The Gaussian Integral // Solved Using Polar Coordinates
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- Опубліковано 29 лип 2024
- The gaussian integral - integrating e^(-x^2) over all numbers, is an extremely important integral in probability, statistics, and many other fields. However, it is challenging to solve using elementary methods from single variable calculus. In this video we will see how we can convert it to multivariable calculus and then use tricks from multivariable calculus - in this case converting to polar coordinates - to solve this single variable integral. The crazy thing is that this integral ends up being in terms of pi, and if you didn't know about the polar trick you might wonder why pi shows up here at all! This proof is due to Poisson.
The previous video on double integration in polar: • Double Integration in ...
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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.
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
Awesome video the display is really nice and clear. I love the graphs helping to visualise the integrals.
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.
Another brilliant and clearly explained video! Thanks for posting.
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!
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.
Just discovered your channel today Dr. Trefor - awesome. Subscribed already.
Awesome, thank you!
Really good explaination, thanks Dr. Bazett!!
Thank you! This helped make sense of verifying the pdf of a normal distribution!
This finally makes sense, thank you so much! Greetings from a physics student from Austria! 😊
Another master class. My brain thanks you!
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
Using the Gamma function is my favourite method to solve this
Great explanation. Thank you
Nice work!
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.
this is a brilliant short video
I am new to your channel and I have watched all day today. It is awesome. You are awesome.
Welcome aboard!
Very helpful video!!! I was trying to use complex analysis, but it didn't quite work out as expected :/
amazing explanation prof. tnx alot
You are an amazing prof I wish you were my instructor
Thanks bro... really helped me understand this...already liked and subscribed :)
Thanks for the sub!
Very cosy methodprofessor thanks.
made it really easy to understand thanks
You are showing the beauty of mathematics 🥰🥰
thank you so much !!!!! this video saved me!
You make it simple.... 👍
Great teacher 💪👍
Wow, the way is cool. And the way you teach is very clear.
Thank you! 😃
Nice idea 💡
that's insane and amazing mathmatica tirck.
I'm waiting for wonderful topics in Vector Calculus
GENIUS
Nice video!
How did you convert the bound of integration ? I mean how to write that mathematically ?
thank's a lot. Very good explanation.
Glad it was helpful!
I want to understand how u defined the limits for theta...
Sir,I need help here,if u can.
nice trick, i didnt know of it :)
Sir , can you made a video on centre of gravity....multiple integral??
Why the (r)dr was not transferred to udu ? As the extra r is there once you change to polar
??!!
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.
My personal favorite
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..
Thanks man!
No problem!
awesome!
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)
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 !
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
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! 🙂
Awesome
How can you change the variable
Great!!!!!
You're the best
Respect from Pakistan!
Just pure respect!!!
Thank you!!
is there a way to compute the gaussian integrale without going in the polar coodinate ?
Infinite series.
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.
squareroot of pi is also the sidelength of a square with the same area as a unitcircle.
So what's your _least_ favourite integral then?
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
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.
What technique you use for this kind of animation
I make everything in MATLAB
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. 🙏🏻👍🏻👍🏻
no music, perfect
♥️♥️SUPERB💝💝♥️
Sir love from the ❤️ 💙 💜 💖
I still wonder,why you allowed to merge the squared integral
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❤
now what happens if we differentiate the root of pi
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.
SIR I am still confused why the LIMIT OF THETA is 0 to 2pi and why not 0 to pi /2
If I did that mighty hand waving in a test I guess my grade would be bad indeed.
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.
Super
Super super
🙏🙏🙏
king
Is this how Gauss solved it originally?
Wwwwwoooooowwww thank youuu
Wow this was 3b1b kind of beautiful
This is used as the main way of proof that the pdf of the gaussian distribution integrates to 1
❤
HALLELUJA 💖💖💖
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
it is best if your vedio suported by animation
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
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.
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...
When is The Gaussian Integral used in real world applications? 🌎
Three analysis classes and I still can't solve it, worst integral
im sorry you cant change the variable and still say " this is squared"
"goewsian" 🙃🙃
Gaussian not gausssssian
Ain't understanding really
Gauss was one of the best matematicans, but worst social persons.
Too talkative!!!