A Wallis-type product for e.
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- čas přidán 11. 09. 2024
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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.)
If Pippenger is your professor, you could ask him?
@@Alex_Deam He was when I was in college, but that was a long time ago.
@@mostly_mental ah fair
Seeing as Sterling’s Formula is my favorite math result, I definitely want to see the video!
Agree
Agreed! Having studied physics, I really like Sterling's approximation. It takes problems in statistical mechanics that would be completely impractical to calculate and turns them into something you can do with a handheld calculator.
6:30 it should say not 2^(2^(n-2)) terms but 2^(2^(n-2)) total pulled out for those confused
Thank you I was very confused
I was completely flummoxed, thanks!
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.
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).
There's a known wallis-type product for e^gamma but it seems like we don't know of a wallis product for gamma
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!
Astonishing to learn that pi and e are so closely related. What's the intuition behind that?
If you take your glasses off, all infinite series are the same.
You can see the connection in many simpler formulas, starting with Euler's identity.
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!
Yes for the Sterling formula video!
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?!
I haven’t seen a proof of Sterling’s approximation in years. Please upload one!
He made a video some time ago
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
First and second terms of the sequence, not sentences.
@@megauser8512
what do you mean my friend
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
This is a decent video if you're interested: czcams.com/video/JsUI40uSOTU/video.htmlsi=WD8HfGKfzFLpJtV_
The e comes from the fact that the Gamma function is an extension of the factorial to the real and complex numbers. And the Gamma function uses e in its definition. The recovery of the "discreteness" of the factorial comes from using integration by parts on the integral definition of the gamma function, where the unique properties of e^x allow the extraction of the falling power for each step.
Would love to see the approximation video!
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?
You are right. But why powers of 2 are used? Because e is a simplex, that's why, while pi is a sphere.
2D example. e/pi=2.72/3.14=0.866=sin(60)=1/1.155. This is are "diametr of 2D sphere /length of 1D edge of inscribed 2D simplex".
3D example. This is are " "diametr of 3D sphere /area of 2D triangle of inscribed 3D simplex".
Length of 1D edge of inscribed 3D simplex in a unit sphere equals 1.63299932. Use Heron's formula Area= sqrt(3)/4 *1.63299932^2=1.155.
I wonder if it's the grouping with the parentheses being a little misleading for the Wallis example up top. My conceptual understanding of the Wallis product is that an appropriate grouping would be each "group of 4" (two numbers in the numerator and two in the denominator) at that fixed size. Whereas the exponents for e keep doubling the size of the grouping.
But since Pi also shows up in Stirling's formula, maybe they're more related like you're saying
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?
The same happens with infinite sums, the sum of rationals is rational, but only a finite sum. In fact every real number is the infinite sum of rationals (it's one of the definitions of the reals as Cauchy sequences). This is indeed counterintuitive, but this is only because we think of infinite sums as a type of sum. In fact it's better to think of infinite sums as just limits (of the partial sums). Then we see that all the strange behaviour happens because limits don't preserve all the properties reals can have (eg limit of positive terms need not be positive)
Irrational numbes are in a way defined as the "missing limits" of sequences of rational numbers (drastically oversimplified of course). So this is actually to be expected.
yes to the Sterling's Approximation video!
Looks like a product of geometric means.
perfect.well done
Very nice video, but it's Stirling, with an "i". :-)
This was thorough bonkers.
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'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).
That was tough! Thank you for nice presentation!
Stirling, not Sterling.
That was a big one.
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.
Shifting the index or reordering a partial(finite!) product before taking a limit is always valid...
I want to see the sterling video
I think it's Stirling, not Sterling.
Beautiful and yet aweful at the same time
That was intense. Pretty sweet result though.
do a pushup everytime he says "two" or "square"
it looks like a half decent IQ question: what is the bracketed number raised to the 1/128th power?
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
رحلة شاقة لكن الوصول مريح
Is this the Futuna product?
an approximation kinda takes the point out or
So good
Why are we allowed to replace the terms with their approximations using sterling’s formula?
You basically multiply and devide the term with his approximation (so you dont change anything), now , the apporaximation say that in the limit n!/sterling is 1 so you left with sterling alone
@@bluelemon243 ah the fraction is one in the limit I see
Doing that assumes the original expression itself converges though, which one would have to show
This was one of the longest problems you ever did
awesome one
Amazing
Great!
Please avoid mixing lower case and upper case N in the same formula. There are so many other letters you could use instead.
It's pretty standard. It works fine as long as you write them differently as Michael does.
@@D.E.P.-J. It is standard, but the problem comes in when he is reading the formulae. I agree with the other poster that I'd prefer to use different letters simply for "production value" clarity.
@@D.E.P.-J. it doesn’t work when you read the formula aloud, unless you specifically say “small n” and “capital n” each time, which Michael didn’t.
Don’t make formulas harder to read and to check than they need to.
There’s a pretty good reason that math is commonly communicated through writing/reading instead of speaking/listening, Imagine going to a math class and not being shown anything visually, just the lecturer reading out all the expressions and formulae for the entire class.
@@divisix024 indeed.
In any case, both when writing and when speaking, I was taught to avoid unnecessary sources of confusion. It’s a form of courtesy to the reader / listener, and it helps the writer / lecturer to avoid mistakes.
:)