So many factorials!!!

[@blackpenredpen]

What?!!! This series has the sum in a closed form?!

[@djvalentedochp]

man where are your videos?

[@VaradMahashabde]

Hello!

[@HeyKevinYT]

DJ VALENTE DO CHP If you check his instagram you’ll see he got a surgery recently (appendicitis I think), so he needs time

[@djvalentedochp]

@@HeyKevinYT thanks for the information 👍

[@ffggddss]

Get well soon, bprp!! Fred

[@jacemandt]

What's so amazing about these videos is how these results at first look (to me) well beyond the ability of a mid-level college math major (that was me in college, but I never studied stuff like this), yet they frequently use just basic first-year calculus, applied creatively.

[@anastasissfyrides2919]

13:32

[@elie.makdissi]

🤣

[@den1fednu]

27:57 ?

[@Manuel-pd9kf]

This video is gonna get alot of views, I can feel it

[@holyshit922]

If we want to do this in other direction 1/sqrt(1-x^2) - by binomial expansion integrate to get arc sin(x) Use Cauchy product to get series for arcsin(x)/sqrt(1-x^2) but we would have to deal with double sum

[@pikupal8996]

Sir can you do a series of lesbegue integration and measure theory , functional analysis?

[@shanmugasundaram9688]

A nice convergence of summation of proper fractions.The definite integration of sine power function is mysteriously connected to the summation.Very interesting.

[@timurpryadilin8830]

Very claasic video my Michael. Excellent!

[@williamchurcher9645]

Just an integration tip I thought I would share: when changing variables, you can multiply the integrand by the modulus of the derivative and make the bounds of the integral ordered, ie the smaller bound on the bottom. The reason for this: when doing change of basis in multivariable calculus, we use the modulus of the Jacobian. If you just use the one dimensional version, you get what I just described. I personally find it easier to work with, but it's up to you.

[@OH-pc5jx]

Yes - unless the derivative is zero within the range, in which case you’re ✨ trouble ✨

[@OH-pc5jx]

Usually a sign of a bad substation tho so I don’t think it comes up too often

[@williamchurcher9645]

@@OH-pc5jx yes well change of basis theorem doesnt hold if the derivative is zero anywhere. I suppose you may be able to split the domain of the integral before substitution, but it may get messy ;)

[@OH-pc5jx]

William Churcher yeah in 1D you can get around it with a bit of thought but in nD I think I’d just use a different substitution

[@goodplacetostop2973]

28:41

[@armanrasouli2779]

:)

[@hoodedR]

Woah I just realised what a long video that was

[@nothowtung7372]

username checks

[@vaxjoaberg9452]

15:06 was not a good place to stop

[@tobiasgorgen7592]

Michael, you have a typo in your thumbnail. the factorial in the thumbnail is inside the brackets

[@sauravthegreat8533]

I saw that at first and then thought “infinity, how is this video so long”

[@ffggddss]

There are two factorial signs in the expression in the thumbnail; one inside parentheses, the other outside. Both are correct. The summand is just 1/C(2n,n). Fred

[@sauravthegreat8533]

ffggddss no He changed the thumbnail, previously it was (2n!) which is completely different from (2n)!

[@ffggddss]

@@sauravthegreat8533 I see... Well that *does* make a difference. Fred

[@duncankoepke7499]

I would love a video about the dominating convergence theorem

[@OH-pc5jx]

Very nice! Wouldn’t know where to start without the hints ngl

[@shanmugasundaram9688]

A nice convergence of summation of proper fractions plus one.The definite integration of sine power function is mysteriously connected to the summation.Very interesting.

[@fmakofmako]

Yes please to the video on dominated convergeance theorem.

[@filipchris245]

15:05 Nice clothing change!

[@kushsinghal1998]

Hey I can't wait for the video on the dominated convergence theorem. I'm struggling with it quite a bit

[@QmcometdudeShardMaster]

As always, a wonderful video. Thank you for the great math content!

[@gnomeba12]

Would love to see some videos on some of the more sensitive convergence tests like Gauss's test

[@bsuperbrain]

Beautiful. How did you find these three lemmas?

[@jonathangrey6354]

Please a video on the dominated convergence theorem!

[@vh73sy]

The result can be expressed as ²F¹[1,1;0.5;(1/2²)] F is the generalized hypergeometric function wolfram notation Hypergeometricpfq[{1,1},{0.5},0.25]

[@jimskea224]

But ²F¹ is the usual "original" (Gauss's) hypergeometric function. It's only really "generalised" if the indices are different from 2 and 1. So much so that the indices 2 and 1 are usually omitted in this case and one simply write F(1,1; 1/2;1/4)

[@ТимофейБакшеев-ь8р]

Oh my God! It’s awesome! Good job

[@fartoxedm5638]

I think you would better mentioned that formula of infinite sum which you used is only worth for convergent sums of course it is arcsin so x is between -1 and 1 however it was not really obvious

[@VerSalieri]

You are starting to remind me of Sami Hamiyyee.... my favorite professor.. Thank you.

