The Stirling approximation for the factorial would make this really quick! x! ≡ Γ(x+1) ~ sqrt(2πx) (x/e)^x as x -> ∞ So x! / x^x = O(x^1/2 e^-x) as x -> ∞ -> 0 as x -> ∞ , since e^-x is beyond all orders in x^k as x -> ∞ (and nice thing is this clearly holds for non-integer x via the gamma function, so no need to worry about the factorial of a non-integer aspect in a continuous limit)
Yep that’s true, you can also use the AM-GM to get an upper bound of [(x+1)/2]^x for x! but this is overkill compared to the simple method presented here 😂
@@maelhostettler1004 I’ve done exactly that so I am satisfied, not sure about everyone else though 😭 Also proved the Stirling series to the next order term using Laplace’s Method and And Watson’s Lemma, in fact: x! ≡ Γ(x+1) ~ sqrt(2πx) (x/e)^x * [ 1 + 1/(12x) + 1/(288x^2) + … ] as x -> ∞ The first term of the series is the commonly known “Stirling Approximation”, which in itself is extremely accurate for large x so the other terms aren’t really needed, but still it’s very interesting! The point is to use what we already know - there’s no point in not using the Stirling Formula given it’s been proved and is now a common result. It’s like saying you can’t use the squeeze theorem, because u don’t know the proof of it. The result I showed is better anyways as it shows the asymptotic behaviour of the limit function at large x precisely, and not just the limiting result at infinity. And arguably, the Stirling series is not hard to show - just need a few integration by parts and clever substitutions. The foundations are set in stone
@@maelhostettler1004very important observation! We often think some proofs are faster or simpler when they actually require some more advanced techniques, that require some longer or more difficult proofs. The beauty of mathematics is to prove apparently difficult statements with elementary techniques, trying to not overcomplicate things.
What I immediately looked at was how the numerator and denominator are defined. The numerator is 1*2*3*4...x. .., the denominator is defined x*x*x*x... Clearly, the denominator is getting bigger faster than the numerator, so the limit will be zero.
Always a joy to watch these as a calc student who is beyond bored by my textbook's bland problems. Your passion is contagious, and you help me realize how beautiful math is. Thank you for these videos.
Wow. I haven't done this kind of math since 1972 when in Physics program at university. Long unused but not totally forgotten. You have a wonderful teaching style, far better than the "Professors" that I had at the time. I love the logic and reasoning that allows such seemingly difficult problems to be solved. Thanks very much.
Honestly, great presentation. I understood from beginning to end. It's never always clear how creative logic can be applied when using inequalities. This example using the squeeze theorem to demonstrate how to rewrite the question in a form that looks much more digestible is priceless. Thank you.
Bro, I just found this channel, and this is really great stuff. This wasnt new, yet very plainly explained. Great to see that the math content creators are not 100% whitebread
For positive integer X, we know that X! = X * (X-1) * (X-2) * .... * 1, totalling X terms, we know X^X is X * X * X... * X, X terms, so X!/X^X is X/X * (X-1)/X * (X-2)/X * ... * 1/X We know that the first term is 1, the second term onwards all the way to 1/X is less than 1, so the original function is always less than 1 for any positive integer X. As X approaches infinity, the last term (1/X) approaches zero, so the original function must also approach zero.
@@BossDropbearWhat would be more interesting is proving the same for the gamma function (generalizing factorial from integers to reals). Intuitively the ratio probably still continuously decreases, but I haven't tried to prove that it does.
Such a beautiful chalkboard-writing 🤗 It really helps digesting the content since it is all so clearly readable and nicely ordered. Compared to others (*cough*Borcherds*cough*) this should have waaay more views and subs! Thanks for the great content!
I've been out of school for ~15 years and i don't use anything more advanced than basic algebra for my current job. Coming back to these concepts is so much fun and so interesting. And you're such a great teacher too!
This is a nice demonstration of the kind of fundamentals that mathematicians use frequently that many students don’t really encounter. I do bounds and rate stuff quite frequently and there’s always a bunch of little tricks that I use to get things into a nice form that aren’t really “advanced” but also aren’t exactly easy. You need to have a good mathematical awareness for this kind of stuff.
For every x >= 2, x! is smaller than x^x BECAUSE x^x = x•x•x•x…x•x (x times) x! = x•(x-1)•(x-2)…(2)•(1) The terms of x! are getting farther away from x, so x^x would in a way reach infinity faster, so the expression is like (small infinity)/(big infinity). This is more easily seen as 1/(infinity) or just 0. *Also multiplying out x! gives some polynomial with leading coefficient one: x^x - x^x + (xC2)x^(x-2) + … This means the degree of the numerator is smaller than the degree of the denominator, so the limit is zero. (x choose 2)x^(x-2) + … ------------- x^x I think..?
