people below average IQ: just compare term by term. people with average IQ: NOOOOOOO YOU MUST USE STIRLING APPROXIMATION people above average IQ: just compare term by term.
@karimalramlawi72289 ай бұрын
Flammy : "so we are going to go with sandwich theorm" Also Flammy : (Continues to go with squeeze theorm)
@PapaFlammy699 ай бұрын
xDDD I noticed that too when editing :'D
@davidemmanuel94188 ай бұрын
Same thing, aren't they?
@karimalramlawi72288 ай бұрын
@@davidemmanuel9418they are the same ,but it's funny After he committed to call it sandwich theorm he continues to call it squeeze theorm
@highviewbarbell7 ай бұрын
Sandwich theorem uses bread and squeeze theorem involves sauce dude@@davidemmanuel9418
@raytheboss46509 ай бұрын
the beginning meme ☠
@cubekoss75479 ай бұрын
Guac and balls
@rachellua54819 ай бұрын
The super guac guac 9000
@nate0___9 ай бұрын
that guac is longer how 😭
@themanthemyththelegenda8 ай бұрын
guac and ball torture
@creatermania8 ай бұрын
💀💀
@ich68859 ай бұрын
I guess the physicist's answer would be the forall x: x!/x^x => lim = 0 and for the computer scientist it will be 0.000000000000000047385
@Harmonicaoscillator9 ай бұрын
No, a physicist’s answer would be the “intermediate” level where he uses sterling’s approximation for a factorial
@ilangated9 ай бұрын
Cool video! Now do it with an epsilon-delta proof
@PapaFlammy699 ай бұрын
breh 3:
@Me-pt8sv9 ай бұрын
nah do the neighborhood version
@hhhhhh01753 ай бұрын
it's epsilon-N since x is approaching infinity. N = ceil(1/epsilon) works by inspection
@adarshiyer48059 ай бұрын
Very nice 👌. I assume x is an integer, otherwise the factorial doesn't work. Of course, if the functional limit works over the integers, then it does for any sequence which approaches infinity.
@proutmobile12289 ай бұрын
you can generalize it with integrals
@grantarneil81429 ай бұрын
Yeah, n! = Γ(n+1). And Γ(n+1) is defined for all n ε C\Z^-. So as long as you're not computing the factorial of a negative integer (for which it diverges), you're all good,
@wulli_9 ай бұрын
You probably meant x to be a natural number, since i am not sure how to define the factorial on negative integers. In this case there are no applicable non-integer sequences anyway, so the second statement is true, but pointless. However, the same limit can be shown, when extending the factorial to the gamma function. Regarding your second statement in general, while the converse is true, since the real sequences contain the integer sequences, it is wrong. As a counter-example, let f(x) = sin(pi*x)*exp(x), then for all integers z we have f(z) = 0 and therefore any limit over an integer sequence is 0. Now take the sequence of half integers a_n = n + 1/2 we get f(a_n) = exp(a_n) if n is even and f(a_n) = -exp(a_n) if n is odd, which does not converge. It does work for monotonic functions f (like the one from the video when restricted to positive numbers and extended using the gamma function), since for a real sequence a_n that tends to infinity, f(a_n) can be bounded by f(floor(a_n)) and f(ceil(a_n)), where floor(a_n) and ceil(a_n) are integer sequences which also go to infinity.
@CAG29 ай бұрын
I think you can just say x! / x^x
@abelhivilikua87358 ай бұрын
@@grantarneil8142 how about the factorial of fractions?
@windowstudios459 ай бұрын
X^x is always greater than X! Same number of numbers (X) But instead of counting up (factorial), the exponent replaces the smaller numbers with the biggest number (also X) Therefore, since x^x also increases at a faster rate, the function is an ever-shrinking fraction. This means that it will eventually approach zero as X gets bigger and bigger.
@tianyuema47979 ай бұрын
Actually, x^x is not always greater than x!. 0
@justinkim5379 ай бұрын
That condition itself is not strong enough, the fact that the fraction is shrinking does not necessarily imply that the limit is zero. Forall x>2, 0.5x < x-1, 0.5x grows slower than x-1, and 0.5x/(x-1) is decreasing on x, but the limit of 0.5x/(x-1) is 1/2.