[@mxminecraft9410]

What's the second tool called And iis there any other way of proving it ?

[@alejandrojimenez108]

How did you get this? Like seriously it seems so arbitrary but boom everything works and fits in perfectly

[@matthias7790]

27:56 was that a burp?

[@peytonglass745]

please do a proof of the dominated convergence theorem!!!

[@pikupal8996]

He have also used the differentiation theorem of power series.He should also do a video of that.

[@noway2831]

How would one approach the sum of (2k choose k) * (-4)^(-k) from k=0 to infinity? I got to the sum from the integral of e^(-x) erf(sqrt(x)) dx from x=0 to infinity. That integral has a surprisingly simple closed form, and I imagine its evaluation is quite elegant. The sum I have verified is correct, and I obtained it through the taylor series for e^x and some gamma function identities.

[@mrmathcambodia2451]

So good solution, I like this video .

[@nontth5355]

Do a video about catalan number please

[@willianmarconbicaio6125]

Amazing!! Shouldn't the result be pi/9/3^0.5 + 4/3 though? I can't figure where the 2 in the numerator comes from

[@sergiokorochinsky49]

The general case is: Sum[(n!)^p/(q n)!,{n,0,Infinity}]=pF(q-1)[1,1,...,1;1/q,2/q,...,(q-1)/q;q^q] where pFq is the Hypergeometric function. The particular sum in the video is 2F1[1,1;1/2;1/4]

[@mrflibble5717]

Excellent! Michael would you do a presentation on the Dominated Convergence Theorem, also what do you recommend as a good reference for detail on it? rgds, Rod

[@iridium8562]

19:02 but by writing 1-x^2 as sqrt(1-x^2)^2 you are assuming that x =< 1, right..?

[@AmitBentabou]

X=1/2

[@demenion3521]

Even from the very beginning, the condition |x|

[@michaelempeigne3519]

not really since the integral is from 0 to 1.

[@FrankDelVecchio]

Can you evaluate the integral log(x)/(1-x^2) from 0 to 1 by means of a contour?

[@birdboat5647]

a lot of prep pays off

[@gardenmenuuu]

Sir its great

[@ramanakv3272]

First definit integral can be easily derived by walleys method than induction

[@xuwei0126]

Hello, I want ti know how did you find those tools? Because if I have to solve this problem from nothing how can I find those tools?

[@aswinibanerjee6261]

Write the fraction as a beta function then change the order of sum and beta integral. Then do the sum first (which will be an easy geometric series) Then do the integral

[@geometrydashmega238]

Thank you for your comment. I thought about gamma functions at first when seeing the problem but I had forgotten about beta. I tried it as you said and indeed, a bit long to compute but very easy approach

[@pacojacomemaura2129]

Very great idea! Using Beta function properties, I obtain that the general term of the series is B(n,n)*n/2. One has to be careful, because Beta funcion isn't defined in B(0,0), and is necessary to pull apart the first term, 1, from the rest of the series. Then, using the Beta funcion definition and Lebesgue's dominated convergence theorem, I exchange the integral symbol for the sumation symbol. Now, the series inside the integral is the derivative of a geometric series (the term is (n+1)(t(1-t))^n, from n=0 to infinity). This series is equal to 1/(1-t(1-t))^2 when t \in (0,1). So the initial series is equal now to 1+1/2 \int_0^1 1/(1-t(1-t))^2dt. This integral isn't funny at all, but can be done and gives the same result obtained by professor Penn.

[@goblin5003]

Suggestion: evaluate the same sum but instead of (2n!) in the denominator, put (2n+1)!

[@hjdbr1094]

Could you prove that (2n)!!/(2n-1)!!~sqrt(πn) please?

[@yossefswelam265]

15:02 magic

[@fredericmonrasividiella7394]

for me it is interesting to listen to the whole demonstration one day and then (one day, two days later) go back to listening to it 1.5 times faster. complete understanding. thank you Mr. Penn!

[@urumomaos2478]

Me: doing the exercise Michael: and thats a good place to stop Me: okay michael senpai :3 i will stop 4 u uwu

[@steve2817]

Factory-al.

[@ethanbeachy6593]

His answer and the way I did it numerically in Matlab gives a consistent result. I wouldn't have gotten it analytically though... This is really good!

