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Possible bug with Cauchy Loss #142

@theartful

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@theartful

This is currently not really a bug, but thought I would share either way.

The implemented cauchy loss formula is:

$$\rho\left(r^2\right) = ln\left(1 + \frac{r^2}{t^2}\right)$$

Instead of:

$$\rho\left(r^2\right) = t^2 \cdot ln\left(1 + \frac{r^2}{t^2}\right)$$

Mathematically I don't think this has any effect, since the residuals/jacobians will just be scaled differently.
But in my use case, I was using multiple loss functions, and a threshold schedule for some of them. Other loss functions approximate r^2 below the threshold, but this implementation has a different much smaller scale (the bigger the threshold, the smaller the scale), which made its contribution negligible.

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