Stirling's Approximation - Close Encounters

[@vwcanter]

This is a lot better than the other videos /lectures on this subject. Because you don't appeal to other, more advanced tools, before getting started. And also, starting with an "eyeball" estimate is incredibly helpful. The drawing with the bars of the reimann sum is incredibly helpful. Because the limits of the integrals are right on it, something that seems random in the other lectures.

[@mostly_mental]

I'm glad you liked it. Thanks for watching!

[@victor1978100]

I found this channel accidentally, searching for "square wheel maths", and I see this is a very good channel.

[@mostly_mental]

Glad you like it. Thanks for watching!

[@swarnendupachal2579]

Superb, very helpful. Keep making videos like this.

[@mostly_mental]

Thanks for watching. I'm glad you liked it.

[@namh]

interesting, thank you for your simple explaination

[@mostly_mental]

Thanks. I'm glad you liked it.

[@abdlazizlairgi9690]

Thank you so much for the video it's so helpful

[@mostly_mental]

Thanks for watching. I'm glad you liked it.

[@manswind3417]

Another well-explained and informative video, thanks for this! (tho I wanted to go a bit deeper in those bounding integrals in section 3 :D) A small error at 5:30 though: the graph you drew below log(n) to find a lower bound should ideally start from 1 not 0; so taking integral log(x)*dx from 1 to n over 0 to n would yield a better/tighter lower bound on our log sum.

[@mostly_mental]

Good catch. And I had originally planned to go into more depth on the bounds, but it ended up being a lot of fiddling with technical details, and that would have interrupted the flow too much.

[@manswind3417]

@@mostly_mental Hmm okay nvm, I understand. Thanks anyway.

[@dziugaschvoinikov4440]

What if we used logarithm with different base?

[@mostly_mental]

Logarithms in different bases only differ by a constant factor, so not much would change. Say we did this with base 2. In the step where we integrate (around 5:30), we would pick up a factor of log_2(e). Then when we get rid of the logs (6:58), we would end up with 2^(-n log_2(e)) instead of e^-n, but by logarithm rules, those are really the same thing. So we'd arrive at the same formula.

[@swarnendupachal2579]

Sir, I need to calculate log(n!), What is the exact formula for this?

[@mostly_mental]

Unfortunately, there isn't a nice exact formula for it. That's one of the reasons we need approximations like this one. But taking the log of Stirling's formula gives (n + 1/2) log(n) - n + log(2pi)/2, which is a pretty good estimate. And if you need something with a bit more precision, there are some other estimates like Ramanujan's approximation (www.johndcook.com/blog/2012/09/25/ramanujans-factorial-approximation/) which you can work with.

[@swarnendupachal2579]

@@mostly_mental thanks

[@swarnendupachal2579]

@@mostly_mental I think it will be 'ln' in place of 'log' in this formula

[@mostly_mental]

@@swarnendupachal2579 When you see log without a base in math, it's usually assumed to be the natural log (base e). In computer science, it's usually base 2, and in the natural sciences, it's base 10. In this case, it doesn't actually matter, since you'll get the same formula no matter which base you use.

[@swarnendupachal2579]

@@mostly_mental Thanks for the clarification.