It’s not quite that simple. For instance, consider the functions f(n) = (1/2 + 1/2ⁿ) n and g(n) = n, and look at f/g. The numerator decreases over time approaching n/2 while the denominator is always n. Notice that f is always moving “farther away” from the denominator, but their ratio is approaching 1/2 and not zero.
honestly your presentation is so intuitive and awesome that i would want to have you as my calculus teacher. no joke youre actually on par with 3blue1brown, if not beyond, when it comes to visual learning like this. i commend the phenomal work here.
My initial guess by looking at it is that it will approach 0. Because breaking it apart it will be a lot of factor terms that start with finites (1,2,3...) on the top and infinite on the bottom. Leading up to factors that approach 1.
I thought it too. Because x! is slower growing than x^x, thus even for small intigers making patern: Let x=3 3!/3³ = 6/27 = 2/9 Let x=4 4!/4⁴ = 24/256 = 3/32 Since 2/9 > 3/32, we can say this fuction tends to go to zero.
Haha great video! Studied calculus 20 years ago, I can still follow you... I'm glad I put the effort into learning it at the time! Thanks for the video, you made it look easy!
According to MatCad, that question isn’t expressed in the correct way. You should define whether the limit is being approached from below, from above, or from both sides of infinity. Remember, it has been proven that different sizes of infinity exist
Fun fact: in Italy we call the squeeze theorem, the "teorema dei 2 carabinieri". Now, the carabinieri are technically policemen so the allegory is that two policemen heading somewhere are dragging with them the central function which is some kind of prisoner! 🙂
Alternative method: The sequence a(n) = n!/n^n is decreasing since a(n+1) = a(n) * (n/n+1)^n < a(n). The sequence also has a lower bound since every term is positive. Therefore the sequence converges by monotone convergence. Notice that the limit as n approaches infinity of (n/n+1)^n = (1 - 1/n+1)^n is e^-1 = 1/e. So now lim a(n+1) = lim a(n) * (n/n+1)^n = 1/e * lim a(n). But lim a(n+1) = lim a(n) so this implies lim a(n) = 0.
You are absolutely right in case of natural numbers whereas we take limit of function over real numbers. According to your definition of factorial, it is not even defined on real numbers so that leads everything that you have done to be actually completely no sense.
You consider the series sum_n^{infty} n!/n^n, use the quotient test, conclude that the series converges and deduce the limit of the sequence of summands to be zero. Note that when you write x!, you would typically mean the (continuous) Gamma-function, which coincides with a factorial only for integers x.
Nice presentation, and clear explanation! I was pretty sure 'by inspection' (sorta guess) that the limit was 0, but couldn't prove it. Now I can! Wheee! Thanks.
I don't think "the squeeze theorem" is easy, but your neat and gentle explanation makes me understand this theorem. Even I'm not native speaker of English. Thank you very much.
man i tell you.. i dont get many new things that i listen to at the first time.. but in this case, i understood it at the first time. thank you man you are great :D!
I want to share an alternate approach i had for this.I am not the best at expressing my thoughts in word so apologies in advance. Basically in x!/x^x ,since x^x has x terms and including 1(even though it doesnt change anything) x! has x terms. so we can re write it as x/x * x-1/x * x-2/x .... 2/x * 1/x Now the first term is 1. Second term can be written as x -1 /x = x/x - 1 /x ,and when x approaches infinity,this approaches one. For the last few terms ,its just constant divided by x,when x goes to infinity they approach 0. So in the end we have 1 * 1 * 1 ... 0* 0 = 0 I hope i didnt accidentally get the wrong answer with the right steps
Here is another way : assume x is a positive integer (otherwise, we use the gamma function and it’s - nearly - the same story). Let f be the fonction x -> x ! / x^x, from N to Q, and A (x) = f (x+1) / f(x). Then A (x) = f (x+1) / f(x) = (x+1) ! x^x / x ! (x + 1)^(x + 1) = x^x / (x + 1)^x = (1 + 1/x)^(-x) = exp (-x ln (1+1/x)). Using equivalents, we see that lim A (x) is 1/e when x -> ∞. Then, for x big enough, A (x) < 1/2 and f(x) < 1/2^x, which proves lim f (x) = 0. Thanks for your videos !
