I know this sounds like a silly question but how did you get to understand such situations in math? Was simply reading the textbook helpful/did you need to read or procure other resources? How did you find them and know the difference between what would be helpful vs what would not? Thanks a lot, fascinating video as always.
when you said the jacobian could be shown in a different video, I got sad cause that's what I wanted to see lol otherwise, this was cool to watch :) I've used the gaussian integral a bit recently in my physics classes, but never really knew where it came from. Also, sqrt(pi)/2 is (1/2)! I assume they're related.
The matrix was likely used to compute the Jacobian. The determinant of the Jacobian matrix yields the scale factor of the linear transformation essentially.
That's probably calculating the Jacobian, the scale factor of different units in integral (dA = dxdy = |J| drdTheta). So in multivariable integral, we can change the original units (dx dy) into whatever other units we feel convenient (dr dTheta in this case). We can treat dr and dTheta as vectors, dx and dy as linear combinations of them (dx = a*dr + b*dTheta, dy = c*dr + d d*Theta) using partial derivatives (a = partial x/partial r, etc.). Then we get dA(change of area, or the area of parallelogram formed by dx and dy) in terms of drdTheta, by putting the 2 vectors into a matrix and calculate its determinant (i.e. ad - bc for a 2x2 matrix). By x=rcosTheta and y=rsinTheta, you can calculate the partial derivatives and the determinant to get r or -r depending on whether you calculate partial r or partial Theta first, since an area must be positive, we take the abs value of the results (|-r| = r). Therefore, we get dA = dxdy = rdrdTheta. Using the same method, you can change cartesian coordinates into cylindrical and spherical systems in triple integrals as dV = dzdydx = rdzdrdTheta = Rho^2 sinPhi dRhodThetadPhi as well, just remember to change the integral bounds using our new terms (in this video, [-inf,inf],[-inf,inf] -> [0,2pi],[0,inf] ).
Great explanation. One wouldn’t even need to know multivariable calculus to understand this.
I agree
Would be interesting to recollect how Jacobian is introduced, and how it's inferred in general
Very smart. The result sqrt(pi) is related to circle or polar coordinates. Coordinates conversion simplifies the solution.
Very elegant solutions. Congratulations!!!
Your explanations are perfect!
The Best! I'm a Brazilian fan!
I remember when I thought this technique smacked of trickery. Then I learned of Feynmann's technique and the world stopped making sense altogether.
We shall visit Feynmann soon 😆
best explanation video so far however i think it would have been better to show change of variables more in depth using jacobian
good job! Gauss would have loved this
Very smart. Great explanation.
For the first person tanks for an other video ..teacher
Bravo !!!!!
Very good, Sir.
Great explanation and blackboard technique!
Respect! Multumesc din Romania!
This is really the way to do it !
Ur greatest funny mathematican
You are so so great guy!! Thank you :))
Thanks for arranging this.
I know this sounds like a silly question but how did you get to understand such situations in math? Was simply reading the textbook helpful/did you need to read or procure other resources? How did you find them and know the difference between what would be helpful vs what would not? Thanks a lot, fascinating video as always.
Great videos!
when you said the jacobian could be shown in a different video, I got sad cause that's what I wanted to see lol
otherwise, this was cool to watch :)
I've used the gaussian integral a bit recently in my physics classes, but never really knew where it came from.
Also, sqrt(pi)/2 is (1/2)! I assume they're related.
Great explanation!
One thing that is passed without saying explicitly is that Z is an integer in this derivation. This is therefore not a general definition
Bless You!
Recently is solved by Double integral and transfer it to polar ordinance and no need use IxI ?
Great❤
What does e have to do with e^-infinity like any value to ^-infinity is 0
Does that mean I have to watch like 2 other videos just to understand the full concept?
Yoooooo.... super cool
Is the derivative of pi equal e?
Prove that dx.dy=rdr.d(Theta)
why is -1/2 -1=1/2 i dont know anyone tell me
tootoo good.
I saw a sped up proof of this and it was going into matrices and all sorts, has anyone come across that before?
The matrix was likely used to compute the Jacobian. The determinant of the Jacobian matrix yields the scale factor of the linear transformation essentially.
That's probably calculating the Jacobian, the scale factor of different units in integral (dA = dxdy = |J| drdTheta).
So in multivariable integral, we can change the original units (dx dy) into whatever other units we feel convenient (dr dTheta in this case). We can treat dr and dTheta as vectors, dx and dy as linear combinations of them (dx = a*dr + b*dTheta, dy = c*dr + d d*Theta) using partial derivatives (a = partial x/partial r, etc.).
Then we get dA(change of area, or the area of parallelogram formed by dx and dy) in terms of drdTheta, by putting the 2 vectors into a matrix and calculate its determinant (i.e. ad - bc for a 2x2 matrix).
By x=rcosTheta and y=rsinTheta, you can calculate the partial derivatives and the determinant to get r or -r depending on whether you calculate partial r or partial Theta first, since an area must be positive, we take the abs value of the results (|-r| = r). Therefore, we get dA = dxdy = rdrdTheta.
Using the same method, you can change cartesian coordinates into cylindrical and spherical systems in triple integrals as dV = dzdydx = rdzdrdTheta = Rho^2 sinPhi dRhodThetadPhi as well, just remember to change the integral bounds using our new terms (in this video, [-inf,inf],[-inf,inf] -> [0,2pi],[0,inf] ).
Из 3,14 а частицы начинают с 2.
thanks
How can I integrate (1/(1+e^x))dx
Multiply both the numerator and the denominator by e^(-x), equivalent to multiplying by 1. After that, a substitution will be very obvious.
ruclips.net/video/hUIfJt4k0Hg/видео.html
@@znhait You're right, a substitution did become obvious.
dx dy = | ∂(x,y)/∂(r,t) | dr dt . Matrix J:
∂x/∂r = cos(t) = J(1,1)
∂x/∂t = -rsin(t) = J(1,2)
∂y/∂r = sin(t) = J(2,1)
∂y/∂t = rcos(t) = J(2,2)
==> | ∂(x,y)/∂(r,t) | == | detJ |
= | cos(t)*rcos(t) - (-rsin(t) sin(t)) |
= | rcos²(t) + rsin²(t) | = r
==> dx dy = r dr dt