Super! Very nice analogy to the Taylor expansion for single variable.

The T for the transposed matrix is missing in the final formula

@MrRenanwill

2 жыл бұрын

Some books do not put the T in the formula.

@MightySapphire

Жыл бұрын

This is a critical mistake as you cannot get a scalar if you don't transpose your delta (x-xo).

The final expression is so neat, and simply beautiful

I love your vids man keep on going

@ryanmike9833

8 жыл бұрын

Agree

Simply beautiful

Good video! thanks!

I was really expecting this to build into a kind of Taylor expansion of multi-variable functions. Does that exist?

@thatdude_93

6 жыл бұрын

that's exactly what that is :) but only to 2nd order terms. there is also a formula for a taylor approximation generalised for n dimensions and up to the nth order term. it's pretty interesting :)

beautiful math

btw this is a truncated Taylor series. You can keep going for n > 2 for greater than quadratic approximations.

Brilliant!!!!!!!

You've proven this for the case for two variables Q(x,y). What if there are n variables Q(x1,x2,x3..xn) ? Is there a more formal way to derive this?

@henrylang7556

Жыл бұрын

The definition'd be the same independently of the amount of variables

Thank you very much. So I guess cubic and higher order approximations would start involving order 3 and higher tensors?

@zairaner1489

7 жыл бұрын

Their called "multilinear functions of order n" in my course, but these are somwhat the same

@martingutlbauer9071

7 жыл бұрын

+ raphael schmidpeter I think duckquickly meant not to increase n (the order of dimensions) but to increase the order of approximation to cubic and higher approx. I am interested too how such approx. would look like in vector/matirx form.

Thank you very much for this, it helped me a lot. However, I find a bit of difficulty extending this to a vector of multivariable functions, i.e. f1, f2, .., fn. I wish to express all of their taylor expansions in vector form and then put the result in a vector of functions. Long story short, I can't write the Hessian of a vector of functions.

@zairaner1489

7 жыл бұрын

Just do it for every single of our f1, ..., fn. Their just realvalued functions you can approximate like shown in the video (especially you can write down their hessian) and then yo can put all of these approximations into a vector

Mistake at 7:15, Grant didn't put in the transpose symbol

At first, x0 is a scalar, then suddenly it turns into a vector without any explanation. The notation is confusing.

At the end, he forgot to write that the first term in the vectorized quadratic form is transposed. Anyways, Sal you're a lifesaver.

@rohitsanjay1

6 жыл бұрын

Polymath That wasn't Sal. That was Grant from 3Blue1Brown

@louis9116

6 жыл бұрын

he is not Sal, he is Grant

You could also write this as: Q(x) = f(x0) + grad f(x) (x-x0) + C q(x-x0) where C would be a large matrix with constants and q(x-x0) would contain all permutation of quadratic terms. How can you find C?

beautiful math