I may suggest another way of reasoning. First we can expand e^(x^2) using its Taylor series and multiply it by the constant x^n. In such a way we would get n*integral(sum(x^(2k+n)/k!)). Since the convergence of the sum is uniform (dominated by sum(1/k!)) we can switch the integral and the sum so that we get sum(n/(k!*(2k+1+n)))=:sum(f(n)). Since sup(abs(f(n)))

A more direct way is to realize that n x^n dx converges to a point mass at x = 1. Then e^(1^2) = e.

@spamspam4117

3 ай бұрын

Exactly! Was just about to comment that nx^n converges to a delta function centered at 1, making the whole limit trivial... to a physicist :) This may sound like sacrilege to a rigorous mathematician :D

@ALX112358

2 ай бұрын

​@spamspam4117 Functional analysis has stuff far weirder than the delta function.

Can you make a video about the question mark function? The function that is monotonically strongly increasing yet has derivative 0 almost everywhere?

@ 8:14 The numerator should be -2n+1, not 4n+1.

@homerthompson416

3 ай бұрын

Yeah Penn's argument just wrong here

@krisbrandenberger544

3 ай бұрын

It would've been easier to just compare the values of a_n and a_(n+1) by evaluating the integrals and comparing fractions. We find a_n=n/(n+1) and a_(n+1)=(n+1)/(n+2). Then, putting everything over a common denominator, we see that the numerator of a_n is bigger than the numerator of a_(n+1), which implies a_n

@nahuelcaruso

10 күн бұрын

The plot of the integrant f(x)=[(n+1)x^(n+a)-n x^ n]e^(x^2) is quite tricky, it is not straightforward proof through inequalities. I have proven (a_(n+1)-a_n)>0 using the Taylor expansion of e^(x^2) and interchanging the integral, then the sum reached is a numeric sum with positive terms.

12:09 we don't need to do any more checks once we guarantee that such an N exists, because if a(n) < 3 for all n ≥ N, then a(0) < a(1) < ... < a(n-1) < a(n) < 3 so the whole sequence is bounded

@schweinmachtbree1013

3 ай бұрын

plus he didn't even use the fact that the sequence is bounded _by_ 3; just that it is bounded, which follows from it being eventually bounded

nx^n converges in distribution to a Dirac centered on 1. Hence the integral converges towards the value of e^x2 at 1, which is e. I’m not sure whether the rigorous reasoning would be shorter, but 1. I guess it should, 2. It gives overall a much more visual interpretation of the final result 3. You can the generalize the result for any function f replacing e^x2

At 12:10, why do you take the expression on the second-to-last line, where it looks like if n->inf, a_n 0 ), and then turn it into 3/e*e = 3?

@UNIverso.Matematica

2 ай бұрын

Actually if a number is less than e, it is also less than 3 😁. The claim was to proof that a_n

Once we have the recursive formula at 4:39, it's quite easy to get a closed form for a_n by induction and derive the limit

I think the resolution is unnecesaryly complicated: first of all the sequence is obviously bounded by e, and doing this integration-by-parts trick a_n equals en/(n+1)-a_{n+1}/n. Then, using the boundedness of the sequence and the squeeze theorem the result inmediately follows.

If we rewrite the integral as int(D(x^n)(xe^x^2) and we integrate by parts, we get: I=e-0.5int(x^ne^x^2) as n grows to infinity, the later integral clearly goes to zero, because 0inf. So I=e.

@egoreremeev9969

3 ай бұрын

It looks like you integrated both multipliers, although you need to integrate one and differentiate the other So it's either int D(x^n)(xe^(x^2)) dx = 1/2 e^(x^2) * D(x^n) |^1_0 - int D^2(x^n)*1/2*e^(x^2) dx = n/2 * e - 1/2 int n(n-1) x^(n-2) e^(x^2) dx or int D(x^n)(xe^(x^2)) dx = xe^(x^2) x^n |^1_0 - int x^n * D(xe^(x^2)) dx = e - int x^n (1+2x^2)e^(x^2) dx Still though your argument holds as in the second one the integral goes to zero too.

@Megathescientist

3 ай бұрын

​@@egoreremeev9969 You're absolutely right, my original intention was to integrate D(x^n) and then differentiate xe^(-x^2), but I mistakenly integrated xe^(-x^2) as well. Fortunately, as you correctly pointed out, my overall argument remains valid despite the erroneous calculation. I really appreciate you catching my mistake and taking the time to clarify the proper steps of integrating D(x^n) and differentiating xe^(-x^2). Thank you for your feedback!

for boundedness yo can just estimate the integral, then you get a_n

THANKS PROFESOR!!!!!!, VERY INTERESTING !!!!!!

