yea, pretty much most math books. "as the readers should verify", "it is indeed trivial and shall be left as an exercise to the readers", "the argument will be outlined in exercise (insert number)"
You have transformed the integral into polar coordinates. This is a classic example of how sometimes complex cartesian coordinate integrals can be simplified in polar coordinates.
I saw Professor Christine Briner from MIT used polar coordinates, double integrals and change of variables to evaluate the Gaussian integral. Now, I see you find the volume under the bell curve by summing tubes(hollow cylinders) whose radii varies from 0 to infinity. I like the geometrical approach to finding the volume of each tube. I find the visual aids intuitive. I think this is a fresh and intuitive way of evaluating the Gaussian integral. Thanks.
Came back to this video after taking calc 3 and seeing you explaining the gaussian integral without getting into jacobians and double integral mess is absolutely stunning
This video randomly popped up on my feed and I am glad that it did. Have been out of touch of mathematics for about 8 years. Brought back a lot of memories.
i find this is the best way to understand what's going on and what is the concept of doing such things with this problem. thanks for making this video.
One of my math teachers at university showed this exact method when I studied Applied Mathematics (which was basically "math used for all kinds of real-life things", so it was like a big test of all the students' previous knowledge in other courses), and I was blown away by it.
Really great video! Thank you so much for this, I had a really hard time understanding how to integrate wave functions that included this exact integral. Thank you so much!!
So in the integral of `e^(-x²)` the simple lack of `·x` is what makes it (almost) impossible to solve? And the whole idea of translating the problem to polar coordinates it what helps to bring that `·x` (or `·r` in this case) back?
Exactly, by multiplying it with itself, but with a different variable, you can turn it into this integral over the entire plane of the function f(x,y) = exp(-x^2-y^2). It is then indeed a smart idea to do a coordinate transform to polar coordinates, doing this transformation then changes the shape and infinitesmal area of the d-bit (no clue what it's called, the dx dy) by an amount according to the jacobian of the transformation (look it up, it's the determinant of a transformation matrix and can be used for any set of coordinates you'd want), this jacobian just happens to be r for polar coordinates, which then makes the entire integral evaluable.
@@NoobLord98 Alternatively, think of dxdy=dA, a tiny area in the xy-plane caused by changes in x and y. In polar coordinates, a change in r and θ will cause a tiny area of approximately rdθdr which will be equal to dA=dxdy in the limit as they become smaller and smaller. So dxdy=rdθdr
Fantastic It is like... you have to use properties of complex numbers to end out with the error funcion That was very funny to think!!! Thanks a lot for the video!!!
OMFG, I hope i could get that video when I started to learn Calculus ... Special Functions(Non-elemental) were such a pain for and after all I just used abstract rules to get them, but omg this interpretation of Gaussian Integral is awesome...
The substitution at 8:44 is spooky, I mean, it comes out so smoothly. It's easy to see that this integral comes down to pi but man I'm still scratching my head on that.
it wasn't necessary tho. You just need to know that integral of f'(x)*e^(f(x)) = e^(fx). So in this case you have -pi * integral of -2x * e^(-x^2) which is -pi*e^(-x^2)
I always knew to do this by converting to polar form integrating from 0 to 2pi and then 0 to infinity, but this explains where the extra r comes from very well! thank you so much
I stopped at 4:07 and tried doing the problem with stacked circles and it gave a very easy answer. Was kinda disappointed that you would go on and do it with cylindrical shells, but it works both ways.
I have seen a couple of videos where this integral is solved by substituting x^2 + y^2 = r^2 and letting dy dx = r dr d(theta) and another one where instead of polar coordinates stuff it goes by substitution y = xt. This is the first time I have seen a visual approach and it's really really good... Much appreciated :slightly_smiling_face:
Thank you. This was helpful. Even though I came with somewhat of an understanding of both the Gaussian function and Integrals, my background isn't math-heavy. This was very pleasant.
Could someone explain why both the integrals are equated to 'I' at 3:10 How does one know beforehand that the integral will not be a function of x or y?
Actually they both are DEFINITE integrals, which means that thay will spit out a number and not a function. Changing variables in the same integral does not affect its value.
Because x^2+y^2 is not a function. They are independent values in the x-y plane defined by the property of a parabola. If you assign a value to x^2+y^2 like r^2 this will be the equation of a circle. If you integrate or differentiate these they are defined by their partial derivative (or integral). By this property they are separable, because in therms of the other intergral they are constants (independent).
