Why does pi show up here? | The Gaussian Integral, explained 

Which software you use to add audio and small videos clips together 😐??

are you inspired by 3 blue 1 brown?? The video is really good

You mean e^-(x²+y²)?

Que programa usas para las grafica?

I'm dumb. Isn't e^-(x^2+y^2) a 2D function? f:R^2->R

Max Haibara Why is an astronaut doing panini in the corner

@@cmbaz1140 and what is Black Hitler doing here?

Ha ha ha

I can hear that voice !!

*"why is it", not "why it is"

Its cute until your professor never taught it and expects you to come up with it when he hands you a midterm

Applied maths is still inferior to pure maths

@@johnle1723 Nah i think thats better than arranging oranges in 42 dimensions.

Ok

@@Uri131 yes, right!

Yea it is very simmilar, also he uses the same framework for visualization as 3blue1brown

also check out reddit.com/r/manin. There are a lot of people making cool stuff with it

@@uzairm3816 Sorry, there aren’t any communities on Reddit with that name.

@@Guztav1337 oops I meand r/manim not manin. Manim stands for math animation

Chudna........

Watched this vid before and after taking calc 3. I had a lot better understanding now

Right yeah, I agree. If I could go back and change anything, this would be the first thing I would

3Blue1Brown has a video on that.

Everything is circle, so π must be everywhere.

刻进DNA的台词

r/im14andthisisdeep

You're welcome! I'm glad it was helpful :)

I got 5 likes, and 5 is such a milestone, it turned blue!

Glad you enjoyed it!

Thanks for this comment, it addresses the exact thing that bothered me: what makes this argument work for Gaussian but not other solids of revolution. However, I feel like either I'm not understanding something, or then your explanation isn't comprehensive. To me, a dependence on theta seems like something a function that is non-symmetric around the origin would have. However, there's a lot of solids of revolution that are perfectly symmetrical around the origin, and thus don't have any dependence on theta. It seems that another property of the Gaussian, the fact that the shape of it's "slope" exp(-x^2) happens to fit exactly the substitution trick at 4:23, plays also role here.

@@DDranks I think you're just looking at it from a different angle. The construction used here is not to take a function f(x) and to consider its solid of revolution f(r). Instead, we consider a sort of "Cartesian product" version of this function f(x)f(y). For the Gaussian this happens to be the same as the solid of revolution of f(x). This equivalence of the "Cartesian product" f(x)f(y) and solid of revolution f(r) for the Gaussian is precisely the property I described in the OP. This is what allows us to simplify the integral and extract information about the 1D integral of f(x). You can say this is special from two angles. The angle I took in the OP is to say: let's look at all functions f(x) and consider f(x)f(y). For which functions does the theta dependence drop out and we can simplify things? The answer is the Gaussian. The other angle, which you took, is to start out with a solid of revolution f(r). This is already simplified and if you integrate this over R2, you will get some answer involving pi (provided the integral exists). However, in general this does not give you much information about the integral of any 1D function. From this angle it's a bit harder to formulate a question for which "the Gaussian" is the answer, which is why I took the other angle in my OP. But you are correct that this is an equally valid way of looking at this problem.

No problem!

Awesome! Thank you!

Thanks!

No no. There was no explicit reasoning for the presence of pi, but it can be somehow inferred. In my understanding, pi is there because of the rotational invariance of the function, the very property that allows the integral to be separated when you rotate it. The 2D case (2D in the domain) is the one that feels intuitive, because pi is defined over a two-dimensional object (the circle), so in that case it is easy to see how given the rotational symmetry the volume can be interpreted as an amalgamation of cylinders - which would inevitably bring over the factor of pi to the calculation. But in higher (and well, the one lower) dimensional case it would behave like the higher dimensional equivalents of that cylinder, i.e. higher powers of pi - a little bit like the factor of pi in the volume of a hypersphere. Well, that's what I took from it. Haha

I can't come up with an intuitive answer, but if you use cylindrical coordinates it just makes sense

I see it like this: when I see the curve of exp(-(x^2+y^2)), I pretend I am looking down at it from a bird's eye view and I see that the graph is radially symmetrical, which implies circles and thus pi.

