Noticing that the limit is actually an integral is beautiful. I've never seen that done before. Not having to use any advanced theorems, just basic calculus methods is a pure joy.
Great approach via the integral, I ended with the same result using Stirling's approximation; only to have a rogue scalar of \sqrt{2 \pi} alongside e^{-1}.
@@Itskev7 if you replace n! with (2*pi*n)^1/2*(n/e)^n it simplifies to lim n -> inf ((1/e)*(2*pi*)^(1/2n)) and you can show what (2*pi*)^(1/2n) goes to 1 so the result is 1/e
Beautiful! A really nicely presented example of the beauty of mathematics. Incidentally, 1/e is the unemployment rate in the limit when a large number of employers randomly chooses to hire a worker from among the same large number of workers. Euler's constant shows up everywhere, like a mathematical Zelig.
Cauchy's theorem on limits works fine on this one. Take a_n = n!/(n^n), it's a sequence of positive numbers. The theorem says: Let a_n be a sequence of positive members. Limit of a_(n+1) / a_n exists. Then lim (a_n)^(1/n) = lim a_(n+1) / a_n. These conditions are met, so we can use it. After cancelling those factorials and n+1 in powers the expression leads in limit to 1/e. You even do not need integral aparatus for it.
First you have to prove that that series converges, which is another large problem, then you say use the stirling approximation but then we don't even need this at all, so its not practical at all.
@@sovietwizard1620 It s not necessary. Rephrasing the theorem, we have b_n that converges to b. Then, we look at the series c_n = (b_1*..*b_n)^(1/n). The theorem states, c_n converges, and it converges to b. NOT that if c_n converges, then it converges to b. See the proof, We are literally constructing n so big that abs(c_n-b)
Nice job! To integrate y=ln(x) from 0 to 1, turn the function on its head. Just integrate x=e^y from minus infinity to zero, which comes to 1, and take the negative of that, which is -1.
So... series gave birth to limits which gave birth to derivatives and integrals.. Now we have this limit problem that can be expressed as a series problem which is, in fact an integral problem... Maths is like a thriller movie: whatever the starting situation is, you never know which character is the turning point in the end. I love thrillers.
谢谢老师。我越看越惊艳！ Although my math knowledge is limited (due to my own laziness, nowadays there are no excuses to learn), I really enjoy your videos and actually learn a lot... And of course get amazed almost everytime. Math is extremely beautiful and rewarding.
It has been long time ago i studied the mathematic but I've gotten 2 book for it , the of them it was specifically for mathematics students , the second book just high school book for maths classe
We can also use the Stirling's approximation. When n tends to infinity, n! ~ sqrt(2πn)*(n/e)^n Which means (n!/(n^n))^(1/n) ~ (2πn)^(1/2n)*(1/e) ~(2π)^(1/2n)*n^(1/2n)*(1/e) So the limit is 1*e^0*1/e = 1/e
Yeah this has a more "geometric" feel, whereas the other using addition is "arithmetic". My thinking is like the difference between geometric and arithmetic mean -- the structure of the two are analogous, where geometric just takes each operation a step up in their order.
im studying very advanced math in my university and i had my doubts watching the whole video but man i really loved it. i wish my professors would have your enthusiasm. it really is a beautiful limit! and i think students should be shown these kinds of examples to see the true power of infinitesimal calculus.
Amazing solution! Another much easier way to work out this limit is to use Stirling’s formula, which states that n! = (n/e)^n • sqrt(2•pi•n) • (1+epsilon(n)), where epsilon(n) goes to zero when n goes to infinity. Using this formula, you almost immediately get the answer 1/e.
je suis vraiment étonné du nombre de gens qui regarde la vidéo alors que l'on vie dans un monde ou les maths sont de plus en plus mis en retrait (en france en tout cas). Je trouve cela vraiment bien et merci pour la résolution.
I find it really fun that you're showing us all of these examples that 0^0 isn't always equal to 1. That it depends upon the function that you're taking the limit of that takes a 0^0 form at the limiting value. Also find it really fun that virtually every single problem you come across that can be solved in multiple ways, your method almost invariably (pun intended) involves an integral at some point. You love your integrals, don't you? ;)
My first reaction was to use Stirling's Approximation for n!. Returns 1/e pretty quickly. (n!/n^n)^(1/n) ~ ((n/e)^n * sqrt(2*n*pi)/n^n)^(1/n) (n!/n^n)^(1/n) ~ (1/e) * sqrt(2*n*pi)^(1/n) (n!/n^n)^(1/n) ~ 1/e
Im afraid you can't, the very first step is invalid. The fact g(n)~h(n) does not imply g(n)^f(n) ~ h(n)^f(n), take, for example, n ~ (n+1), but 1=n^(1/n) does not ~ (n+1)^(1/n)
@@szymonantoniak1772 It applies for f(n)=1/n. g(n)~h(n) means g/h -> 1 so (g/h)^f -> 1^0=1, nothing nasty... Also your example is not good, (n+1)^(1/n) -> 1 as well...