[@soundsleep4119]

Enough Sound explanation....but may I get the same series by the help of fourier series? May I get a periodic function which may help me?...eagerly waiting for your reply.... From India ♥️

[@rontiemens2553]

In your title you forgot the rejoinder, "... so little time!!!". Seriously, you put out great content.

[@RickyKwokMath]

Funny, usually when factorials appear in infinite series, e usually shows up. This time it's pi.

[@sergiokorochinsky49]

Try the numerator without the square... Sum[(n!)/(2 n)!,{n,0,Infinity}]

[@ffggddss]

Remarkable result! Remarkable that you can even get a result! Especially seeing what it takes to get it!! Might I ask how you came across this result? 2π/(9√3) + 4/3 = 1.736399858718715077909795168364923... Finally, kudos for your teaching style. It is really exemplary! Fred

[@VaradMahashabde]

Did anyone else notice when he changed shirts after the second tool?

[@vh73sy]

In general Sum( (n!)^b / (b n)! ), n=0 to inf for b>=2 can be expressed as the generalized hypergeometric function p F q b F b-1 [1, ... (b times) ... ,1 ; 1/b, 2/b, ... , (b-1)/b ; 1/(b^b)]

[@ramanakv3272]

The teacher has to tell where term Wise differentiation ,integration is applicable

[@riadsouissi]

Nice problem. Did it differently though (after looking at the video, there are some similarities) - defined sum y(x) but adding x^(2n-1) to the sequence and starting at n=1 (so our final sum = y(1)+1). - after some manipulation and derivation, I get a differential equation y'(4-x^2)-3xy=2 with y(0)=0 - Using integration factor, I get in the end y(x) = 2(4-x^2)^(-3/2) * integral(sqrt(4-t^2)dt, t=0..x). Which can be integrated with simple substitution t=sin(u). - End result y(x) = 1/(4-x^2)^(3/2) * (4*arcsin(x/2)+x(4-x^2)^(1/2). Substituting x=1, I get same result.

[@Notthatkindofdr]

That's close to how I did it too, though your equation was slightly simpler than mine.

[@diegohcsantos]

Amazing! Could you explain how did you found this ODE? Also, why add x^(2n-1) instead of x^n?

[@pappaflammyboi5799]

@Michael Penn Your answer is wrong, it's: pi/(6×sqrt(3)) + 4/3

[@mohamedelkhmissi3504]

15:05

[@nicholashernandez4367]

Hello, I have a question. What is (319!)! ?

[@coycatrett2303]

Dat burp doe

[@General12th]

15:06 Everything changed when the red-shirt nation attacked.

[@Jack_Callcott_AU]

Isn't math(s) amazing. One never knows what will turn up in series. Here we have pi again!

[@jotaro6390]

Nice

[@stedis7259]

Hello Michael! Where is this problem from? Was it from a math competition?

[@khiemngo1098]

Nice problem and thanks for sharing this video! By the way, there's a minor mistake in that the derivative of arcsin(x) should be 1/sqrt(1 - x^2).

[@maxblack493]

This remind me the beta function.

[@vh73sy]

the way it's written on the post cover (2n!) is wrong, it leads to divergence. The right notation is (2n)! as it appears on the board in the video. 2 x n! is not (2n)!

[@gvomet1]

It seems so strange to obtain a sum as the result ....

[@CarlosFloresP]

13:33 burp xd

[@Walczyk]

this one is cute, reminds me of quantum mechanics prroblems

[@tobiasgorgen7592]

Even though the answer is a closed form. The fact that is a sum leaves me... Unfulfilled

[@schweinmachtbree1013]

you're welcome to take a common denominator lol, unless you would also be unfulfilled by the resulting sum in the numerator xD

[@tobiasgorgen7592]

@@angelmendez-rivera351 it definitely is, don't get me wrong. By the way Michael prefaced the video I expected the sum to be straight up pi or maybe pi ^ k or something. Having a closed form be it convoluted as it may be is always amazing starting with such weird sums

[@PriyaGupta-sr1iy]

7th comment

[@parameshwarhazra2725]

13:33 Michael penn remain silent for 2sec straight giving respect to Chadwick Boseman

[@parameshwarhazra2725]

@Adam Romanov are you Russian

[@IoT_]

@@parameshwarhazra2725 по-моему, нет.

[@parameshwarhazra2725]

@@IoT_ what did you say please tell me in English

[@IoT_]

@@parameshwarhazra2725 I said, in my opinion he's not а Russian, but allegedly he knows it a bit.

[@parameshwarhazra2725]

@@IoT_ oh okay