Let's get Product-ive: Integrating one infinite boi
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In today's episode of Papa Flammy's integral week: Integrating an infinite product involving trigonometric functions. Enjoy the ride =)
Twitter: / flammablemaths
Visit my website! =)
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When exactly did i start watching integral solving recreationally?
I’m asking myself the same question:)
It’s weirder when you’re not even majoring in mathematics
It's much weird when you're still in high school 🤣
@@user-fungus relatable
@@micomrkaic same here
at this point the product function is just used for ridiculous questions such as these
0:28 Oh shit... he's onto us.
Classic German humor at the start
Oi, this is german humor. It's no laughing matter
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.
Yeah,I have used this way ! Great .
I'd guess he didn't think of it and decided to pursue the solution he thought of instead.
i did with the same approch !! and yeah the product term becomes much simpler .
Because I didn't remember this identity.
Such a beautiful infinity sign 4:30
i pictured a burned hotdog when i thought of the phrase
"infinite tan boi"
Es ist immer wieder spannend, deine Videos bei TH-cam anzusehen... Danke sehr, Schnuckel-Mathematiker! und Grüße aus Brasilien :)
"Cute twink." Suddenly the channel name makes sense.
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!
Could ur titles and intros get any cringier? Love the math skillz tho
The intros are hilarious. Cringy but hilarious
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.
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.
@@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
@@ezioauditore7378 1 year later, how’s your calculus skill?
particularly your ability in the symbolic evaluation of integrals
8:29 "... ONE THING I DONT KNOW WHY IT DOESN'T EVEN MATTER HOW HARD YOU TRY"
wow, beautiful trigonometry right there!
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 .
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
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)
"Boi", always cracks me up xD
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.
”Infinity boy.”
William Tachyon *boi
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.
i love this guy
I have a feeling a bunch of clever tricks will be used.
Spicy integral evaluates to a simple function. Damn clever boi
Just finished the video. What an elegant solution. Thank you! :D
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
"That was quite easy."
If you say so.
Wow that was quite the knarly integral. You know your trig super well, man.
Beautiful proof ! I love maths and the way you explain it !
You forgot to put absolute value around sin(x) when integrating cot(x) :D
Friends, if you dont love him and his parents, then see your cardiologist because you dont have a heart !!
I had to search for the definition of twink in the urban dictionary:))
Great video by the way.
7:05 , “so working with finite things is way easier than working with infinite things” huh coulda fooled me.
Amazing! I love you flammy boi! Keep it up.
I appreciate your accent it makes it more agreeable and acceptable. Please forgive me for being a Deuschverderber.
"seek n square" flammy says.😂
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
I lost you when you stated talking about "telescopic functions", but that's likely attributable to my limited education.
You could also use the double angle tan formula to get a telescoping product.
This madman still reads the comments on this
that's a damn to be given
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.
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.
came for math, stayed for the memes
@@PapaFlammy69 \(πωπ)/
Flammy Jens what's the name of the music in the beginning of video
First 40 seconds earned my like. That aside brilliant video ;)
Calculus and a cute twink what an amazing combination.
Alex Schiffer I thought he is straight? I’m confused
Flammable Maths you are cute though and clearly smart so anybody would be lucky with you lol
there is an illness in your mind
Incredible video! Love it!
Papa Flammy this ist a sehry neiß Koßein formula. Very gut
Splitting the infinite product needs some conditions man
Telescoping series was pretty cool
A simplification: 1 - tan2(a) = 1-sin2(a)/cos2(a) = (cos2(a) - sin2(a)) / cos2(a) = cos(2a)/cos2(a)
keep it up from morocco
We ain't callin em functions here we call em Bois 💀
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....
You read my mind; I do only come for the cute twink
This was not deep math but trig skill. Both of which are enjoyable.
This reminds me of an old MIT Integration bee problem
I wonder if these big integrals are constructed by reverse process or if you can really figure out such amount of steps
Nice one 😊
good job man love ur videos , i learn i lot from u.
"that was quite easy"
this is so beautiful
This channel is so gooooodd.
this was my first papa flammy video ever
This was a fucking amazing integral boi
@@PapaFlammy69
You forgot to take absolute value at last!
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.
This was beautifulll
More infinite products bruh please!
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 :)
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?
Thank you
bloody legend
I hyper love my infinite boiis.
14:00 you can make your life much easier by using the Taylor series expansion for limits like these.
Is anyone able to tell me where he got this integral from??
Can you find the area between the functions sqrt(X) and ln(X + 1) + 1 from 0 to their intercept
Please
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.
This was one spicy Boi
Amazing video sir
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
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.
**calls himself twink**
Most bestest calculus video maker
Hey math guy prove this:
Every Irrational number can be written as a ratio between two p-adic numbers
Back in the day when his handwriting was still understandable
12:00 you could use instead
LIM (sin f(x))/f (x) = 1
f (x)->0
But it.s in the special limits list!
My maths teacher,aparently :)))
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?
What so now we can log negative values?
Yes
Why did you change your i into a k at 5:44 ?
Do more Putnam problems!!!!
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 :)
Short boi disguise as long boi
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) ?
nice solution
that is a bad boy!!!!
I always fall for that too but you missed the absolute value on the ln when integrating the final thing
2cos^2(a)-1 is also cos(2a)