It's really interesting how the integration of the original function between 0 and 1 ends up being equal to the infinite discrete sum of the same function starting from 1. I'm not sure why but it just feels fascinating that something like this exists.
I got the same feeling. There must be some interesting symmetry. On the other hand, I also feel like the answer isn't any more helpful than the original question 🤣
I can think of the Riemann integral from the shape of the function and the intervals of the integral and series, but I can’t quite come up with a way to express the Riemann sum properly.
@@letao12 the power in answer is negative and it is sum rather than integration... it can be calculated by a computer easily with some approximation so I think it is very helpful answer
@@letao12 if you learn integral properly in calculus, you'll know it's not really that surprising, considering integral comes from limit to infinity of the sum of the function.. This is why we learn calculus, the study of limit, we must never forget the origin of derivative and integral, that they are all just limits..
I think it’s “those who stop learning stop living”. Not to be nitpicky but changing “those” to “if” makes it sound like you’re suggesting suicide if they stop learning😂
I subscribe to a few other maths channels..all of them just show the right way(s) of getting to the ans..here, we get to understand WHY a particular way won't work, & how to get around/through/above the block..much appreciated!
Man I aspire to understand math this well one day. I don’t know how to do any of this but the way you work and alter the math so intricately is beautiful
I appreciate you all's scientific curiosity, but what is the point of computing the numerical value of a converging series if you can't prove its convergence, can't write it in simple terms usually involving natural numbers and usual constants and without having a manual method of solving the series.
The video felt very interactive because instead of directly showing us the solution, you walked us through the problems by showing us the various ways you tried to approach the problem.
very beutifully done. i just love the way you put up an act of a few fumble here and there - fumbling like any average student would. Please do a video on tests for convergence. the answer for this integral converges to approx 1.29129. i would have preferred if you had finished off the video with a quick evaluation of the infinite sum - may be calculating 5 or 6 terms to show how quickly this converges. from a student perspective she is going to demand the value of the infinite sum at the final answer. ofcourse if i were the techer i would have said “thats left as an exercise”😅😅😅
I’ll be honest I have no idea why someone would ever want to learn how to do these kind of integrals, as I don’t see a reason to use them anywhere in real life problems, but I recognise your math skills to be a thousand times better than mine, and your videos to be a lot helpful to get ready for Math exams at uni, so, kudos.
4:03 😂😂😂 I'm dead. I havent laughed that hard in a math video in a long time. Hahahaha 😆 😂 😆 But for real... I hate this problem... sometimes I wish math was easier.
Thank you for another wonderful video and what a beautiful and fascinating result. As an alternative approach. I had the idea of trying to compute the Riemann sum for the integral: Lim as n tends to infinity of the sum from r = 1 to n of 1/n*(r/n)^~(r/n) directly. But so far I have not been able to do this. I also had the idea of proving the result you found by considering the difference between this sum, re expressed in the form: lim as n terns to infinity of the sum from r = 1 to n of r^-r and the above Riemann sum This is potentially easier I think, by establishing an upper bound, in terms of n, on the mod of this difference, and then showing that this bound is 0 in the limit as n terms to infinity. But so far I have not been able to do this either. I would be interested if you or others know if either of these alternative approaches can be made to work for this particular problem. Again, than you for your blessed and inspirational videos ❤
There must be a continuation, you must evaluate the sum whether it converges or diverges, which it is converging. The irony of this answer that it resembles the original sum so that if we try to evaluate it by method of integration we end up with the same integral.
No, the integral is n! since: Γ(z) = (z-1)! = ∫[0 to ∞] t^(z-1) e^(-t)dt i.e. Γ(z+1) = z! = = ∫[0 to ∞] t^z e^(-t)dt The power of the integrand is itself shifted in the definition of the Γ function
I have not done calculus in over 40 years .This is beyond what I am currently capable of doing.I am College level Algebra I couldn’t pass pre-calculus / Trigonometry right now .
