Fixed minus sign errors in momentum in branch fix-momentum-2

Author: nickmcgreivy

chapter_optimization/momentum.md

@@ -258,11 +258,11 @@ $$h(\mathbf{x}) = \frac{1}{2} \mathbf{x}^\top \mathbf{Q} \mathbf{x} + \mathbf{x}
 
 This is a general quadratic function. For positive definite matrices $\mathbf{Q} \succ 0$, i.e., for matrices with positive eigenvalues this has a minimizer at $\mathbf{x}^* = -\mathbf{Q}^{-1} \mathbf{c}$ with minimum value $b - \frac{1}{2} \mathbf{c}^\top \mathbf{Q}^{-1} \mathbf{c}$. Hence we can rewrite $h$ as
 
-$$h(\mathbf{x}) = \frac{1}{2} (\mathbf{x} - \mathbf{Q}^{-1} \mathbf{c})^\top \mathbf{Q} (\mathbf{x} - \mathbf{Q}^{-1} \mathbf{c}) + b - \frac{1}{2} \mathbf{c}^\top \mathbf{Q}^{-1} \mathbf{c}.$$
+$$h(\mathbf{x}) = \frac{1}{2} (\mathbf{x} + \mathbf{Q}^{-1} \mathbf{c})^\top \mathbf{Q} (\mathbf{x} + \mathbf{Q}^{-1} \mathbf{c}) + b - \frac{1}{2} \mathbf{c}^\top \mathbf{Q}^{-1} \mathbf{c}.$$
 
-The gradient is given by $\partial_{\mathbf{x}} h(\mathbf{x}) = \mathbf{Q} (\mathbf{x} - \mathbf{Q}^{-1} \mathbf{c})$. That is, it is given by the distance between $\mathbf{x}$ and the minimizer, multiplied by $\mathbf{Q}$. Consequently also the velocity  is a linear combination of terms $\mathbf{Q} (\mathbf{x}_t - \mathbf{Q}^{-1} \mathbf{c})$.
+The gradient is given by $\partial_{\mathbf{x}} h(\mathbf{x}) = \mathbf{Q} (\mathbf{x} + \mathbf{Q}^{-1} \mathbf{c})$. That is, it is given by the distance between $\mathbf{x}$ and the minimizer, multiplied by $\mathbf{Q}$. Consequently also the velocity  is a linear combination of terms $\mathbf{Q} (\mathbf{x}_t + \mathbf{Q}^{-1} \mathbf{c})$.
 
-Since $\mathbf{Q}$ is positive definite it can be decomposed into its eigensystem via $\mathbf{Q} = \mathbf{O}^\top \boldsymbol{\Lambda} \mathbf{O}$ for an orthogonal (rotation) matrix $\mathbf{O}$ and a diagonal matrix $\boldsymbol{\Lambda}$ of positive eigenvalues. This allows us to perform a change of variables from $\mathbf{x}$ to $\mathbf{z} \stackrel{\textrm{def}}{=} \mathbf{O} (\mathbf{x} - \mathbf{Q}^{-1} \mathbf{c})$ to obtain a much simplified expression:
+Since $\mathbf{Q}$ is positive definite it can be decomposed into its eigensystem via $\mathbf{Q} = \mathbf{O}^\top \boldsymbol{\Lambda} \mathbf{O}$ for an orthogonal (rotation) matrix $\mathbf{O}$ and a diagonal matrix $\boldsymbol{\Lambda}$ of positive eigenvalues. This allows us to perform a change of variables from $\mathbf{x}$ to $\mathbf{z} \stackrel{\textrm{def}}{=} \mathbf{O} (\mathbf{x} + \mathbf{Q}^{-1} \mathbf{c})$ to obtain a much simplified expression:
 
 $$h(\mathbf{z}) = \frac{1}{2} \mathbf{z}^\top \boldsymbol{\Lambda} \mathbf{z} + b'.$$