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.
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.
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.
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
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.
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,
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.
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.
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).
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
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?
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...
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)
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.
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
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.
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
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.
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
"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.
@@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
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
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.
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 !✌
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.
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)|
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-> ∞
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)
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.
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.
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.
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
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.
Flammy : "so we are going to go with sandwich theorm"
Also Flammy : (Continues to go with squeeze theorm)
xDDD I noticed that too when editing :'D
They are synonymous
Same thing, aren't they?
@@davidemmanuel9418they are the same ,but it's funny After he committed to call it sandwich theorm he continues to call it squeeze theorm
Sandwich theorem uses bread and squeeze theorem involves sauce dude@@davidemmanuel9418
the beginning meme ☠
Guac and balls
The super guac guac 9000
that guac is longer how 😭
guac and ball torture
💀💀
Cool video! Now do it with an epsilon-delta proof
breh 3:
nah do the neighborhood version
it's epsilon-N since x is approaching infinity. N = ceil(1/epsilon) works by inspection
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
No, a physicist’s answer would be the “intermediate” level where he uses sterling’s approximation for a factorial
Did it with using log and the harmonic series approximation!
Lovely question!
definitely buying a couple hoodies papa flammy thanks for the discount code
The support is highly appreciated, thank you so much!!!!
Funny thing is i was teached about sandwich in highschool
awesome! :)
@@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
@@tererere3877mood
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.
Actually, x^x is not always greater than x!. 0
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.
The noises are the greatest part of these videos!
This is such a beautiful explanation, thank you
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
Did it the “PhD way” when looking at the thumbnail thinking it was gonna be the beginner way 😂
Some content on measure theory? That'd be fun! or some topology, there's a lot of really nitty results out there!
Fun fact: Avocado means Testicle in one of the Aztec languages which makes the meme even more fun.
:^)
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.
you can generalize it with integrals
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,
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.
I think you can just say x! / x^x
@@grantarneil8142 how about the factorial of fractions?
I thought the pro one would be an epsilon-N proof and the squeeze theorem one would be the intermediate one :(
Me too
My expectations were too high
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.
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).
Damn. That was a clean argumentation.
Not my experience in anal class
Beginner was dirty though ;)
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
Ya if its treated as seq then ratio test for limit of seq can be invoked directly
This is great!
The last one was really impressive!
sandwich theorem ftw. also, the avocado looks a lot like a geoduck.
I though you would use gamma function to express x! and them operate the limits, maybe even using L’Hospital.
Very useful
0:22 the inner pigdog awaked.
Im truely sorry for you.
oink oink
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
Unsure if it's too next level for a general math audience, but can you talk about concepts from the Langlands program?
5:50 - "e to the x grows like this. It's a fast fuckin' boi."
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?
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...
You can also solve it by doing the series ratio test
But what if you have X!/X^(X-1) ?
Will it leave you with analitis, if you follow that class?
Where did u get the chalkboard from? Im looking to pick one up :)
this guy is like if shitposting was an integral part of math
$1 limit vs. $5 limit vs. $100 limit Buzzfeed
👍
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
I like to think that 0
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)
what the hell with the avocado, i am going to warch a math content, brooo lmao...
:'D
how about using gamma function? Can that be a valid approach?
ngl, that avocado made me dirty 💀
:^)
u can form an integral too!
The first and last solution is only correct for x in positive integer numbers
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.
Is making it an integral wrong?
no, you can also do that!
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
My initial guess is 0 because x^x blows up faster than x!
i love ur tshirt
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.
I was going to say that too
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
Bro! Growth factor test and you’re done lmao
Hy don't you make a video about Fourier transform
Already made quite some lol
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.
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
"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.
did not expect whatever tf was the intro
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
07:39
I want to be able to do this
The beginner way is wrong. if x grows, then the number of factors also grows. a small estimate would be correct here.
As mentioned, it is a >heuristic< approach that utilizes that the denominator is outgrowing the numerator
factors not terms, right?
@@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
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.
@@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.
i didnt catch that code papa flammy could you say it again?
fbyhsghevklisvkrhkebg7hhjhrg
internal screeching intensifies
😂😂😂
How come pro was harder than PhD
!⁻ˣ
breh
Ich liebe dich
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
As mentioned, it's a heuristic approach only.
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.
L hospital and gamma function would be cool
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 !✌
I think his way will work if one can prove that in the case of x! being defined as the Gamma function, when 1
😂
The pro method wasn't rigorous enough...
True rigor lies in using the epsilon delta definition😂😂
I think a PhD would say it's equal to 0 trivially and not provide any further proof
Leave as an exercise for the students. 😂
X^X = X.X…X, X times only works with integers. What happens when X -> ♾ but isn’t an integer ?
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
Sandwich
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.
Wait,you don't need to prove this,this is just obvious
I cannot tell if he is Australian or German ;D
Do you guys learn about limits before University? 💀
Ye, a bit
if dont have number, it isnt math. it is english
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)|
it's rigorous as they are asymptotically equal
How is Stirling's formula not rigorous?
@@aozora9059 It's using a much more advanced theorem to prove something simple, so it's basically circular reasoning.
it's not even in the slightest circular, what do you mean?...
@@danielspivak3926Since when do need to prove this limit in order to derive Stirling formula?
how is striling intermediate lmfao
Your introduction, wackooooo
But how do REAL alpha males evaluate it?
with ligma
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-> ∞
Pressshshhsh
you forgot the wolfram alpha method. disappointing
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)
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.
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.
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.
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
Love that this video releases 1h after my analysis final 🫠
rip ;_; I hope it all went well!
You meant your anal final