solving an Impossible Integral... | Bernoulli's Integral
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- Опубликовано: 12 сен 2024
- 🎥 The Infamous Bernoulli Integral Solved! 📚🔥
Have you ever wondered if the impossible can be achieved? In this video, I tackle the legendary Bernoulli integral, also known as the integral of x^x, and provide a detailed solution. This complex math problem has puzzled mathematicians for years, but today, we break it down step-by-step. Join me on this incredible math journey and see how we conquer the impossible!
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Suggestion for future videos:
Do not write the plus sign like a "t", specially if you are going to use the variable t.
Noted !
@ 10:02 I think it should be Gamma(n+1) not Gamma(n-1)
Yes you are right, I made an error. Does not effect the rest of the video though
the gamma function is incredibly cool! love the videos!
It really is!
@@JagoalexanderBit it's just something you know or don't so therefore just a contrivance righg can't younsoove without it?? Thanks for sharing.
Good work dude!! This is one of my favorite integrals and i really liked your explanation on how to solve it. Hoping for more to come!
BTW, your voice is very soothing
Thank you so much!
0:37 I'll give you a second[INSTANT JUMP CUT]Right!
hey, today i have found your channel and watched the x^{-x} video. you explain really well, and in a manner that everyone can grasp easily. keep the good work!
Awesome, thank you!
Honest tears filled up inside me when I saw the integral 💀💀💀 😭😭😭
If someone had told me to squash that into the gamma function I would have thrown up, very well done
Wow, dude my brain was trippin and feelin fine just then. What a deep dive.
Hey! I was a bit confused on the very last part. Could you please explain how you rounded the 'sum of alternating inverse squares' to roughly 0.783431. From my limited understanding, the sum should approach (pi^2)/12, which is approximately 0.822467. Thanks for the awesome video either way!!
It isn't the sum of inverse squares. If you look carefully each term is n^-n not n^-2
@@Jagoalexander ohhhhhh. Thank you so much!
Johann solved it prior to the luxury of e, they were also seeking the sum of 1/k^2 where a student named Euler came to the rescue. Great job!
Interesting!
Good. Jako student bardzo interesowałem się tą funkcją. Super metoda całkowania.
Such a good explanation thanks brother. keep it up
Thanks man
The last result can be rewritten as
Sig(n=1->inf)[(-1)^(n+1)/n^n]
amazing thank you so much!! 😊
No problem 😊
Meanwhile the Pi function is crying in the corner . JUSTICE for PI function😝. anyways , great vid
What is the pi function
@Jagoalexander it's just Gamma(x+1) . it's cooler imo 😁but noone mentions it for some reason
@@gitboyyy i also like the pi function. it more directly corresponds to factorial cuz theres no offset. the gamma function is nice in other places too but i feel like it takes too much of the spotlight
Bravo Lilin ! Grazie !
4:36
Just, thank you!!
So many RUclipsrs skip this step now... I'm tired of writing comments with proper proof :)
You're welcome mate
n=ʌ t=+ ∞=o makes it a little confusing but enjoyable as well
great job. keep going
The n looks like an among us
Maybe it is? 👀
Sus@@Jagoalexander
Is requirement 1 really met? I'm pretty sure that term approximates 0 with n->inf.
I was searching for a video like this... I was wondering, how could you solve the same integral but by the method by partial integration you mentioned... i'd like to see the demonsttation made in that way, thanks!😁
Isn’t limit n->infinity of fn(x)=0 for the first condition of the Dominating Convergence theorem?
It doesn’t converge to xlnx
Yes, I think f(x) and fn(x) were mislabelled.
DCT is used for swapping a *limit* with an integral, and so in the case of a summation, the relavent sequence of functions is the sequence of partial sums.
Take f(x) to be the whole integrand e^(xlnx) and fN(x) to be the sum from 0 to N of (xlnx)^n/n!. Then as N->infinity, fN pointwise converges to f (this is just the convergence of e^x). So we’re good to go!
@@maxchemtov3482 thanks for the explanation
@@maxchemtov3482 Excellent comment. We can also use g(x) = e^|x log(x)| (and not simply e, as wrongly explained in the video) as the dominating function to apply the DCT. I used the absolute value because log(x) is negative between 0 and 1.
(x ln x)^n/n! converges to x ln x? I call bullshit.
Hello dude, Nice vidéo.. what app did u use in this video?
Goodnotes on iPad
1-1/4+1/27-1/256+1/3125-1/6^6...
Slight mistake. You are correct about factor n! In the term, but it is equivalent to gamma(n+1) not gamma(n-1) as you wrote down in your proof.
Thank you, was just a mistake when I was remembering my method !
What happens if u change the integration boundaries, for example the integral of x^x from 0 to 2 ?
Could you prove that the last part converges ?
It can be proven using the ratio test, have a go!
The alternating series test also makes is super clear that it converges imo
No tendría que ser mínimo mayor a 1?, ya que en 1 la función vale 1. Pregunto desde mi propia ignorancia
f(x)=x^x; f(0)=1 ; f(1)=1; 1x1> S f(x) dx
Ойй, сорри, а Х может быть комплексным числом? i, с действительной составляющей. Не досмотрел, некогда.
нет, «i» не может быть X, так как этот интеграл определен для действительных чисел, а не мнимых.
good math fun...
I think this is wrong. n! = Gamma (n+1), not Gamma (n-1).
My bad, I meant to write n+1, solution is still valid and works though
👏👏👏🔝
Great video!
Круг.
Circle 👍
5:00 let y=x lnx, you say that lim[n to inf] (y^n/n!) = y. I don't believe that, I think it's 0. Wolfram alpha too.
Here's the intuition:
In the expression (y/1)(y/2)...(y/n), nearly all factors (those with n>y) are smaller than 1 and they get arbitrarily small as n goes to infinity.
So now given your f_n(x), now we have that f(x)=0. That's certainly integrable and therefore it's all fine, it does not affect the rest of the proof.
Thank you
@@JagoalexanderI have another question, when using the dominated convergence theorem, shouldn't you prove that the sum converges, rather than just "(xln(x))^n/n!"? If anything, this further proves that this expresion should converge to 0, for the sum to converge to some number (as necessary but not sufficient condition). It is not a correction, I have genuinely never seen applying dominated convergence with summation, so I could be wrong.
Kann man nicht einfach so integrieren 1/ x +1 * x ^ x +1
nein. versuche dein ergebnis abzuleiten - und vergiss nicht das x im Exponent von x^x...
d/dx (1/(x+1) x^(x+1))=
x^x * (1 - x/(x+1)^2 + x lnx)
Bro we cannot expand it about x = 0.
Explain?
x^x is not defined at x=0 and we cannot even find limit of this function at x=0.
lim x→0 x*lnx = 0, so lim e^(x*lnx) = e^0 = 1. So, lim x→0 e^(ln(x^x)) = x^x = 1.....
0^0 is an indeterminate form obviously, but that only means that _x^y does not exist for x,y=0 in a defined fashion_ , like it exists *uniquely* while x,y=1. But that doesn't affect the existence of the limit x→0 x^f(x) (y=f(x)...); the limit could be checked for existence or calculated from itself (if existed).
@@krishnamaity5056 Pls check differentiability at x = 0, since it's a must condition for expansion.
@@salmankhan2910the limit is 1, I believe. Use l’ hospital on ln(x^x) = x ln x = lnx/(x^-1) first. This-> 0. So x^x -> 1. So the integral is improper but with an extra step or two I think his solution still works.