If we're assuming that x is a whole number, so that x! is defined, I would have written it with n, rather than x. Sometimes this doesn't make a difference, but sometimes it does. For example, the sequence sin(πn) is just 0, 0, 0, ... so it has limit 0. But the limit of the function sin(πx) (where x is a real number) as x approaches infinity does not exist.
A simple way is to use the definition of factorial and exponents x! = x * (x - 1) * (x - 2) ... 2 * 1 x^x = x * x * x * x ... (x times) Note when we compute x!/x^x, the first x matches, but all the remaining terms in the numerator will be less than the remaining terms in the denominator as we let x go out towards ♾️. Which means that the denominator grows much faster than the numerator thus the limit approaches 0
my explanation of why the limit equals zero, not even near as rigorous was that x^x grows in magnitude quicker than x! which was easily proved by taking a few exmples, so as x tends to infinity the numerator value is much smaller than the denominator, and becomes closer and closer to zero compared to the denominator as x increases even more. so the limit is zero over infinity which is zero.
I like using logarithms in these situations to see how it works out. In this case, x is a positive integer. ln( x!/x^x) = ln( [1*2*3*…*x] / x^x) = ln( 1/x 2/x 3/x … x/x) = ln(1/x) + ln(2/x) + … + ln(x/x) ≤ ln(1/x) Which tends to -∞ as x->∞. So the original limit tends to -∞ as well. Exponentiating we get the original limit is 0.
If you talk about integer x You can do that by splitting into 1/x * 2/x * ... * x/x there are x terms, all of them ∞)(1/x * 2/x * ... * x/x) = 0*0*...*1 = 0 Some people have shown other methods too, all of them worth reading.
Im always waiting for a moment when this guy start rapping and its not comming, which makes me feeel a little bit awkward, but i really love the way you explain things, good job !
Artificial time inflating. For example (x^(x-1))/(x^x) can be simplified much faster: (x^(x-1))/(x^x) = (x^(x-1))/(x*x^(x-1)), then x^(x-1) cancel out ans 1/x is left.
I got here by videos I think are aimed at middle schoolers. Even if the title screen made me feel I must be missing something in my basic understanding. I was always annoyed by the presentation and ended up just looking at the comments to check I was correct - and yes always was. However this is much more interesting. Made me actually think and I even learned (or with my memory might even be reminded of) something. I had the right answer but I certainly didn't write in down and do it rigourlessly I and if I had written it down I wouldn't have got marks for the working out section (basically I just went straight from x!/x^x to 1/x and didn't do any "bracketing" for the squeeze). Good video. I look forward to finding more from you in my feed.
Consider the limit as x approaches infinity of x!/(x^x). In mathematics it's understood that x signifies a limit in the real numbers in this context. If we pretend this is a limit over the natural numbers n instead, then n! and n^n are docile. Then for a given n consider n!/(n^n). One can readily see there are n terms in both the numerator and denominator giving a product of n fractions. Each fraction is less than 1; in addition one fraction is 1/n. Apply the squeeze theorem with 0 and 1/n. After doing some research i've concluded that taking this limit over the real numbers is beyond the scope of a calculus two course due to the complexity of the functions involved; in addition Prime Newtons seems to be thinking of x! and x^x as functions of the natural numbers in his argument which is erroneous.
Solution without any calculation: x^x grows much faster than x!, because if you expand those Terms, for every term of x^x, you have a smaller term (except for the first, which is equal) in x!. As such, the Limit tends to 0, as the denominator becomes much larger than the numerator.
We could simply say that by mathematical induction 1 x 2 x 3 x 4 x 5 x . . . x n-1 x n = n ! is necessarily less than n x n x n x n x n x . . . x n-1 x n = X^X since, while there are n multipliers in each product, each multiplier in the denominator for the expression lim[X→ ∞] (X! /X^X) is less than or just equal to n. The significance of this disparity between the corresponding multipliers in X! and in X^X grows as the number of multipliers in X! and X^X itself grows, so that, for increasingly larger X, X! /X^X necessarily tends to zero.
I'm but a grammar school mathematician, but I would say: as x gets larger, x^x grows a lot faster than x!, so you can already tell, without doing calculations, that x! / x^x approaches 0.
This one is simple. x^x has x*x…*x, while x! Has the same number of multiplicands, only the first (or last) is x, every subsequent multiplicand is smaller than x. Therefore, x^x grows faster than x!, so the limit approaches 0.
