Very good. I electrical engineering grad school 1979, my math professor solved this integral for us. I was fascinated with it since. The key is the dx dy = dA =r dr d(theta). In polar form the integrated possesses antiderivative. Great example of coordinate transformation being useful. In one coordinate system a problem which is very difficult becomes *easy peasy* in another coordinate system.
just 3 minutes into the video, and i'm already in love with the clean color commented animations. Keep up the good work. Your setting the bar for others.
You have an awesome way of teaching, thanks. It is fascinating how cool is geometry sometimes when compared with calculus. It is also worth noting the physical meaning of 'sqrt pi' where calculus also touches statistics!
I did this out of my textbook today, and by the looks of it I got it right without using any outside sources to help. So happy. Thanks for the explanation with the visuals backing up my initial intuition
Yes, beautiful. So straightforward when you know how. Nicely presented, part way through I remembered this from years ago but enjoyed it to the end. Like watching a row of dominos topple, inevitable but satisfying.
Interestingly, the result can be interpreted as the square of the Gaussian curve area equal the area of a unit circle and that makes the pi so much more meaningful.
I haven’t learned about polar coordinates yet but I did notice that having x^2 + y^2 has some resemblance to radius squared from origin on xy plane, so I found that if k = e^(-x^2), x = sqrt(-ln(k)), so I^2 can also be written as pi* int_0^1(-ln(k)dk) by volume of revolution which very easily just becomes pi. I’m guessing this polar coordinates method and volume of revolution is just slightly different ways of saying the same thing, pretty cool
My method of solving it is. convert it to the multivariable version. Then imagine it as infinitely many cylinder. then add up those cylinders. the radius of the cylinder is sqrt(-ln x) (the inverse of e^-x^2). Adding up them is just pi*r^2. where r is the function. So its just intergrating pi (sqrt(-lnx))^2. then You'll get -pi*-1=pi. Then take the sqrt of it
There’s another video claiming that Laplace solved the Gaussian integral without needing to switch coordinate systems. However, the nuts and bolts all look the same. The claim is that a parameter, t, avoids it. However, r is also just a parameter, not a function of theta, when it comes to converting dxdy to the tiniest area in the polar coordinate system. Start with S=rtheta. Differentiate with respect to theta, treating r as a parameter now. dS=rdtheta. Now multiple both sides by dr. You get dA = rdrdtheta = dydx. Took me a while to work it out starting with length of a sector of a circle, which is where my intuition starts.
Also, the area under the curve 1/(x²+1) is exactly π, which I think is cool. This is something more trivial though, I suppose. The antiderivative for 1/(x²+1) exists and is well known to be arctan(x). But now I wonder, why is the derivative of arctan(x) equal to 1/(x²+1)?
There are many of us who could not solve this equation, but have enough background to enjoy watching the solving. To ignore us is to leave a large fraction of potential subscribers on the sideline. You may think that the content of a prior video is all you need to refer to. That is not true. When I watch a video, I want you to take me from start to finish. Don’t shorten your video simply because you made a video about the process previously.
Check his channel. This video is part of a full course on multi variable calculus. Expecting him to cover all the fundamental ideas again in this video is like showing up to a class for the first time on the 15th day and being mad you don’t know what’s going on
I just came across this. Thanks for helping me understand this! I especially liked how you showed the 3D function and linked it to polar coordinates with spherical symmetry.
I think those who have worked with 3D modelling and rendering have a more intuitive grasp on these things. Especially if they used something like POV-Ray, which is a fully scripted and Turing complete modelling language.
Hey, thanks for your video Prof. It was all clear to me, except for one step. What theorem did tou use in order to justify combining the multiplication of the two single variable integrals to one double integral?
I had a question- In some beta or gamma integrals, after substituting some variable as sin(theta) or cos(theta) we get intehrals in terms of theta. So, in that case, integrating from 0 to 2pi, gives the integration value 0 if the answer is in terms of sine. So to get non-zero answer, we need to break integral according to symmetry as 4×integral(0 to pi/2). Why do we get 2 different answers then? Shouldn't the answers be same, if we are equating one thing to another for solving.
Well I guess due to discontinuity of that particular func^ at some points..so we need to split them...one example - integral (dx/(2+sin2x) limits 0->2π..