@@3snoW_ was there not different equivalencies? I remember you had ~ representing equivalence in the substraction of limits and ~~ equivalency on their quotient. I don't know, but 20 years after my "limits" times, that problem begs for Stirling (in fact the solution presented reminds too much to the traditional proof of strilings equivalence)
You can calculate the integral between zero and one of ln(x) by integrating with respect to y. Alternatively: You can calculate the integral between zero and one of ln(x) by considering the fact that ln(x) is the inverse function of e^x, therefore y=ln^x is a reflection in the line y=x of y=e^x, so you can just calculate the improper integral between -infinity and 0 of e^x, and this area will have the same magnitude as the integral we want to calculate. I say magnitude because the e^x integral = 1 because all e^x values are greater than zero. Both methods work (they are basically the same thing anyway)
Here is a much simpler solution: It is well-known that if for a sequence a_n of positive numbers the limit of a_{n+1}/a_n exists then so does the limit of the n-th root of a_n, and both are equal. For the obvious choice of a_n, here we have the ratio of interest equal to (n/n+1)^n which converges to 1 over e. The end. And the result referred to above is more elementary than information on Riemann integral.
We can demonstrate it using "D'Alambert's quotient limit implies Cauchy's radication limit", but we lost a brillant method and application of Riemann's sums. Thank you for an excellent video. King regards from Argentina
Very nice use of recognizing the series was equivalent to an integral. I had started the problem using the gamma function, and restricting to integral values, but your solution was much more elegant. Nice job.
I have vague memories of going from infinite product to an integral back in Highschool. Watching your video gave me a wonderful aha moment as the recollection hit me! Thank you!
This is awesome, I tried doing some algebra to this equation but got no where, the way you did this is dope! I first did n! ^(1/n) / n^n^(1/n) = n! ^(1/n) /n^(n*1/n) = n! ^(1/n) /n then i watched what you did lol.
I too immediately saw that (n^n)^(1/n) is simply n, so the equation reduces to limit ((n!)^(1/n))/n as n goes to infinity. This does not have the 0^0 problem, so it should have been an easier way for him to solve.
Am I the only one who solved this using stirling's approximation for n factorial? It makes the solution super easy: Sqrt(2pi*n)*(n/e)^n < n! < (e^(1/12n))*sqrt(2pi*n)*(n/e)^n If you replace the n! In expression in the limit with these two expressions the n^n bit cancels out leaving some expression taken to the power of 1/n which clearly converges to 1 as n goes to infinity, multiplied by a constant 1/e. Placing the expression in the limit between the two in a chain of inequlities and taking the limit of the upper bound and the limit of the lower bound, the result clearly goes to 1/e and by the sandwich rule we get that the limit we are evaluating must go to the same limit as well. This is so much quicker than what was shown here. However, his solution is still a neat one and I get why he would prefer to show it rather than the quick and easy route I took, which may seem like trickery to someone unfamiliar with this inequality.
I did! I have a physics background, and Sterling's approximation is always invoked in statistical mechanics courses. The other thing is to cancel out the exponents in the denominator, since (n^n)^(1/n) is just n. So ln(L) ~ (1/n)*[nln(n) - n] - ln(n) = ln(n) - 1 - ln(n) = -1
I clicked this video because i was bored and expected it to be clickbait, but boy, was this a brilliant limit. Had me captured for the full 15 minutes.
In India in jee adv examination for IIT entrance this type of q is asked Very similar q was in jee adv 2018 and jee adv 2017 and many earlier also. We study this topic as "limit as a sum" .
An alternative, and perhaps easier solution would just be to use a conclution of the stolz-scesaro theorm ("L-Hopital rule for sequances") which is that for any sequance an, if the limit of (a(n+1)/an) converges, then the limit of n'th root of (an) converges into the same value. So all we have to do is just calculate the (a(n+1)/an) limit (which is super easy) and we are done. (This may sound complecated for those who aren't familier with this theorm but it's a very very useful one for calculating complecated limits and series Although I have to admit that bprp's solution is freaking awsome!).
I was getting depressed I was stressed I started physics 2 months ago calculus is so much difficult I've been all day trying to solve this limit and I found your video about that that's incredible that's brilliant thank you that's the most beautiful thing I've ever seen in this period u reminded me why I love this shit thank you very much ❤️❤️
You can do the integral pretty quickly, you can realize that geometrically it is the exact same as the area from -infinity to 0 on y=e^x , which of course is very quick to do as its just e^0 - e^(-infinity) = 1 - 0 = 1, and then since its below the y axis its negative. Maybe its not rigorous though?
Graphing is a powerful method in reasoning, the solution to problems in calculus. He had to evaluate the integral in the end ( integration by parts ) to get xlnx -x + C as the answer. Evaluating the upper limit of integration: 1*(ln(1) -1 ) + C which is 1*( 0 - 1) + C = -1 + C. Now the lower limit of integration: 0 * (ln(0) -0 ) + C = 0 + C. Subtracting from the lower limit of integration the upper limit of integration: ( -1 + C ) - (0 + C) = -1 .