Before watching it, it seems obvious that the limit is 0. The numerator is the product of all the numbers between 1 and x. The denominator is the product of x with itself x times. The denominator obviously grows much faster as x grows. So the limit is 0 as x approaches infinity. After watching, I see the squeeze theorem would be the formal way to show the same thing.
x!/x^x = x/x * (x-1)/x * (x-2)/x * ... * 3/x * 2/x * 1/x = 1 * (1-1/x) * (1-2/x) *... * 3/x * 2/x * 1/x Every factor is smaller than or equal to one, so it's enough if just one of them tends to 0 for the whole product to tend to 0. Last factor is easy: 1/x -> 0 as x -> inf. Therefore the whole expression tends to 0.
Thank you making a pretty straight forward question so complicated. x factorial = x.x-1……3.2.1 Now take out x from each term x factorial = x^x(1.1-(1/x).1-(2/x)…2/x.1/x) Hence numerator and denominator in question cancel out and limit is equal to 0
I'm pretty sure another way that you could do it is: x!=x*(x-1)*(x-2)...*1 and x^x=x*x*x*x...x (with "x" amount of x's) Since x decreases for x! and x^x doesn't change for each x, x^x takes more power over x!, therefore, it's pretty much simplified to 1/x^n (where n is any positive integer), and as we know, lim x->oo 1/x^n is equal to 0
for x^[(x-1)]/x^x at 8:54 we can use exponent rules and say that this is equal to x^(x-1-x) and that's equal to x^-1 = 1/x btw the exponent rule im using is (x^y)/([x^z)=x^(y-z)
It's amazing how passionate you are about teaching, thank you!
Also "those who stop learning stop living" hit me hard.
This channel is a true gem.
The Stirling approximation for the factorial would make this really quick!
x! ≡ Γ(x+1) ~ sqrt(2πx) (x/e)^x as x -> ∞
So x! / x^x = O(x^1/2 e^-x) as x -> ∞
-> 0 as x -> ∞ , since e^-x is beyond all orders in x^k as x -> ∞
(and nice thing is this clearly holds for non-integer x via the gamma function, so no need to worry about the factorial of a non-integer aspect in a continuous limit)
Yep that’s true, you can also use the AM-GM to get an upper bound of [(x+1)/2]^x for x! but this is overkill compared to the simple method presented here 😂
however the proof of stirling involve Wallis Integral and general properties of equivalents... not that ez
@@maelhostettler1004 I’ve done exactly that so I am satisfied, not sure about everyone else though 😭
Also proved the Stirling series to the next order term using Laplace’s Method and And Watson’s Lemma, in fact:
x! ≡ Γ(x+1) ~ sqrt(2πx) (x/e)^x * [ 1 + 1/(12x)
+ 1/(288x^2) + … ] as x -> ∞
The first term of the series is the commonly known “Stirling Approximation”, which in itself is extremely accurate for large x so the other terms aren’t really needed, but still it’s very interesting!
The point is to use what we already know - there’s no point in not using the Stirling Formula given it’s been proved and is now a common result. It’s like saying you can’t use the squeeze theorem, because u don’t know the proof of it. The result I showed is better anyways as it shows the asymptotic behaviour of the limit function at large x precisely, and not just the limiting result at infinity.
And arguably, the Stirling series is not hard to show - just need a few integration by parts and clever substitutions. The foundations are set in stone
@@scottparkins1634I don’t think this is overkill, this a nice solution.
@@maelhostettler1004very important observation! We often think some proofs are faster or simpler when they actually require some more advanced techniques, that require some longer or more difficult proofs. The beauty of mathematics is to prove apparently difficult statements with elementary techniques, trying to not overcomplicate things.
In Italy we call it “The Cops Theorem” because the two external functions are like cops carrying the middle function to their same limit (prison).
Same in France except we don't say cops but "gendarme" which is different but for the sake simplicity let's just say they're a kind of cop.
same in hungary
Same in Ukraine, we call it: "Теорема про двух поліцейських" - Theorem about two cops
In romania, we call it the "claw criterion"
@whitedemon6382 My favorite so far!
What I immediately looked at was how the numerator and denominator are defined. The numerator is 1*2*3*4...x.
.., the denominator is defined x*x*x*x...
Clearly, the denominator is getting bigger faster than the numerator, so the limit will be zero.
Yes this one is very simple.
Yup! I quickly came to the same conclusion.
It’s still not obvious that the limit tends to 0 and not some constant in (0,1) - that requires proving
@@adw1z pretty obvious to me, especially when observing x! And x^x using my old friend Desmos. For x>0, x^x clearly blows away x!.
why does the denom increasing faster than the numerator mean the limit is 0?
I'm Japanese, and I'm not good at English so much, but your explanation is very easy to understand for me.
Thank you! and Excellent!
Glad to hear that!
