This isn't a Circle - Why is Pi here?
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- Опубліковано 23 вер 2024
- This famous bell shaped curve has a pretty famous result. It's not exactly clear why the circle constant pi is showing up in this normal distribution curve. Let's investigate this Gauss Curve (Gaussian Distribution) and get to the bottom of why Pi is here!
#math #brithemathguy#integral
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
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I honestly believed this guy was writing backwards until I realized you can just flip the video :p
You mean he's not left-handed?!
Don't reveal the magic
@keith Surely then he'd have to write backwards so that when he flips it we can read it. It could work if he's in face cam mode, with a screen in front of the camera and then flips that? ....is that what you mean?
That’s why all of his T-shirts are textless. If I were in his place, I would go to order custom t-shirts with inverted text to take deception to the next level.
@Joshua Millet nope. what if instead of a word he just drew a single straight horizontal line. get it?
I'm literally taking an upper-level mathematical statistics course right now and immediately saw that equation and thought "Wait..... isn't that the pdf of a standard normal distribution???"
Well spotted! Best of luck with your class!
That's not standard normal
@@user-se2pl5hd5s correct. Standard normal (and any probability distribution function) means your total area under the curve is 1, this video shows that is not the case. You have to divide it by root 2pi
@@conmattang8492 There have been attempts and redefining the standard normal to have variance 1/2*pi, which would take care of the pesky normalizing constants. It hasn't stuck, though.
Almost. The pdf of the standard normal has a 1/2 in the exponent. Of course the function is also normalized so that the integral is 1. Other than that, you're right.
Damn that was clean. What a smooth transition to pi
Glad you liked it!
Funny story: When I was in college, I took a course designed for people who wanted to be math professors. One of the presenters showed us (old tech alert) a calculator that you could put on the glass surface of an overhead projector. It would even integrate functions, something I hadn't seen done on any calculator at the time. For fun, I asked the presenter to have it integrate e^(-x^2). She didn't even understand why I was asking for that particular function until the professor explained that it didn't have a closed-form antiderivative. She dutifully typed it in and it responded (whatever the constant is) times erf(x). I protested, "That's cheating!"
what's erf(x) ?
@@omeraydindev The "error function," which is used in statistics. It's a scalar multiple of the antiderivative that I asked for.
I feel you, but do you also shout thats cheating if somebody tells you the integral of 1/x is ln?
@@andik70 lnx is the desired result, and is a totally different operation to 1/x. erf(x) is literally just his original input. It would be like if you asked me:
"What is the derivative of y = x^2?"
and I flatly responded:
"d/dx(x^2)"
@@andik70 Fair point, but no. The logarithm function was identified in 1614, a quarter century before Isaac Newton was even born. (A better rejoinder would have been to
"Did you expect a pie to fall out of here?" THAT'S WHAT SHE SAID.
Yessssss...
Nice.
You're the Bob Ross of math
BobTheRossGuy
XD
That's high praise
So damn right xD
Figuring this out on our own was a homework problem in the calculus course I took last semester
Brutal
I would stuck at the very first step
That is a standart homework for multivariable calculus students.
I often deal with Gaussian curves in many dimensions as an electrical engineer with signal processing. The really cool part is that (given that the covariance matrix is diagonal), the pdf is constant over an N-D sphere! I think that is a beautiful way to connect the two concepts.
I think there are some confusing elements here. The y domain does not "become" the theta domain, it's more of a double substitution (x,y)->(r,theta) and maybe explain more why it is that dxdy=r*dr*d(theta) from the jacobian etc.
yeah...that part there lost me
both dxdy and rdrdθ are small changes in area, but in different coordinate systems
i just thought it was θ or arclength scales with r, so dxdy = dr(rdθ)
At least for the rdrd(theta) for now its better to just say don't worry about it until you get to jacobians later on in Calc 3
I did this shit in physics and it's pretty non-rigourous there, but I bet they'd flay me alive the moment I step into Calc 3
Great presentation! We see, again, why Pi is one of "holiest" numbers in mathematics. Cheers
Have a great day!
@Certyfikowany Przewracacz Hulajnóg Elektrycznych 355/113
@@certyfikowanyprzewracaczhu33909/3
@@certyfikowanyprzewracaczhu3390π
As soon as you started talking about using y in the second integral I saw it.... that is so cool
A few comments from a statistician:
Nice walkthrough of the polar coordinates trick. I will link this video for my class!
