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.
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.
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
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).
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
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...
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
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.
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?
@@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 flawed because you can only split products of its individual limits, if the terms are finite, with x approaching infinity thats not the case.
"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.
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)
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
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
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.
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 !✌
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)
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 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.
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.
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.
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
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.
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).