👌🏻👌🏻
Always a joy to watch these as a calc student who is beyond bored by my textbook's bland problems. Your passion is contagious, and you help me realize how beautiful math is. Thank you for these videos.
You should use aops books if you want more interesting problems / more of a challenge
What’s calc? UC Berkeley? They should have something way much harder than this
From one math teacher to another, you are a great teacher.
Wow. I haven't done this kind of math since 1972 when in Physics program at university. Long unused but not totally forgotten. You have a wonderful teaching style, far better than the "Professors" that I had at the time. I love the logic and reasoning that allows such seemingly difficult problems to be solved. Thanks very much.
Honestly, great presentation. I understood from beginning to end. It's never always clear how creative logic can be applied when using inequalities. This example using the squeeze theorem to demonstrate how to rewrite the question in a form that looks much more digestible is priceless. Thank you.
Bro, I just found this channel, and this is really great stuff. This wasnt new, yet very plainly explained. Great to see that the math content creators are not 100% whitebread
That is the most beautiful and satisfying limit demonstration I’ve ever seen
For positive integer X, we know that X! = X * (X-1) * (X-2) * .... * 1, totalling X terms, we know X^X is X * X * X... * X, X terms, so X!/X^X is X/X * (X-1)/X * (X-2)/X * ... * 1/X
We know that the first term is 1, the second term onwards all the way to 1/X is less than 1, so the original function is always less than 1 for any positive integer X. As X approaches infinity, the last term (1/X) approaches zero, so the original function must also approach zero.
Exactly. Not sure how this is a 10 min video.
@@BossDropbearWhat would be more interesting is proving the same for the gamma function (generalizing factorial from integers to reals). Intuitively the ratio probably still continuously decreases, but I haven't tried to prove that it does.
Yeah I did the same thing, but It's interesting to see other approaches.
I really like your use and explanation of the squeeze theorom
I never took Calc and I understood everything you said. You are marvelous.
That's big brain
Such a beautiful chalkboard-writing 🤗 It really helps digesting the content since it is all so clearly readable and nicely ordered. Compared to others (*cough*Borcherds*cough*) this should have waaay more views and subs!
Thanks for the great content!
I've been out of school for ~15 years and i don't use anything more advanced than basic algebra for my current job. Coming back to these concepts is so much fun and so interesting. And you're such a great teacher too!
I've been out of high school for 51 years (calc 1 and 2) and college for 44 (diff eq) and couldn't agree more!!
And that's sad. Research, programming with luck and teaching are the only accesible jobs where you can apply advanced maths.
This is a nice demonstration of the kind of fundamentals that mathematicians use frequently that many students don’t really encounter. I do bounds and rate stuff quite frequently and there’s always a bunch of little tricks that I use to get things into a nice form that aren’t really “advanced” but also aren’t exactly easy. You need to have a good mathematical awareness for this kind of stuff.
For every x >= 2, x! is smaller than x^x BECAUSE
x^x = x•x•x•x…x•x (x times)
x! = x•(x-1)•(x-2)…(2)•(1)
The terms of x! are getting farther away from x, so x^x would in a way reach infinity faster, so the expression is like (small infinity)/(big infinity). This is more easily seen as 1/(infinity) or just 0.
*Also multiplying out x! gives some polynomial with leading coefficient one: x^x - x^x + (xC2)x^(x-2) + …
This means the degree of the numerator is smaller than the degree of the denominator, so the limit is zero.
(x choose 2)x^(x-2) + …
-------------
x^x
I think..?
This was my immediate intuition as well, the denominator "grows" faster so the result should approach zero as we approach infinity
Yep.
This is not a mathematical proof lmao.
@@wiilli4471did they say it is?
It’s not quite that simple. For instance, consider the functions f(n) = (1/2 + 1/2ⁿ) n and g(n) = n, and look at f/g. The numerator decreases over time approaching n/2 while the denominator is always n. Notice that f is always moving “farther away” from the denominator, but their ratio is approaching 1/2 and not zero.
I’ve never seen such a best teacher. Thank you so much.
honestly your presentation is so intuitive and awesome that i would want to have you as my calculus teacher. no joke youre actually on par with 3blue1brown, if not beyond, when it comes to visual learning like this. i commend the phenomal work here.
..what?
My initial guess by looking at it is that it will approach 0. Because breaking it apart it will be a lot of factor terms that start with finites (1,2,3...) on the top and infinite on the bottom. Leading up to factors that approach 1.
I thought it too. Because x! is slower growing than x^x, thus even for small intigers making patern:
Let x=3
3!/3³ = 6/27 = 2/9
Let x=4
4!/4⁴ = 24/256 = 3/32
Since 2/9 > 3/32, we can say this fuction tends to go to zero.