Since I wondered about this myself: the reason that you cannot simply take any base b for the initial logarithm operation and then find a solution of 1/b is that the processes are only the same until the integration step. You can pull out the limit for many values of b (where logb(x) is continuous with x>0). You can always pull the exponent of the argument into a coefficient; this is a basic property of logarithms. Then another basic property is used to turn the problem into a series. However, at this point, it is demonstrated that the series limit is equal to the integral from 0 to 1 of ln(x), which is equal to -1. It is still correct to say that that limit finds the integral of logb(x) from 0 to 1 of any b (for which the function is continuous from 0 to 1), but the answer will not be -1 if we use another base b! Recall what we are finding: this limit (which is also an integral) is equal to ln(L), or log2(L), or logb(L). Which integral we find depends on which logarithmic base we used first; log2(L) does NOT equal the integral from 0 to 1 of ln(x), but of log2(x), which is a different number (-1 is only so clean because e is special). Then, we should raise our base b to the value of this integral, to solve for L. That number, if you multiply it by e, gives you 1; that means it is 1/e. This is the part where e comes in whether you like it or not! Since our original limit only had one answer, we can use different bases to get it, but the combination of integrating those bases and then raising them to the power of that integral always gives us 1/e. You can try this with base 10: the integral from 0 to 1 of log10(x) dx is -0.4342944.... Raising 10 to this power gives us 1/e, which you can prove by multiplying by e to receive a number very close to 1.
Never noticed despite watching BPRP videos for some time. Chinese calligraphy? Isn't horizontal stroke the first and most basic symbol in Chinese language? If I could remember my first lessons...
Did it quite in a same way and then checked your video. The caveat is to show that the Integral(log(x)dx) = xlog(x) - x has a limit at {x>0 and x->0}. xlog(x) = log(x) / (1/x) and by L'Hopital's rule you take the derivative and get (1/x)/(-1/x^2) as x-> 0. The limit is 0 so xlog(x) - x from 0 to 1 is (1*log(1)-1)-0 = -1 so log((n!/n^n)^(1/n))->-1 and we get your result (n!/n^n)^(1/n) -> 1/e as n->Infinity. Cool result. Thank you.
I'm 17 and solved this in 2 minutes through summation and integration..saw that you used same method ..Felt good. Did many problems like these will i was 16 for sharpening calculus..
Cool video. Another way to get the answer is if you know Stirling's approximation: ln(n!) goes to n ln(n) - n as n goes to infinity. So here we have (1/n) ln(n!/n^n) = (1/n)(ln(n!) - ln(n^n)) = (1/n) (ln(n!) - n ln(n)) goes to (1/n)(n ln(n) - n - n ln(n)) = (1/n)(-n) = -1. Rewriting it as the integral of ln(x) is super clever, though! I think that also works as a way of deriving Stirling's approximation.
Commented before I saw the pinned comment on Stirling's approximation :). I knew that one from statistical physics, so it's the first thing I thought when I saw the log of n!
No need to integrate by parts. You can change from a dx integral to a dy integral, and it becomes straightforward. integral(ln(x) dx, from 0 to 1) = - integral(exp(y) dy, from -infinity to 0).
@@wydadiyoun Informal proof: draw the graph. Formal proof: switch to a 2-d integral of 1 dx dy over the desired area, change order of integration, do integrals.
Since the integral from 0 to 1 of ln(x) is the area unther the x line you know that that is the area of the function e^x from - infiniti to 0 since a funtion and its inverse are simetric along the line y=x, and the integral from e^x from - infinity to zero is simply e^0 - e^(- infinity) = 1, since the integral of the ln(x) is below the x axis the integral is -1
It's really surprising that 0^0 can be like literally anything. Like in the step: ln(L) = lim as n goes to inf (1/n × [ln(1/n) + ln(2/n) +....+ ln(n/n)] We know that the limit of a product is the same as finding individual limits of the numbers in the Multiplication, and then finding the product of them. lim as n goes to inf (1/n) is basically 0. => ln(L) = 0 × lim as n goes to inf (ln(1/n) + ln(2/n) +...+ ln(n/n)) => ln(L) = 0 => L = e^0 => L = 1 But after the integration technique, we get the answer 1/e Really interesting. Amazing video as always!
taking natural log on both sides and interchaging log an limit on right hand side is not permitted unless the limit exists..by composite functions therorem.
looking at it and thinking about it for 15 seconds, i feel like it should go to 0, but in my head i know it’s somehow gonna go to e edit: well, close enough
I think there is an easier way to do this using basic root properties. And the limit of: n^(1/n) = 1. You break up the root into two pieces. The top and bottom because. (a/b)^(1/n) = a^(1/n) / b^(1/n) the bottom term then becomes just n. The top term then becomes 1 because: (a*b*c)^(1/n) = a^(1/n) * b^(1/n)*c^(1/n) then doing the limits of each of those separately we see that it becomes: 1*1*1... so now you are taking the limit of: 1/ n
Using Cauchy d'Alembert's criterion, we can easily calculate the limit of (n + 1)! / (n+1)^(n+1) * n ^ n / n!. This is equivalent to (n / (n+1))^n, which by adding and subtracting one becomes (1 - (1 / (n + 1)))^n. See that - 1 / (n+1) converges to 0, therefore we can simply raise this to the - (n+1) power and divide again by -(n+1), this new series converging to e to the power of limit as n goes to infinity of n/(-n - 1), which goes to -1. Therefore, as X(n+1)/Xn goes to 1/e, the n-th root of n!/n^n goes to 1/e.
Limits involving factorials often do, since e can be expressed as lim 1/1! + 1/2! + 1/3! ... + 1/n! The original expression is also a bit similar to another limit that approaches e, (1+1/n)^n
Maybe, but sometimes things which seem intuitive in maths end up doing what you wouldn't expect them to. I _think_ you're right and that will work, but perhaps not.
