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 Жыл бұрын
I'm glad you liked it. Thanks for watching!
@victor1978100 Жыл бұрын
I found this channel accidentally, searching for "square wheel maths", and I see this is a very good channel.
@mostly_mental Жыл бұрын
Glad you like it. Thanks for watching!
@swarnendupachal25793 жыл бұрын
Superb, very helpful. Keep making videos like this.
@mostly_mental3 жыл бұрын
Thanks for watching. I'm glad you liked it.
@namh3 жыл бұрын
interesting, thank you for your simple explaination
@mostly_mental3 жыл бұрын
Thanks. I'm glad you liked it.
@abdlazizlairgi96903 жыл бұрын
Thank you so much for the video it's so helpful
@mostly_mental3 жыл бұрын
Thanks for watching. I'm glad you liked it.
@manswind34172 жыл бұрын
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_mental2 жыл бұрын
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.
@manswind34172 жыл бұрын
@@mostly_mental Hmm okay nvm, I understand. Thanks anyway.
@dziugaschvoinikov4440 Жыл бұрын
What if we used logarithm with different base?
@mostly_mental Жыл бұрын
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.
@swarnendupachal25793 жыл бұрын
Sir, I need to calculate log(n!), What is the exact formula for this?
@mostly_mental3 жыл бұрын
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.
@swarnendupachal25793 жыл бұрын
@@mostly_mental thanks
@swarnendupachal25793 жыл бұрын
@@mostly_mental I think it will be 'ln' in place of 'log' in this formula
@mostly_mental3 жыл бұрын
@@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.