sRGB's linear segment to avoid infinite slope - why?

Question

In this article about sRGB (https://en.wikipedia.org/wiki/SRGB) is stated, that the gamma transformation has a linear portion near zero, to "avoid having an infinite slope at K = 0, which can cause numerical problems". I'd like to know what's the problem with that.

Answer 1

There are two answers, as is usual for gamma. The modern variant is:

The problem is that with an infinite slope you need "infinite" resolution (many bits of storage) in order to arrive at a linear representation that is invertible to gamma-encoded without loss. In other words, it allows for a small lookup table to produce an invertible linear encoding (8bit -> 10 bit -> 8 bit).

The numerical problem is most easily understood on the first step (8 bit -> 10 bit). With an infinite slope near zero, you need a much bigger encoding range to stay faithful and reversible, i.e. you'd need more than 16 bit (assuming integer coding, halfs do not have this problem).

The linear equivalent of #010101 or 1/255th with square (gamma = 2.0) coding is 1/(255*255)th. You would need 16 bits to represent that faithfully, and using 2.2 not 2.0 as an exponent would make it worse. These quite small numbers are just a corollary of the coding function, and in practice you don't need much resolution in a lightness range that is, roughly, black. So the linear segment helps coding by not wasting resolution to detail around black (or near zero).

The slightly older answer, taken from

http://www.poynton.com/notes/colour_and_gamma/GammaFAQ.html#gamma_correction

is that in some equipment this linear segment will be less sensitive to noise. This is probably mostly true of the analog signal path.