Very clear. And I love your cap. It suits you!
Haha great video! Studied calculus 20 years ago, I can still follow you... I'm glad I put the effort into learning it at the time! Thanks for the video, you made it look easy!
X!=1*2*3*4*5.....*X, then the ratio is 1/x* 2/x*.....x/x. Each one is
Can you clarify your notation please?
According to MatCad, that question isn’t expressed in the correct way. You should define whether the limit is being approached from below, from above, or from both sides of infinity. Remember, it has been proven that different sizes of infinity exist
Fun fact: in Italy we call the squeeze theorem, the "teorema dei 2 carabinieri". Now, the carabinieri are technically policemen so the allegory is that two policemen heading somewhere are dragging with them the central function which is some kind of prisoner! 🙂
That's a beautiful allegory.
In french too!!
We call it "theoreme des gens d'armes" 😂
One of the best videos I've ever seen. Fascinating problem solved in a fundamental yet brilliant way!!! Keep up the great work
It's an amazing approach! Thank you from Russia!
In India, we call it the Sandwich Theorem which makes sense, and also we some times refer it to as Squeeze Play Theorem
Alternative method:
The sequence a(n) = n!/n^n is decreasing since
a(n+1) = a(n) * (n/n+1)^n < a(n).
The sequence also has a lower bound since every term is positive.
Therefore the sequence converges by monotone convergence.
Notice that the limit as n
approaches infinity of
(n/n+1)^n = (1 - 1/n+1)^n
is e^-1 = 1/e.
So now
lim a(n+1) = lim a(n) * (n/n+1)^n
= 1/e * lim a(n).
But lim a(n+1) = lim a(n) so this implies
lim a(n) = 0.
Good stuff, mate. Super clear discussion. Making complex concepts easy is a gift. I see why you're on the way to your million sub=scribers.
This was easily done by inspection. Every numerator term (x-1), (x-2) etc will be over X. This product will go to zero as X goes to infinity.
Does anyone else feel that there is sleight of hand in using
Electrical Engineer here - my degree was like a deep dive into maths which I loved. Love your passion for maths snd teaching.
Thank you you opened a way in my mind in maths section the way of your solving is very logic and good
Good Job Professor
You are absolutely right in case of natural numbers whereas we take limit of function over real numbers. According to your definition of factorial, it is not even defined on real numbers so that leads everything that you have done to be actually completely no sense.
your explanation and english are both very clear and understandable. As an old mentor of engineering math. i appreciated you so much.Thank u so much.
just a clarification, the gamma function can be defined for negative non integers and can also be lower than 1 for positive numbers
True. I realized what I said but it was too late 😢
You consider the series sum_n^{infty} n!/n^n, use the quotient test, conclude that the series converges and deduce the limit of the sequence of summands to be zero.
Note that when you write x!, you would typically mean the (continuous) Gamma-function, which coincides with a factorial only for integers x.
| x!/x^x |< E, for any E > 0 by the archimedian principle you can always find x such that the inequality holds for any arbitrary E.
Nice presentation, and clear explanation! I was pretty sure 'by inspection' (sorta guess) that the limit was 0, but couldn't prove it. Now I can! Wheee! Thanks.
Good presentation style very soft easy to follow voice, simple explainations.
The fact that you could figure that out without even writing it down makes me happy
I was very much enjoying the video, but the outro got you a new subscriber. You have a very theatrical & charismatic way of talking. I love it!
Nicely explained. Thank you.
I don't think "the squeeze theorem" is easy, but your neat and gentle explanation makes me understand this theorem. Even I'm not native speaker of English. Thank you very much.
man i tell you.. i dont get many new things that i listen to at the first time.. but in this case, i understood it at the first time. thank you man you are great :D!
Glad to hear that!
Really enjoyed watching your teaching style
Math is so amazing! I'm a school teacher in Brazil and I love it. Thanks.❤
I want to share an alternate approach i had for this.I am not the best at expressing my thoughts in word so apologies in advance.
Basically in x!/x^x ,since x^x has x terms and including 1(even though it doesnt change anything) x! has x terms.
so we can re write it as
x/x * x-1/x * x-2/x .... 2/x * 1/x
Now the first term is 1.
Second term can be written as x -1 /x = x/x - 1 /x ,and when x approaches infinity,this approaches one.
For the last few terms ,its just constant divided by x,when x goes to infinity they approach 0.
