Multiple Variables Analysis, where the function is not a curve. The function is a Surface. I love it. I would definitely like to be so smart like you. 😊😊😊
Another great video. Also, it is very cool of you that you gave credit to the site you borrowed the example from. You are one of the best teachers on youtube.
It is basically the same idea as exchanging sum orders for the array or double sequence (1, -1, 0,0,0,0...), (0,1,-1,0,0,0,0...), (0,0,1,-1,0,0,0,0,0...), ...... If you stack sequences and sum vertically, you get 1 left over, but horizontal sums are all 0. To do this with integrals on a finite region, the infinite amount of cancellation has to be squeezed in the domain and then stretched in the range to balance it all out. This could be done stepwise with step functions over a 2d domain, but it is satisfying to use elementary function examples instead.
In polar you get r^(-2)*(1/4)*sin(4*theta), if my work checks out. This counterexample is more convenient on a Cartesian disc than a rectangle. The rays (that is, f dr) integrate to +- infinity except where sin(4*theta)=0, which leads to ***infinite cancellation in the double integral***. If you integrate d theta first, it looks like it's all 0, but it really isn't so clear-cut. Perhaps the first rule of integral convergence is to see if ***the integral of the absolute value of your function*** is finite, and if not, you are dealing with something that requires careful definitions, such as principal value integrals or other such headaches...
i realized this counter example too .i say that this means that not only is fubini invalid under , changes of coordinates are also invalid. i think for it to be valid, at least one of the integrals has to agree with the original integrals in the x or y world.
At 15.40 you made a mistake: It should be y cubed over u cubed. The 1/2 term comes from antidifferientiating. Confused me for about 10 minutes. Still An amazing video
You're right; I found the same trick, and putting into the fraction the correct term and integrating, I found that the integral diverges (I'm non a monster in computation, so please check), but anyway the integration give a very different result from the first one
What do you mean by "each slice has to be integrable"? Every conceivable slice of the domain has to be integrable? Not only the 2 slice examples A(x) and B(y)? E.g. in polar coordinates over the quarter circle of radius 1, the result is infinity*0 = rubbish...
@@drpeyam That can not be enough, since for fixed x not equal to zero, f(x,y) is integrable in y with A(x)=x/(2*(x^2+1)^2) which is everywhere defined. For x=0 we have f(0,y)=0 which is trivially integrable. Similarly vice versa for B(y). I would guess that every conceivable slice has to be finite.
do things like change of coordinates with the jacobian also break when fubini's theorem doesn't work like when the integral of the absolute value function is divergent.
@@drpeyam actually i think it does break because you can get even more different answers than just interchanging y and x. for example change the integration reigon given in this counter example to the portion of the unit circle in the first quadrant, that is if we consider ∫_{0}^1∫_{0}^{√(1-x^2)}(xy(x^2-y^2))/(x^2+y^2)^3dydx=∫_{0}^1x/2*∫_{0}^{√(1-x^2)}(2y(x^2-y^2))/(x^2+y^2)^3dydx ∫((2y)(x^2-y^2))/(x^2+y^2)^3∂y,u=x^2+y^2,y^2=u-x^2,∂u=2y∂y ∫(x^2-(u-x^2))/(u)^3∂u=∫(2x^2-(u))/(u)^3∂u=x^2∫2u^(-3)∂u-∫u^(-2)∂u= u^(-1)-x^2u^(-2)+g(x)=(x^2+y^2)^(-1)-x^2(x^2+y^2)^(-2) plugging in both endpoints for ∫_{0}^{√(1-x^2)}(2y(x^2-y^2))/(x^2+y^2)^3dy gives us ((x^2+1-x^2)^(-1)-x^2(x^2+1-x^2)^(-2))-(x^(-2)-x^2*(x^-4))= 1-x^2 ∫_0^1 x/2(1-x^2)dx=∫_0^1 x/2-x^3/2dx=x^2/4-x^4/8]_0^1=1/4-1/8-0=1/8.also by the way the integral with dxdy is -1/8. by the way here it turns out that the integral of the absolute value is infinity too. however changing to polar coordinates for this integral gives different results than both the integrals given above. this integral in that world would be ∫_0^{π/2}∫_0^1(r^2cos(t)sin(t)(r^2cos^2(t)-r^2sin^2(t))/(r^2)^3 rdrdt= ∫_0^{π/2}∫_0^1(r^2cos(t)sin(t)(r^2cos^2(t)-r^2sin^2(t))/(r^2)^3 rdrdt= ∫_0^{π/2}∫_0^1(r^5cos(t)sin(t)(cos(2t))/(r^2)^3 drdt = ∫_0^{π/2}∫_0^1(cos(t)sin(t)(cos(2t))/r drdt.if we integrated with respect to r first, this integral would be undefined, because we get ∫_0^1 (cos(t)sin(t)(cos(2t))/rdr which is divergent, since cos(t)sin(t)cos(2t)ln(1)-cos(t)sin(t)cos(2t)ln(0) which is a massive sad face,if we integrate