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.
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.
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!
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
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.
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'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!
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.
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 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 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.
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.
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 !
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!
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.
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
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 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.
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.
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 !
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.
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.
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)
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. :-)
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
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?
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.
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?
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.
great video, but you can find that solution without having to do any calculations. we can logically deduce that x! increases at a slower rate x^x, and so we know that, as x approaches infinity, we get a small number over a very big number, or zero.
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.
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.
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.
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?
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!
👌🏻👌🏻
That is the most beautiful and satisfying limit demonstration I’ve ever seen
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!
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.
From one math teacher to another, you are a great teacher.
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!
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
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.
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?
I never took Calc and I understood everything you said. You are marvelous.
That's big brain
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.
Does anyone else feel that there is sleight of hand in using
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'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.
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.
Electrical Engineer here - my degree was like a deep dive into maths which I loved. Love your passion for maths snd teaching.
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.
I’ve never seen such a best teacher. Thank you so much.
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.
One of the best videos I've ever seen. Fascinating problem solved in a fundamental yet brilliant way!!! Keep up the great work
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.
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?
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 !
I really like your use and explanation of the squeeze theorom
It's an amazing approach! Thank you from Russia!
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 😢
Very clear. And I love your cap. It suits you!
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.
Hello, I am in the third year of studying mathematics. I really enjoyed solving the example and your teaching method. Thank you
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!
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.
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
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" 😂
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.
Nice solution and explanation.
You could make it even simpler by stating: 0 < x!/x^x ≤ 1/x
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.
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
Man, you're great! Greetings from Brazil
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?
Good Job Professor
Thank you you opened a way in my mind in maths section the way of your solving is very logic and good
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.
Very well explained, thanks you very much, greetings from Perú.
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!
Isnt this just growth rate of functions?
Ln(x) < x^n < n^x < x! < x^x
Or something like that?
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.
The fact that you could figure that out without even writing it down makes me happy
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.
Math is so amazing! I'm a school teacher in Brazil and I love it. Thanks.❤
Encountered precisely this function in the wild as the likelihood according to an empirical distribution of the sample that was used to construct it.
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!
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.
In India, we call it the Sandwich Theorem which makes sense, and also we some times refer it to as Squeeze Play Theorem
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!
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)
Thank you for your passion. Very well presented proof.
6:40 factorial of any number between 0 and 1 is less than 1.
Unmatched teaching, even knowning the theorem I would never had though in using it.
Amezing teaching style sir. I impressed you.😊😊
Bravo beautiful introduction keep it coming brother !!!
Great explanation and use of the squeeze theorem.
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. :-)
Really enjoyed watching your teaching style
Nicely explained. Thank you.
X!/(X^X) = (X-1)!/(X^(X-1)) = Product (a) where a
Very clever and well done . Enjoyed watching it simplified .
Good presentation style very soft easy to follow voice, simple explainations.
Thanks, I've been looking for that for a long time!
| 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.
another very nice math example...
Everyone: (Impressed by this dude or their ways to solve this question)
Me: He has really nice hand writing.
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
Beautifully Explained. Thanks
When i see the problem, i am sure the limit is equal to zero. But i cant prove it like that. İts nice solution.
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.
At 5:07, how is x!
Great video. As you say "never stop learning" and I don't (74 years old now!)
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
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.
Refreshing my maths ans learning English at the same time... superb!
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.
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?
Very cool! Thank you for sharing this
Best use of the squeeze theorem I've seen in a very long time.
Very clever. I thought I'd be too ignorant to understand this but I'm pretty sure I got it! Thank you for that
Fantastic explanation! So therefore x^x grows faster than x! is what we can deduce?
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.
that is amazing limit problem. Please show for more solving like this, I'm really love no L'hpital rule.
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’re a great teacher.
great video, but you can find that solution without having to do any calculations. we can logically deduce that x! increases at a slower rate x^x, and so we know that, as x approaches infinity, we get a small number over a very big number, or zero.
You have a delightful voice! I'd listen to you read an audiobook
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.