So in the end we have 1 * 1 * 1 ... 0* 0 = 0
I hope i didnt accidentally get the wrong answer with the right steps
Here is another way : assume x is a positive integer (otherwise, we use the gamma function and it’s - nearly - the same story).
Let f be the fonction x -> x ! / x^x, from N to Q, and A (x) = f (x+1) / f(x).
Then A (x) = f (x+1) / f(x) = (x+1) ! x^x / x ! (x + 1)^(x + 1) = x^x / (x + 1)^x = (1 + 1/x)^(-x) = exp (-x ln (1+1/x)).
Using equivalents, we see that lim A (x) is 1/e when x -> ∞. Then, for x big enough, A (x) < 1/2 and f(x) < 1/2^x, which proves lim f (x) = 0.
Thanks for your videos !
If we're assuming that x is a whole number, so that x! is defined, I would have written it with n, rather than x.
Sometimes this doesn't make a difference, but sometimes it does. For example, the sequence sin(πn) is just 0, 0, 0, ... so it has limit 0. But the limit of the function sin(πx) (where x is a real number) as x approaches infinity does not exist.
Very well explained, thanks you very much, greetings from Perú.
For the squeeze theorem part, all you need on the lower bound is 0. X!:x^x is greater than 0 as all factors are positive.
A simple way is to use the definition of factorial and exponents
x! = x * (x - 1) * (x - 2) ... 2 * 1
x^x = x * x * x * x ... (x times)
Note when we compute x!/x^x, the first x matches, but all the remaining terms in the numerator will be less than the remaining terms in the denominator as we let x go out towards ♾️. Which means that the denominator grows much faster than the numerator thus the limit approaches 0
Great explanation and use of the squeeze theorem.
my explanation of why the limit equals zero, not even near as rigorous was that x^x grows in magnitude quicker than x! which was easily proved by taking a few exmples, so as x tends to infinity the numerator value is much smaller than the denominator, and becomes closer and closer to zero compared to the denominator as x increases even more. so the limit is zero over infinity which is zero.
Beautifully Explained. Thanks
I like using logarithms in these situations to see how it works out. In this case, x is a positive integer.
ln( x!/x^x)
= ln( [1*2*3*…*x] / x^x)
= ln( 1/x 2/x 3/x … x/x)
= ln(1/x) + ln(2/x) + … + ln(x/x)
≤ ln(1/x)
Which tends to -∞ as x->∞. So the original limit tends to -∞ as well. Exponentiating we get the original limit is 0.
If you talk about integer x
You can do that by splitting into
1/x * 2/x * ... * x/x
there are x terms, all of them ∞)(1/x * 2/x * ... * x/x) = 0*0*...*1 = 0
Some people have shown other methods too, all of them worth reading.
Very clever and well done . Enjoyed watching it simplified .
What an excellent teacher!
Im always waiting for a moment when this guy start rapping and its not comming, which makes me feeel a little bit awkward, but i really love the way you explain things, good job !
Someday!
@@PrimeNewtons Im glad You're not taking it in bad way friend!
Bravo beautiful introduction keep it coming brother !!!
Thank you for your passion. Very well presented proof.
Artificial time inflating.
For example (x^(x-1))/(x^x) can be simplified much faster: (x^(x-1))/(x^x) = (x^(x-1))/(x*x^(x-1)), then x^(x-1) cancel out ans 1/x is left.
I can seemingly just tell because I know x^x must grow at a faster rate. x! is like a half version of x^x
Excellent explanation! Thank you very much for sharing you knowledgement.
I got here by videos I think are aimed at middle schoolers. Even if the title screen made me feel I must be missing something in my basic understanding. I was always annoyed by the presentation and ended up just looking at the comments to check I was correct - and yes always was.
However this is much more interesting. Made me actually think and I even learned (or with my memory might even be reminded of) something. I had the right answer but I certainly didn't write in down and do it rigourlessly I and if I had written it down I wouldn't have got marks for the working out section (basically I just went straight from x!/x^x to 1/x and didn't do any "bracketing" for the squeeze).
Good video. I look forward to finding more from you in my feed.
Consider the limit as x approaches infinity of x!/(x^x). In mathematics it's understood that x signifies a limit in the real numbers in this context. If we pretend this is a limit over the natural numbers n instead, then n! and n^n are docile. Then for a given n consider n!/(n^n). One can readily see there are n terms in both the numerator and denominator giving a product of n fractions. Each fraction is less than 1; in addition one fraction is 1/n. Apply the squeeze theorem with 0 and 1/n. After doing some research i've concluded that taking this limit over the real numbers is beyond the scope of a calculus two course due to the complexity of the functions involved; in addition Prime Newtons seems to be thinking of x! and x^x as functions of the natural numbers in his argument which is erroneous.