The Gaussian distribution isn't too common for actual empirical things. For example, raw scores on IQ tests aren't really Gaussian; the tests are transformed to be Gaussian via the probability integral transformation. Heights too aren't Gaussian. Even in a restricted population such as adult females, heights tend to be somewhat skewed. What is usually Gaussian? Sampling distributions of well-behaved estimators. (In a sense, "well-behaved" actually means "has a Gaussian sample distribution in a reasonable sample size". In other words, the Central Limit Theorem holds.) It also works out tolerably well as an error distribution in regression, but that comes after systematic effects have been removed, and a lot of modern statistics involves using the Gaussian as a building block to make better distributions that fit more empirical phenomena.
An even deeper reason the pi shows up is because of Stirling's approximation. The original derivation of the Gaussian was by Abraham de Moivre, who used it to approximate probabilities of the binomial and sought something that would be easier to calculate than the intractable combinatoric terms. That would make a great video, too.
In think in practice the Gaussian integral (aka erf) is actually approximated by a series, although as I recall it's not a Taylor series, but a Hermite polynomials---don't quote me on that. I suspect that in Ye Olde Days they used a power series to do it.
I haven't even learned about integrals in school. Why am I watching this?
Sounds rather interesting, though.. I'll come back when I know about it.
Check 3blue1brown’s calculus playlist if you’re interested, it’s a great introduction to higher math :)
Calculus is really interesting. I was upset that I didn't learn it in high school or my first time in college, so I taught myself the basics. Then I had to take four of the classes when I went back to college.
Basically, a derivative of a function is another function that tells you the "slope" or rate of change of the original function. The derivative of x^2 is 2x, so for any value of x, the function has a slope of double x. You can check that for any value of x by seeing what the slope would be from that to the point where x is very slightly greater. For instance, at x = 2, y = 4, and at x = 2.001, y = 4.004001. If you divide the difference in y by the difference in x, you get extremely close to 4 as the derivative predicts. (The derivative is exact, but since we're going from one value of x to another, our method of checking it isn't exact.)
The antiderivative or indefinite integral is the inverse of the derivative, so just as 2x is the derivative of x^2, x^2 is the integral of 2x. (You usually add a constant for indefinite integrals, but that's not important.) The definite integral finds the area under a function down to y = 0, and you just subtract the indefinite integral with the lower bound of x substituted from the indefinite integral with the upper bound substituted. So if you wanted to find the area under the curve of 2x from 0 to 1, it would be (1)^2 - (0)^2, or just 1. This is easily checked because the function, when graphed, forms a triangle with a width of 1 and a height of 2, and the area of a triangle is 1/2 b h.
If you're curious, they're used a lot in physics for going from one thing, such as distance moved, to the rate change of that thing, velocity in this case; and in biology for population growth; and all kinds of other things.
Maybe this will give you and a few other people what he's trying to find, even if you don't know the rules.
@@chitlitlah Thank you for this explanation. The first part I already knew, though. But this is the pretty much all the knowledge I have about that at the moment.
I did (or rather do) learn that in School. Here in germany we have a different education system; so there is no colledge or something.
And yeah, im interested in that because of physics. Though I'm still in school, I like to learn about much more complicated concepts like the theory of relativity or quantum mechanics.
I learn the integration.it's beautiful
Same and it will be years until i do get taught it but i know integrals well so i will anyway
Im taking integral calculus in college rn and our teacher told us this integral would be the last one we'd integrate at the course, but yesterday she made us demonstrate it comverges. It always impresses me how creative you gotta get to solve some math problems.
Slowly learning claculus and understanding these types of videos better and better is the most satisfying thing ever
I am really really trying to understand this, but I have not taken any form of a Calculus class yet (not even precal) and it is very difficult.
But I will continue watching these videos until I understand. We persevere.
You should watch the series on calculus from 3Blue1Brown if you haven't already. It only covers the very basics, but exceptionally well.
You should definitely either take a course or teach yourself the math that this is based on. Otherwise you don't understand a lot of the motivations for the things he does
Copied and pasted from another reply I just made.