Thank you for the explanation, Dr. Trefor. I would like to ask what would have changeg if instead of integrating from -inf --> inf, you integrated from say xo --> inf, where xo is a a point in the curve. Cheers
That's an interesting question, but the answer doesn't exist unfortunately, in other words this integral from xo to infinity can be obtained only numerically and not algebraically
This is an example of a non-elementary function, in other words, there is no writeable combination of sines, cosines, polynomials, logarithms, or exponents that can give your answer in general. It can only be solved given bounds through non-elementary methods (like this trick or feynmans trick or laplace transforms)
I came across a really abstract way to solve this integral, to obtain an ODE from two different integrals of multi valued function and using Fenyman’s Integration Technique of differentiating under the integral sign to obtain a differential relation
Thanks a lot lot Sir, Watching this Video ; I instantly becomes yours subscriber. I had seen multiple videos, but I didn't get it whole. I am student of +2 class from India, Use of polar coordinates is not there in our curriculum; But help me to provide yours video regarding polar coordinates in description; Thus to enjoy this fun. 🙏🏻👍🏻👍🏻
Your way of presenting the content in a vsual ways witg animations is great 💚💚💚💚. You make hard things easy. Can you please make a video series on complex analysis?
I have taken Calc 3 up to double integrals and I just have one question. Why can we assume that e^-x^2 and e^-y^2 are the same? Is it just because in the xz/yz plane they are the same?
Unfortunately no. Otherwise we could find the cumulative distribution function in elementary functions, and not need to define the erf(x) function, or use infinite series to evaluate it.
I didn't understand how you can split double integral by multiplying two integral and vice verse.Iwill be glad ifyou can write an explanation abot this step. Thanks. MUSTAFA AKYOL
Given I = ∫ e^(-x^2) dx. Make a copy of I, and change the variable to y: I = ∫ e^(-y^2) dy Multiply it with itself squared: I^2 = ∫e^(-x^2) dx * ∫e^(-y^2) dy The same way that we can pull constants out of an integral, we can add constants back in to the integral. Thus: I^2 = ∫(∫e^(-y^2) dy)*e^(-x^2) dx Since the differential terms are really just implicitly multiplied by the integrand, we can relocate dy to the end, and collect the integral signs at the beginning: I^2 = ∫∫ e^(-y^2) *e^(-x^2) dx dy Consolidate the exponents: I^2 = ∫∫ e^(-x^2 - y^2) dx dy Then transform to polar coordinates to carry it out.
The bell shape created by e^(-x^2) is so beautiful as that created by other base constants instead of 'e'. At excel drawing ,eg, 12^(-x^2), gives that shape, with less variance, centered at 0 and same maximum value 1. Sure integrating this with the same tricks showed by Dr. Bazett, will obtain other finite razonable constant . So is very interesting explain why Gauss uses base 'e' instead of any other number (except the obvious not useful like base 1 or 1/2 or other between 1 and 0). Tradition ? Because 'e' is the "natural" base although irrational? Advantages of this? ...
The calculus is more elegant when you use base e. Using another base ultimately is the same thing as having a constant grouped with -x^2 in the exponent, because B^x in general is the same thing as e^x. Thus, B^(-x^2) is the same thing as e^(-ln(B)*x^2). Let L=ln(b). When you integrate ∫∫B^(-x^2) * r dr dθ, your inner integral will become: ∫B^(-r^2) * r dr = ∫e^(-L*r^2) * r dr = -1/(2*L) * ∫e^(-L*r^2) * (-2*L*r) dr = -1/(2*L) * ∫e^(u) * du = -1/(2*L) * e^u + C= -1/(2*L) * e^(-L*r^2) + C Apply r=0 to infinity, recall L: 1/(2*ln(b)) Then evaluating the outer integral from 0 to 2*pi, we get: pi/ln(b) If you use base e, then all of those L's and ln(b) terms will equal 1, which greatly simplifies this. The actual Gaussian distribution contains constants all over the place, to adjust it for mean, standard deviation, and to force it to have an area of 1. Knowing to put sqrt(pi) in the leading constant, to normalize its total area, is an application of this knowledge.
@@DrTrefor do you know about manim lib?? its amazing for making animation for math ...its made by 3blue1brown... github.com/3b1b/manim/tree/master/ its in python btw .....
Very good. I electrical engineering grad school 1979, my math professor solved this integral for us. I was fascinated with it since. The key is the dx dy = dA =r dr d(theta). In polar form the integrated possesses antiderivative.
