A Controversial Maths Video.

Рет қаралды 10,387

Flammable Maths

Күн бұрын

Пікірлер: 78

@DatuxGames 6 ай бұрын

Most polygons are just finitely sided circles

@PapaFlammy69 6 ай бұрын

brave

@genos1458 6 ай бұрын

Damn 😂

@angeldude101 6 ай бұрын

This also means that polygons have circumferences and that attempting to correct a past word choice was pointless.

@shy_dodecahedron 6 ай бұрын

Bruh

@joelklein3501 5 ай бұрын

Your mama's ass is just two finitely faced hemispheres

@HugoLHL 6 ай бұрын

Lines are just circles of infinite radius so polygons are just collections of arcs of circles but also circles are collections of arcs of circles qed

@opensocietyenjoyer 6 ай бұрын

all of them are just boundaries of convex sets

@gustavgnoettgen 6 ай бұрын

@@opensocietyenjoyerWhat if they're concave?

@thes7274473 6 ай бұрын

This was one of the first things I independently discovered back in high school. It was one of the early things that helped establish my love of mathematics. I realized that a regular n-gon can be cut into n isosceles triangles and so I derived the relationship between the apothem (radius of the incircle) and the area and found it to be A = a² n tan(π/n). And the limit as n→∞ of n tan(π/n) is π. I later discovered that the perimeter is also given by p = 2n tan(π/n). For an interactive demonstration, see Desmos graph 6ijb6uc1yu

@Kloppin4H0rses 6 ай бұрын

Suuuurrrreee buddy. Sure

@himanbam 6 ай бұрын

@@Kloppin4H0rses No high schooler has ever taken a calculus class / liked doing math in their free time

@friender6886 6 ай бұрын

By the end of senior year in high school, I too discovered this problem. I am now a first-year math major. How cool! Being in calculus at the time, I also realized that the integral of our result yielded the formula for the area of a circle as well.

@symphonyofsolidarity 6 ай бұрын

Sameee :)

@Mystery_Biscuits 6 ай бұрын

What a coincidence! I was just using this idea to prove the curved surface area of a cone formula (pi r l) from the limit of an n-gon based pyramid. Very cool!

@PapaFlammy69 6 ай бұрын

niceeee :)

@winteringgoose 6 ай бұрын

"We're going to start off with a square." *draws a kite* Love the vids, keep up the good work!

@threepointone415 6 ай бұрын

Good to know

@maxhagenauer24 6 ай бұрын

If you have a polygon and increase the number of sides, the perimeter will approach the circumference of a circle of the same radius and you can use this to compute digits of pi. 2^n * sqrt(2 - sqrt( 2 + sqrt (2 + ...) with n many square roots, plug in huge numbers for n to get a better approximation for pi.

@advaitkamath8442 6 ай бұрын

Aren't infinte sided polygons are called aperigons, they have the exact same properties as a circle, with the same area, the only difference is that they have interior angles which eachone 180 degrees, while a circle has none

@soupisfornoobs4081 6 ай бұрын

Apeirogons, from the greek word for infinite "Apeiro" (Άπειρο). And they're very cool

@Filipnalepa 6 ай бұрын

With world apeirogon you just reminded me of mind bending clip of jan Mislai about regular polihedra. Starts with platonic solids, somewhere in the middle we land in honeycomb apeirogon to end in some stellated net of pointy helixes.

@advaitkamath8442 6 ай бұрын

@@Filipnalepa I watched that one too!

@youtuberdisguiser6075 6 ай бұрын

Not really, i agree the distance converges, but some points on the border of the circle will never be touched, since the border of the circle is uncountable yet the points of the infinigon are only countable infinite. For example lets go with aquidistant n-gon, the point on the border at the angle sqrt(2) will never be exactly touched by any n-gon. You could say you take a setunion integral, but that's cheating. And not n gon convergence

@FaerieDragonZook 6 ай бұрын

The difference is that, for a circle, every point on the circumference is a radial point, and there are uncountably many points on the circumference, whereas an apeirogon only has a countably infinite number of radial points. Also, polygons are kinky ;), circles aren't.

