thank for your comment, i know which video is "last video". :D

Thank you, for "last video" was "Jacobian prerequisite knowledge" in the playlist I'm watching ...

@@joluju2375 kzfaq.info/get/bejne/jNOWh7uHrJ-3YIk.html

you rock yourself.

You are of great help! THX

I Knew it!

Dude!! that’s “Grant Sanderson” from 3blue1brown.

@@jjqerfcvddv bruh...thats the joke

@@sindhiyadevimaheshwaran3738 not everyone knows my friend - we must convert the vast unwashed into Grant's mathematical following

BRUH I LEGIT JUST THOUIGHT ABOUT THAT

Michael L He is...or so I hear

thats awesome i was just looking on his channel for this very topic

He IS 3B1B

Yeah, you are right. This does sound like 3Blue1brown

I think that's because it is ( ͡° ͜ʖ ͡°)

That is only true if the local linear approximation is still valid at further distances. Let me explain some more: The columns in the matrix track where the points (x_o+dx, y_o) and (x_o, y_o+dy) are mapped with respect to where (x_o, y_o) is mapped. For example, f(x_o+dx, y_o) - f(x_o, y_o) is the distance between (x_o+dx, y_o) and (x_o, y_o) after being mapped. This is simply our Jacobian times (dx, 0). Since the second entry is zero, we only recover (df_1/dx) + (df_2/dx), which by the chain rule is simply the total derivative of f with respect to x. Likewise, we can show the Jacobian times (0, dy) is the total derivative of f with respect to y. So the Jacobian matrix is mapping *differential* vector quantities that are in the direction of our original basis vectors. We can think of these differential vectors dx and dy as our new basis! But if we choose a basis that is very small, we better make sure our transformation returns a number that isn't very small too. This is why we can imagine "normalising" by the small quantities "dx" and "dy" in the bottom of our matrix. In a normal transformation matrix, we know the denominator is simply "1". But since our function isn't actually linear, we do not have the luxury of using such a simple basis. We can only act on small vectors accurately, so we re-scale :)

no you havent. the jacobian matrix is much deeper than this . he is just touching the tip of the iceberg.

@@maxpercer7119 Then please do point in the direction where I can gain an even deeper understanding.

@@aryamanpatel8250 you could say... please point him in the locally linear direction so he can arrive at his deeper destiantion ;)

@@maxpercer7119 ma vai a cagare

change in dx results in change in df and it has two components which is the first column in jacobian matrix ,similarly for dy

It is 3Blue1Brown voice!

He's not considering that translation, he's just considering the linear transfromation around (-2,1). It's like when we are on earth, we don't consider that earth is moving when we are doing some physics calculations.

At about 1:20, you can see he selects -2,1 on the original matrix. That selected point moves to near -1,0 after the various partial transformations performed throughout the video.

Did you ever find out?

Yes it indeed is. Here is the link to the entire playlist which is called "Multivariable Calculus": kzfaq.info/sun/PLSQl0a2vh4HC5feHa6Rc5c0wbRTx56nF7

Because we are looking for the ratio of how much the axis is stretched or squeezed

The reason is that in the approximation the Jacobian is multiplied by the vector [delx,dely]. If the vector was [1,1] you'd be correct that it should be just delf1 and delf2. Think that the approximation (taking some liberties with notation) is dF = J*dX, where F is the vector of the function and X is the vector with the variables.

He has made a project named "manim" for these animations. Check it out!

It's a playlist dude

Same problem here. I hoped the next video "Computing a Jacobian matrix" would clarify that and answer this question, but nope. What is missing here is a real example of how the use of the jacobian matrix would give a satisfying solution to a problem otherwise too complicated. So far, the best I can understand is that the Jacobian matrix can simplify determining what's happening to the *neighborhood* of the point by using only linear functions, but I can't imagine a situation where I would need that. Finally, my best bet is that I missed something important.

It depends on how you see the input space, that is, if it's filled with vectors (things that can be added to each other and scaled by numbers) or dots (simple pairs of numbers that cannot be added or scaled). If you think about vectores, then it's a transformation, like those you see in Linear Algebra, but this time they are not necessarily linear. If you think the function is mapping dots to vectors, then it's a vector space. But I think that Grant's point in this course is that those are two complementary ways of seeing the same thing, it's just that the transformation has this agile nature of taking vectors from one place to another, while vector spaces are more static.

Abs(x) at 0; 1/x at 0 etc.

@@PfropfNo1 can you help a bit, I have a doubt, is all the entries in jacobian matrix represent : Change in output space divide by change in input space ? Considering the jacobian matrix the mapping of basis in input space must be transformed in to the entries of jacobian matrix. But I did not get how delf1/delx,dlef1/dy.... are obtained ?

Same guy. Khan academy has different teachers, one of whom is the same guy as from 3b1b.

Locally linear = differentiable. If it's not differentiable at a certain point, this means that it can't be locally approximated by a linear transformation, and vice versa.

@@Cessedilha Thanks a lot!

Only one row*

Harish D Keep in mind that when using this matrix, we’re only focusing on local points surrounding the point we originally focused on, not the grid as a whole. The fact that we’re taking partial derivatives automatically encapsulates this idea of locality. Also, the origin in this example moves because the matrix transformation isn’t linear.

I bet it's proprietary

github.com/3b1b

Perhaps my explanation to Daneil C above may help! We are mapping small changes (dx, 0) and (0, dy) to small changes of f using the chain rule! J (dx, 0) = df_1/dx + df_2/dx But this is just the total derivative of f with respect to x by the chain rule. Likewise for (0, dy) = total derivative of f with respect to y. So, locally, we know how far we would move from the point we are evaluating if we took small steps.

sounds right

Yup, 1x1 Jacobian Matrix is essentially a derivative of a univariate scalar function, 1 x m Jacobian Matrix is the transposed gradient vector of a multivariate scalar function. Cool beans.

Any possible linear transformation of x and y can be conceptually represented as shown in the video by the matrix (with a-f being constants): [ ax+by+e] [ cx+dy+f ] (As should be expected, these are just equations for lines.) What happens if you apply the Jacobian to this matrix? It reduces to precisely the linear transformation matrix that's normally used to transform (x,y) points: [ a b ] [ c d ] Why is this so? Why is it just constants? ... Because the Jacobian expresses how much a transformation is "changing things locally", and a _linear_ transformation changes the entire transformation space in exactly the same way (which is why lines stay parallel, and whatnot). In other words, it does not vary; it stays constant. It is comprised entirely of uniform scaling and shearing (and potentially translating). In short, the reason the (general, i.e., unevaluated) Jacobian shown in the video varies from point to point is _because_ the functions selected for the transformation were *not* linear (sine and cosine). If they were linear, the resulting matrix would have simply been full of constants.

The Jacobian matrix is a linear approximation. For a linear transformation (matrix multiplication), the Jacobian would be the linear transformation itself. Kind of what happens in 1-d derivation, when multiplying a constant by x the derivative is the constant itself.

local linearity is true at every point

@@li_chengliang for any function?

Well, each of the partial derivatives will give you a function that tells you the rate of change of one function with respect to another, and when we evaluate it at a specific point, its going to tell us what that change was. That's the important part, and he said it, that we have to evaluate it and it will just turn into a matrix with numbers in it instead of functions.

What do you mean howto ? This isn't some simple cooking tutorial or something this is academic teaching

@@That_One_Guy... I didn't mean anything, that is how KZfaq categorized the video.

paricin469 Nerd

am here

Justin Ward nonintellectuals…

it is 3b1b! :)

You don't understand anything or if u say nothing it means u understand :P

Wow