Great example of coordinate transformation being useful. In one coordinate system a problem which is very difficult becomes *easy peasy* in another coordinate system.
Interesting thanks for sharing
We can also solve Gaussian integral by Laplace transform,but this method is really cool,I like this more, thnk u professor
Laplace transform is so much more elegant, I don't know what you're on about.
just 3 minutes into the video, and i'm already in love with the clean color commented animations. Keep up the good work. Your setting the bar for others.
Awesome video the display is really nice and clear. I love the graphs helping to visualise the integrals.
You have an awesome way of teaching, thanks. It is fascinating how cool is geometry sometimes when compared with calculus. It is also worth noting the physical meaning of 'sqrt pi' where calculus also touches statistics!
I did this out of my textbook today, and by the looks of it I got it right without using any outside sources to help. So happy. Thanks for the explanation with the visuals backing up my initial intuition
Just discovered your channel today Dr. Trefor - awesome. Subscribed already.
Awesome, thank you!
This finally makes sense, thank you so much! Greetings from a physics student from Austria! 😊
Using the Gamma function is my favourite method to solve this
Yes, beautiful. So straightforward when you know how.
Nicely presented, part way through I remembered this from years ago but enjoyed it to the end. Like watching a row of dominos topple, inevitable but satisfying.
Another brilliant and clearly explained video! Thanks for posting.
I am new to your channel and I have watched all day today. It is awesome. You are awesome.
Welcome aboard!
You are showing the beauty of mathematics 🥰🥰
Interestingly, the result can be interpreted as the square of the Gaussian curve area equal the area of a unit circle and that makes the pi so much more meaningful.
Wow, the way is cool. And the way you teach is very clear.
Thank you! 😃
Respect from Pakistan!
Just pure respect!!!
Thank you!!
Thanks bro... really helped me understand this...already liked and subscribed :)
Thanks for the sub!
I haven’t learned about polar coordinates yet but I did notice that having x^2 + y^2 has some resemblance to radius squared from origin on xy plane, so I found that if k = e^(-x^2), x = sqrt(-ln(k)), so I^2 can also be written as pi* int_0^1(-ln(k)dk) by volume of revolution which very easily just becomes pi. I’m guessing this polar coordinates method and volume of revolution is just slightly different ways of saying the same thing, pretty cool
You can use gamma function
x²=t
2xdx=dt
And then the integral convert in
Gamma(1/2)
Which is √π
QED
Respect from india❤
This is circular reasoning because Gamma(1/2) exact value is found via Gaussian integral
My method of solving it is. convert it to the multivariable version. Then imagine it as infinitely many cylinder. then add up those cylinders. the radius of the cylinder is sqrt(-ln x) (the inverse of e^-x^2). Adding up them is just pi*r^2. where r is the function. So its just intergrating pi (sqrt(-lnx))^2. then You'll get -pi*-1=pi. Then take the sqrt of it
Really good explaination, thanks Dr. Bazett!!
this is a brilliant short video
Very helpful video!!! I was trying to use complex analysis, but it didn't quite work out as expected :/
There’s another video claiming that Laplace solved the Gaussian integral without needing to switch coordinate systems. However, the nuts and bolts all look the same. The claim is that a parameter, t, avoids it. However, r is also just a parameter, not a function of theta, when it comes to converting dxdy to the tiniest area in the polar coordinate system. Start with S=rtheta. Differentiate with respect to theta, treating r as a parameter now. dS=rdtheta. Now multiple both sides by dr. You get dA = rdrdtheta = dydx. Took me a while to work it out starting with length of a sector of a circle, which is where my intuition starts.
Another master class. My brain thanks you!
Also, the area under the curve 1/(x²+1) is exactly π, which I think is cool. This is something more trivial though, I suppose.
The antiderivative for 1/(x²+1) exists and is well known to be arctan(x). But now I wonder, why is the derivative of arctan(x) equal to 1/(x²+1)?
2:12 , sir what was the logic of that step
that's insane and amazing mathmatica tirck.
Thank you! This helped make sense of verifying the pdf of a normal distribution!
You make it simple.... 👍
You are an amazing prof I wish you were my instructor
There are many of us who could not solve this equation, but have enough background to enjoy watching the solving. To ignore us is to leave a large fraction of potential subscribers on the sideline. You may think that the content of a prior video is all you need to refer to. That is not true. When I watch a video, I want you to take me from start to finish. Don’t shorten your video simply because you made a video about the process previously.