@pneujai9 ай бұрын
I thought the pro one would be an epsilon-N proof and the squeeze theorem one would be the intermediate one :(
@karimalramlawi72289 ай бұрын
Me too My expectations were too high
@MClilypad9 ай бұрын
For the epsilon-N proof, we can use a variant of sterlings approximation which provides a strict upper bound on the factorial, with which we can computer a strict upper bound on the fraction. From there it is not too hard to find a lower bound on N such that the fraction is less than epsilon by inverting the lower bound.
@WhosParx9 ай бұрын
definitely buying a couple hoodies papa flammy thanks for the discount code
@PapaFlammy699 ай бұрын
The support is highly appreciated, thank you so much!!!!
@tererere38779 ай бұрын
Funny thing is i was teached about sandwich in highschool
@PapaFlammy699 ай бұрын
awesome! :)
@tererere38779 ай бұрын
@@PapaFlammy69 not really, noone really understood it then and we had this on a test, at least when it was introduced on Uni i wasnt totally clueless
@Sir_Isaac_Newton_9 ай бұрын
@@tererere3877mood
@mr.inhuman79329 ай бұрын
Damn. That was a clean argumentation.
@bertilhatt9 ай бұрын
Not my experience in anal class
@Rafau859 ай бұрын
Beginner was dirty though ;)
@gonzalezm2448 ай бұрын
Did it the “PhD way” when looking at the thumbnail thinking it was gonna be the beginner way 😂
@KAI-wn1pg8 ай бұрын
Did it with using log and the harmonic series approximation! Lovely question!
@diogeneslaertius33659 ай бұрын
Fun fact: Avocado means Testicle in one of the Aztec languages which makes the meme even more fun.
@PapaFlammy699 ай бұрын
:^)
@__christopher__8 ай бұрын
Actually you only need the upper bound, as the expression is quite obviously positive and therefore 0 works as the lower bound (BTW, you also need that x^x is positive for your division, not just that it is nonzero, because you are dealing with inequalities).
@b0ngdon8839 ай бұрын
you can prove by induction that the limit holds for the succession {a_n}n = n!/n^n. I guess that is good enough to also prove the limit of this video, since it looks like you treated x as a natural number when you wrote x! = x(x-1)(...)*2*1
@Aman_iitbh9 ай бұрын
Ya if its treated as seq then ratio test for limit of seq can be invoked directly
@ianweckhorst32009 ай бұрын
Also, I was thinking about a good way to really compare the two, I first thought a sum would work, but that would just be infinity, then I thought, weirdly not directly of integrals, but some sort of infinite average, but after that I realized that given my own experimentation, an infinite mean of that nature is just a simplified and halved version of an integral, don’t ask me why it’s halved, I just know it is from bs on desmos, Symbolab and a weird visual programming app called oovium for which I’m trying to define all the things that didn’t come pre installed with it, like calculus and various combinatorial functions, and as I recently found out gcf and lcm are not included, which means that I will likely do that first
@zyxzevn9 ай бұрын
But what if you have X!/X^(X-1) ?
@V-for-Vendetta019 ай бұрын
sandwich theorem ftw. also, the avocado looks a lot like a geoduck.
@neutronenstern.9 ай бұрын
0:22 the inner pigdog awaked. Im truely sorry for you. oink oink
@ANunes069 ай бұрын
5:50 - "e to the x grows like this. It's a fast fuckin' boi."
@dfcastro9 ай бұрын
I though you would use gamma function to express x! and them operate the limits, maybe even using L’Hospital.
@ianweckhorst32009 ай бұрын
Although back on point, I think the integral might actually be e at infinity, but I have nothing to go off other than desmos and sloppy guesswork
@andreapaolino59059 ай бұрын
seems to me that the beginner version should only apply to sequences of natural numbers, right? I mean, the expressions x! and x^x are not _really_ defined in the same way for real numbers. For example: expressions like pi^pi or pi! do not naturally translate into phrases like "multiply pi by itself pi times" so on and so forth...
@beastbum9 ай бұрын
$1 limit vs. $5 limit vs. $100 limit Buzzfeed
@bramss9999 ай бұрын
Is making it an integral wrong?
@PapaFlammy699 ай бұрын
no, you can also do that!
@aimsmathmatrix9 ай бұрын
Some content on measure theory? That'd be fun! or some topology, there's a lot of really nitty results out there!
@Happy_Abe9 ай бұрын
The noises are the greatest part of these videos!
@symmetricfivefold9 ай бұрын
ngl, that avocado made me dirty 💀
@PapaFlammy699 ай бұрын
:^)
@dAni-ik1hv9 ай бұрын
this guy is like if shitposting was an integral part of math
@a17waysJackinn9 ай бұрын
what the hell with the avocado, i am going to warch a math content, brooo lmao...
