Let's get Product-ive: Integrating one infinite boi

[@user-zj8jn3hs6f]

When exactly did i start watching integral solving recreationally?

[@micomrkaic]

I’m asking myself the same question:)

[@thomaslopez8395]

It’s weirder when you’re not even majoring in mathematics

[@user-fungus]

It's much weird when you're still in high school 🤣

[@manamritsingh969]

@@user-fungus relatable

[@Nothingtonnobodson]

@@micomrkaic same here

[@SteamPunkLV]

at this point the product function is just used for ridiculous questions such as these

[@NoNTr1v1aL]

0:28 Oh shit... he's onto us.

[@blazep5881]

Classic German humor at the start

[@dylantellez1758]

Oi, this is german humor. It's no laughing matter

[@AlexiLaiho227]

i pictured a burned hotdog when i thought of the phrase "infinite tan boi"

[@blueberrypoptart2424]

Such a beautiful infinity sign 4:30

[@oliverinspace9252]

Why not use the identity 1-tan^2(x) = 2tanx/tan(2x) instead? This simplifies the product directly to ∫(1/x) lim(2^i * tan(x/2^i)/tan x) dx which can also be evaluated using l'H rule in the same way as shown in the video.

[@fengshengqin6993]

Yeah,I have used this way ! Great .

[@desertrainfrog1691]

I'd guess he didn't think of it and decided to pursue the solution he thought of instead.

[@madgodyt2391]

i did with the same approch !! and yeah the product term becomes much simpler .

[@decreasing_entropy3003]

Because I didn't remember this identity.

[@cupcakesandrose]

"Cute twink." Suddenly the channel name makes sense.

[@IlTrojo]

Very interesting as always. Just one little thing: at about 8:00 you ought to have also shifted from k to k-1 on top of the big pi - no harm done as you were going to take a limit soon afterwards, but it could have cost you the loss of some constant. Cheers!

[@glukhas]

Es ist immer wieder spannend, deine Videos bei KZbin anzusehen... Danke sehr, Schnuckel-Mathematiker! und Grüße aus Brasilien :)

[@ezioauditore7378]

Dude, I'm 16 and even don't understanding a thing on this video, just basic trigonometry, I got surprised because this seems impossible for me to solve but you nailed it so easy. Hope I can get to this level some day.

[@Vibranium375]

Bro u just need a bit of practice, hard work and a bit of interest. It has been 3 years so I think seeing ur passion you might already have reached this level.

[@ezioauditore7378]

@@Vibranium375 wow, I didn't even remember this comment. And yeah, I've improved so much since then, not so much in Calculus, but in math in general I did. Thanks for the words

[@maalikserebryakov]

@@ezioauditore7378 1 year later, how’s your calculus skill? particularly your ability in the symbolic evaluation of integrals

[@echopotato]

8:29 "... ONE THING I DONT KNOW WHY IT DOESN'T EVEN MATTER HOW HARD YOU TRY"

[@achalanand2213]

Take a simple math problem keep multiplying and dividing by same quantity and then use some properties to expand them, - tadaaaa - a mind bending math problem .

[@tomsxe]

I hope no one has removed their headphones after they've heard "if you want to support me a bit more, take a look ..." at 17:52 when you pointed down

[@NotLegato]

about the pythagorean identities: you can derive them all from the unit circle by drawing a triangle with all the trig functions on it and so you won't have to memorise as much.

[@hopethisnameisntok]

VERY NICE! I really enjoy. Its great for when you leave uni and start working with something that does not involve pure maths. 7:54 don't forget the upper limit, the last term should be for x/2^k-1 altough it makes it all much more smooth as you did and in the limit nothing matters

[@hunghinsun2123]

At 13:12, instead of using L'Hôpital's rule, actually you can simply use the standard result of the limit of sin(x)/x being equal to 1 as x tends to 0.

[@impossiblemission4ce]

I had a lot of fun doing this one. I'm horrible at trigonometry, so I made a cheat sheet starting with Euler's formula and going from there, deriving another form of tan^2(x) and the double angle formulas and such. I think I'll have to keep this question ready, for a long trip.

[@joeyazbeck849]

Could ur titles and intros get any cringier? Love the math skillz tho

[@soyvjdexter]

The intros are hilarious. Cringy but hilarious

[@alexcollins9983]

You could also use the double angle tan formula to get a telescoping product.

[@Phi1618033]

"That was quite easy." If you say so.

[@paulg444]

Friends, if you dont love him and his parents, then see your cardiologist because you dont have a heart !!

