二范数的Taylor展开
考虑一个方程f(c)=∥a−bc∥f(\mathbf{c})=\Vert \mathbf{a}-\mathbf{bc}\Vertf(c)=∥a−bc∥,其中a∈RN,b∈RN×N,c∈RN\mathbf{a}\in \mathbb{R}^N,\mathbf{b}\in\mathbb{R}^{N\times N},\mathbf{c}\in \mathbb{R}^Na∈RN,b∈RN×N,c∈RN。需要对该方程应用Taylor展开至一次项,有
f(c)=f(0)+∇f(x)∣x=0Tc=∥a∥+∇f(0)∣Tc
f(\mathbf{c}) = f(0)+\nabla f(x)|_{x=0}^T \mathbf{c}=\Vert\mathbf{a}\Vert+\nabla f(0)|^T \mathbf{c}
f(c)=f(0)+∇f(x)∣x=0Tc=∥a∥+∇f(0)∣Tc
探讨∇f(0)∣Tc\nabla f(0)|^T \mathbf{c}∇f(0)∣Tc,有
f(c)=∥a−bd∥=(a−bc)T(a−bc)∇f(c)=12∥a−bc∥2(a−bc)(−1)bT∇f(0)=−12∥a∥2abT=−abT∥a∥
\begin{aligned}
f(\mathbf{c}) &= \Vert \mathbf{a} -\mathbf{bd}\Vert=\sqrt{(\mathbf{a-bc})^T(\mathbf{a-bc})}\\
\nabla f(\mathbf{c})&=\frac{1}{2\Vert \mathbf{a-bc}\Vert}2(\mathbf{a-bc})(-1)\mathbf{b}^T\\
\nabla f(0)&=-\frac{1}{2\Vert\mathbf{a}\Vert}2\mathbf{ab}^T=-\frac{\mathbf{ab}^T}{\Vert \mathbf{a}\Vert}
\end{aligned}
f(c)∇f(c)∇f(0)=∥a−bd∥=(a−bc)T(a−bc)=2∥a−bc∥12(a−bc)(−1)bT=−2∥a∥12abT=−∥a∥abT
所以带入有
f(c)=∥a∥+∇f(0)∣Tc=∥a∥−baT∥a∥c⇒f(c)∥a∥=∥a∥2−baTc⇒(f(c)−∥a∥)∥a∥=−baTc \begin{aligned} f(\mathbf{c})&=\Vert \mathbf{a}\Vert+\nabla f(0)|^T \mathbf{c} \\ &= \Vert\mathbf{a}\Vert - \frac{\mathbf{ba}^T}{\Vert \mathbf{a}\Vert}\mathbf{c}\\ \Rightarrow f(\mathbf{c})\Vert \mathbf{a}\Vert &=\Vert \mathbf{a}\Vert^2-\mathbf{ba}^T\mathbf{c}\\ \Rightarrow \left(f(\mathbf{c})-\Vert \mathbf{a}\Vert\right)\Vert \mathbf{a}\Vert &=-\mathbf{ba}^T\mathbf{c}\\ \end{aligned} f(c)⇒f(c)∥a∥⇒(f(c)−∥a∥)∥a∥=∥a∥+∇f(0)∣Tc=∥a∥−∥a∥baTc=∥a∥2−baTc=−baTc