Noticing that the limit is actually an integral is beautiful. I've never seen that done before.
Not having to use any advanced theorems, just basic calculus methods is a pure joy.
This approach is called a Riemann sum.
@@skycocaster Queridos nobles amigos Profesores, alumnos, conocidos y amigos de este sencillo canal, ¿qué impacto causaría en el Universo de las Matemáticas, al afirmar que algunos números citados no son primos? y los primos gemelos no existen?
¿Cómo sería la "teoría" de Riemann, donde estos números no son primos?
2; 19; 41; 59; 61; 79; 101; 139; 179; 181; 199; 239; 241; 281; 359; 401; 419; 421; 439; 461; 479; 499; 521; 541; 599; 601; 619; 641; 659; 661; 701; 719; 739; 761; 821; 839; 859; 881; 919; 941; 1019; 1021; 1039; 1061; 1181; 1201; 1259; 1279; 1301; 1319; 1321; 1361; 1381; 1399; 1439; 1459; 1481; 1499; 1559; 1579; 1601; 1619; 1621; 1699; 1721; 1741; 1759; 1801; 1861; 1879; 1901; 1979;
Rafael Morais Magão
@@SidneySilvaCarnavaleney I don't speak Spanish
It's called limit as a sum. We study it in class 12 in India.
@@priyanjala5975 no one asked
Great approach via the integral, I ended with the same result using Stirling's approximation; only to have a rogue scalar of \sqrt{2 \pi} alongside e^{-1}.
There shouldn't be a (2*pi)^1/2 coefficient. If you do it properly, you should just get 1/e
@@user-wx8bm1pg1d Following the OP method yes, but I fail to see how the coefficient vanishes using Stirling's Approximation.
@@Itskev7 if you replace n! with (2*pi*n)^1/2*(n/e)^n it simplifies to lim n -> inf ((1/e)*(2*pi*)^(1/2n)) and you can show what (2*pi*)^(1/2n) goes to 1 so the result is 1/e
did the same, no im relieved my method works
I cry a little inside every time I see him use a pen that isn’t Black or Red ...
Dealwiththe Bob
:)
So that represents the tricky part.
I get excited. "We must break out the forbidden markers for this one."
wooooowww..math..
I really love you and your math....
Beautiful! A really nicely presented example of the beauty of mathematics. Incidentally, 1/e is the unemployment rate in the limit when a large number of employers randomly chooses to hire a worker from among the same large number of workers. Euler's constant shows up everywhere, like a mathematical Zelig.
What ?
Waking up and watching a math video from the best UA-camr while having breakfast, this feels good.
日西ディエゴ
Thanks!!!
日西ディエゴ
I am glad to hear!!!
It fork for dinner as well. I have the proof of experience.
I just woke up and I am doing exactly this and I can't believe I read your comment.
This is honestly so weird. I just got woken up and the first thing I do is watch this while still in bed.
Cauchy's theorem on limits works fine on this one. Take a_n = n!/(n^n), it's a sequence of positive numbers. The theorem says: Let a_n be a sequence of positive members. Limit of a_(n+1) / a_n exists. Then lim (a_n)^(1/n) = lim a_(n+1) / a_n. These conditions are met, so we can use it. After cancelling those factorials and n+1 in powers the expression leads in limit to 1/e. You even do not need integral aparatus for it.
exactly
First you have to prove that that series converges, which is another large problem, then you say use the stirling approximation but then we don't even need this at all, so its not practical at all.
@@sovietwizard1620 It s not necessary. Rephrasing the theorem, we have b_n that converges to b. Then, we look at the series c_n = (b_1*..*b_n)^(1/n).
The theorem states, c_n converges, and it converges to b. NOT that if c_n converges, then it converges to b. See the proof, We are literally constructing n so big that abs(c_n-b)
Sure but the integral thing was pretty cool
n-factOREO
LOL
loloreo
weerman44
Love it sooooo much!!!
I will differentiate x-facOREO one day!!
I hope you'll take the real x-factOREO hahaha
blackpenredpen is BRILLIANT!
Talking about all of this makes me feel hungry.. need a non-bunny Oreo soon
Nice job! To integrate y=ln(x) from 0 to 1, turn the function on its head. Just integrate x=e^y from minus infinity to zero, which comes to 1, and take the negative of that, which is -1.
Yes,that's what I thought as well
So... series gave birth to limits which gave birth to derivatives and integrals.. Now we have this limit problem that can be expressed as a series problem which is, in fact an integral problem...
Maths is like a thriller movie: whatever the starting situation is, you never know which character is the turning point in the end.
I love thrillers.
Except it's not because there's a simple way to find this limit withot any integrals
@@shacharh5470 what the key word on this simple way?
So true. As said in the movie 'hidden figures' , math is always relatable.
Derivatives and integrals existed before the limit. Infinitesimals were the name of the game. That is unless you are talking about Riemann integrals.
It can be solved in other way too.
I actually cried out "holy shit" when you converted it to an integral. That was real eureka moment for me!
