I am reading this paper which explains the algorithm for the method Positive Matrix Factorization (different from non-negative matrix factorization). The paper can be found on https://doi.org/10.1002/cem.894
In one part of the paper, an objective function is given as follows,.
\begin{align} \mathcal{Q}\left(\mathbf{Q,F}\right) = \sum_{i=1}^{m}\sum_{j=1}^{n}\left(\dfrac{\mathbf{R}_{ij}}{{\sigma}_{ij}}\right) - \underbrace{\alpha\sum_{i=1}^{m}\sum_{k=1}^{p}\log \mathbf{G}_{ik} - \beta\sum_{k=1}^{p}\sum_{j=1}^{n}\log \mathbf{F}_{kj}}_{\text{Logarithmic Penalty}} \nonumber \ \\ + \underbrace{\gamma\sum_{i=1}^{m}\sum_{k=1}^{p} c_i\mathbf{G}_{ik}^2 + \delta\sum_{k=1}^{p}\sum_{j=1}^{n} d_j\mathbf{F}_{kj}^2}_{\text{Regularization}} \label{revisedQ} \end{align}
Main objective is to minimize the objective function using Gauss-Newton Method which needs all the terms in the equation in the form of sum of squares. I quote from the paper,
"Therefore the logarithmic penalty terms need to be approximated by a quadratic function around the current $\mathbf{G}$ and $\mathbf{F}$ values. If the logarithmic function $f(\theta) = \log_{10}(\theta)$ is used in the penalty terms, the quadratic function to approximate its value at a point $\theta_0$ is
\begin{align} g(\theta) = a(\theta - b)^2 \end{align} where, \begin{align} a & = \dfrac{1}{4\ln(10)\ln(\theta_0)\theta_0^2}\\ b & = \theta_0 - 2\theta_0\ln(\theta_0) \end{align}
The equations of $a$ and $b$ are derived by letting the values of first derivative of $f(\theta)$ and $g(\theta)$ equal at point $\theta_0$"
Firstly, I have a doubt, how can one approximate $f(\theta)$ as $g(\theta)$, I tried searching in literature but was not successful. Secondly, If I accept the first point, I am unable to derive the values of $a$ and $b$
Generally, you have the Taylor expansion $f(\theta) \approx f(\theta_0) + f^{'}(\theta_0)(\theta-\theta_0) + \frac{1}{2}f^{''}(\theta_0)(\theta-\theta_0)^2$. Now crunch the data with $f(\theta) = \frac{\ln(\theta)}{\ln(10)}$ and factorize to desired form.
Why one would use $\log_{10}$ in the penalty instead of the natural logarithm is beyond me though as it's just a scaling and just complicates the notation.