Check his channel. This video is part of a full course on multi variable calculus. Expecting him to cover all the fundamental ideas again in this video is like showing up to a class for the first time on the 15th day and being mad you don’t know what’s going on
I just came across this. Thanks for helping me understand this! I especially liked how you showed the 3D function and linked it to polar coordinates with spherical symmetry.
I want to understand how u defined the limits for theta...
Sir,I need help here,if u can.
Very cosy methodprofessor thanks.
I think those who have worked with 3D modelling and rendering have a more intuitive grasp on these things. Especially if they used something like POV-Ray, which is a fully scripted and Turing complete modelling language.
I'm waiting for wonderful topics in Vector Calculus
Hey, thanks for your video Prof. It was all clear to me, except for one step. What theorem did tou use in order to justify combining the multiplication of the two single variable integrals to one double integral?
Fubinis theorem
I think it's the constant factor rule... because both integrals are constants w.r.t. each other.
I had a question-
In some beta or gamma integrals, after substituting some variable as sin(theta) or cos(theta) we get intehrals in terms of theta. So, in that case, integrating from 0 to 2pi, gives the integration value 0 if the answer is in terms of sine. So to get non-zero answer, we need to break integral according to symmetry as 4×integral(0 to pi/2). Why do we get 2 different answers then? Shouldn't the answers be same, if we are equating one thing to another for solving.
Well I guess due to discontinuity of that particular func^ at some points..so we need to split them...one example - integral (dx/(2+sin2x) limits 0->2π..
Great explanation. Thank you
thank you so much !!!!! this video saved me!
Thank you for the explanation, Dr. Trefor. I would like to ask what would have changeg if instead of integrating from -inf --> inf, you integrated from say xo --> inf, where xo is a a point in the curve. Cheers
That's an interesting question, but the answer doesn't exist unfortunately, in other words this integral from xo to infinity can be obtained only numerically and not algebraically
This is an example of a non-elementary function, in other words, there is no writeable combination of sines, cosines, polynomials, logarithms, or exponents that can give your answer in general. It can only be solved given bounds through non-elementary methods (like this trick or feynmans trick or laplace transforms)
made it really easy to understand thanks
Sir , can you made a video on centre of gravity....multiple integral??
thank's a lot. Very good explanation.
Glad it was helpful!
Nice work!
Great teacher 💪👍
I came across a really abstract way to solve this integral, to obtain an ODE from two different integrals of multi valued function and using Fenyman’s Integration Technique of differentiating under the integral sign to obtain a differential relation
or you could do the way laplace did it
Nice idea 💡
Thanks a lot lot Sir,
Watching this Video ; I instantly becomes yours subscriber. I had seen multiple videos, but I didn't get it whole.
I am student of +2 class from India, Use of polar coordinates is not there in our curriculum;
But help me to provide yours video regarding polar coordinates in description;
Thus to enjoy this fun. 🙏🏻👍🏻👍🏻
Why when you have I^2 can you turn that into a double integral (if of course it is convergent, which it is)?
My personal favorite
Your way of presenting the content in a vsual ways witg animations is great 💚💚💚💚. You make hard things easy. Can you please make a video series on complex analysis?
@@DrTrefor Ha ha Ha but thanks You again Sir !
squareroot of pi is also the sidelength of a square with the same area as a unitcircle.
Sir love from the ❤️ 💙 💜 💖
I have taken Calc 3 up to double integrals and I just have one question. Why can we assume that e^-x^2 and e^-y^2 are the same? Is it just because in the xz/yz plane they are the same?
now what happens if we differentiate the root of pi
Thanks man!
No problem!
GENIUS
is there a way to compute the gaussian integrale without going in the polar coodinate ?
Infinite series.
How did you convert the bound of integration ? I mean how to write that mathematically ?
amazing explanation prof. tnx alot
So what's your _least_ favourite integral then?
no music, perfect
Why the (r)dr was not transferred to udu ? As the extra r is there once you change to polar
??!!
SIR I am still confused why the LIMIT OF THETA is 0 to 2pi and why not 0 to pi /2
nice trick, i didnt know of it :)
What technique you use for this kind of animation
I make everything in MATLAB
brain storming but how dx and dy and x and y are matching and making a polar couple. with what major sense. out of scope of my head
I still wonder,why you allowed to merge the squared integral
If I did that mighty hand waving in a test I guess my grade would be bad indeed.