Basically, a derivative of a function is another function that tells you the "slope" or rate of change of the original function. The derivative of x^2 is 2x, so for any value of x, the function has a slope of double x. You can check that for any value of x by seeing what the slope would be from that to the point where x is very slightly greater. For instance, at x = 2, y = 4, and at x = 2.001, y = 4.004001. If you divide the difference in y by the difference in x, you get extremely close to 4 as the derivative predicts. (The derivative is exact, but since we're going from one value of x to another, our method of checking it isn't exact.)
The antiderivative or indefinite integral is the inverse of the derivative, so just as 2x is the derivative of x^2, x^2 is the integral of 2x. (You usually add a constant for indefinite integrals, but that's not important.) The definite integral finds the area under a function down to y = 0, and you just subtract the indefinite integral with the lower bound of x substituted from the indefinite integral with the upper bound substituted. So if you wanted to find the area under the curve of 2x from 0 to 1, it would be (1)^2 - (0)^2, or just 1. This is easily checked because the function, when graphed, forms a triangle with a width of 1 and a height of 2, and the area of a triangle is 1/2 b h.
If you're curious, they're used a lot in physics for going from one thing, such as distance moved, to the rate change of that thing, velocity in this case; and in biology for population growth; and all kinds of other things.
Maybe this will give you and a few other people what he's trying to find, even if you don't know the rules.
@@chitlitlah Thanks a lot for that explanation, it cleared up a lot of things for me as I have not yet been able to find an explanation as easy to understand as yours.
W mindset
This is so beautifully done ✅ I love math even more now !!!
I'm very glad to hear it!
@@BriTheMathGuy math confuses me especially this math cus i don't understand it yet but it's fun
I thought in stats we just let that area equal 1. The chance something will appear under that circle is 1.
The reason the area is 1 in statistics, is that there is an intentional factor with sqrt(pi) in the denominator of the formula that defines the normal distribution curve, in order to force its area to equal 1. And area under the curve has to equal 1, because a probability density function by definition has to add up to 1 over the entire domain. That's due to the fact that it is 100% likely that the variable will be somewhere within the domain of all possibilities.
*Me in my math class at 9 am*
"Calculus is hard"
*me at 3am watching this guys video"
"Yep that seems simple"
Just in time. I have probability and statistics course this sem and I learned about normal distribution
Very cool!
I don't take calc 3 until next semester but everything you said in this video made perfect sense! Been watching your stuff a ton lately and am loving it. Cheers!
Great to hear! Thanks so much and have a nice day!
yep. the only real thing i had to take his word for was the jacobrian constant
Having recently finished calculus in 3D, I started having flashbacks when I saw that polar coordinates could be substituted in
Another Amazing Integral!
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Why does Wolfram Alpha say it's 1/2 * sqrt(pi) ?
@@Polleke123456 If you type in the indefinite integral it might display that (make sure bounds are -infinity to infinity)
*me who is a math nerd and is very confused by bell curves but is still only is in 9th grade and doesn't understand integrals*
@@kyokajiro1808 ah yes. Keep working
@@kyokajiro1808 exact. same. situation.
As a EE and mathematician I love these videos. Let' all count like EE's. 0, 1, 2, e, 3, Pi, 4, 5....
I think you might have double counted a few times with one of your numbers.
@@WindsorMason pi is 4. Fight me :P
@@cmelton6796 normally it's e=π=3, but when you cut particularly big slices it's π=4.
OMG! this is amazing!
Really glad you think so!
Your explanations and examples are very clear and well presented. I suggest, however, that you allow more vertical space when doing the actual computations. Perhaps you could stand "in the middle" and have the problem being worked on on one side of you, and the scratch work on the other?
Keep up the great work!
I really need to know how these glass screen videos are done. Do you actually write backwards? Is there an optical technique involved?
I write normally and flip the video horizontally using video editing software :)
@@BriTheMathGuy thanks. Do you teach math at college level? I like how you explain mathematical concepts. Are you more of a pure or applied mathematician?
@@michaeltamajong4659 I teach at the community college level and I prefer pure math over applied.
@@BriTheMathGuy amazing! I do prefer applied math, but i find pure math concepts so elegant and interesting.