with respect to t first ∫_0^{π/2}cos(t)sin(t)cos(2t)/rdt=(∫_0^{π/2}(2cos(t)sin(t)cos(2t))/rdt)/2=(∫_0^{π/2}(sin(2t)cos(2t))/rdt)/2=(∫_0^{π/2}(sin(4t))/rdt)/4=([-cos(4t)/4]_0^{π/2})/r)/4=0. since thats true we would get 0 if we integrated with respect to t first. i say that alone invalidates changes of coordinates under the case of the integral of the absolute value being infinity.for changes of coordinates to be valid, we should have that at least one of the new integrals agrees.since neither result here agrees with either of the integrals in the x and y world,this means you cannot just change coordinate systems willy nilly like that. if dealing with integrals in this case to calculate either integral here.what did you have in mind with your statement that changes of coordinates don't break under the integral of the absolute value being infinity.
Who else appreciates Dr. Peyam's neat and legible handwriting? 👍
Studying fubini-tonelli right now for my probability course you are of tremendous help
Multiple Variables Analysis, where the function is not a curve. The function is a Surface. I love it. I would definitely like to be so smart like you. 😊😊😊
Another great video. Also, it is very cool of you that you gave credit to the site you borrowed the example from. You are one of the best teachers on youtube.
It is basically the same idea as exchanging sum orders for the array or double sequence (1, -1, 0,0,0,0...), (0,1,-1,0,0,0,0...), (0,0,1,-1,0,0,0,0,0...), ...... If you stack sequences and sum vertically, you get 1 left over, but horizontal sums are all 0.
To do this with integrals on a finite region, the infinite amount of cancellation has to be squeezed in the domain and then stretched in the range to balance it all out. This could be done stepwise with step functions over a 2d domain, but it is satisfying to use elementary function examples instead.
Nice comment!
Lotta fun, worked the details, good example for not working blindly and taking a result for granted.
Your channel is great, I wish you the best of luck!
dx ^ dy = -dy ^ dx. Do some videos on differential geometry ;)
It is like christmas then you two make a shared video :-)
Mistake at 15:42: second term should be y^3/U^3 dU. Is this why the final answers are different?
Thanks, but it’s not why the answers are different
@@drpeyam 2 mistakes cancelled each other out!
So fubinis theorem is wrong here? Or the function doesn't comply to the conditions for fubini to work?
The latter one :)
In polar you get r^(-2)*(1/4)*sin(4*theta), if my work checks out. This counterexample is more convenient on a Cartesian disc than a rectangle. The rays (that is, f dr) integrate to +- infinity except where sin(4*theta)=0, which leads to ***infinite cancellation in the double integral***.
If you integrate d theta first, it looks like it's all 0, but it really isn't so clear-cut. Perhaps the first rule of integral convergence is to see if ***the integral of the absolute value of your function*** is finite, and if not, you are dealing with something that requires careful definitions, such as principal value integrals or other such headaches...
i realized this counter example too .i say that this means that not only is fubini invalid under , changes of coordinates are also invalid. i think for it to be valid, at least one of the integrals has to agree with the original integrals in the x or y world.
15:41 should be u^3 because the 2s cancel
same question
true, fortunately his message to us holds :D
Nice, very useful for probability
Great video!!
I think most textbook on Calculus do give counterexample of the Fubini's Theorem. It is just the matter of who would read it
At 15.40 you made a mistake: It should be y cubed over u cubed. The 1/2 term comes from antidifferientiating. Confused me for about 10 minutes. Still An amazing video
You're right; I found the same trick, and putting into the fraction the correct term and integrating, I found that the integral diverges (I'm non a monster in computation, so please check), but anyway the integration give a very different result from the first one
Very good content for understanding fubini's thm and integrals!!👍
Explain the nontrivial zeros of zeta Riemann by Riemann Siegel method
Can you do a proof on the general case of the implict function theorem? And also differential forms
What do you mean by "each slice has to be integrable"? Every conceivable slice of the domain has to be integrable? Not only the 2 slice examples A(x) and B(y)?