Its a Riemann sum
I realized much sooner since I just learned them in school 😁
谢谢老师。我越看越惊艳！
Although my math knowledge is limited (due to my own laziness, nowadays there are no excuses to learn), I really enjoy your videos and actually learn a lot... And of course get amazed almost everytime. Math is extremely beautiful and rewarding.
sjmarel 謝謝收看跟支持！！！
sjmarel 希望你也好好加油
blackpenredpen 我希望你的UA-cam频道增长很多。Sorry for my rusty Chinese, but I haven't used it in quite a time. Jiayou!
Very few people teach themselves mathematical skills outside of a course. It's much more fun with other students and direct contact to tutors.
It has been long time ago i studied the mathematic but I've gotten 2 book for it , the of them it was specifically for mathematics students , the second book just high school book for maths classe
We can also use the Stirling's approximation.
When n tends to infinity, n! ~ sqrt(2πn)*(n/e)^n
Which means (n!/(n^n))^(1/n) ~ (2πn)^(1/2n)*(1/e) ~(2π)^(1/2n)*n^(1/2n)*(1/e)
So the limit is 1*e^0*1/e = 1/e
Yeah but thats less elegant
That's way more practical, less elegant though.
Or use the fact that if n>=3,
Exp(1-n)
I like your funny words magic man
That approximation is just another example of why we need to be using tau :)
I love how happy this guy is just to do the maths and share it. It makes me smile. 😀
Man, I love your channel. It's been 35 years since I did any proper maths, but thankyou, you have reminded that mathematics is awesome.
The instant I saw this problem, I figured it would involve Euler's number. It even looks similar to the definition of *e.*
Yeah this has a more "geometric" feel, whereas the other using addition is "arithmetic".
My thinking is like the difference between geometric and arithmetic mean -- the structure of the two are analogous, where geometric just takes each operation a step up in their order.
im studying very advanced math in my university and i had my doubts watching the whole video but man i really loved it. i wish my professors would have your enthusiasm. it really is a beautiful limit! and i think students should be shown these kinds of examples to see the true power of infinitesimal calculus.
Yes buddy
Math is just love;)
You are studying "very advanced math"? Is that code for being a PhD student or for being a professor?
@@MrCmon113 lol
I've been using this approach since an year but today's the first time when I truly understood it. Thank you so much.♥️🙏🏻😊
brilliant question and solution
Bruce
Thank you!
Amazing solution! Another much easier way to work out this limit is to use Stirling’s formula, which states that
n! = (n/e)^n • sqrt(2•pi•n) • (1+epsilon(n)), where epsilon(n) goes to zero when n goes to infinity.
Using this formula, you almost immediately get the answer 1/e.
but that's an approximation
Mind. Blown. That was amazing.
What an absolutely brilliant solution! This is one of the coolest limits I've ever seen, thank you so much for showing this to the world!
This man is an incredible mathematician. He makes me go "oh fuck how did I miss that, it seems so obvious now".
This is one of the most brilliant and beautiful solutions I saw you doing.
Holy cow. How do you relate that to a definite integral from 0 to 1?! That's pure genius omg.
je suis vraiment étonné du nombre de gens qui regarde la vidéo alors que l'on vie dans un monde ou les maths sont de plus en plus mis en retrait (en france en tout cas). Je trouve cela vraiment bien et merci pour la résolution.
I find it really fun that you're showing us all of these examples that 0^0 isn't always equal to 1. That it depends upon the function that you're taking the limit of that takes a 0^0 form at the limiting value.
Also find it really fun that virtually every single problem you come across that can be solved in multiple ways, your method almost invariably (pun intended) involves an integral at some point. You love your integrals, don't you? ;)
I was thinking the same thing regarding 0^0. And that’s exactly why 0^0 is indeterminate.
maths is beautiful, by far the coolest limit i've seen
My first reaction was to use Stirling's Approximation for n!. Returns 1/e pretty quickly.
(n!/n^n)^(1/n) ~ ((n/e)^n * sqrt(2*n*pi)/n^n)^(1/n)
(n!/n^n)^(1/n) ~ (1/e) * sqrt(2*n*pi)^(1/n)
(n!/n^n)^(1/n) ~ 1/e
Im afraid you can't, the very first step is invalid. The fact g(n)~h(n) does not imply g(n)^f(n) ~ h(n)^f(n),
take, for example, n ~ (n+1), but 1=n^(1/n) does not ~ (n+1)^(1/n)
@@szymonantoniak1772 the limit is still the same though, lim (n+1)^(1/n) = 1
I got 1/e instantly because I remember long ago accidentally making e with the x root of (x^x / x!).
@@szymonantoniak1772 It applies for f(n)=1/n. g(n)~h(n) means g/h -> 1 so (g/h)^f -> 1^0=1, nothing nasty... Also your example is not good, (n+1)^(1/n) -> 1 as well...
@@3snoW_ was there not different equivalencies? I remember you had ~ representing equivalence in the substraction of limits and ~~ equivalency on their quotient. I don't know, but 20 years after my "limits" times, that problem begs for Stirling (in fact the solution presented reminds too much to the traditional proof of strilings equivalence)
bro this is the best ad for brilliant ive seen on youtube so far
The moment you said you couldn't differentiate n! I was like: actually there is... then you mentioned the Gamma-function and I was happy
You can calculate the integral between zero and one of ln(x) by integrating with respect to y.