That is simply amazing, I subscribed on the spot.
hey bro make video on partial differentiation and euler's theorem , composite function, jacobian, taylor and maclurin on ane and two variable, multiple integral(double integral,triple integral)
You're a great teacher
You can understand geometrically the origin of that \sqrt{\pi} if you just notice that when you calculate I^2, you are essentially calculating the volume of a solid of revolution (the surface of a half of the gaussian curve rotated around the y axis), hence I^2 is proportional to \pi according with the second Pappus-Guldinus theorem: en.wikipedia.org/wiki/Pappus%27s_centroid_theorem
This was on my multivariable calc exam, now its in my recommended
Take the successive rows of Pascal's Triangle: (1); (1,1); (1,2,1); (1,3,3,1); (1,4,6,4,1)... If you present each row as a bar graph centered on the Y-axis, as the rows progress the result will look more and more like a normal curve!
You should make a video on how Laplace solved this without polar coordinates!
I remember this in a multivariable calculus test. This was the question that stumped me during the test, but I asked a couple people about it and they told me about it. I tried it on my own after the test....and I felt a bit stupid. :P
don't feel stupid my guy i failed a test recently because i didn't use a hyperbolic substitution for in integration question and got completely stuck. it literally said to use a hyperbolic substitution in the question. i retried the question when we got our papers back and solved it in minutes we all do silly stuff sometimes
Couldn’t you make the theta boundary from 0 to pi, because the curve has no area under the x axis
No, that is irrelevant. The range of the function has nothing to do with the y-coordinate.
Didn't understand a lot since i'm still learning calculus but still fascinating.
Explained clearly
Making maths more accessible to everyone good job 👍
Glad you liked it
Flipping the image is pure genius
The u substitution is not needed, just multiply by -2 inside of the integral and -1/2 outside of it.
If I'm reading what you're saying correctly: You'd still be doing a u-substitution to evaluate the integral afterwards, you would just be leaving the work unwritten. Similar to when someone has two binomials multiplied together and didn't actually show using the distributive property; instead going straight to the the expanded and sinplified form. They're equal, but how you get to the end result is still using that step, stated or not.
It surprises me how many times PI can appear, in a different equation than those related to circles and spheres.
If you generalize, the integral of t^(-x^2) is sqrt(ln(t)*pi). Cute.
This is one of the most beautiful demonstrations in math.
Glad you think so!
pretty cool. I was able to find that the integral between A and B, both A and B being large numbers, of (e^(-(x^2)/2pi))/rt2pi)^2 was equal to pi
It's easier to write it as (e^1/π)^-x²
Any large numbers? Such as A = a googol and B= a googol + 1/(a googol)
Every time Pi shows up somewhere unexpected: That little bastard!
I should point out that the IQ is defined by its distribution. Pointing it out as such would be like circular reasoning.
another thing is get the integral and then multiply by pi. its sorta like the disc method
Well done! I would appreciate it if you clarify this small issue. Maybe you find out the absolute value of I, do you have to prove that I is equal to absolute value of itself before ending the video. You can improve your technique by using Laplace transform and would be solved faster and shorter.
As soon as I saw x squared plus y squared I expected pi to show up.
Random distributions in real life tend to approximate the normal distribution because things in real life tend to be caused by the sum of many random things. The distribution of the sum of many random things approximates the normal distribution.
A somewhat similar curve is 1/(1+x^2), which has the area pi.
Also sech(x) integrated over R has area π. It's because sech(x)=1/cosh(x)=cosh(x)/cosh²(x)=cosh(x)/(1+sinh²(x)) which takes the form du/(1+u²) and sinh(±∞)=±∞
@@xinpingdonohoe3978 That's real nice tho. You should suggest an integral based on that to professor Penn! :D
Thanks for a great video again!
Glad you enjoyed it!
Fun fact:
derivative of arctan (radians) of x
is equal to
1/(x^2 + 1)
I love how this problem utilizes the most advanced forms of single-variable calculus, even showing a perfect transition into multivariable Calculus viz-a-viz the multiplication of two single integrals.
Calculus is so damned sexy.
When we use the normal distribution on quality, the formula is actually more complicated. Something to the effect of 1 ÷ [sigma * sqrt(2pi)] * e ^ [-1/2 * (x - mu)^2 / sigma^2]
Is e ^ (-x^2) somewhat more fundamental?
@@tddupaid That makes a lot of sense, thanks !
I’m wondering do you write backwards on glass to present the video?
