How can Stirling's approximations be used to simplify a log equation?

[UniPhysics90]

This is from a worked example involving using Stirlings approximations.

I have the log equation, where c<<1:

S=K(Nln(N)-cNln(cN)-N(1-c)ln(1-c))


In the next line of the example, this is simplified to

S=-kNcln(c)



I've tried a few ways of getting to this, but have had no success. I've tried making the last term 0, but still don't get the right answer.

Any help would be great :) Thanks

 

[Mute]

First off, I think you're missing a factor of N inside the last logarithm. I.e., your expression (aside from the Boltzmann constant) is

$$N\ln N - cN\ln(cN) - N(1-c)\ln(N(1-c))$$

From there a bunch of stuff should cancel, leaving you with

$$-Nc\ln c - N(1-c)\ln(1-c)$$.

Using the fact that c is small, you can show that the first term is much larger than the other term, hence you can neglect the other term.

 

[Redbelly98]

It looks like they are using

ln(cN) = ln(c) + ln(N)​

and then neglecting the ln(N) terms relative to -ln(c), i.e. N « 1/c.

UniPhysics90, from the context of what you are reading, does that seem a reasonable assumption?