It’s easier too. If you know the gaussian formula, you can moltiplicate and divide that expression for sqrt(2*π*1/2), and drag out the numerator from integral. Now the integral represents the area of a gaussian with mean=0 and variance=1/2 that’s equal 1 and the amount out of integral is sqrt(2*π*1/2), that’s actually sqrt(π). Great video anyway
5:51 we were supposed to find volume under the curve, therefore we should have taken solid cylinder rather than taking hollow cylinder. And if you had taken hollow cylinder then you should find volume under hollow cyling including it's volume but you had just find volume of cylinder itself and not volume under the cylinder. Why???
He used tubes (with 0 volume) instead of solid cylinders (with >0 volume) because by integrating an area you get a volume. It’s a bit confusing and if you want to do it with solid math than you do it like this: ∫ ∫ f(x,y) dx dy (with x and y going from -inf to +inf) = ∫ ∫ f(r,θ) * J dr dθ (with r going from 0 to +inf and θ going from 0 to 2*pi) J is the Jacobean coefficient from cartesian to polar, in this case J = r This is a bit to complicated for the video but voila
I would also like to add that if you take the 3D image at 3:57, rotate it 90 degrees so that the tip of the cone is pointing towards you and imagine that the point of the cone is a stack with smoke steadily streaming out from against a steady wind, that cone would be a concentration profile of the smoke as it travels from the tip of the stack to infinity. We have much to credit Gauss for.
Oh dammit. This is exactly what i needed to answer a question on a paper that was due to 2 days ago. Glad that I understand it now, even if its a little late. Good video!
The law of non-contadiction : nothing is it's opposite, is important technical apparatus for underwriting the operational - institutional * - possibility of equal opportunity, because it functions like a complementary epistemological standard to the ideosyncratic faculties - emotion, inuition, revelation, faith, or some combination of these - for percepto-reason and it's value is illustrated by the Normal Probability Density Function, which returns a unique solution for it's input, however is derived from the equation for a circle, which is a relation with more than one possible output : instead of defining the square root of number representative syntax < . < for " . " multiplied by as < opposite to > , conventionally it is - < and < for " - " negative. * The rules and conventions guiding behaviour.
I was literally just in a lecture for 3D calculus and they introduced polar 3d integrals for the intersecting areas and volumes of sheets and cylinders.
I didn't see any error function video on youtube which depicts closer picture of the phenomenon than yours. It is the best one the approaching perfection asymptote.
Can someone please explain why if you plot f(x,y) at 3:51 you get a 3 dimensional surface rather than a 2-d plane. I thought a third z coordinate would need to be part of the function in order to achieve that shape
Going to three dimensions allows you to transform the function into polar coordinates, where integration is more manageable. It works because the function in the y-axis is equal to the function in the x-axis. The solution is as elegant as it gets...that it gives a finite solution is critical for statistical applications (where the sum of probabilities over a range of out comes is exactly unity).
why do u use hollow cylinders instead of just normal cylinders at 5:50? since u are trying to find the volume under the 2d surface, aren't u adding up infinite number of solid cylinders. (sorry if my qn doesn't make sense I dont major in math...)
Hey, thanks for the video, this was an incredibly well-worded demonstration. Can you by any chance tell me what program you used to graph the 3d graph? That would come in pretty handy
I love how the video makes you feel smart by hinting at where it’s going so that you figure a lot of it out on your own.
Lies again? Dark X
yea, pretty much most math books. "as the readers should verify", "it is indeed trivial and shall be left as an exercise to the readers", "the argument will be outlined in exercise (insert number)"
"Might ring a bell"... I see what you did there
hehehehehehe
*Laughs in Bell curve*
No word for this great video.
Spectacular.
According to you, there are actually 7 words.
@@David_F97 ok
But you said no word.
That 7 words doesn’t include
Dude integrated so hard that his voice changed at 9:35
"I talked to Barzini."
I am blown away by the way you simplified it. Pure awesomeness.
I mean he didnt invent anything its in every calc textbook about integral calculus right
@@slender1892 stfu bit*h , you probably scored 0 in maths 😂😂
You have transformed the integral into polar coordinates. This is a classic example of how sometimes complex cartesian coordinate integrals can be simplified in polar coordinates.