Solution without any calculation:
x^x grows much faster than x!, because if you expand those Terms, for every term of x^x, you have a smaller term (except for the first, which is equal) in x!.
As such, the Limit tends to 0, as the denominator becomes much larger than the numerator.
讲的非常好，点赞支持。
We could simply say that by mathematical induction
1 x 2 x 3 x 4 x 5 x . . . x n-1 x n = n !
is necessarily less than
n x n x n x n x n x . . . x n-1 x n = X^X
since, while there are n multipliers in each product,
each multiplier in the denominator for the expression lim[X→ ∞] (X! /X^X)
is less than or just equal to n.
The significance of this disparity between the corresponding multipliers in X! and in X^X grows as the number of multipliers in X! and X^X itself grows,
so that, for increasingly larger X,
X! /X^X necessarily tends to zero.
I'm but a grammar school mathematician, but I would say: as x gets larger, x^x grows a lot faster than x!, so you can already tell, without doing calculations, that x! / x^x approaches 0.
This was a nice way to show not only how to use the squeeze theorem, but also why it works
Amezing teaching style sir. I impressed you.😊😊
Dude, I've never seen you before, one minute into the video and I can see how passionate you are about math, I love it dude! Have a great day
Unmatched teaching, even knowning the theorem I would never had though in using it.
We can easily say that the limit is 0 within 10s coz x^x will be greater than x!
Coz exponential function outperforms multiplication by miles
Literally I used to think that squeeze theorem is useless! Thanks for this video!
Claude 2 used Stirling's approximation for the factorial function:
x! ≈ sqrt(2πx) * (x/e)^x
I've never heard of it but it apparently worked.
This one is simple. x^x has x*x…*x, while x! Has the same number of multiplicands, only the first (or last) is x, every subsequent multiplicand is smaller than x. Therefore, x^x grows faster than x!, so the limit approaches 0.
Before watching it, it seems obvious that the limit is 0. The numerator is the product of all the numbers between 1 and x. The denominator is the product of x with itself x times. The denominator obviously grows much faster as x grows. So the limit is 0 as x approaches infinity.
After watching, I see the squeeze theorem would be the formal way to show the same thing.
For non-integer values of x, one cannot say that x! = x*(x-1)*(x-2)*...*1 so, yes, one will need to use the Gamma function.
Hello, I am in the third year of studying mathematics. I really enjoyed solving the example and your teaching method. Thank you
Man, you're great! Greetings from Brazil
X!/(X^X) = (X-1)!/(X^(X-1)) = Product (a) where a
x!/x^x = x/x * (x-1)/x * (x-2)/x * ... * 3/x * 2/x * 1/x = 1 * (1-1/x) * (1-2/x) *... * 3/x * 2/x * 1/x
Every factor is smaller than or equal to one, so it's enough if just one of them tends to 0 for the whole product to tend to 0.
Last factor is easy: 1/x -> 0 as x -> inf.
Therefore the whole expression tends to 0.
Thank you making a pretty straight forward question so complicated.
x factorial = x.x-1……3.2.1
Now take out x from each term
x factorial = x^x(1.1-(1/x).1-(2/x)…2/x.1/x)
Hence numerator and denominator in question cancel out and limit is equal to 0
I'm pretty sure another way that you could do it is:
x!=x*(x-1)*(x-2)...*1
and
x^x=x*x*x*x...x (with "x" amount of x's)
Since x decreases for x! and x^x doesn't change for each x, x^x takes more power over x!, therefore, it's pretty much simplified to 1/x^n (where n is any positive integer), and as we know,
lim x->oo 1/x^n is equal to 0
Why the duce am watching it instead of going to bed and why is it so satisfying to watch. Great job. Thank you 😊
Just want to say I absolutely love your videos! Your energy and enthusiasm are so captivating and really makes me appreciate mathematics much more.
Superb sir..wonderful teaching...thank you
.
Very nice. Well explained and clear handwriting
Best use of the squeeze theorem I've seen in a very long time.
Nice solution and explanation.
You could make it even simpler by stating: 0 < x!/x^x ≤ 1/x
I havent heard about the squeeze theorem before, im not on that level yet i guess, but thats actually super useful. Thank you for this video.
for x^[(x-1)]/x^x at 8:54 we can use exponent rules and say that this is equal to x^(x-1-x) and that's equal to x^-1 = 1/x btw the exponent rule im using is (x^y)/([x^z)=x^(y-z)
Great video. As you say "never stop learning" and I don't (74 years old now!)