Fascinating video about the process but I'm not quite sure what we achieved - given the form of the solution looks so very similar to the original integral 🤔
You just did something called discretisation. You essentially converted an integral of a continuous function 1/x^x to a sum of the same function of n+1 where n is an integer.
So, we go from one over x to the x, and the integral from 0 to 1 of that equals the sum of k=1 to infinity of k to the minus k, which is one over k to the k. But what does this approximate? You've rewritten a finite integral into an infinite sum (of the same function), but that's only one step.
It’s more than half a century since I last studied maths, but I’m still a bit wary of your answer. I think you need to show that that series actually exists and is well defined. Unfortunately I can’t remember the conditions for convergence.
On simpler terms..... Is ut not just the discrete identity of the continuous integral? May be nowadays with cimputational capability the continuous integral is reduced to just a notation.....
I agree with @misteribel that you replaced one riddle with another one. And solving that one, gives the first again. The only thing (okay, a great find) you showed is that some finite integral of x to the power -x can be replaced by an infinite sum of more or less the same function. What I missed in this video is what in fact is the meaning or consequence of this result.
Thank you, this was a perfect presentation, congratulations! One question still remains: Is there a closed expression for sum_(k=1)^infinity(k^(-k)) ? It can easily numerically be computet but the question would be, if this number can be expressed as a multiple of pi , e or whatever. Would be very interesting to know!
At 15:15, I don't see why e^(n+1) is moved out as a constant. Then, the answer becomes immediately, sum n=0 to n=inf e^(n+1). Use of "r" seems overkill.
Hey newtons, I’m a 10 year old learning calculus, I know a lot (not like a whole college course) I’ve started Calculus 3, So I need help and my exams are there too. everything’s to me is easy. I started in February of my advanced mathematics learning when I was 9.
I'm a little bit confused. He started with an integral[0,1] of x^(-x) and ended up with a sum, that basically is sum[1,infinity] of x^(-x) 🤔 what's the clue?
Thanks for the great illustration, however I'm not sure what has been achieved here, all I can notice that the original integral is replaced by the sum of the similar function, which is basically the integration 🤔 Not sure if I'm seeing the full picture here!
@oniondeluxe9942 and, I just read the Wikipedia articles and got their sources and tracked down texts tbh. They are in most analysis books probably. Fubini's was in My calc 3 text book, but Patrick JMT has a video on it if I remember correctly
14:36 how is flipping the integral by the negative sign even a step? They means the same, both mean the definite integral from 0 to infinite, I don't understand at all
It feels like I am watching a Mathematical opera when ever I watch your videos. The drama! The suspense! Bravo 👏
Meanwhile I feel like I'm being tied down and mathematically sodomized
It's really interesting how the integration of the original function between 0 and 1 ends up being equal to the infinite discrete sum of the same function starting from 1. I'm not sure why but it just feels fascinating that something like this exists.
I got the same feeling. There must be some interesting symmetry.
On the other hand, I also feel like the answer isn't any more helpful than the original question 🤣
I can think of the Riemann integral from the shape of the function and the intervals of the integral and series, but I can’t quite come up with a way to express the Riemann sum properly.
@@letao12 the power in answer is negative and it is sum rather than integration... it can be calculated by a computer easily with some approximation so I think it is very helpful answer
@hammondkakavandi7738 the power in the original integral is negative also man.
1/(x^x)
=
x^(-×)
@@letao12 if you learn integral properly in calculus, you'll know it's not really that surprising, considering integral comes from limit to infinity of the sum of the function..
This is why we learn calculus, the study of limit, we must never forget the origin of derivative and integral, that they are all just limits..
"Never stop learning.If you stop learning. Stop living..." I appreciate you very much.. Nice explanation and nice question..
I think it’s “those who stop learning stop living”. Not to be nitpicky but changing “those” to “if” makes it sound like you’re suggesting suicide if they stop learning😂
@@ranae6566 learning means not only studies..