[@AFIyingKiwi]

At 3:23, why didn't you directly turn 2cos^2(a) - 1 into cos(2a) since that also is a double angle formula for cosine.

[@affrokilla]

Love your videos, always explained very well. I would like to see some maths involving machine learning (solving Support Vector Machine and gradient descent for example)

[@mijoo5685]

En el minuto 13:30, por qué se puede usar L’Hôpital si k es una variable no continua? Osea, k es un número natural... en ese caso no se podría derivar o si me equivoco entonces por qué se podría?

[@williamtachyon2630]

”Infinity boy.”

[@gregoriousmaths266]

William Tachyon *boi

[@kono152]

crystal clear explanation, though i wish you calculated the second product too since that result kinda confused me, but ill go ahead and do it myself

[@tomvanmoer8202]

"Boi", always cracks me up xD

[@omarifady]

You forgot to put absolute value around sin(x) when integrating cot(x) :D

[@RkMrn101]

i love this guy

[@FitR_MusicProductions]

7:05 , “so working with finite things is way easier than working with infinite things” huh coulda fooled me.

[@justwest]

wow, beautiful trigonometry right there!

[@poutineausyropderable7108]

I have a feeling a bunch of clever tricks will be used.

[@nicolasbaghdassarian2119]

Beautiful proof ! I love maths and the way you explain it !

[@sergioh5515]

Just finished the video. What an elegant solution. Thank you! :D

[@jiaming5269]

A simplification: 1 - tan2(a) = 1-sin2(a)/cos2(a) = (cos2(a) - sin2(a)) / cos2(a) = cos(2a)/cos2(a)

[@alexvasilachi9558]

came for math, stayed for the memes

[@alexvasilachi9558]

@@PapaFlammy69 \(πωπ)/

[@insouciantFox]

Imagine seeing a problem/solution with this in reverse... [Part way way through] So we now have cot(x) in our expression. We can’t use it as such, but a product representation might work. Recall from your notes that cot(x) == 1/x *prod[k=1,inf](1-tan^2(x/2^k)) From this it is obvious that....

[@jemcel0397]

Spicy integral evaluates to a simple function. Damn clever boi

[@neutrino5695]

I had to search for the definition of twink in the urban dictionary:)) Great video by the way.

[@thomasblackwell9507]

I appreciate your accent it makes it more agreeable and acceptable. Please forgive me for being a Deuschverderber.

[@aniketeuler6443]

Flammy Jens what's the name of the music in the beginning of video

[@UrasSomer]

This madman still reads the comments on this

[@nashweekendcovers]

"seek n square" flammy says.😂

[@noway2831]

I lost you when you stated talking about "telescopic functions", but that's likely attributable to my limited education.

[@yvesdesille78]

Why did you change your i into a k at 5:44 ?

[@sibsbubbles]

Wow that was quite the knarly integral. You know your trig super well, man.

[@shivanshnigam4015]

We ain't callin em functions here we call em Bois 💀

[@ryderpham5464]

You read my mind; I do only come for the cute twink

[@almanahulzilnicdesuceava5379]

12:00 you could use instead LIM (sin f(x))/f (x) = 1 f (x)->0

[@almanahulzilnicdesuceava5379]

But it.s in the special limits list!

[@almanahulzilnicdesuceava5379]

My maths teacher,aparently :)))

[@Bollibompa]

Why can we use L'Hôpital's when the values k take are discrete? Shouldn't we use Stolz-Cesàro? I remember pondering this when deriving sinc(x) from the infinite product of cos(x/2^n). On the other hand I can't find a case where expanding from discrete to continuous would be a problem unless the argument is quite nasty.

[@subhadeepsarkar5606]

that's a damn to be given

[@alielhajj7769]

Splitting the infinite product needs some conditions man

[@RaspingBubbles6]

Can you find the area between the functions sqrt(X) and ln(X + 1) + 1 from 0 to their intercept

[@RaspingBubbles6]

Please

[@nicholasheilig3694]

Amazing! I love you flammy boi! Keep it up.

[@omgopet]

14:00 you can make your life much easier by using the Taylor series expansion for limits like these.

[@bimbumbamdolievori]

I wonder if these big integrals are constructed by reverse process or if you can really figure out such amount of steps

[@alexschiffer6237]

Calculus and a cute twink what an amazing combination.

[@jc8384]

Alex Schiffer I thought he is straight? I’m confused

[@jc8384]

Flammable Maths you are cute though and clearly smart so anybody would be lucky with you lol

[@maalikserebryakov]

there is an illness in your mind

[@janderson2709]

14:46 'don't forget the x from before' Which x? From where?

