Partial Derivatives of Energy

For convenience we repeat the definition of raw Energy (3.16) here:

$$\sigma_{Er}(\vec{X})=\sum_{i<j}\frac{\left(d_{ij}(\vec{X})-\hat{d}_{ij}\right)^2}{\hat{d}_{ij}^2}$$ (9.1)

and that of the Euclidean distance (3.1) between points $\vec{x}_i=\left(x_{i1},\ldots,x_{ip}\right)^T$ and $\vec{x}_j$ in the p-dimensional MDS configuration $\vec{X}=[x_{ia}]$:

$$d_{ij}(\vec{X})=\left\Vert\vec{x}_i-\vec{x}_j\right\Vert=\sqrt{\sum_{a=1}^p\left(x_{ia}-x_{ja}\right)^2}$$ (9.2)

 
The First Partial Derivative of the Euclidean Distance

$$\frac{\partial d_{ij}(\vec{X})}{\partial x_{ka}} \frac{1}{2d_{ij}(\vec{X})}\frac{\partial d_{ij}^2(\vec{X})}{\partial x_{ka}}$$ =

$$\frac{1}{2d_{ij}(\vec{X})}\sum_{b=1}^p2(x_{ib}-x_{jb})\frac{\partial (x_{ib}-x_{jb})}{\partial x_{ka}}$$ =

$$\frac{x_{ia}-x_{ja}}{d_{ij}(\vec{X})}\frac{\partial (x_{ia}-x_{ja})}{\partial x_{ka}}$$ =

$$\frac{x_{ia}-x_{ja}}{d_{ij}(\vec{X})}(\theta_{ik}-\theta_{jk})$$ = (9.3)

where is the Kronecker delta

 
The First Partial Derivative of Energy

$$\frac{\partial \sigma_{Er}(\vec{X})}{\partial x_{ka}}=2\sum_{i<j}... ...)-\hat{d}_{ij}}{\hat{d}_{ij}^2}\frac{\partial d_{ij}(\vec{X})}{\partial x_{ka}}$$

$$2\sum_{i<j}\frac{d_{ij}(\vec{X})-\hat{d}_{ij}}{d_{ij}(\vec{X})\hat{d}_{ij}^2}(x_{ia}-x_{ja})(\theta_{ik}-\theta_{jk})$$ =

$$\sum\left\{\begin{array}{l@{\quad\Rightarrow\quad}l} k=i<j\neq k&... ...)\\ k\neq i<j\neq k&0\\ k=i<j=k&\textnormal{contradiction} \end{array}\right.$$ =

$$2\sum_{l>k}\frac{d_{kl}(\vec{X})-\hat{d}_{kl}}{d_{kl}(\vec{X})\ha... ...rac{d_{kl}(\vec{X})-\hat{d}_{kl}}{d_{kl}(\vec{X})\hat{d}_{kl}^2}(x_{la}-x_{ka})$$ =

$$2\sum_{l\neq k}\frac{d_{kl}(\vec{X})-\hat{d}_{kl}}{d_{kl}(\vec{X})\hat{d}_{kl}^2}(x_{ka}-x_{la})$$ = (9.4)

$$2\sum_{l\neq k}\frac{x_{ka}-x_{la}}{\hat{d}_{kl}^2}-2\sum_{l\neq k}\frac{x_{ka}-x_{la}}{d_{kl}(\vec{X})\hat{d}_{kl}}$$ =

 
Second Partial Derivatives of Energy

$$\frac{\partial^2 \sigma_{Er}(\vec{X})}{\partial x_{ka}^2} 2\sum_{l\neq k}\frac{\theta_{kk}-\theta_{lk}}{\hat{d}_{kl}^2}-2\s... ...ac{\partial\displaystyle\frac{x_{ka}-x_{la}}{d_{kl}(\vec{X})}}{\partial x_{ka}}$$ =

$$2\sum_{l\neq k}\frac{1}{\hat{d}_{kl}^2}-2\sum_{l\neq k}\frac{1}{\... ...isplaystyle\frac{\partial d_{kl}(\vec{X})}{\partial x_{ka}}}{d_{kl}^2(\vec{X})}$$ =

$$2\sum_{l\neq k}\frac{1}{\hat{d}_{kl}^2}-2\sum_{l\neq k}\frac{d_{k... ...})^2}{d_{kl}(\vec{X})}(\theta_{kk}-\theta_{lk})}{d_{kl}^2(\vec{X})\hat{d}_{kl}}$$ =

$$2\sum_{l\neq k}\frac{1}{\hat{d}_{kl}^2}-2\sum_{l\neq k}\frac{d_{kl}^2(\vec{X})-(x_{ka}-x_{la})^2}{d_{kl}^3(\vec{X})\hat{d}_{kl}}$$ = (9.5)

$$\mathop{\forall}_{a\neq b}\frac{\partial^2 \sigma_{Er}(\vec{X})}{\partial x_{ka}x_{kb}} 0-2\sum_{l\neq k}\frac{1}{\hat{d}_{kl}}\frac{\partial\displaystyle\frac{x_{ka}-x_{la}}{d_{kl}(\vec{X})}}{\partial x_{kb}}$$ =

$$-2\sum_{l\neq k}\frac{0-(x_{ka}-x_{la})\displaystyle\frac{(x_{kb}... ...lb})}{d_{kl}(\vec{X})}(\theta_{kk}-\theta_{lk})}{d_{kl}^2(\vec{X})\hat{d}_{kl}}$$ =

$$2\sum_{l\neq k}\frac{(x_{ka}-x_{la})(x_{kb}-x_{lb})}{d_{kl}^3(\vec{X})\hat{d}_{kl}}$$ = (9.6)