1 2 (Δ X) T \times \nabla 2 f (X) \times Δ X

$f(X+\Delta X)=f(X) + \nabla ^Tf(X)\times \Delta X+\frac{1}{2}(\Delta X)^T\times \nabla ^2f(X)\times \Delta X$
令

f(X+ΔX)=0→ $f(X+\Delta X) = 0 \to$

Δ X = - (\nabla 2 f (X)) - 1 \times \nabla f (X)

$\Delta X = -(\nabla^2f(X))^{-1}\times \nabla f(X)$

ΔX $\Delta X$ 是X逼近解的方向。

F = x T A x + b T x + c x m a x = - 0.5 A - 1 b

$F = x^TAx +b^Tx + c\\ x_{max} = -0.5A^{-1}b$

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