My method of choice was to substitute e^(x^2) with its Maclaurin series, change the order of integration and summation, integrate, and take the limit afterwards.

Wow, that was a challenge for me to follow. I wonder how long it took to do this the first time that it was proven….

As a faster non rigourous way , a_n ≈a_n+2 as n tends to infinity so put an+2=an in the recursion relation and solve for a_n(which is actually only for n-> inf)and you get e

6:48 Wouldn't you need to prove first that e

Your videos are great

Interesting that for all values of x not including 1, all points eventually tend to zero, yet the integral tends to a nonzero value. So switching the limit and integral wouldn’t be valid here since limiting the function first would have a zero result yet

My first thought is lim[n -> ∞] ∫ [0, 1] n*x^n*f(x)dx = f(1), so the limit is f(1) = e^1 = e. I am not sure if this identity is true or not but it gives the right anwser for this promblem.

@user-oe5eg5qx4c

3 ай бұрын

I am wondering. What is the limit when f(x)=sin((1-t)^-1)?

This was a_n awesome limit! 👍

how about t=x^n and DCT?

@anshumanagrawal346

3 ай бұрын

Yup, that's what I did

@spiderjerusalem4009

2 ай бұрын

what's dct?

@anshumanagrawal346

2 ай бұрын

@@spiderjerusalem4009 Dominated Converge Theorem. You can look it up if you haven't seen it before. It is a theorem about Lebesgue Integrals but can be applied to Riemann Integrals as well since Lebesgue reduces to Riemann in case the latter exists

@vladimirlucic1276

2 ай бұрын

@@spiderjerusalem4009 Lebesgue Dominated Convergence Theorem.

@vladimirlucic1276

2 ай бұрын

@@spiderjerusalem4009 Dominated Convergence Theorem

Does anybody know why one cannot use the Lebesgue Dominance Convergence Theorem here? The integrand (n*x^n*exp(x²)*1_[0,1](x) ) is measurable for all n, where 1_[0,1](x) is the indicator function on [0,1] evaluated at x. Also the integrand converges pointwise to 0, except for at x=1. The important assumption of an integrable majorant: |n*x^n*exp(x²)| is continious on [0,1], so it has a maximum C there. So it follows |n*x^n*exp(x²)*1_[0,1](x)| ≤ C*1_[0,1](x), where the latter is integrable. But obviously this doesn't work, as one would get 0 as the result?

@Nico-kd9yh

3 ай бұрын

That's because the constant C depends on n, in order to use DCT it is necessary to have a C (or an integrable funcion C(x)) such that |n*x^n*exp(x²)*1_[0,1](x)| ≤ C(x)*1_[0,1](x) for all n natural

@Hello-jm1zw

3 ай бұрын

@@Nico-kd9yh Oh yeah thank you, that's why it doesn't work, now I understand!

@leonardolazzareschi9347

3 ай бұрын

To you use correctly the dominated convergence theorem you can observe that the integral is equal to the integral from 0 to 1 of (n+1)*(x^n)*(e^(x^2))dx - integral from 0 to 1 of (x^n)*(e^(x^2))dx. Now, using integration by parts, you can note that the first integral is equal to e - integral from 0 to 1 of 2(x^(n+2))*e^(x^2)dx. Finally, the integrals of (x^n)*e^(x^2) and of 2(x^(n+2))*e^(x^2) converges both to 0 using the DCT, because a majorant is 2e^(x^2) (remember that 0

nx^n acts like a dirac function with the spike a x=1, is there a way to make this into a shorter yet formal proof?

@jay_sensz

3 ай бұрын

Showing that the integral of the function n*x^n over the interval 0

@ilayyundler8857

3 ай бұрын

Thank you!

I think the easiest way to do it is by bounding above and below as n * e / ( n + 3 ) The limit is straightforward.

He didn't need to use the recursive sequence to prove boundedness, just use the original definition. e^x^2

@DrR0BERT

3 ай бұрын

OMG, I came here to write that exact statement. Sometimes there is a much simpler way.

@StanleyDevastating

3 ай бұрын

i think you meant: a_n < 3n\int{0}^1 x^n dx. But yes your way is simple and the way Michael Penn uses doesn't work either.

Huzzah!

e.

E

First??