E.g. in polar coordinates over the quarter circle of radius 1, the result is infinity*0 = rubbish...
Just A(x) and B(y) for every x and y
@@drpeyam That can not be enough, since for fixed x not equal to zero, f(x,y) is integrable in y with A(x)=x/(2*(x^2+1)^2) which is everywhere defined. For x=0 we have f(0,y)=0 which is trivially integrable. Similarly vice versa for B(y). I would guess that every conceivable slice has to be finite.
Thank you very much for the example.
do things like change of coordinates with the jacobian also break when fubini's theorem doesn't work like when the integral of the absolute value function is divergent.
I don’t think so
@@drpeyam actually i think it does break because you can get even more different answers than just interchanging y and x. for example change the integration reigon given in this counter example to the portion of the unit circle in the first quadrant, that is if we consider ∫_{0}^1∫_{0}^{√(1-x^2)}(xy(x^2-y^2))/(x^2+y^2)^3dydx=∫_{0}^1x/2*∫_{0}^{√(1-x^2)}(2y(x^2-y^2))/(x^2+y^2)^3dydx
∫((2y)(x^2-y^2))/(x^2+y^2)^3∂y,u=x^2+y^2,y^2=u-x^2,∂u=2y∂y
∫(x^2-(u-x^2))/(u)^3∂u=∫(2x^2-(u))/(u)^3∂u=x^2∫2u^(-3)∂u-∫u^(-2)∂u=
u^(-1)-x^2u^(-2)+g(x)=(x^2+y^2)^(-1)-x^2(x^2+y^2)^(-2)
plugging in both endpoints for ∫_{0}^{√(1-x^2)}(2y(x^2-y^2))/(x^2+y^2)^3dy gives us ((x^2+1-x^2)^(-1)-x^2(x^2+1-x^2)^(-2))-(x^(-2)-x^2*(x^-4))=
1-x^2
∫_0^1 x/2(1-x^2)dx=∫_0^1 x/2-x^3/2dx=x^2/4-x^4/8]_0^1=1/4-1/8-0=1/8.also by the way the integral with dxdy is -1/8. by the way here it turns out that the integral of the absolute value is infinity too.
however changing to polar coordinates for this integral gives different results than both the integrals given above.
this integral in that world would be ∫_0^{π/2}∫_0^1(r^2cos(t)sin(t)(r^2cos^2(t)-r^2sin^2(t))/(r^2)^3 rdrdt=
∫_0^{π/2}∫_0^1(r^2cos(t)sin(t)(r^2cos^2(t)-r^2sin^2(t))/(r^2)^3 rdrdt=
∫_0^{π/2}∫_0^1(r^5cos(t)sin(t)(cos(2t))/(r^2)^3 drdt
= ∫_0^{π/2}∫_0^1(cos(t)sin(t)(cos(2t))/r drdt.if we integrated with respect to r first, this integral would be undefined, because we get ∫_0^1 (cos(t)sin(t)(cos(2t))/rdr which is divergent, since cos(t)sin(t)cos(2t)ln(1)-cos(t)sin(t)cos(2t)ln(0) which is a massive sad face,if we integrate with respect to t first
∫_0^{π/2}cos(t)sin(t)cos(2t)/rdt=(∫_0^{π/2}(2cos(t)sin(t)cos(2t))/rdt)/2=(∫_0^{π/2}(sin(2t)cos(2t))/rdt)/2=(∫_0^{π/2}(sin(4t))/rdt)/4=([-cos(4t)/4]_0^{π/2})/r)/4=0. since thats true we would get 0 if we integrated with respect to t first.
i say that alone invalidates changes of coordinates under the case of the integral of the absolute value being infinity.for changes of coordinates to be valid, we should have that at least one of the new integrals agrees.since neither result here agrees with either of the integrals in the x and y world,this means you cannot just change coordinate systems willy nilly like that. if dealing with integrals in this case to calculate either integral here.what did you have in mind with your statement that changes of coordinates don't break under the integral of the absolute value being infinity.
thanks a lot for the nice example!
What if the function is bounded
Awesome :D !
when are you going to make its-a-me fubini into a tshirt like use the chen lu and all. that should be a t shirt.
15:44 Last 2 is wrong
unlisted?
Yeah