@@ranae6566 I literally liked his version better. Bro was like if you stop learning I will be personally looking for you
I really like that sir shows it's OK to back track & re-think when we reach a road block in solving
Yes, it's very reassuring
He sees the future that we have never, as of yet, seen, then he backtracks.
Yes I agree, but I think this is very common in calculus especially.
I subscribe to a few other maths channels..all of them just show the right way(s) of getting to the ans..here, we get to understand WHY a particular way won't work, & how to get around/through/above the block..much appreciated!
It’s so beautiful. One of the best problems I’ve ever seen. Everything just worked out so beautifully and it uses my favorite, the gamma function!
As an ex calculus private teacher i appreciate your expression so much.Your english and explanation is so clear.
Clear and detailed explanation of the steps taken to tackle the problem , thank you , opera writer !
Excellent teacher! It is so refreshing to experience mathematics taught well. His enthusiasm and knowledge makes the difficult easy.😊
Wow, did not expect that is so complicated.
I just loved this and your whole approach. As a 74 year old UK guy who took his BSc in 1971, I am indeed still learning. Thank you!
I thought your comment was good to see that at your age you are still learning . I am sure the education system has changed since you were at uni.
You do a very good job explaining the solution.The result is really nice.
That was spectacular! Beautifully done!
One of your BEST videos that I have seen of yours. I really enjoyed it.
This teacher is absolutely awesome. I am really a fan of his way to explain. Perfect !
It really feels like a Sophomore’s Dream!
Man I aspire to understand math this well one day. I don’t know how to do any of this but the way you work and alter the math so intricately is beautiful
I HAVE NEVER BEEN SO PROUD OF MYSELF USING THE SAME EXACT WORKING OUT IN THE VIDEO
Really good video! Even though I'm not that good with math, I find you videos really understandable!
Excellent explanation! You are even better than some math professors!🎉
Haha. That's hard to prove.
Watching this video was mesmerizing! Now, I want to know what that infinite series converges to.
1.29129
@@PrimeNewtons Thanks! I'm 79 and still learning!
do a spreadsheet, it converges very fast, after some six terms to 1.291286
I appreciate you all's scientific curiosity, but what is the point of computing the numerical value of a converging series if you can't prove its convergence, can't write it in simple terms usually involving natural numbers and usual constants and without having a manual method of solving the series.
@@BartBuzz Same here..i am 70.
amazing!! beautiful result!
The video felt very interactive because instead of directly showing us the solution, you walked us through the problems by showing us the various ways you tried to approach the problem.
very beutifully done. i just love the way you put up an act of a few fumble here and there - fumbling like any average student would. Please do a video on tests for convergence. the answer for this integral converges to approx 1.29129. i would have preferred if you had finished off the video with a quick evaluation of the infinite sum - may be calculating 5 or 6 terms to show how quickly this converges. from a student perspective she is going to demand the value of the infinite sum at the final answer. ofcourse if i were the techer i would have said “thats left as an exercise”😅😅😅
Well the ending was a Revelation! Thanks for sharing!
most intellectual 20 minutes & 36 seconds of my life. Thanks
This is one of those videos that I don’t have enough math background to understand what is going on but my fascination with math keeps me watching
Hey PN, you’re getting a lot of attention from other channels lately. It’s well deserved brother, god bless you and your work.
Pleasant videos! I'll have to spend some hours to understand all the details here, but I think I'll set aside an evening for just that!
That made my Sunday evening pleasant.
This Professor is genius
I’ll be honest I have no idea why someone would ever want to learn how to do these kind of integrals, as I don’t see a reason to use them anywhere in real life problems, but I recognise your math skills to be a thousand times better than mine, and your videos to be a lot helpful to get ready for Math exams at uni, so, kudos.
Much love from Australia. What a cliffhanger hahaha. Thanks
You are a great teacher, congrats!
4:03 😂😂😂
I'm dead. I havent laughed that hard in a math video in a long time. Hahahaha 😆 😂 😆
But for real... I hate this problem... sometimes I wish math was easier.
