Chain Rule Surprise

In this video, I present a neat result I have noticed when calculating some partial derivatives! Is it a coincidence? I'm not sure! Enjoy!

[botondosvath]

In the generalized spherical coordinate system (an orthogonal curvilinear system), you will have n-1 angles and the euclidian distance (r) as coordinates. In our case, f(x_1, ..., x_n)=r, so it does not depend on any angle. This means that the gradient will have only one component, in the r direction, with magnitude (df)/(dr)=1, so it will have norm 1.

[blackpenredpen]

Wtf is the chain rule? It’s the Chen Lu man 🤦‍♂️

[replicaacliper]

The function is just the distance from the origin of an n dimensional space. The fastest way to move away from the origin is through a ray passing through the origin - the magnitude of the gradient being 1 tells you that the change in the distance from the origin is exactly equal to the distance you move along that line (if you move 1 unit in that direction, your distance from the origin increases by 1). Tada.

[Kentrosauruses]

I like that you thank us for watching before the video starts

[eliyasne]

Of course the gradient is 1, its the gradient of the distance function (from the origin) which is necessary in the direction of the position vector and it must be 1 since a change in the position along the gradient is a direct addition to the distance function.

[kroyboy]

Doesn't have to be /sqrt(/sum(x_i)²)). It'll work for the magnitude of a vector in n-dimensional space in any p-adic norm

[Zonnymaka]

It's an hypercone. No surprise the steepest positive direction is constant and equal to one (all the coefficients are 1)

[Gamma_Digamma]

This kind of reminds me of finding the length of a curve formula but here the curve is the distance from origin function itself

[thedoublehelix]

For n=2, f is just a cone and it is obvious from the lines that make up the cone that the gradient vectors will have a length of one since the incline of a cone is just 1. For n=1 it is abs of x which is sorta like a 2d cone. Anyway it makes sense than an n-dimensional cone would have unit gradient vectors because that's how cones work. Their slope curvature doesn't change ig. Also once you work it out in 1/2 dimensions you could try and use induction f_n = sqrt((f_(n-1))^2 + (x_n)^2) and use the same argument. That sorta helped me understand it

[ohanneskamerkoseyan]

In spherical coordinates,
grad(r)=r unit vector
Your expression is equivalent to:
sqrt(grad(r) • grad(r))
which is equal to 1.

[tylershepard]

This almost seems like a definition of a unit vector. You take the gradient with respect to each coordinate of the lengths of your coordinate system. Interesting.

[jamesbra]

please please do videos on legendre differential equation

[TheAvokela]

Since f=|x| grows linearly away from the origin, we have a) grad f = gx for some real valued g and b) f = x.grad f (Taylor). From a) and b) we get g=1/|x| so that |grad f|=1, which of course is trivial to see in spherical coordinates as already pointed out by other commentators.

[paulkohl]

This video reminds me of the Eikonial equation with the speed function set to one. The deep reason for the video probably has to do with Lie Group Theory related to these equations. Quadratic forms, from an algebraic point of view, are an island of easily discoverable results resting above a sea-ledge that demarcates where the abyssal realms of higher-order -- hard to figure out -- forms, begin and end. Love the video. Open-ended, audience participation on this one :) Noice!

[stormhoof]

Hi, I’d like to briefly outline a different method that more explicitly uses Chen Lu

let A=R^n
let B=[0, inf)
let tr be TRANSPOSE function
let g: B to B by g(x)=sqrt(x)
let h: A to B by h(x)=trans(x)*x

So f: A to B by f=g•h

Let’s also define the object we want to compute as s(x)
s: from A to B by s=f•tr(f’)

Note that for x in B, g(x)^2=x
Note that for x in A, c in R, h(cx)=c^2*h(x)

anyway we’ll need derivatives
g’(x)=1/(2*sqrt(x))
h’(x)=2*tr(x)

By Chen Lu, f’=(g’•h)*h’.
So we can revise our s(x) as
s=g•h•tr((g’•h)*h’).

Now there are two ways forward to showing that s(x) is identically one:

1.i. Use the chain rule to show that s’(x) is identically zero
1.ii. Compute s for a particular value of x to get one
Or
2. Directly compute that s(x) results in one

Either way you have used Chen Lu. I think method 2 here is cleaner

[frozenmoon]

I think a relevant restriction is that x1 +... xn > 0, otherwise I think this may cause some contradiction.

[joaogabriel]

I don't know the meaning of this, but it happens because of the form of the gradient in spherical coordinates. Grad (f)= df/dr in the radial direction plus some other derivatives in the theta and phi directions. If f is such as you defined, f=r, then the only non zero component is in the radial direction and It is equal to 1 (dr/dr). Obviously the formula of the grad that I mentioned is for 3 dimensions, but it can be generalized to n dimensions via hiperspherical coordinates.

[FURY]

Is video reversed by saying thanks for watching on first ...!??

[Matlockization]

I picked on this guy ia an alpha male, no shit ! I good at things like that.

[dolev]

1:40 - Usually this is the point when I fast forward.
But since you suggested the viewers to do so, I got curious and watched all the spiel.
On hindsight, I should've listened to your advice...