Alternatively:
You can calculate the integral between zero and one of ln(x) by considering the fact that ln(x) is the inverse function of e^x, therefore y=ln^x is a reflection in the line y=x of y=e^x, so you can just calculate the improper integral between -infinity and 0 of e^x, and this area will have the same magnitude as the integral we want to calculate. I say magnitude because the e^x integral = 1 because all e^x values are greater than zero.
Both methods work (they are basically the same thing anyway)
Here is a much simpler solution: It is well-known that if for a sequence a_n of positive numbers the limit of a_{n+1}/a_n exists then so does the limit of the n-th root of a_n, and both are equal. For the obvious choice of a_n, here we have the ratio of interest equal to (n/n+1)^n which converges to 1 over e. The end. And the result referred to above is more elementary than information on Riemann integral.
This needs to be a video of it's own. You seem to be saying something but there's a lot to unpack.
@@TheSoleYankeeI remember this theorem...
We can demonstrate it using "D'Alambert's quotient limit implies Cauchy's radication limit", but we lost a brillant method and application of Riemann's sums. Thank you for an excellent video. King regards from Argentina
You pretty much explained how Stirling’s approximation works. Excellent video
Very nice use of recognizing the series was equivalent to an integral. I had started the problem using the gamma function, and restricting to integral values, but your solution was much more elegant. Nice job.
I wouldn't have let you pet me any longer if you had said IN FRONT OF ME, that I was not as fluffy as the other one, aswell..
AndDiracisHisProphet
Awwww loll. I realized that's what happened to the cat Oreo...
Love how excited he is about solving and explaining math exercises, keep it going
This same question came in our Jee Advanced Mock tests last day and today I came across it on youtube..😂 What a coincidence
I have vague memories of going from infinite product to an integral back in Highschool. Watching your video gave me a wonderful aha moment as the recollection hit me! Thank you!
Why am i watching this before going to bed like wtf
To get sweet dreams
My teacher give us the same limit six years ago and I used Riemann integral to solve it,this video bring me some great memories.
This video made me look at math from a whole new perspective! What a beautiful question
That was such an elegant solution! The way it simplified was so satisfying :D
Thanks!
Great video! I used Stirling’s formula after doing the logarithm of the limit instead, that way you can transform ln(n!) to n*ln(n)-n.
This is awesome, I tried doing some algebra to this equation but got no where, the way you did this is dope! I first did n! ^(1/n) / n^n^(1/n) = n! ^(1/n) /n^(n*1/n) = n! ^(1/n) /n then i watched what you did lol.
I too immediately saw that (n^n)^(1/n) is simply n, so the equation reduces to limit ((n!)^(1/n))/n as n goes to infinity. This does not have the 0^0 problem, so it should have been an easier way for him to solve.
isn't it
This math problem is brilliant!
you can see how as he slowly gets closer to the solution his smile gets bigger and bigger and his facial expression over flows in excitement XD
Makes me feel humble. i would never have got that in a gazillion years.
How is this guy so entertaining. It's actually insane.
Am I the only one who solved this using stirling's approximation for n factorial? It makes the solution super easy:
Sqrt(2pi*n)*(n/e)^n < n! < (e^(1/12n))*sqrt(2pi*n)*(n/e)^n
If you replace the n! In expression in the limit with these two expressions the n^n bit cancels out leaving some expression taken to the power of 1/n which clearly converges to 1 as n goes to infinity, multiplied by a constant 1/e. Placing the expression in the limit between the two in a chain of inequlities and taking the limit of the upper bound and the limit of the lower bound, the result clearly goes to 1/e and by the sandwich rule we get that the limit we are evaluating must go to the same limit as well. This is so much quicker than what was shown here. However, his solution is still a neat one and I get why he would prefer to show it rather than the quick and easy route I took, which may seem like trickery to someone unfamiliar with this inequality.
I did! I have a physics background, and Sterling's approximation is always invoked in statistical mechanics courses.
The other thing is to cancel out the exponents in the denominator, since (n^n)^(1/n) is just n.
So ln(L) ~ (1/n)*[nln(n) - n] - ln(n) = ln(n) - 1 - ln(n) = -1
I clicked this video because i was bored and expected it to be clickbait, but boy, was this a brilliant limit. Had me captured for the full 15 minutes.
In India in jee adv examination for IIT entrance this type of q is asked
Very similar q was in jee adv 2018 and jee adv 2017 and many earlier also.
We study this topic as "limit as a sum" .
we does same question in class illustrations
Ah yes! Limit as a sum! We did that back when I was in Highschool some 15 years back. Your comment brought the memory back.
An alternative, and perhaps easier solution would just be to use a conclution of the stolz-scesaro theorm ("L-Hopital rule for sequances") which is that for any sequance an, if the limit of (a(n+1)/an) converges, then the limit of n'th root of (an) converges into the same value. So all we have to do is just calculate the (a(n+1)/an) limit (which is super easy) and we are done.
(This may sound complecated for those who aren't familier with this theorm but it's a very very useful one for calculating complecated limits and series
Although I have to admit that bprp's solution is freaking awsome!).
Solved this while preparing for IITJEE
Exactly
It is quite a fundamental problem in which the limit changes to integral.
You need to be able to solve this so fast
Exactly me too
You are expected to solve this in under 4 mins though.