Do also read about cylindrical coordinate system
@Kappa Chino laughs in confocal eliptical coordinates.
Hey can you please explain why was the height taken as the function itself ( -e power R2 squared) m
One integral is used for area, two for volume, what’s the use of triple integral?
@@danipent3550 density
That is stunningly beautiful. One of the best mathematical explanations I’ve ever seen. Well done, sir
This is easily the best video explaining the Gaussian Integral I have ever seen!
This is no random math as your channel name suggests. The geometrical proof is always the most elegant way. I can't skip ads for you. Great content!
I saw Professor Christine Briner from MIT used polar coordinates, double integrals and change of variables to evaluate the Gaussian integral. Now, I see you find the volume under the bell curve by summing tubes(hollow cylinders) whose radii varies from 0 to infinity. I like the geometrical approach to finding the volume of each tube. I find the visual aids intuitive. I think this is a fresh and intuitive way of evaluating the Gaussian integral. Thanks.
thx a lot man. finally a good explanation
no kidding this is probably the best explaination i found on youtube
I have watched a number of clips on the Gaussian Integral, but I like the practical way this has been explained.
Came back to this video after taking calc 3 and seeing you explaining the gaussian integral without getting into jacobians and double integral mess is absolutely stunning
I'll just echo what others have typed; this is probably the best explanation on math I've ever had.
Probably the best video on TH-cam explaining this problem. Thank you!
This video randomly popped up on my feed and I am glad that it did. Have been out of touch of mathematics for about 8 years. Brought back a lot of memories.
THIS IS BEAUTIFUL! This is the best video I’ve seen that talks about this integral.
you have an entertaining teaching style (like a short movie), to the point that I was excited about what's next 🍿🍿🍿
This video is AWESOME! I think they skipped this in all my calc classes in college. He makes it so intuitive. I'll def be coming back to this channel
i find this is the best way to understand what's going on and what is the concept of doing such things with this problem. thanks for making this video.
One of my math teachers at university showed this exact method when I studied Applied Mathematics (which was basically "math used for all kinds of real-life things", so it was like a big test of all the students' previous knowledge in other courses), and I was blown away by it.
I listened 9:35 to 9:36 at 25% speed, and I enjoyed it very much!
ouuuhhhhfff paaaaaaaiiii
Really great video! Thank you so much for this, I had a really hard time understanding how to integrate wave functions that included this exact integral. Thank you so much!!
I am not good at English but this video is one of greatest explanation videos I have ever seen. Thank you very much
So in the integral of `e^(-x²)` the simple lack of `·x` is what makes it (almost) impossible to solve?
And the whole idea of translating the problem to polar coordinates it what helps to bring that `·x` (or `·r` in this case) back?
Exactly, by multiplying it with itself, but with a different variable, you can turn it into this integral over the entire plane of the function f(x,y) = exp(-x^2-y^2). It is then indeed a smart idea to do a coordinate transform to polar coordinates, doing this transformation then changes the shape and infinitesmal area of the d-bit (no clue what it's called, the dx dy) by an amount according to the jacobian of the transformation (look it up, it's the determinant of a transformation matrix and can be used for any set of coordinates you'd want), this jacobian just happens to be r for polar coordinates, which then makes the entire integral evaluable.
@@NoobLord98
Alternatively, think of dxdy=dA, a tiny area in the xy-plane caused by changes in x and y. In polar coordinates, a change in r and θ will cause a tiny area of approximately rdθdr which will be equal to dA=dxdy in the limit as they become smaller and smaller. So dxdy=rdθdr
Sehr raffinierte Transformation, die das Problem entscheidend vereinfacht. Das könnte man sogar in einem motivierten Leistungskurs bringen! Super!
pff i love how you just turned one integral into the same integral+another
An extremely well done video for an extremely beautiful integral.
this channel is like sent from god, great video! you deserve a lot more!
Absolutely brilliant.... never seen such an explanation. Thanks a lot.
Fantastic
It is like... you have to use properties of complex numbers to end out with the error funcion
That was very funny to think!!!
Thanks a lot for the video!!!
OMFG, I hope i could get that video when I started to learn Calculus ... Special Functions(Non-elemental) were such a pain for and after all I just used abstract rules to get them, but omg this interpretation of Gaussian Integral is awesome...
