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.
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.
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.
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!
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 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.
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!
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!
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.
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.
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
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!
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.
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
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.
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! 🙂
You don't need to use any theorem. Just rewrite x^x as x.x.x...x and X! as X(x-1)(x-2) .... 2.1. and it is not hard to see that the fraction is smaller than 1/X which tends to zero at the infinity. But we usually compare n^n together with n! Multiplied by a^n a>1. Which really makes sens. And to do so we know that if n is sufficiently large then after some fixed integer N, we look for the limit of α(n)/α(n+1). And it should be equal to one of three values ∞, 0 or some integer.
Well, the limit as x goes to 0 is 1, due to the Taylor polynomial of e^x, where e^0 = 1 and has a 0^0/0! term that equals to 1, meaning that its inverse, 0!/0^0 is also equal to 1.
Absolutely great Mr. Newton. However for positive integers we can easily see by inspection that the numerator will have the highest power of X as (X-1), whereas the denominator is X^X- that will be simply an expression with 1/X ---> infinity gives the answer = 0
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.
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.
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.
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
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!
👌🏻👌🏻
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.
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.
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!
That is the most beautiful and satisfying limit demonstration I’ve ever seen
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.
I never took Calc and I understood everything you said. You are marvelous.
That's big brain
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'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.
Good Job Professor
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!
One of the best videos I've ever seen. Fascinating problem solved in a fundamental yet brilliant way!!! Keep up the great work
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.
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.
It's an amazing approach! Thank you from Russia!
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.
Just want to say I absolutely love your videos! Your energy and enthusiasm are so captivating and really makes me appreciate mathematics much more.
Beautifully Explained. Thanks
Does anyone else feel that there is sleight of hand in using
Thank you for your passion. Very well presented proof.
Very clear. And I love your cap. It suits you!
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.
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
I really like your use and explanation of the squeeze theorom
Really cool video! We touched on this concept in calc but never really used it, it’s nice to see it applied
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!
Bravo beautiful introduction keep it coming brother !!!
Very clever and well done . Enjoyed watching it simplified .
Amezing teaching style sir. I impressed you.😊😊
Really enjoyed watching your teaching style
Superb sir..wonderful teaching...thank 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
Electrical Engineer here - my degree was like a deep dive into maths which I loved. Love your passion for maths snd teaching.
Nicely explained. Thank you.
Great explanation and use of the squeeze theorem.
So smooth. Thank you.
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 😢
Your way of teaching is just amazinggg
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.
Excellent explanation. Thank you.
Very well explained, thanks you very much, greetings from Perú.
I loved your explanation!
Unmatched teaching, even knowning the theorem I would never had though in using it.
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!
Thank you you opened a way in my mind in maths section the way of your solving is very logic and good
Thank you, Sir!
讲的非常好，点赞支持。
The fact that you could figure that out without even writing it down makes me happy
Literally I used to think that squeeze theorem is useless! Thanks for this video!
Beautiful! 🤩. Thx!
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.
Awesome approach! 😎
Thanks for the clear explanation.
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.
Excellent explanation! Thank you very much for sharing you knowledgement.
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.
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
This was a nice way to show not only how to use the squeeze theorem, but also why it works
Very cool! Thank you for sharing this
It's very good demonstrations! I like it!
Very good. Thanks 🙏
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
Clear and well -presented.
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.
Best use of the squeeze theorem I've seen in a very long time.
Nicely done. You’re a great teacher.
Thank you, Sir! Great explanation!
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" 😂
Great presentation!
Thank you sir!
Excellent teaching. Rare.
You don't need to use any theorem. Just rewrite x^x as x.x.x...x and X! as X(x-1)(x-2) .... 2.1. and it is not hard to see that the fraction is smaller than 1/X which tends to zero at the infinity. But we usually compare n^n together with n! Multiplied by a^n a>1. Which really makes sens. And to do so we know that if n is sufficiently large then after some fixed integer N, we look for the limit of α(n)/α(n+1). And it should be equal to one of three values ∞, 0 or some integer.
In this case it's obvious that x can't be negative, but in the general case how is the < ruled out to give just the = case?
Hello, I am in the third year of studying mathematics. I really enjoyed solving the example and your teaching method. Thank you
Congratulations for 100K🎉
Great teaching passion!
A pleasure to watch! Tx u!
Beautiful proof and great explanation. Instant subscribe.
Math is so amazing! I'm a school teacher in Brazil and I love it. Thanks.❤
Excellent job, I really enjoyed it.
great presentation!!
Very clever. I thought I'd be too ignorant to understand this but I'm pretty sure I got it! Thank you for that
Very nice. Well explained and clear handwriting
You have a delightful voice! I'd listen to you read an audiobook
That was excellent, thank you!
Well, the limit as x goes to 0 is 1, due to the Taylor polynomial of e^x, where e^0 = 1 and has a 0^0/0! term that equals to 1, meaning that its inverse, 0!/0^0 is also equal to 1.
Great job, Professor
Beautiful. ❤
Absolutely great Mr. Newton.
However for positive integers we can easily see by inspection that the numerator will have the highest power of X as (X-1), whereas the denominator is X^X- that will be simply an expression with 1/X ---> infinity gives the answer = 0
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.
Nicely done!
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.