I still say the only real way to prove Laplace's Method is to use rigorous mathematics, but, either way, you did a great job. (+1). I am rooting for your channel to get a surge in subscribers. You produce high quality content for sure. Can't wait to see the next video! (assuming you're going to continue to produce more)
Thanks for these videos. Your explanations are very clear and easy to follow. I'm currently about to start college, but enjoy looking into complex mathematics for fun. Your videos allow me to learn new things and broaden my understanding of math and engineering.
It made me sad seeing the sqrt(2pi n) getting ignored in the final formula. Also, I've noticed that Gamma(x+1) is unreasonably well approximated by sqrt(2pi(x+1/6))(x/e)^x. Why is so good in absolute difference for x < ~6, and very good in relative difference for all x?
It is because by the binomial series sqrt(x+1/6)=sqrt(x)sqrt(1+1/(6x))=sqrt(x)(1+1/(12x)+...) which approximates the asymptotic series of the gamma function. See mathworld.wolfram.com/StirlingsSeries.html
If x is sufficiently close to x0, then (x-x0) to a high power is negligible. As a numerical example, suppose x0 = 1 and x = 1.01. Then, (x-x0)^3 = (1.01-1)^3 = (0.01^3) = 10^(-6) which is very very small. As the power goes from 3 to 4 to 5 etc, (x-x0) gets even smaller when x and x0 are sufficiently close. That's why I can ignore the higher order terms. Hope that helps!
Why is this channel so underrrated
You can always share my work on social media and let your friends know about it to make it slightly less underrated!
keep it up .. compact concise and clear explanation .. Thanks for the effort
I still say the only real way to prove Laplace's Method is to use rigorous mathematics, but, either way, you did a great job. (+1). I am rooting for your channel to get a surge in subscribers. You produce high quality content for sure. Can't wait to see the next video! (assuming you're going to continue to produce more)
Wow this is the only video that also derives the sqrt(2pin) part. Thank you so much!
I knew of stirling's approx for the gamma function but never have I seen it derived... that was awesome, great videos!
Thanks for these videos. Your explanations are very clear and easy to follow. I'm currently about to start college, but enjoy looking into complex mathematics for fun. Your videos allow me to learn new things and broaden my understanding of math and engineering.
Thank you for making this amazing video
simply superb
Very clear. Thanks!
simply amazing!!!
love your lecture thanks
7:07 ah when I derived stirlings formula that point went straight over my head!
Amazing video, but can someone explain why we take the exponential of f(x)?
Thank you! We might want to take the exponential of f(x) for an integral we choose to find; could be a multitude of things!
Amazing, amazing , amazing.
Great
It made me sad seeing the sqrt(2pi n) getting ignored in the final formula.
Also, I've noticed that Gamma(x+1) is unreasonably well approximated by sqrt(2pi(x+1/6))(x/e)^x. Why is so good in absolute difference for x < ~6, and very good in relative difference for all x?
It is because by the binomial series sqrt(x+1/6)=sqrt(x)sqrt(1+1/(6x))=sqrt(x)(1+1/(12x)+...) which approximates the asymptotic series of the gamma function. See mathworld.wolfram.com/StirlingsSeries.html
@@gergonemes88 thanks!
Could You derive normal distribution just from binomial distribution after that? Best regards
When you say that M is large, how 'large' it actually must be? How can determine the largeness of M?
There's no specific cutoff, but M is large is taken to mean as M approaches infinity.
Hi! can you please tell me the software and hardware you are using for the making of this video?
3:15 may i know how you can cancel out all higher order terms ?
If x is sufficiently close to x0, then (x-x0) to a high power is negligible. As a numerical example, suppose x0 = 1 and x = 1.01. Then, (x-x0)^3 = (1.01-1)^3 = (0.01^3) = 10^(-6) which is very very small. As the power goes from 3 to 4 to 5 etc, (x-x0) gets even smaller when x and x0 are sufficiently close. That's why I can ignore the higher order terms. Hope that helps!
3rd veiw 1st to comment
First time in my life
Beautiful voice
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