The substitution at 8:44 is spooky, I mean, it comes out so smoothly. It's easy to see that this integral comes down to pi but man I'm still scratching my head on that.
it wasn't necessary tho. You just need to know that integral of f'(x)*e^(f(x)) = e^(fx).
So in this case you have -pi * integral of -2x * e^(-x^2) which is -pi*e^(-x^2)
[2024] Greetings from Curaçao, an Island Nation in The Caribbean.
Great clear video explanation.
Bravo! Best explanation of this integral ever!!!
Very easy-to-digest explanation of a very complex topic. Excellent video!
The best intuitive and clear video ive seen on this topic!
One of the best explanations I've ever seen...well done!
Thank you for providing with detailed explanation not just formulas, great job!
First time I understood this integral. Incredible video
WOW THIS IS AMAZING THIS IS SO CLEAR!!!!!
This is so beautiful. Amazing explanation, thank you
Fantastic video. Kept me entertained with great explanation and graphical representation.
1:25 "might ring a bell" nice pun haha
I am stunned and amazed with your explanation
Finally i can understand this without polar things. Thank Youuuu. Very Logical and easy to understand. Big much thanks
this is a very great video that helped me understand the gaussian integral and I hereby recommend it to everyone
Super explanation, this should have more likes and views.
I always knew to do this by converting to polar form integrating from 0 to 2pi and then 0 to infinity, but this explains where the extra r comes from very well! thank you so much
In a more mathematical way, the 2pi*r is the jacobian of the substitution and you can calculate it solving the determinant of the jacobian matrix.
A neat solution with a remarkable result...and with profound real world applications.
Beautiful! Congratulations for such a wonderful video and demonstration.
I stopped at 4:07 and tried doing the problem with stacked circles and it gave a very easy answer. Was kinda disappointed that you would go on and do it with cylindrical shells, but it works both ways.
I have seen a couple of videos where this integral is solved by substituting x^2 + y^2 = r^2 and letting dy dx = r dr d(theta) and another one where instead of polar coordinates stuff it goes by substitution y = xt. This is the first time I have seen a visual approach and it's really really good... Much appreciated :slightly_smiling_face:
Thank you. This was helpful. Even though I came with somewhat of an understanding of both the Gaussian function and Integrals, my background isn't math-heavy. This was very pleasant.
You explained it very well
Best explanation I've seen for this integral. Well done sir.
โอ้วพระเจ้าจ้อดดดด
คุณเป็นคนไทยรึนี่ นั่งดูจนจบ สำเนียงโครตเทพ วิชาการ เทพสัด สุดๆครับ ดีใจ กับ อนาคตของชาติจริงๆ
Could someone explain why both the integrals are equated to 'I' at 3:10 How does one know beforehand that the integral will not be a function of x or y?
Actually they both are DEFINITE integrals, which means that thay will spit out a number and not a function. Changing variables in the same integral does not affect its value.
Because x^2+y^2 is not a function. They are independent values in the x-y plane defined by the property of a parabola. If you assign a value to x^2+y^2 like r^2 this will be the equation of a circle. If you integrate or differentiate these they are defined by their partial derivative (or integral). By this property they are separable, because in therms of the other intergral they are constants (independent).
Thank you. You just earned a subscriber. Looking for more such great content.
It’s easier too. If you know the gaussian formula, you can moltiplicate and divide that expression for sqrt(2*π*1/2), and drag out the numerator from integral. Now the integral represents the area of a gaussian with mean=0 and variance=1/2 that’s equal 1 and the amount out of integral is sqrt(2*π*1/2), that’s actually sqrt(π). Great video anyway
I have watched a lot of videos , but understood nothing. But you explained it perfectly!!
Fantastic explanation of a somewhat abstract idea!
Just spectacular! Well explained and visualized. Speechless
The f ing best integral ever
my jaw dropped when it came out to be square root of pi!
Wow.. great presentation sir
Perfect explanation. Big thanks from Germany 👌👌
5:51 we were supposed to find volume under the curve, therefore we should have taken solid cylinder rather than taking hollow cylinder. And if you had taken hollow cylinder then you should find volume under hollow cyling including it's volume but you had just find volume of cylinder itself and not volume under the cylinder. Why???