@jack002tuber 6 ай бұрын

I did this last March on PI day. I calculated a/2 by splitting the triangle and made two right triangles and solved for a/2 using law of cosines. Try it, I think it's easier.

@ricardoparada5375 6 ай бұрын

I love seeing these arguments. Great video Flammy ma boi

@Happy_Abe 6 ай бұрын

I don’t think using l'hospital would’ve been illegal here since a limit in a continuous variable x converging implies that it converges in the discrete variable of naturals n. The other way doesn’t work so we can’t use limits in n for limits in x but we can use limits in x for limits in n so we can use l'hospital for the x limit and then conclude it for this n limit as well.

@destinlol277 6 ай бұрын

Let's go another video

@Katzeblow 6 ай бұрын

My heart is boiling blood already, I'm shocked by the thumbnail 😢

@erikhaag4250 Ай бұрын

15:34 "Because we're dealing with positive n, the absolute value is going to cancel out" I would've used "idempotent" instead of "cancel out"

@jasurmakhkamov 6 ай бұрын

Feels great to understand this stuff/see the solution just about before it happens. Thanks papa flammy for making us more smarter :D

@TheGildedMackerel0 6 ай бұрын

I came up and did this problem a while ago in high school. very cool!

@avinashbabut.n4123 6 ай бұрын

I love your videos. They make me enjoy math.

@emmettnelson7260 6 ай бұрын

I think polygons are a key to alot of things in math, I just used them to prove that sin^2(a+c) = sin(a)sin(b) + sin(c)sin(d) when a+b+c+d = pi

@douglasfalter2337 6 ай бұрын

Papaflammy papapaflammy6969 love your videos…we need a video/series on numerical methods ❤️❤️ atmospheric science rulez btw ✨

@wjalp 6 ай бұрын

I love your videos papa! Looking forward for the next video!

@koenth2359 6 ай бұрын

It's somewhat easier with a = 2rsin(π/n), so lim na= lim 2nr sin (π/n) = 2πr, using that lim sin x = x as x approaches 0.

@joelklein3501 5 ай бұрын

Alternative approache for the "cheatung" part sqrt(1-cos(2π/n)) = sqrt(1-cos(2π/n))*sqrt(1+cos(2π/n)) /sqrt(1-cos(2π/n)) = sqrt(1-cos²(2π/n))/sqrt(1+cos(2π/4n) = sqrt(sin²(2π/n))/sqrt(1+cos(2π/n)) = |sin(2π/n)|/sqrt(1+cos(2π/4)) for n >1 , 0 1, cos(2π/n) > -1 So we can conclude sqrt(1-cos(2π/n)) = sin(2π/n)/sqrt(1+cos(2π/n)) In conclusion L[ n*sqrt(2)*r*sqrt(1-cos(2π/n)) ] = L[ n*sqrt(2)*r*sin(2π/n)/sqrt(1+cos(2π/n)) ] = L[ r*sqrt(2)/sqrt(1+cos(2π/n)) ] * L [ n*sin(2π/n) ] = r * L[ 2π * sin(2π/n)/(2π/n) ] = 2πr

@viewerfan 6 ай бұрын

Aperigon: 2 plus 2 plus 2 plus 2 plus 2 plus 2 plus 2 plus 2 Numberblock 18: AHH!!!

@tszhanglau5747 6 ай бұрын

I derived that the area of regular polygons approach that of circles myself

@John-zx2ss 6 ай бұрын

What about substituting the Taylor series of cos instead of the double angle formula

@gustavgnoettgen 6 ай бұрын

It's like saying that all reality happens in my head. Well, sort of. And this polygon is sort of a circle. And it sort of is not.

@ivnsaw7422 6 ай бұрын

A circle is an infinitely cornered polygon without any sides

@Bobbel888 6 ай бұрын

Infinitely-Sided, already the interior .... :) :) :),

@neutronenstern. 6 ай бұрын

its easier with a non regular polygon. you can always draw your polygon in a way, such that the edges are perpendicular to the radius. And then its just pythagoras. And if ypu take the inscribed polygon and compare it to the outscribed one,you will see, that they dont differ for n->∞.

