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.
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.
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.
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.
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.
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!
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.
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!
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.
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.
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!
Very cool explanation! Thank you. In looking at this problem, in my mind, I figured the denominator (x^x) would approach infinity faster than the numerator (x!). With that thought, I “guessed” the limit would be zero. Is there any logic to that thought process?
An alternate approach is to use the Limit of a product is the product of the limits. Thus the limit becomes ( lim x/x ) ( lim x-1/x )( lim x-2/x) ... (lim x/2)(lim x/1 ) . The first limit as x goes to infinity is one, the rest in the line of products is zero. Therefore the limit is zero.
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
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.
Shouldn't you introduce Stirling formula ? I have an application: Consider a set E = { x_1 ... x_N } ; N times, we randomly choose an element of E, and note it (and replace it in E). What's the probabibility to get every element of set E (once, since we perform N drawings) ?
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.
That's it. The way you teach is really great. You allow (as it should be) students to think before you pronounce the answer, and that's the way we work at my Academy. I look up to you. :-)
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.
Dude, where are you from? Your voice is so smooth and soothing. Can you just narrate my whole life? I can feel my blood pressure dropping as I listen to you.
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
Interesting premise of the video. Seems like product of k/x for k from 1 to x, and x going to infinity? So that must be 0, right? To prove, fix an epsilon and choose an x large enough so that 1/x < epsilon. Done because just the first term is less than epsilon, and all other terms are less than 1. Right?
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
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.
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
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!
👌🏻👌🏻
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?
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.
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.
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.
From one math teacher to another, you are a great teacher.
That is the most beautiful and satisfying limit demonstration I’ve ever seen
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.
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!
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!
I never took Calc and I understood everything you said. You are marvelous.
That's big brain
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?
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.
One of the best videos I've ever seen. Fascinating problem solved in a fundamental yet brilliant way!!! Keep up the great work
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
Just want to say I absolutely love your videos! Your energy and enthusiasm are so captivating and really makes me appreciate mathematics much more.
I’ve never seen such a best teacher. Thank you so much.
It's an amazing approach! Thank you from Russia!
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!
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
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.
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.
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.
Very clear. And I love your cap. It suits you!
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.
So smooth. Thank you.
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?
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.
Really cool video! We touched on this concept in calc but never really used it, it’s nice to see it applied
Thank you for your passion. Very well presented proof.
Unmatched teaching, even knowning the theorem I would never had though in using it.
Very clever and well done . Enjoyed watching it simplified .
Bravo beautiful introduction keep it coming brother !!!
Thank you, Sir!
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.
This was a nice way to show not only how to use the squeeze theorem, but also why it works
Literally I used to think that squeeze theorem is useless! Thanks for this video!
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.
Amezing teaching style sir. I impressed you.😊😊
Isnt this just growth rate of functions?
Ln(x) < x^n < n^x < x! < x^x
Or something like that?
Thank you you opened a way in my mind in maths section the way of your solving is very logic and good
I really like your use and explanation of the squeeze theorom
Excellent explanation! Thank you very much for sharing you knowledgement.
Best use of the squeeze theorem I've seen in a very long time.
Beautifully Explained. Thanks
Beautiful! 🤩. Thx!
Great explanation and use of the squeeze theorem.
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!
I loved your explanation!
Beautiful proof and great explanation. Instant subscribe.
Excellent explanation. Thank you.
You have a delightful voice! I'd listen to you read an audiobook
Awesome approach! 😎
Thanks for the clear explanation.
Very cool explanation! Thank you. In looking at this problem, in my mind, I figured the denominator (x^x) would approach infinity faster than the numerator (x!). With that thought, I “guessed” the limit would be zero. Is there any logic to that thought process?
Yes. Almost every limit that converges to 0 as x goes to infinity is predictable. The problem is the algebra.
A pleasure to watch! Tx u!
Your way of teaching is just amazinggg
Very clever. I thought I'd be too ignorant to understand this but I'm pretty sure I got it! Thank you for that
Clear and well -presented.
Very cool! Thank you for sharing this
An alternate approach is to use the Limit of a product is the product of the limits. Thus the limit becomes ( lim x/x ) ( lim x-1/x )( lim x-2/x) ... (lim x/2)(lim x/1 ) . The first limit as x goes to infinity is one, the rest in the line of products is zero. Therefore the limit is zero.
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?
Thank you, Sir! Great explanation!
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
Great job, Professor
Very well explained, thanks you very much, greetings from Perú.
im glued to your channel now
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.
Really enjoyed watching your teaching style
Nicely explained. Thank you.
Math is so amazing! I'm a school teacher in Brazil and I love it. Thanks.❤
Shouldn't you introduce Stirling formula ?
I have an application: Consider a set E = { x_1 ... x_N } ; N times, we randomly choose an element of E, and note it (and replace it in E). What's the probabibility to get every element of set E (once, since we perform N drawings) ?
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.
That's it. The way you teach is really great. You allow (as it should be) students to think before you pronounce the answer, and that's the way we work at my Academy. I look up to you. :-)
Is the squeeze theorem the same as the sandwich theorem?
I remember using something similar to prove sinx/x is 1 at x -> 1.
Hello, I am in the third year of studying mathematics. I really enjoyed solving the example and your teaching method. Thank you
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.
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 😢
Superb sir..wonderful teaching...thank you
.
Thank you sir!
That was excellent, thank you!
It's very good demonstrations! I like it!
Fantastic explanation! So therefore x^x grows faster than x! is what we can deduce?
Never seen a chalkboard so clean. Makes me want to get some chalk and write something out. Hagoromo?
Beautiful. Subscribed!
Dude, where are you from? Your voice is so smooth and soothing. Can you just narrate my whole life? I can feel my blood pressure dropping as I listen to you.
Nicely done!
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
Beautiful. ❤
Interesting premise of the video. Seems like product of k/x for k from 1 to x, and x going to infinity? So that must be 0, right?
To prove, fix an epsilon and choose an x large enough so that 1/x < epsilon. Done because just the first term is less than epsilon, and all other terms are less than 1. Right?
The fact that you could figure that out without even writing it down makes me happy
讲的非常好，点赞支持。
Man, you're great! Greetings from Brazil
Nice solution and explanation.
You could make it even simpler by stating: 0 < x!/x^x ≤ 1/x
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