I was getting depressed I was stressed I started physics 2 months ago calculus is so much difficult I've been all day trying to solve this limit and I found your video about that that's incredible that's brilliant thank you that's the most beautiful thing I've ever seen in this period u reminded me why I love this shit thank you very much ❤️❤️
You can do the integral pretty quickly, you can realize that geometrically it is the exact same as the area from -infinity to 0 on y=e^x , which of course is very quick to do as its just e^0 - e^(-infinity) = 1 - 0 = 1, and then since its below the y axis its negative.
Maybe its not rigorous though?
You are allowed to do this, it works as long as your transformation of variables is linear (your transformation is x=y and y=x).
You're totally allowed to do that, you just need to integrate along y instead of x.
Busted_Bullseye just becomes a little bit weird when u suddenly have to integrate on dy
I think your argument is 100% rigorous but I'm not a specialist.
100% rigorous. Just turn your head 90 degrees to look at the area. Or just use integration by parts on the logarithm.
beautiful. just beautiful. very elegant. reduces to a right Riemann sum after all.
Is not a rieman sum, thas require a bounded function
II'm Olli 8
Taking AP BC Calculus has given me a whole new side of math to love, we haven’t gotten to integrals yet but this limit made me excited for it
I remember solving this myself. It literally took weeks to find the solution
Graphing is a powerful method in reasoning, the solution to problems in calculus. He had to evaluate the integral in the end ( integration by parts ) to get xlnx -x + C as the answer. Evaluating the upper limit of integration: 1*(ln(1) -1 ) + C which is
1*( 0 - 1) + C = -1 + C. Now the lower limit of integration: 0 * (ln(0) -0 ) + C = 0 + C.
Subtracting from the lower limit of integration the upper limit of integration:
( -1 + C ) - (0 + C) = -1 .
Since I wondered about this myself: the reason that you cannot simply take any base b for the initial logarithm operation and then find a solution of 1/b is that the processes are only the same until the integration step. You can pull out the limit for many values of b (where logb(x) is continuous with x>0). You can always pull the exponent of the argument into a coefficient; this is a basic property of logarithms. Then another basic property is used to turn the problem into a series.
However, at this point, it is demonstrated that the series limit is equal to the integral from 0 to 1 of ln(x), which is equal to -1. It is still correct to say that that limit finds the integral of logb(x) from 0 to 1 of any b (for which the function is continuous from 0 to 1), but the answer will not be -1 if we use another base b! Recall what we are finding: this limit (which is also an integral) is equal to ln(L), or log2(L), or logb(L). Which integral we find depends on which logarithmic base we used first; log2(L) does NOT equal the integral from 0 to 1 of ln(x), but of log2(x), which is a different number (-1 is only so clean because e is special).
Then, we should raise our base b to the value of this integral, to solve for L. That number, if you multiply it by e, gives you 1; that means it is 1/e. This is the part where e comes in whether you like it or not! Since our original limit only had one answer, we can use different bases to get it, but the combination of integrating those bases and then raising them to the power of that integral always gives us 1/e. You can try this with base 10: the integral from 0 to 1 of log10(x) dx is -0.4342944.... Raising 10 to this power gives us 1/e, which you can prove by multiplying by e to receive a number very close to 1.
Thanks for the explanation! I was wondering about this as well.
I had a big smile when you made the connection between the series and a Riemann sum.
This question was once asked in
IITJEE
When they have subjective paper
I had solved this
This has to be one of the most beautiful math ive ever seen
10:00 and elsewhere
Anyone else notice that, when writing "f", he does the horizontal stroke first and then the ∫ ?
Never noticed despite watching BPRP videos for some time.
Chinese calligraphy? Isn't horizontal stroke the first and most basic symbol in Chinese language? If I could remember my first lessons...
That's so cool and flipping amazing !
Did it quite in a same way and then checked your video. The caveat is to show that the Integral(log(x)dx) = xlog(x) - x has a limit at {x>0 and
x->0}. xlog(x) = log(x) / (1/x) and by L'Hopital's rule you take the derivative and get (1/x)/(-1/x^2) as x-> 0. The limit is 0 so xlog(x) - x from 0 to 1 is (1*log(1)-1)-0 = -1 so log((n!/n^n)^(1/n))->-1 and we get your result (n!/n^n)^(1/n) -> 1/e as n->Infinity. Cool result. Thank you.
what a creative solution. I love it!
I'm 17 and solved this in 2 minutes through summation and integration..saw that you used same method ..Felt good.
Did many problems like these will i was 16 for sharpening calculus..
the tittle must be "most emotional moments of 21st century" or smt...
Cool video. Another way to get the answer is if you know Stirling's approximation: ln(n!) goes to n ln(n) - n as n goes to infinity.
So here we have (1/n) ln(n!/n^n) = (1/n)(ln(n!) - ln(n^n)) = (1/n) (ln(n!) - n ln(n)) goes to (1/n)(n ln(n) - n - n ln(n)) = (1/n)(-n) = -1.
Rewriting it as the integral of ln(x) is super clever, though! I think that also works as a way of deriving Stirling's approximation.
Commented before I saw the pinned comment on Stirling's approximation :). I knew that one from statistical physics, so it's the first thing I thought when I saw the log of n!