@bachvaroff 6 ай бұрын

~12:22 ← one can also apply the Cesàro-Stolz theorem from that point on…

@jimnewton4534 6 ай бұрын

Is a perimeter a number or a set of edges?

@PapaFlammy69 6 ай бұрын

a number

@mighty8357 6 ай бұрын

What I take from this video is that perimeter and circumference are the same thing!

@PapaFlammy69 6 ай бұрын

KInda ye, that's probably why I messed up the terms before XD

@ChrisVenus 6 ай бұрын

Feels like you might have saved yourself a whole lot of working by just bisecting your triangle to give a right triangle and using sin(pi/n) = a/2r . so a = 2rsin(pi/n) . Then you take n lots of that to get 2rn*Sin(pi/n) and then solve the limit as you did...

@shy_dodecahedron 6 ай бұрын

Well if 0.999...=1 it's not all that strange. Like there's a reason mathematicians avoided infinity for a long time, - it doesn't follow normal rules which makes it weird.

@billevans7488 3 ай бұрын

Except it isn't - it can keep getting closer to 1 as you allow more decimal positions but it will never get there. For it to be 1 you would have to be able to add 0.000...1 to it but you can't terminate an infinite number of zeroes with a 1. That expression is simply 'defined' to be 1 as a matter of utility just as some consider it to be mathematically useful to define 1/infinity as 0. But then that would mean than EVERY value less than 1 but greater than 0 divided by infinity must also be 0 which, of course, is absurd. And what about 1.1 divided by infinity? Is that now somehow greater than zero. The fallacy here is that you can't mix absolute values and non-existent values except in the imagination. 'What if we consider 1 divided by infinity to be equal to 0?'.

@Rafau85 6 ай бұрын

Did I hear you cannot use l'Hospital for a sequence? You can do it. There is nothing wrong about it, if you can extend the sequence to a function.

@encounteringjack5699 6 ай бұрын

Each side length of a circle is epsilon. One over omega. Some infinitely small quantity. But wait! Unpopular opinion, there is only one size of infinity. This means the idea of infinitely large and infinitely small quantities no longer make sense. Thus, a circle is not an infinitely sided polygon. However, it can be approximated as a polygon, getting an exact answer for the limit as the polygons side lengths go to zero. Precisely getting the circumference of the circle even though there are lengths involved, since they go to zero.

@billevans7488 6 ай бұрын

@12 min - LOL! Sorry but 1 divided by infinity is NOT zero. It may be infinitely close (as close as you like) to zero but it never quite gets there. You are perpetuating a fantasy similar to the one about 0.9999 ad infinitum being equal to 1 - which it isn't - they are ALWAYS different no matter what decimal place you wish to examine. Things that are always different can never really be the same. But it can be, and is, useful to assume they are the same. Limits are a 'convenient' mathematical concept that are useful in understanding. A circle with a 'rational' radius will have an irrational circumference because you multiple by Pi, an irrational number. Likewise a circle with a 'rational' circumference will have an irrational radius. As far as I know no one has ever proven/shown that a circle with an irrational radius can have a rational circumference. Put another way the result of multiplying two irrational numbers is NEVER, and can not be, rational. Yet square roots (e.g. square root of 2) are useful conveniences. I think a lot of the confusion arises because even mathematicians forget the difference between the contrived, and logical, world of mathematics and the physical world. In the physical world there is no such thing as an 'exact' distance. How big is an atom? an electron, a proton? There is no such thing as a physical ruler that is exactly 12 inches long because physical things don't have exact dimensions.

@monishrules6580 6 ай бұрын

"Infinitely sided polygon" mfs when i tell them about a oval

@MyikeOndreov 6 ай бұрын

Brooo, you are good

@Gabriel-tf5ws 6 ай бұрын

you said that using L'Hospital is not very clean, because we are dealing with discrete values. But if i'm not mistaken, to show that sin(x)/x approches 1 you also use L'Hospital and x is pi/n and therefore also discrete, so you run into the same problem

@PapaFlammy69 6 ай бұрын

Noone ever uses L'Hospital to prove this limit lol