@PapaFlammy699 ай бұрын
:'D
@PragmaticAntithesis9 ай бұрын
Noob solution: Prove by induction that x^(x-1)>=x! for all x>=1. This is left as an exercise for the reader. Rewrite problem as lim(x->inf): (x!/x^(x-1))/x. Let f=x!/x^(x-1). 0inf): f/x. This is finite/infinity, so the limit is 0.
@zakyy_178 ай бұрын
An other good method is to considere the serie sum(fact(n)/n^n) then prove its convergence using the d’Alembert method then use the necessary condition of series convergence (Un converge => lim(n->infinity)Un = 0). Am i correct?
@koenth23599 ай бұрын
Will it leave you with analitis, if you follow that class?
@liamw69767 ай бұрын
This is such a beautiful explanation, thank you
@JSSTyger8 ай бұрын
My initial guess is 0 because x^x blows up faster than x!
@zxyjulzeeeks8 ай бұрын
The beginner way seems incorrect. The product rule of limits applies for fixed and finitely many terms. Here the number of terms itself is growing. You can construct 0=1 type ‘proofs’ using this
@PapaFlammy698 ай бұрын
As mentioned, it's a heuristic approach only.
@garrikwolfe8 ай бұрын
I appreciate that you have helped solved some problems brought on by my Calculus courses. However , I find it distracting that you refer to Analysis as Anal and I'm having trouble watchimg
@somedud11408 ай бұрын
"Pro" is beautifully simple! basically lim->♾x!/x^x ♾x^(x-1)/x^x = lim->♾1/x=0. and of course all of these limits are >=0, because both numerator and denominator are positive. So it's 0♾x!/x^x♾x!/x^x=0.
@manasawalaaamir9 ай бұрын
the Beginner way seems flawed because you can only split products of its individual limits, if the terms are finite, with x approaching infinity thats not the case.
@RichardBuckman8 ай бұрын
I was going to say that too
@armstrongtixid68738 ай бұрын
Can't you use AM-GM here? AM of the values 1 through x = 1/2(x+1). By AM-GM (1/2*(x+1)) > (x!)^(1/x) (1/2*(x+1))^x > x! and given that we have the value < 2^-x * ((x+1)/x)^x ==> value < 2^-x (1+1/x)^x ==> lim as x goes to infinity: 2^-x * e and this goes to 0.
@_anonymousxd8 ай бұрын
You can also solve it by doing the series ratio test
@SpennyBoi9 ай бұрын
i didnt catch that code papa flammy could you say it again?
The beginner way is wrong. if x grows, then the number of factors also grows. a small estimate would be correct here.
@PapaFlammy699 ай бұрын
As mentioned, it is a >heuristic< approach that utilizes that the denominator is outgrowing the numerator
@karelvanderwalt36259 ай бұрын
factors not terms, right?
@markerena22749 ай бұрын
@@PapaFlammy69yes, but you cannot split the limit into a product of multiple limits if there are an infinite amount of terms in the product It can lead to the wrong result in many cases
@ExplosiveBrohoof9 ай бұрын
A more rigorous heuristic approach would be to just take the product of the limits 1/x and x!/x^(x-1) instead of the infinite product.
@DrR0BERT9 ай бұрын
@@PapaFlammy69 It's still a bit of a stretch to use that approach. I would have a major issue with that if one of my students gave that as a solution.
@오성연-j6l8 ай бұрын
how about using gamma function? Can that be a valid approach?
@risedown52029 ай бұрын
How about we just use lim(as x -> inf) of An = lim(as x ->inf) A(n+1)/An ?(An is the general term of the series (An), where n belongs to N or any "variation" of N)
@epsilia36119 ай бұрын
sin(2pi*n) has a limit at +inf, but sin(2pi*x) doesn't. That's why I think there is a lack of argumentation in the so called "PhD" way. When you say "x!=x(x-1)...(2)(1)", this equality holds for x being a positive integer, poggers. And if you want to show it to an anal class, you definitely don't want to make use of the gamma function, which I suppose wasn't talked about here in that regard. So without this tool, I suppose you really need an argument like : "Let (u_n) be a sequence over N such that u_{n+1}=n*u_n. Then u_{n+1}/u_n = n, so u_{n+1}/u_n > 1 for all n > 2, which proves how (u_n) strictly increases after some n. Plus, u_n = n! for all n > 2. So lim u_n = +inf, so """lim x! = +inf""" as well" I don't know exactly how the end of what I just wrote would need to be justified, but I hope people in the comments can find it out if they know better about it than me ! I thought, to do the same thing with the x^x function, that we could be ordering quantities on intervals of size 1, with monomials, appearing with the ceiling and floor functions. What do you think ? All things considered, a very interesting video, I appreciated !✌
@Noam_.Menashe9 ай бұрын
I think his way will work if one can prove that in the case of x! being defined as the Gamma function, when 1
@saadbenalla36789 ай бұрын
Hy don't you make a video about Fourier transform
@PapaFlammy699 ай бұрын
Already made quite some lol
@BigFloppyHat9 ай бұрын
This is great!
