A Wallis-type product for e. 

That's a really pretty result, and the proof is very clear. But how on earth did Pippenger come up with that formula? (Also, really cool to see one of my professors in the wild.)

Seeing as Sterling’s Formula is my favorite math result, I definitely want to see the video!

6:30 it should say not 2^(2^(n-2)) terms but 2^(2^(n-2)) total pulled out for those confused

Very interesting. Also, by dividing the Wallis product of pi by the product formula of e it is possible to define a neat product formula for pi/e.

Awesome exposition. Huge potential for muddle here

23:42 Michael,yes! It is a really good idea about Sterling's approximation. I would like to see how to get this useful fact!

Yes for the Sterling formula video!

I'd like to know if there's something similar for gamma (the Euler-Mascheroni constant)... and if finding such an infinite product would enable a proof of gamma's irrationality (and possibly also transcendentality).

Here is another strange equation: Binomial(j,j/e) ~ 1.4027, j ~ 1.2954 , Slope ~ 1.4142 Anyway, hard to understand, but nice beautiful pattern with exponents. Great video!

Astonishing to learn that pi and e are so closely related. What's the intuition behind that?

I haven’t seen a proof of Sterling’s approximation in years. Please upload one!

Absolutely astonishing. I don't think I've ever seen anything like a relation between pi and e that is quite so simply "arithmetic" outside of complex analysis. The fact that the change of the exponent from 1 to 1/(2^n) takes us from pi to e (or, more precisely, e/2, but that's still in the "world of e") makes one consider the range of exponentiated "a-sub-n" over "b-sub-n" products in general, for an arbitrary domain of exponents. Call that domain E of exponents "E "-- Do other values for E yield other "important numbers," like but other than e or pi? What about E=1/n? Or maybe E=1/n! ? Or E=e^(-n)? E: Most enticingly: What is the class of functions E=f(x) for E over the reals that has both pi an e in its range for special cases of x? Does it yield phi (golden mean) as well?!

Would love to see the approximation video!

perfect.well done

That was tough! Thank you for nice presentation!

yes to the Sterling's Approximation video!

Can you help me solve this problem? {A_n+2}+{A_n}={a_n}×{a_n+1} Write this recursive sequence in terms of the first and second sentences of the sequence

I was waiting for e emerging from the definition and ended up just being dropped as another formula (the approximation) so I'll wait for that video hoping the connection does come up from the ground instead of being taken for granted

So, if we take finite products that end after each parenthesized set (doubled to offset the 2 in the denominators), they increase to π in one case and increase a little more slowly (because we're including 2^(n-1)-th roots instead of the whole thing) to e in the other case. In some sense, that makes e a version of π, when seen as limits, where you're taking care not to increase the products quite as fast for e as you do for π. There must be a deeper reason here that e and π are related in this way. Is this a version of their relationship that derives from Euler's formula e^(πi)=-1? Or is this a different relationship entirely? Maybe the answer is to wait for Michael's video on Stirling's formula, since that also expresses a relationship between the constants in terms of factorials?

This was thorough bonkers.

That was a big one.

Very nice video, but it's Stirling, with an "i". :-)

Looks like a product of geometric means.

It is interesting that pi is given by the product of ratios, yet pi is not rational. Presumably this is because it is an infinite product (although this seems to be counter-intuitive)? I know that an infinite sum can converge to a limit that is outside of the space of the terms in the sum (if it is not in a Hilbert space), but the product of rational expressions (as in this example) has a numerator and denominator that are both integers, so the product should have the form of a rational expression, no matter how many terms are multiplied?

So good

It's almost nauseating to see how similar the expression for e is to Wallis infinite product for pi They seem to be completely non related constants🤷‍♀️ and well here we are....

I would like a video on Sterling's approximation

I wonder if it is legal to reindex n, because ln(P) is conditional converge, such that rearrange its terms gives different result.

That was intense. Pretty sweet result though.

Stirling, not Sterling.

I want to see the sterling video

do a pushup everytime he says "two" or "square"

awesome one

Great!

I've got a 'personal' question (unrelated to the video). Are the videos still being edited by the same person as those a few weeks ago? I believe their name was Stephanie? I feel like the editing is less present in the recent videos (not saying that's a bad thing though, I just want to know).

Plaease upload the Stirling approx video! We always use it in physics and statistics (especially for calculate log(n!)) but noone explain why it's true

Amazing

Beautiful and yet aweful at the same time

an approximation kinda takes the point out or

This was one of the longest problems you ever did

Why are we allowed to replace the terms with their approximations using sterling’s formula?

I think it's Stirling, not Sterling.

رحلة شاقة لكن الوصول مريح

Is this the Futuna product?

it looks like a half decent IQ question: what is the bracketed number raised to the 1/128th power?

Please avoid mixing lower case and upper case N in the same formula. There are so many other letters you could use instead.

:)