what a difficult limit!!
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Probability method: provided limit is the probability that random variable with Poisson distribution with parameter n is less than n. Central Limit Theorem implies that the given limit equals Ф(0) = 1/2
Wow that’s beautiful
Thank you
Yeah! I had this problem in a probability theory homework last week and that was the solution lol.
I knew beforehand this probability approach, but also notice the integrals Mr.Penn studied are just the remainder in the Taylor's formula in its integral form,, so either way we get a more understandable solution.
You meant with parameter k less than n and not n less than n as you wrote!
@@heliumfrancium8403 Not really?
At 17:17, small typo: (x-1)^n should be (1-x)^n. That would have broken the end result.
If you pull the e^(-n) inside the summation, the summand is just the pdf of the Poisson distribution with mean n. As n gets large, Poisson converges to normal distribution with mean n. So summing from 0 to n is like finding P(k
You just copied the math stack exchange answer
I think in essence that's what he's doing, but rigorous.
Saying one distribution family approaches another is wishy washy and imprecise. In physics that's acceptable, but this is a math channel.
As n tends to infinity the poisson distribution IS a normal distribution.
You can proof it starting with a binomial distribution, find the convergence and the correction, see that as the limit of n going to infinity the correction goes to zero.
Since the poisson distribution is a subcase of the binomial you get that the reasoning with the poisson distribution is exact and rigorous. It's a standard exercise in probability courses.
Since one has knowledge about the Poisson distribution why not use it? You can call it a trick but it's correct, exact and rigorous.
The video is clearly been made assuming the viewer doesn't have further knowledge. This makes the video enjoyable for everyone even without knowing much about probability.
He doesn't need to kill the exercise, just make it interesting and maybe teach, to at least someone, something new.
@@MooImABunny
True, but the proof is relatively straightforward using moment generating functions. Another way to think of it… poisson(n) can be thought of a sum of n iid poissons(lambda=1).As n goes to infinity, sum of n iid variables converges to normal via the standard CLT
@@MooImABunny How is saying that a sequence of distributions converges to another distribution wishy washy? Were you just referring to the fact that entine4 did not write out the entire argument in their comment? Or am I missing a technical detail?
My first thought when I saw this was “Poisson distribution” lol
Appeared as Problem E2723 in American Mathematical Monthly in 1978. Related to Szasz generalization of Bernstein polynomial approximations.
@ 18:09 I think it should be 1-x not x-1
If we set λ as the value of the limit, i think it's easy to show that λ=1-λ, thus λ = 1/2. The idea is to add and subtract the sum from m to infinity of n^k/k! so you have that λ = lim(1 - e^(-n) * sum (m, +oo, k^m/m!)) and with substitution of the summation variable το generate again the Σn^ν/ν! with ν from 0 το Infinity), thus having λ = 1-λ, leading το λ = 1/2
Was not expecting that, crazy limit
Yes, 1/2 feels a bit unexpected. 1 would be believable. So would 0. But it's right in the middle.
18:20
No
7:15 hold on, that can't be right, you can set x=1. Then f(x) = 0, but the RHS is a_n*e-½, which is clearly positive...
In the next line you say we want to decrease the value so we shrink the bounds of integration to [0, n^-¼], which is valid.
I think the claim should be that f(x) ≥ a_n e^(-x²/2) on that new range.
Personally I would have made use of the central limit theorem
Beautiful simply
4:10 doesn’t this integral evaluate to sqrt(2pi/n)? In which case the argument results in L \leq 1, not 1/2, so the later squeeze theorem stuff doesn’t work?
..a real marvel of data !!
7:15 hold on, that can't be right, you can set x=1. Then f(x) = 0, but the RHS is a_n*e-½, which is clearly positive...
These are extremely adhoc methods.
Why doesn't SUM(n^k/k!), k=0..n converge to e^n when n->infinty?
Stolz-Cesàro theorem?
The limit should be one. Take another look at the two Gaussian integrals. They were not computed correctly.
Very Nice
3:08 Do not understand HOW we can extend to infinity and still say that the integral will be less. Comparing integrals with same boundaries is ok, but not a different ones
Given that g(x) = e^(...) > 0, if you extend the interval over which you integrate the value of the integral increases. He basically condensed two steps into one: intg-0^1 f(x) < intg_0^1 g(x) < intg_0^inf g(x)
@@ivandebiasi6657 Ah thanks. Forgot we were dealing with an equality and was super confused by that step.
The sum is the beginning of the maclauren series for e^n, so we can rewrite as e^n - SUM(n^k/k!) for k>n. Distributing the e^(-n) we get 1-SUM(e^(-n)n^k/k!). So my guess for the limit is 1, I think the tail will die to zero in the limit. Let's see if I'm right.
Oof. Well, I wasn't too far off! I think with some work I could show the tail bit goes to ½ so the limit also goes to ½.
It's not the McLauren series of e^n since it depends on the upper bound of the sum
@@josephcamavinga9721 I did not say it was the maclauren series, I said it was *the beginning* of the series. That distinction is necessary and important.
I just looked at the store - nowhere does it say "A good place to shop." ... Seriously?
How can I suggest a neat proof of the Euler product formula for the Riemann zeta function? I'm not sure a RUclips comment is the best place
How can anyone understand this?
It is understandable from the solution why he came up with the integral. But those inequalities especially the second one is not easy to find. The first inequality is also not easy. May be if someone is lucky and thinks that this has to involve gaussian intergrals may be motivated to come up with those inequalities. This is clearly extremely adhoc. But nice solution.
Not the most intuitive proof I've ever seen. Kinda hard to follow.
One of the most important parts of education or presentation in general is knowing when to present information so the flow of logic seems consistent and understandable. Why would you ever introduce that random limit at the start and then justify its appearance? It just makes everything way worse to follow
That type of presentation is very common in mathematics. And I don't think it's wrong to do things that way. Of course, always keeping in mind that behind that neat blackboard, there are several pages of frustrated calculations.
You wrote f wrong the 2nd time around (e^(-x) instead of e^x) and we should have a_n^n. Ignoring these, you "proved" e^(-x^2/2) a_n inf, this shows f(x) = e^(-x^2/2). A bound on f which depends on n is highly suspicious, you must have some constraint on x.
exactly. Watching past that makes me think this inequality holds on the shrunk range [0, n^(-¼)]
I've been stuck in the same part, it is clear that a_n->1, but a_n
Fir.... sorry, Second.
This only works if, n! ~ ( ) and also always less than. Is that true?