@emanuelvendramini20459 ай бұрын
I like to think that 0
@memecleave42999 ай бұрын
Where did u get the chalkboard from? Im looking to pick one up :)
@charlievane9 ай бұрын
!⁻ˣ
@PapaFlammy699 ай бұрын
breh
@MCentral80869 ай бұрын
Unsure if it's too next level for a general math audience, but can you talk about concepts from the Langlands program?
@SuperTommox9 ай бұрын
Very useful
@EpicIEO8 ай бұрын
did not expect whatever tf was the intro
@gurkiratsingh7tha9939 ай бұрын
I have solved the riemann hypothesis and the non trivial zeroes are of the form s = (1/2 + ib)*69/69 + 6*9+6+9-69
@tangomuzi9 ай бұрын
The first and last solution is only correct for x in positive integer numbers
@syed33449 ай бұрын
u can form an integral too!
@perseusgeorgiadis78218 ай бұрын
Bro! Growth factor test and you’re done lmao
@afrolichesmain7779 ай бұрын
I think the “intermediate” and “pro” approaches should be switched, since the squeeze theorem is something you learn in an intro to calc class, whereas using stirling’s approximation is not as common.
@qm3chan1c28 ай бұрын
The pro method wasn't rigorous enough... True rigor lies in using the epsilon delta definition😂😂
@ianweckhorst32009 ай бұрын
I am at a weird point in my mind right now in which I’m angry he used absolute values on his shirt instead of sqrt(x)^2 because it would look slightly more complex
@55hzdxlh738 ай бұрын
i love ur tshirt
@robinbfh58939 ай бұрын
Isn't there an implicit assumption here that the limit goes over sequences of natural numbers? If you include real sequences it doesnt cancel as nicely in the beginner way. Squeeze theorem should work tho
@bridgeon75029 ай бұрын
How come pro was harder than PhD
@oni83379 ай бұрын
👍
@uwuowo77759 ай бұрын
Ich liebe dich
@hanuskamenik14119 ай бұрын
I don't think, that the first level was correct in argumation, you said that it is finite multiplication, but I think that it is the opposite
@maximilianmueller47079 ай бұрын
L hospital and gamma function would be cool
@Ludovicusgoertz9 ай бұрын
07:39
@fesslerivan6039 ай бұрын
X^X = X.X…X, X times only works with integers. What happens when X -> ♾ but isn’t an integer ?
@theupson9 ай бұрын
an easy if inelegant line: 0< gamma(x+1)/x^x < gamma( ceiling(x+1))/floor(x)^floor(x) < 4/x and squeeze. apologies for the eye pain from my formatting
@Miaumiau33339 ай бұрын
I think a PhD would say it's equal to 0 trivially and not provide any further proof
@TranquilSeaOfMath9 ай бұрын
Leave as an exercise for the students. 😂
@danielspivak39269 ай бұрын
The intermediate level was unrigorous and the pro level was needlessly complicated. The beginner level, which essentially says |lim(x!/x^x)| = |lim(x!/x^(x-1))||lim(1/x)|
@maxthexpfarmer39579 ай бұрын
it's rigorous as they are asymptotically equal
@aozorah059 ай бұрын
How is Stirling's formula not rigorous?
@danielspivak39269 ай бұрын
@@aozorah05 It's using a much more advanced theorem to prove something simple, so it's basically circular reasoning.
@PapaFlammy699 ай бұрын
it's not even in the slightest circular, what do you mean?...
@user-ky4qs2ib2q9 ай бұрын
@@danielspivak3926Since when do need to prove this limit in order to derive Stirling formula?
@ExplosiveBrohoof9 ай бұрын
Before watching: it looks to me like the limit should be 0, since x!/x^x = (1/x) × (x!/x^(x-1)). 1/x --> 0 and x!/x^(x-1) < 1 for all x, so the product of the limits would yield 0.
