Minimize Logged Sum of Squares? When numerically maximizing the likelihood function it is standard practice to do this indirectly by minimizing the negative log-likelihood. When numerically minimizing the residual sum of squares (RSS), is it therefore recommended to log the RSS before minimizing?
 A: The residual sum of squares already is the negative log likelihood of a Gaussian likelihood function.
Quick and dirty explanation
Assume you model your data $y$ as some true value $f(x)$ plus some error $\epsilon$ like so:  $$y = f(x) + \epsilon.$$ Now assume your error is distributed as zero mean Gaussian, i.e. $error \sim \mathcal{N}(0,\sigma^2)$ then also your measurement for $y_j$ (which is the value of $y$ at point $x_j$) is distributed as Gaussian around mean $f(x)$: $$y_j \sim \mathcal{N}(f_j,\sigma^2), $$
where $f_j = f(x_j)$. This means that the likelihood of observing $y_j$ given $f_j$ and sigma is
$$prob(y_j|f_j,\sigma) \propto \exp\left(-\frac{(f_j-y_j)^2}{\sigma^2}\right).$$ 
Now for a set of observed values $\{y_j\}$ at positions $\{x_j\}$ we have the likelihood
$$prob(y_j|f_j,\sigma) \propto \prod_j\exp\left(-\frac{(f_j-y_j)^2}{\sigma^2}\right)$$ 
assuming the measurements at each position are independent.
If you now take the negative log likelihood of that expression you end up with:
$$L = \sum_j\left(\frac{(f_j-y_j)^2}{\sigma^2}\right) +const.$$ 
which is the RSS expression you want to minimize. This is why it does not make sense to take the log again. Hope that helps.
Reasons for taking the log
The reason for taking the log is this: 
In principle you are interested in maximizing your likelihood $prob(y_j|f_j,\sigma) \propto \prod_j\exp\left(-\frac{(f_j-y_j)^2}{\sigma^2}\right)$. This however is a product of numbers (usually numbers smaller than one if you use normalization). If you want to maximize this expression numerically, it can quickly turn into a nightmare because your computer will quickly reach its precision limits.
Since the log is a monotone function it will produce the same maximum. But also it will reduce the product to a sum and also get rid of the exponentials for you in the process. This is computanionally great and also simplifies analytical calculations greatly. By convention you take the negative log likelihood to make this a minimization problem.
Now think about this: if you take the log again on the RSS expression you will still produce the same optimum. So there is no reason to do it. You might even potentially do yourself a disservice, when RSS goes close to zero because the log diverges rapidly near zero.