He used tubes (with 0 volume) instead of solid cylinders (with >0 volume) because by integrating an area you get a volume. It’s a bit confusing and if you want to do it with solid math than you do it like this:
∫ ∫ f(x,y) dx dy (with x and y going from -inf to +inf)
=
∫ ∫ f(r,θ) * J dr dθ (with r going from 0 to +inf and θ going from 0 to 2*pi)
J is the Jacobean coefficient from cartesian to polar, in this case J = r
This is a bit to complicated for the video but voila
I would also like to add that if you take the 3D image at 3:57, rotate it 90 degrees so that the tip of the cone is pointing towards you and imagine that the point of the cone is a stack with smoke steadily streaming out from against a steady wind, that cone would be a concentration profile of the smoke as it travels from the tip of the stack to infinity. We have much to credit Gauss for.
That video make me remember my first year in college when we were studying electrostatic and electromagnetism 😊😊
Is it the electric field of a uniformly charhed disk?
@@vincentdublin3127 Yes those things XD
Oh dammit. This is exactly what i needed to answer a question on a paper that was due to 2 days ago. Glad that I understand it now, even if its a little late.
Good video!
1:26 "Might ring a bell" I see what you did there
Very Good explanation ! Keep sharing such valuable content !
You put a lot of effort to make this video!! A great work done!!!!!!!!!!
very nice video, i like how you explain things, its like how my mind process things exactly thank you
My first time in your channel. You describe well everything, so I have to subscribe.
Great teacher! Thank you very much. Keep making maths videos...
Doing this is calc 3 was so amazing for me to see
Simply elegant and brilliant!
The law of non-contadiction : nothing is it's opposite, is important technical apparatus for underwriting the operational - institutional * - possibility of equal opportunity, because it functions like a complementary epistemological standard to the ideosyncratic faculties - emotion, inuition, revelation, faith, or some combination of these - for percepto-reason and it's value is illustrated by the Normal Probability Density Function, which returns a unique solution for it's input, however is derived from the equation for a circle, which is a relation with more than one possible output : instead of defining the square root of number representative syntax < . < for " . " multiplied by as < opposite to > , conventionally it is - < and < for " - " negative.
* The rules and conventions guiding behaviour.
How fun was that! Thanks!
Wow! Great illustration!
Amazing, now I know the proof of error function very clearly
I was literally just in a lecture for 3D calculus and they introduced polar 3d integrals for the intersecting areas and volumes of sheets and cylinders.
This is so beautiful and awesome
Truly brilliant and magnificent
I didn't see any error function video on youtube which depicts closer picture of the phenomenon than yours. It is the best one the approaching perfection asymptote.
Can someone please explain why if you plot f(x,y) at 3:51 you get a 3 dimensional surface rather than a 2-d plane. I thought a third z coordinate would need to be part of the function in order to achieve that shape
Going to three dimensions allows you to transform the function into polar coordinates, where integration is more manageable. It works because the function in the y-axis is equal to the function in the x-axis. The solution is as elegant as it gets...that it gives a finite solution is critical for statistical applications (where the sum of probabilities over a range of out comes is exactly unity).
@@hetsmiecht1029 aaaah. It's been while since I've done multi variate in class. Thank you
@@hetsmiecht1029 what is f(x,y,z)? what do triple integrals represent if double integrals are volume
This is such a good explanation btw
why do u use hollow cylinders instead of just normal cylinders at 5:50? since u are trying to find the volume under the 2d surface, aren't u adding up infinite number of solid cylinders. (sorry if my qn doesn't make sense I dont major in math...)
The area inside the cylinders is accounted for by all of the other cylinders which exist inside of that one
Ive only taken Calculus 1 but somehow this all makes sense haha! wonderful video!
1:41 is that right? Variance 1 means a SD of 1 too, right? So is really only 2/3 of the are between -1 and 1?
Nicely done sir👍.
we can also find this integral using Gamma function.
This is a great video.... very understandable. A lot of videos go through what to do, but this really helped to visualize it. Thanks, and great job!
Hey, thanks for the video, this was an incredibly well-worded demonstration. Can you by any chance tell me what program you used to graph the 3d graph? That would come in pretty handy
I think it is too late, but i believe he used Maxima
looks like MATLAB
I love this video and the way you explain
Excellent! A very clear explanation, thank you!
yea exactly what I said. exactly. Thanks for reaffirming my answer
1:36 I guess it can't be considered a distribution unless the area under the curve is 1.
Amazing method to explain complicated expression
Hello, it's great, with which programm did you make this video? I mean the animation of figures?