There are lot of moving parts in this sum -- Also: We will use this result in upcoming videos! 

There's a mistake at 10:07. The m-sum goes from zero to *n*, not infinity.

That was intense yet clear. Thank you professor, I’m going to use this.

For the "Homework problem", it seems that one cannot just use the Newton's formula in Pascal Triangle, because of the nasty (2m+1) denominator. I solved this by replacing 1/(2m+1) with INTEGRAL(x^2m from 0 to 1). From the binomial theorem, this leads to an integral of the form I(n)=INTEGRAL((1-x^2)^n from 0 to 1). after an x = sin Q substitution and integration by parts, this leads to a reduction formula I(n)= (2n)/(2n+1).I(n-1), which yields (2n)!!/(2n+1)!!.

A very cool problem! Thanks Prof Penn!

3:40 1 is probably the most important number in the odd double factorial 😂 10:25 Homework 17:15 Good Place To Stop

I also want to point out that the integral from 0 to 1 of (1-y²)^n can also be evaluated by the substitution x=sin(u) or x=cos(u). This will result in the integral from 0 to π/2 of (sinx)^(2n+1), or equivalently, (cosx)^(2n+1). These can be evaluated either by using the Beta function, or by first deriving a reduction formula for the integral of (sinx)^m or (cosx)^m using integration by parts, and then using the resulting recurrence relation. Or you can use the beta function right off the bat after substituting t=y².

I think that in the derivation of the 2nd tool, the x isn't a dummy variable, so it seems important to point out that the conversion to the geometric series is only valid for abs(x)

0:59 that's no theta, that's the death star.

Holy this is the most satisfying math I've seen in a long time!

A different approach would be this: The series S can be rewritten as 1/4*sum( n!n!/(2n+1)! * x^(n+1)/(n+1)) where x=4. Then taking the derivative with respect to x and replacing the factorial terms by the beta function Beta(n+1,n+1) in its integral form, we get 1/4*sum(integral (x(t)(1-t))^n*dt)) = 1/4*integral(dx/(1-x(t)(1-t))) after noticing that x(t)(1-t) is always between 0 and 1 for x less or equal than 4 and t between 0 to 1. Then integrating with respect to x and setting x=4, we get 1/2*integral(log|2t-1|/(t(t-1))dt). Substituting y=2t-1 and after some basic manipulations, we get 2*integral(-log(y)/(1-y^2)dy) from 0..1. We can use the geometric representation of the denominator which gives 2*sum(1/(2n+1)^2) = p^2/4

love these types of problems that call on so many tools

⭐️ Lately it seems like every video is a tour de force. I enjoy these videos very much. Thank you, Dr. Penn.

For the HW of the formula for the sum identity: Recall that the forward finite difference of a function f is defined to be Δf(x) := f(x+1)-f(x), and the k-th forward finite difference is gotten by applying this difference operator k times. For the k-th difference, I’ll denote it by Δ^k[f](x) := f(x+1). One can prove by induction that the n-th forward finite difference of 1/x is (-1)^n * n! / (x(x+1)…(x+n)), because when taking the forward difference of the (n-1)-th forward differences, one can divide on the outside and multiply the inside by x(x+1)…(x+n), and so then the inside becomes (-1)^(n-1) * (n-1)! * (x - (x+n)) = (-1)^n * n!. By induction, one can prove in general that the formula for the n-th forward difference of a function f(x) is Σ{k=0…n, (-1)^(n-k) * nCr(n,k) * f(x+k)}, where nCr(n,k) is the binomial coefficient. Another way to kind of prove this would be to define the shift linear operator Tf (x) = f(x+1), note that Δ= T - 1, and expand Δ^n = (T-1)^n with the binomial theorem, which works because addition and composition of commuting linear operators satisfies all the properties of multiplication needed for the binomial theorem to apply. After this, we can compute Δ^n[1/x](1/2) using both formulas, and after some simple algebra, we get the result we are after. The second homework problem of showing the series converges would be rather tricky to write in a comment, but you can do it by showing that the n-th term is always 0 for all sufficiently large n. To show this, it suffices to show that for h*ln(N+1) = x - x^2/2 for x>0, then showing that since 1/2n is being subbed in for x to obtain this lower bound, the x^2/2 terms become 1/8n^2, the sum of which is bounded above by a constant (π^2 / 48). After all of these lower bounds are used, the inequality that remains to be proven is that h*ln(N+1) 0 for all sufficiently large N. Taking h = 1/4, N >= exp(π^2/12) is enough to prove the claim.

Just a tip, you should probably start recording your audio in mono. I've noticed a few people in the comments complaining about your voice "Jumping from side-to-side" which makes sense since you have to constantly turn your head to look at the board. The stereo audio doesn't really add much to the video besides being distracting for those of us wearing headphones. Still a great video! Keep up the good work :)

I was waiting for 15 minutes 44 seconds for the upper red line to be erased.

8:30 instead of expansion , we could substitute sinx and get answer by bernoulli's shortcut integral

I feel the mind of Ramanujan behind this one.

Are you implicitly changing the order of limits in the second fact, is that justified simply because everything converges absolutely?

I was wondering where is the dependence of X in the third tool and turns out it didn't get corrected until almost 13 minutes into the video.

How to divide the sum in minute 9:28 into two of them?

11:15 Magician Michael

As currently in many maths manuals, exercises left for the reader are in no way trivial and even more difficult than the principal problem to solve... Pascal Identity doesn't work (and induction over n? Over m?...) There's a 2m+1 in the denominator of the LHS and only n's in the RHS. You cannot get rid of the "m's". You get a bunch of m's and only n's respectively.

I was waiting for the magic that occurs at 13:08 ^^

Can someone explain how (2n)!! can be obtained from induction using Pascal's identity? That's at 10.58.

For x!! to not indicate (x!)! is an irksome abuse of notation. Mathematics has a few of these, for me.

Hi, 0:05 : missing x in the last formula.

Michael, I tried the Maths Induction HW and I found that when I make n=2 for the identity the LHS does not equal the RHS??

Can someone explain how they showed the convergence of this sum? I tried the ratio test, but the limit L was equal to one. I was thinking that maybe you could use extensions to the ratio test where L=1, such as the ones described in the Wikipedia page here: en.wikipedia.org/wiki/Ratio_test

15:33, isn't there a mistake? (2^n)^2 isn't 4^n

Oggi sono arrivato al min 13

Tiny Moving Parts *wink* *wink*

Early!!

Pls, use just 1 microphone. Your voice keeps changing the direction it comes from, which is distracting and annoys a lil bit. But the problem is quite interesting. Nice choice. 😉

Why in hell would anyine use trig at all in this problem instead of just algebra..it's random and unintuitive and out of nowhere..isnt everyone else wondering this??

Why the hell would anyone multiply by y at 5:01?? And the. Do all that elliptical craziness that no one would ever think of??That's just random arbitrary nonsense like so much of math is....it's crazy and frustrating..