You draw your xs so perfectly
Véritablement magnifique ! Merci pour cette brillante résolution.
Your voice is soothing
You earned a subscriber bro. Hats off
I tell you this is wonderful piece of Maths
Great, yet easy presenting approach.I like your channel.
Top quality! Grateful for this teaching!
Never Stop Teaching!
This guy explained math in a very detailed way.
You know your stuff Man.
Keep on.
Thank you for another wonderful video and what a beautiful and fascinating result.
As an alternative approach. I had the idea of trying to compute the Riemann sum for the integral:
Lim as n tends to infinity of
the sum from r = 1 to n of
1/n*(r/n)^~(r/n)
directly. But so far I have not been able to do this.
I also had the idea of proving the result you found by considering the difference between this sum, re expressed in the form:
lim as n terns to infinity of the sum from r = 1 to n of r^-r
and the above Riemann sum
This is potentially easier I think, by establishing an upper bound, in terms of n, on the mod of this difference, and then showing that this bound is 0 in the limit as n terms to infinity. But so far I have not been able to do this either.
I would be interested if you or others know if either of these alternative approaches can be made to work for this particular problem.
Again, than you for your blessed and inspirational videos ❤
WOW!! Outstanding!! I did not foresee a Gamma Function was going to be applied.
a beautiful subject with dashing teacher
Very interesting problem and clear explanation. Also you have such a lovely voice
and such perfect handwriting on blackboard ! A joy to behold.
And the sum from (k=1) to ∞ of [k^(-k)] = 1.29128599706
and thx for the great video
There must be a continuation, you must evaluate the sum whether it converges or diverges, which it is converging. The irony of this answer that it resembles the original sum so that if we try to evaluate it by method of integration we end up with the same integral.
Thats phenomenal sir thank you but what to do with the infinite series
Hooooo , hermoso , me quedé pegada viendo, que lindas son las matemáticas❤
LOVE YOU DUDE
Thx for the dominate convergence theorem
19:10 Isn’t the integral equal to (n-1)!, because it is gamma(n). But previously you established gamma(n+1) as equal to n! and not (n+1)!
I was wondering the same thing
No, the integral is n! since:
Γ(z) = (z-1)! = ∫[0 to ∞] t^(z-1) e^(-t)dt
i.e. Γ(z+1) = z! = = ∫[0 to ∞] t^z e^(-t)dt
The power of the integrand is itself shifted in the definition of the Γ function
Never give up 💪🏻
Excellent.
Can it be integrated if it is indefinite?
I'd approach it via the lambert W function. If that pays off, then your result gives an interesting expansion for W(x)
Not sure that I understood everything but it's awesome!
This man is like the bob ross of calculus!
Thanks a lot for you vidéo from France 🇫🇷
Well explained 👌🏼
Love uuuu ❤
Love uuuu tuuu
Einfach genial!! Und so was von unterhaltsam 🙂 (Genius and best Entertainment!!)
from Morocco thank you for your clear wonderful explanations
Amazing solutions. I felt like I was watching the crucial scene of John Wick. (Last sentence translated.)
God this is epic
I have not done calculus in over 40 years .This is beyond what I am currently capable of doing.I am College level Algebra I couldn’t pass pre-calculus / Trigonometry right now .
Well, you need to prove the uniform convergence of that series to be able to switch integral ans series sum.
"now, can this be easily integrated?... no :("
Fascinating video about the process but I'm not quite sure what we achieved - given the form of the solution looks so very similar to the original integral 🤔
In terms of computing the integral's value? Not much (if anything at all). But the result looks very interesting _because_ of the similarity.
Pretty funny and pretty beautiful.
Very interesting ! Thank you for your solution
Can you teach a full course on calculus from beginning through Cal III?
That is my new goal. I'm working on it
You just did something called discretisation. You essentially converted an integral of a continuous function 1/x^x to a sum of the same function of n+1 where n is an integer.
That was a juicy one! ❤
Sloane's constant ~ 1.29...