No need to integrate by parts. You can change from a dx integral to a dy integral, and it becomes straightforward. integral(ln(x) dx, from 0 to 1) = - integral(exp(y) dy, from -infinity to 0).
sounds logical but any formal proof?
@@wydadiyoun Informal proof: draw the graph. Formal proof: switch to a 2-d integral of 1 dx dy over the desired area, change order of integration, do integrals.
Since the integral from 0 to 1 of ln(x) is the area unther the x line you know that that is the area of the function e^x from - infiniti to 0 since a funtion and its inverse are simetric along the line y=x, and the integral from e^x from - infinity to zero is simply e^0 - e^(- infinity) = 1, since the integral of the ln(x) is below the x axis the integral is -1
I used the Stirling formula :D maybe it is cheating....
So much faster tbh
Yeah it works, much faster
You're content is fantastic!! Your ability to make calculus content feel inherently interesting is rare. I love it!!
Thank you!!
"Imagine you're just petting her and do math"
It's really surprising that 0^0 can be like literally anything.
Like in the step:
ln(L) = lim as n goes to inf (1/n × [ln(1/n) + ln(2/n) +....+ ln(n/n)]
We know that the limit of a product is the same as finding individual limits of the numbers in the Multiplication, and then finding the product of them.
lim as n goes to inf (1/n) is basically 0.
=> ln(L) = 0 × lim as n goes to inf (ln(1/n) + ln(2/n) +...+ ln(n/n))
=> ln(L) = 0
=> L = e^0
=> L = 1
But after the integration technique, we get the answer 1/e
Really interesting. Amazing video as always!
I knew it had something to do with e, I felt it coming....
Any other base for the log would work. Of course, you would get ln after integrating still.
I'm from india . I'm in class 11 broo keep bringing these type of troubled questions so that we can learn more and more from u
although I am bad at English，but I can understand your analysis of this question
Thanks!
@@blackpenredpen 👏👏👏👏👏
This beautiful limit is literally like one of the starting questions from the infamous black book of Kota
taking natural log on both sides and interchaging log an limit on right hand side is not permitted unless the limit exists..by composite functions therorem.
It is applicable since log is a continuous function.
Its a pretty standard problem for students preparing for JEE Advanced (we study this in form of limit as a sum in definite integration)
looking at it and thinking about it for 15 seconds, i feel like it should go to 0, but in my head i know it’s somehow gonna go to e
edit: well, close enough
Zero to the zero power, that's the turning point !
Hai fatto capire benissimo la tua esperienza esilarante. Complimenti per il video!
Sterling's approximation of n! would also do it.
I think there is an easier way to do this using basic root properties. And the limit of: n^(1/n) = 1. You break up the root into two pieces. The top and bottom because. (a/b)^(1/n) = a^(1/n) / b^(1/n) the bottom term then becomes just n. The top term then becomes 1 because: (a*b*c)^(1/n) = a^(1/n) * b^(1/n)*c^(1/n) then doing the limits of each of those separately we see that it becomes: 1*1*1... so now you are taking the limit of: 1/ n
This is truely beautiful
You are excellent!
You always bring a new level of creativity to solve the problems.
OH MY GOSH I MADE THIS LIMIT YWATERDAY BEFORE I WENT TO BED
I am watching because I saw factorial inside limiting function. Thanks for bringing up cool math problems.
Could you solve it with gamma function please?
Using Cauchy d'Alembert's criterion, we can easily calculate the limit of (n + 1)! / (n+1)^(n+1) * n ^ n / n!. This is equivalent to (n / (n+1))^n, which by adding and subtracting one becomes (1 - (1 / (n + 1)))^n. See that - 1 / (n+1) converges to 0, therefore we can simply raise this to the - (n+1) power and divide again by -(n+1), this new series converging to e to the power of limit as n goes to infinity of n/(-n - 1), which goes to -1. Therefore, as X(n+1)/Xn goes to 1/e, the n-th root of n!/n^n goes to 1/e.
how come the number "e" appears so often in limits? can someone explain this to me?
Limits involving factorials often do, since e can be expressed as lim 1/1! + 1/2! + 1/3! ... + 1/n!
The original expression is also a bit similar to another limit that approaches e, (1+1/n)^n
@@3snoW_ Continuously Continuous Interest !
I remember I solved it when I was preparing for jee and the question was roaming in the whole class .That used to be a great feeling.
I correctly guessed the answer by doing an Excel spreadsheet up to n = 143.
Chad
I just plugged in the famous formula for n! namely (2*pi*n)^(1/2) * (n/e)^n, and then it was pretty easy to show the limit being 1/e thankfully! :)
You quite like your 0^0 don't you. Anyway. I have a question for you: can 0^0 ever be a complex number?
Jane
I will have to think about it.
Coming up next, I will have inf vs zero
Of course it can. Just take a limiting form that goes to 1 and "rotate" it 90º in the complex plane. I think it can be done that way.
Maybe, but sometimes things which seem intuitive in maths end up doing what you wouldn't expect them to. I _think_ you're right and that will work, but perhaps not.
Well I haven't tried it out; it's just an idea I'm throwing out there...
It can be argued that all real numbers are also complex numbers...
Man, converting limit to Reimann sum is amazing and fun.
Crazy indeed