It's basically because for the Gaussian exp(-x^2-y^2)=exp(-r^2) which doesn't depend on theta in polar coordinates and allows us to get a pi from the theta integral. The Gaussian is the only function which satisfies the property f(x)f(y)=f(r), which makes it special. See my comment for a more long-winded explanation

@@DavidSmyth666 Exactly. There are lots of distributions that have circular, or, more broadly, elliptical, contours, at least in a very generalized sense of ellipse. These are called, obviously enough, elliptical distributions. en.wikipedia.org/wiki/Elliptical_distribution An example is the T family, of which the bivariate Gaussian is an example (being a limiting case). Of course, these don't factor out quite so nicely so you're left with gamma function terms that, in the limit, only leave pi behind. Multivariate_t-distribution I sort of wonder whether one simply gets used to all these correspondences, but in Calculus classes they sure do seem weird!

Thank you! I use manim: github.com/3b1b/manim

What's funny though?

@@justforfun2238 I mean yeah it's not really "jokey" funny or anything which is why I even mentioned it... just the fact that it's just an equation and his deadpan "where is the circle" where there's obviously no circle, cause, like I said, it's an equation. idk it was just the moment and the way he said it that got me

@@onsewatch yes it's actually kind of funny when something unexpected suddenly popped out XD

I don't see the mistake

@@vcubingx It's rotated around the z axis. At 3:15 you say z and then later y.

@@Gameboygenius Yes, the function e^(-(x^2 + y^2)) is symmetrical with respect to the z axis, but if you rotate the function e^(x^2) around the y axis you get the function e^(-(x^2 + y^2)), which is what I meant to say. Probably should have made it more clear.

@@vcubingx I agree with the previous guy. Rotating e^(-x^2) around the z-axis will give you e^(-y^2). Rotating around the y axis will give you flipped e^(-x^2) -- if rotation is by pi it will flip to negative for example. The cylinder you describe is projecting a circle to the xy-plane, which means the center of the rotation is the z-axis. Hence to make a set of infinitely many cylinders by rotating the e^(-x^2) function, you have to rotate around the z-axis. Please, correct me if I am wrong

@Hamza Omar Because f(x,y) is one dimensional, it gives out one output. Since it has two arguments it is a two dimensional input space, which maps a value on a 1D map by the transformation f(x,y).

@Hamza Omar No, you seem to misinterpret the input, output space with the spaces it is represented in. It is represented in three dimensional space, since it is mapped by two input spaces into an output space. However, the function transforms the input spaces into one output space (note that we give the function two independent values and only get one in return).

@Hamza Omar the function takes in two numbers as inputs and outputs a single numbers. Although this function is 1D, it has a nice visualization in 3D as a surface. A 3D function takes in N variables as inputs and outputs 3 numbers. Something like a parametric curve is the simplest being F:R->R³

What definition it 'dimension of a function' are you using? I would say it should be the dimension if the graph, because a function is set-theoretically defined by its graph

@Spherical Laplacian No, it is not 2D. It is a function of two variables, however the function outputs a scalar.

Thanks!

@@vcubingx Sure. It was succinct, well presented & clear--I loved it, & as a result, I'm going to see your other lectures, too😊

Nice, I wasn't aware that he covered it in his course. Do you have a link? I can't seem to find it.

@@vcubingx It was in this article at the very bottom: www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/double-integrals-a/a/double-integrals-in-polar-coordinates?modal=1

Thanks!

He should have written that it is the volume element dV

Take the hollow cylinder and imagine slicing it vertically (like how you'd slice a cake) and then unroll it. You end up with a cuboid. Its height is the same as the height of the cylinder, which is e^-r^2 (plug in x=r,y=0 to the multivariable function, for example). The length can be imagined as the circumference of the original cylinder, which is 2*pi*r. Finally the depth is just dr, some small shell width that tends to 0. A similar trick can be used to show that the curved surface area of a cylinder is 2*pi*r, by imagining unrolling this area into a rectangle.

Thanks, you're absolutely right. I made this video right before I took statistics, so I wasn't very aware of the CLT and it's applications. Thanks for watching and commenting! I'd like to add that I've always wanted to make a video on the CLT because of how powerful it is, I'm definitely considering it to be my next video.

The explanation is also similar to RandomMathsInc

Yeah that animation was dodgy. Imagine slicing into the cylinder vertically (like you would slice a cake) and then unrolling that. This is what is meant.

@@fahrenheit2101 exactly