@edmundwoolliams12409 ай бұрын
But how do REAL alpha males evaluate it?
@PapaFlammy699 ай бұрын
with ligma
@glafayettegorillo42898 ай бұрын
I cannot tell if he is Australian or German ;D
@bnmy6581i8 ай бұрын
Sandwich
@АннаСивер-г8м9 ай бұрын
Wait,you don't need to prove this,this is just obvious
@alexandresiqueira42198 ай бұрын
if dont have number, it isnt math. it is english
@ernestomamedaliev42539 ай бұрын
The "beginner proof" is just wrong... In 2:37 you say "we have finitely many terms" while you are trying to compute the limit of x going to infinity... You there have infinitely many terms. However, last proof was cool, hehe.
@dariuschitu32547 ай бұрын
The beginner level is wrong, You cannot freely say that the product of the limit is limit of product if you are in a nondeterminate case (which you are) and when you have infinitely many sequences whose product you write down (which you have) without proof. Frankly, how many terms does the product have? This question's answer explains why it is wrong. Not to mention that what you are modelling here is obviously the limit of a sequence, not a function; as such you can trivially apply the ratio test, much simpler than the Théorème des gendarmes (don't know equivalent in English)
@KazACWizard8 ай бұрын
you forgot the wolfram alpha method. disappointing
@ldanielmule89 ай бұрын
Do you guys learn about limits before University? 💀
@PapaFlammy699 ай бұрын
Ye, a bit
@o_27319 ай бұрын
😂😂😂
@jonathanv.hoffmann30899 ай бұрын
😂
@JohnSmith-mz7dh9 ай бұрын
Engineer proof. I just plug in 69!/69^69 and basically get zero. QED Also, just plug in a value of x until the calculator rounds down to 0 due to floating point errors.
@neutronenstern.9 ай бұрын
my way of doing the limit: If x! is equal to the amount of atoms in the sun, then x^x is the number of metric tons in your momma, sooooooo x!/x^x =0 for x-> ∞
@Yougottacryforthis9 ай бұрын
how is striling intermediate lmfao
@zenombereznicki6 ай бұрын
Your introduction, wackooooo
@mircopaul52599 ай бұрын
Pressshshhsh
@ashotdjrbashian96068 ай бұрын
Dear math channel authors, can you, please, stop using the notation x! ? It's simply insulting to see it, because you clearly know that it's nonsense. What the hell means what you started writing at 1:30, x!=x(x-1)...(2)(1) ? If you are assuming that x is a positive integer, then just say it. Otherwise, how do you know that subtracting 1 at a time from x you'll come down to 2 and then 1? Again, I know that you know better, don't mislead your viewers. P.S. I didn't watch after that, maybe you clarified? Even in that case notation x! is unacceptable.
@natebrown28059 ай бұрын
I just did: x! = prod 1 to x of n x^x = prod 1 to x of x x!/x^x = prod 1 to x of n/x limit of products = product of limits forall n, lim (x->infinity) n/x = 0 prod 1 to x of 0 is 0
@florianfister99699 ай бұрын
Love that this video releases 1h after my analysis final 🫠
@PapaFlammy699 ай бұрын
rip ;_; I hope it all went well!
@YoutubeModeratorsSuckMyBalls9 ай бұрын
You meant your anal final
@blbbggins2 күн бұрын
Secondly, could you, an (αηαl)ysis genius, explain your definition of x! for a real x?! For example, (10.4)!=10.4*(10.4-1)*...*(10.4-10)*?? what is next? How do you get to the x!=x*(x-1)*...*2*1? Where, after (10.4-10), do you start the product of ...*2*1 and what is its first factor? (10.4-10)*2*1 or (10.4-1)*3*2*1?? Or what?? Above mentioned shows, that you have NO IDEA of the erroneous meaning of x! as a product!!! The second way of your proof is the RIGHT ONE, but you forgot mentioning that Stirling's formula gives an approximation of the \Gamma(x) function for real x, as the x tends towards positive infinity! And, since for NATURAL values of n holds \Gamma(n+1)=n!, some another (αηαl)ysis genius decided to write x! in Stirlings formula!!! This VERY HARD example is an ELEMENTARY one for those who know Stirling's formula.
@blbbggins2 күн бұрын
First of all, he is a rigorous proof, that one can know math, but be a complete idiot (by using profanities, describing the knowledge many people, before he was born, revealed).