So, we go from one over x to the x, and the integral from 0 to 1 of that equals the sum of k=1 to infinity of k to the minus k, which is one over k to the k. But what does this approximate? You've rewritten a finite integral into an infinite sum (of the same function), but that's only one step.
But the infinite sum is always defined as a limit, which is in this case a certain (finite) constant, I think.
@@zzambezi1959Wolfram Alpha gave the finite answer of:
≈1.29128599706266
It’s more than half a century since I last studied maths, but I’m still a bit wary of your answer. I think you need to show that that series actually exists and is well defined. Unfortunately I can’t remember the conditions for convergence.
On simpler terms..... Is ut not just the discrete identity of the continuous integral? May be nowadays with cimputational capability the continuous integral is reduced to just a notation.....
so int 0 -> 1 x^(-x) = sum 1 -> inf x^(-x) *mindblowing*
The mathematical delinquency in me wants to just set u=x^x even tho i know that is one of the worst things you could do lmao
love it ❤
Really intersting..good job
bro ended with discrete inverse tetration sum while solving continuous inverse tetration sum 🤣🤣🤣
I agree with @misteribel that you replaced one riddle with another one. And solving that one, gives the first again. The only thing (okay, a great find) you showed is that some finite integral of x to the power -x can be replaced by an infinite sum of more or less the same function.
What I missed in this video is what in fact is the meaning or consequence of this result.
It's the non-closed form solution for the definite integral thats much easier to evaluate than the integral by itself.
Wonderful solution by Mathematical manipularon. Thank you, sir. S Chitrai Kani.
Thank you, this was a perfect presentation, congratulations! One question still remains: Is there a closed expression for
sum_(k=1)^infinity(k^(-k)) ? It can easily numerically be computet but the question would be, if this number can be expressed as a multiple of pi , e or whatever. Would be very interesting to know!
At 15:15, I don't see why e^(n+1) is moved out as a constant. Then, the answer becomes immediately, sum n=0 to n=inf e^(n+1). Use of "r" seems overkill.
How?
n+1 is here multiplied by t not in addition
So we cant separate powers of e as you are mentioning
Look carefully
me don't understand anything but just wants to watch it
Wow. My compliments!
Hey newtons, I’m a 10 year old learning calculus, I know a lot (not like a whole college course) I’ve started Calculus 3, So I need help and my exams are there too. everything’s to me is easy. I started in February of my advanced mathematics learning when I was 9.
I'm a little bit confused. He started with an integral[0,1] of x^(-x) and ended up with a sum, that basically is sum[1,infinity] of x^(-x) 🤔 what's the clue?
Beautiful
Thanks for the great illustration, however I'm not sure what has been achieved here, all I can notice that the original integral is replaced by the sum of the similar function, which is basically the integration 🤔
Not sure if I'm seeing the full picture here!
Chapeau 👌🙏👍
Could you do a more elaborate video on when you can swap an integral and a sum, and when you cannot? Preferably with some examples.
Check out Dominated Convergence, monotone convergence, Fubini/Tonelli theorem(s),
@@alexwarner3803 link to video?
@@oniondeluxe9942 can't link on YT. It deletes the comment
@oniondeluxe9942 and, I just read the Wikipedia articles and got their sources and tracked down texts tbh. They are in most analysis books probably.
Fubini's was in My calc 3 text book, but Patrick JMT has a video on it if I remember correctly
@oniondeluxe9942 sry I don't have more info.
14:36 how is flipping the integral by the negative sign even a step? They means the same, both mean the definite integral from 0 to infinite, I don't understand at all
In integration the order of the numbers matter and when you reverse the numbers you get the negative of the value.
Wow! Great job.
{1/(-x+1)}.(x)^(-x+1)
The upper limit
(1/0).(x)^0=infinite
Lower limit
1.x=x=0
Infinite -0=infinite
never in my life did i think i would sit and watch such a crazy integral be solved yet here i am. amazing