[@domenicopiegaia7816]

this was my first papa flammy video ever

[@Rotiiii98]

When you moved the index on that product form 2 to 1 should you haved to move the index on k too? from k to k+1?

[@Jeff-wc5ho]

This reminds me of an old MIT Integration bee problem

[@colorfulcalculus4526]

Incredible video! Love it!

[@mihaiciorobitca5287]

What means those 3 points ,from 2:15

[@aniruddhvasishta8334]

At 6:06 he says the product of the two infinity bois are reciprocals so why can't he cancel out the infinite cosine boi with one of the infinite secant bois?

[@nujabraska]

Telescoping series was pretty cool

[@immersionmusic]

Papa Flammy this ist a sehry neiß Koßein formula. Very gut

[@ericthegreat7805]

Great video, but in my calculus class i was taught an easier way to do limcos(x)sin(x)(1/2k)/(sin(x/2k)) = 1/x * cos(x)*sin(x)* lim(x/2k)/sin(x/2k). Now we know sin(a)/a -> 1 as a -> 0 which does if a = x/2^k, so the limit is just cos(x)sin(x)/x.

[@brayancantero]

Hello, do you explain the change from cartesian to polar coordinates in Gaussian Integral? It's very important. Thanks. Like, explain how to obtain rdrd(theta) from dxdy, thanks.

[@karljoyeux5148]

"that was quite easy"

[@yasuotheunforgiven2839]

keep it up from morocco

[@Joe-bb4yi]

2cos^2(a)-1 is also cos(2a)

[@faithhill9811]

Is anyone able to tell me where he got this integral from??

[@sleindarfeau8657]

Great one boi ! I just wonder why you didn't use the fact that sin(x/2^k)~x/2^k when k->oo which would have given you the 1/x limit quickier that with L'hôpital's rule

[@itsanotheraccount]

**calls himself twink** Most bestest calculus video maker

[@clementboutaric3952]

This was not deep math but trig skill. Both of which are enjoyable.

[@daemonguy2]

What so now we can log negative values?

[@razimograbi2135]

good job man love ur videos , i learn i lot from u.

[@chuayewhui145]

Short boi disguise as long boi

[@CreativeStyled]

This channel is so gooooodd.

[@ranjitsarkar3126]

Back in the day when his handwriting was still understandable

[@pranavva5418]

This was beautifulll

[@soyoltoi]

I hyper love my infinite boiis.

[@teekayanirudh]

First 40 seconds earned my like. That aside brilliant video ;)

[@mokouf3]

Wait...cosine * secant = 1 isn't it? When you did make lim(n→∞)Π(i=1 to n)cos(x/2^(i-1)) = cos(x)*lim(n→∞)Π(i=1 to n)cos(x/2^(i)) Note that you have cos(x) *Πcos(x/2^i) * [Πsec(x/2^i)]^2, isn't it cos(x)*Πsec(x/2^i) ?

[@mokouf3]

You forgot to take absolute value at last!

[@hallowizer440]

17:34 Wait...ln(u) = ln(sin(x))+C? That means C=0, so every arbitrary constant was always equal to 0!

[@xFamous24]

Where can you get problems like this? I would love to solve some of these trick integrals. Do you have a specific source? Danke schonmal :)

[@darmstadtbeste4590]

this is so beautiful

[@anmolempire1197]

Nice one 😊

[@PandaWaffle3]

Do more Putnam problems!!!!

[@curiousminds301]

Amazing video sir

[@bluebears6627]

I always fall for that too but you missed the absolute value on the ln when integrating the final thing

[@lasa18]

Hey math guy prove this: Every Irrational number can be written as a ratio between two p-adic numbers

[@theopapa8232]

Isnt this an integrak from the integration bee at mit?

[@joshuabonet]

This was a fucking amazing integral boi

[@PapaFlammy69]

[@joshuabonet]

@@PapaFlammy69

[@michaelempeigne3519]

what are some properties of pi notation to compute values ?

[@copperfield42]

Michael Empeigne basically same as multiplication, because is just a notation for multiplying many stuff following a certain pattern just like the sigma notation for summation/addition

[@adrianamor8472]

Instead of using L'H rule you could just use the fact that lim x->0 x/sin(ax) = 1/a to prove that lim k->inf 2^-k / sin(x*2^-k) = 1/x . You don't even have to calculate derivatives :)

[@thomasdevos4590]

there is a mutch eazier way using dubbel angle formula of the tangent

[@thomasdevos4590]

then 2tan(x/2^i)/tan(x/2^(i-1))=1-tan^2(x/2^i)

[@anupamdas8541]

Thomas De Vos Exactly!