Nice video!
Hey, where is the url of video that you mentioned in the current one. I couldn't find it in description. Please help, i wanna learn
@@DrTrefor Thanks Prof! 🙂
What if the interval were from a to b, would it be feasible?
Unfortunately no. Otherwise we could find the cumulative distribution function in elementary functions, and not need to define the erf(x) function, or use infinite series to evaluate it.
Thanks Sir
♥️♥️SUPERB💝💝♥️
I didn't understand how you can split double integral by multiplying two integral and vice verse.Iwill be glad ifyou can write an explanation abot this step. Thanks. MUSTAFA AKYOL
Given I = ∫ e^(-x^2) dx.
Make a copy of I, and change the variable to y:
I = ∫ e^(-y^2) dy
Multiply it with itself squared:
I^2 = ∫e^(-x^2) dx * ∫e^(-y^2) dy
The same way that we can pull constants out of an integral, we can add constants back in to the integral. Thus:
I^2 = ∫(∫e^(-y^2) dy)*e^(-x^2) dx
Since the differential terms are really just implicitly multiplied by the integrand, we can relocate dy to the end, and collect the integral signs at the beginning:
I^2 = ∫∫ e^(-y^2) *e^(-x^2) dx dy
Consolidate the exponents:
I^2 = ∫∫ e^(-x^2 - y^2) dx dy
Then transform to polar coordinates to carry it out.
How can you change the variable
Is this how Gauss solved it originally?
I love it!
You're the best
Awesome
This is used as the main way of proof that the pdf of the gaussian distribution integrates to 1
The bell shape created by e^(-x^2) is so beautiful as that created by other base constants instead of 'e'.
At excel drawing ,eg, 12^(-x^2), gives that shape, with less variance, centered at 0 and same maximum value 1.
Sure integrating this with the same tricks showed by Dr. Bazett, will obtain other finite razonable constant .
So is very interesting explain why Gauss uses base 'e' instead of any other number (except the obvious not useful like base 1 or 1/2 or other between 1 and 0). Tradition ?
Because 'e' is the "natural" base although irrational? Advantages of this? ...
The calculus is more elegant when you use base e. Using another base ultimately is the same thing as having a constant grouped with -x^2 in the exponent, because B^x in general is the same thing as e^x. Thus, B^(-x^2) is the same thing as e^(-ln(B)*x^2).
Let L=ln(b). When you integrate ∫∫B^(-x^2) * r dr dθ, your inner integral will become:
∫B^(-r^2) * r dr =
∫e^(-L*r^2) * r dr =
-1/(2*L) * ∫e^(-L*r^2) * (-2*L*r) dr =
-1/(2*L) * ∫e^(u) * du =
-1/(2*L) * e^u + C=
-1/(2*L) * e^(-L*r^2) + C
Apply r=0 to infinity, recall L:
1/(2*ln(b))
Then evaluating the outer integral from 0 to 2*pi, we get:
pi/ln(b)
If you use base e, then all of those L's and ln(b) terms will equal 1, which greatly simplifies this.
The actual Gaussian distribution contains constants all over the place, to adjust it for mean, standard deviation, and to force it to have an area of 1. Knowing to put sqrt(pi) in the leading constant, to normalize its total area, is an application of this knowledge.
This guy's awesome
awesome!
Great!!!!!
🙏🙏🙏
Super super
When is The Gaussian Integral used in real world applications? 🌎
All statistics is based on this result 😅
Wow this was 3b1b kind of beautiful
Super
it is best if your vedio suported by animation
HALLELUJA 💖💖💖
pi shows up all unexpected places..hehe
@@DrTrefor
yes...
math is always cool...and amazing 😁...
@@DrTrefor
do you know about manim lib?? its amazing for making animation for math ...its made by 3blue1brown...
github.com/3b1b/manim/tree/master/
its in python btw .....
@@DrTrefor
you have done really great job sir...
the green screen and the editing 👏👏👏👏👏...
really amazing channel...it deserves more IMO...
Three analysis classes and I still can't solve it, worst integral
king
❤
Wwwwwoooooowwww thank youuu
im sorry you cant change the variable and still say " this is squared"