It is a circle it just melted
😂
I'm taking calk 2. I feel like I learned something but I'm not shoure what it is.
I'm having so many flashbacks to calculus 3. Glad I graduated.
The math there is currently over my head, but I do understand the amazing unexpected result.
You indirectly used Gaussian Integral to solve this. The calculation is actually much simpler when you know the Gauss Error Function lol.
great vid man
I only recently noticed. How do you write in reverse so fluidly?!
How is this type of recording done? Are you using some glass surface?
I think it's Verisatium who said that pi really shouldn't be associated with circle because it appears on far many fields.
Grant Sanderson says otherwise.
There's always a circle hiding somewhere!
@@blackholedividedbyzero I know, but I actually think Grant is wrong here. In modern mathematics, π is not actually defined in terms of circles. It is defined in terms of functions, much how e is defined. And I know Grant has a series of videos explaining how π shows up in certain formulae from a circle. But it is not the case that this is true for every mathematical formula out there.
the most impressive thing here is him being able to write all of this in reverse, he looks to be writing on a glass, writing to the left lol
Thank you so much.👌
I expected you to explain how it is connected to a circle the way a different video compared Zeta of 2 to a circle. Instead, I saw an integral I'd seen before, but that's okay!
If you plot this out on a curve you’ll get a bell curve and the formula to express a bell curve contains pie so with some math you can probably prove this
f(x)=e^-x^2 p - 1 Co-sufficient formula
**shows sine wave** This isn't a circle, why is pi here?
A whole semester of calc 3 and NOW I UNDERSTAND DAMNIT
so whats the actual answer to the title? i mean it seems to be just coming from the polar coordinates switch?? is it that simple or some deeper things ?
I don't understand why the area isn't infinite itself, because the integral goes from negative to positive infinity and the y-values never go to zero. What am I missing here?
A sequence can have a limit, without ever being a constant.
how about x^x = e^(ln(x^x)) = e^(x * ln(x))
lim x ln(x) = 0 when x aproaches 0 so lim x^x = lim e^(x * ln(x)) = e^0 = 1 when x aproaches 0
i don't know is it correct guys
How do you write backwards? Do you write normally on your side and then flip the video?
Yes :)
I didn’t understand that polar coordinate system substitution
Probability means time. For an outcome or events management systems.
Probability means desired result over all possible results.
It's an improper integral you have to use the integral convergence test to check the integral exists before you can use this method
Isn't the integral of e^-x^2 just -1/2x times that?
That was humbling. How in the world did somebody figure out the first step? Maybe this is just a standard trick with the idea of preparing to switch to polar coordinates? Still from my low level math understanding that first step looks like it came from some alien civilization.
I'm not sure what the actual reasoning would have been, but I could imagine sort of working backwards from the end. We know we could integrate this easily if it had an extra factor of x. But how would we transform this into a form with that extra factor? Well, you get that with the radial integral in polar coordinates. But this is a one-dimensional integral--how to relate it to an integral in polar coordinates? Well, if you have an integral in x times a similar integral in y, the exponents add and since the exponent has x^2 in it, that bit becomes x^2 + y^2 and suddenly everything looks very Pythagorean...
This is one of those standard tricks physicists all know because physicists deal with Gaussians so much. I remember never being able to remember the precise formula but knowing I could re-derive it if I needed it.
0:19 The distribution of IQ is not really symmetrical. That would produce a skewed Gaussian curve in reality, unless of course you take log of IQ to force it into symmetry.
I laughed out loud when I saw this title. I am learning Digital Signal Processing right now for my masters program and my undergrad degree was Sociology so needless to say.. I have never used Pi in this way before. 😂
Thank you!!!
Maybe a stupid question but why can't you use u supstitution where u=x² or -x²
Math is awesome and weird at the same time. So many crazy outcomes to problems
Odd coincidence that I happened to notice \Gamma(1/2) = sqrt(\pi) right after watching this video while working on something unrelated. That's the curve e^(-x)/sqrt(x) integrated from x=0 to \infty with respect to x.
This is because of the transformation sqrt(x) |-> x in the integrand.
can anyone please explain how he is writing? It is blowing my mind while making me feel weird
From what I've read on other comments, I believe he's likely writing normally on clear glass using his right hand, and then the entire video is inverted (hence he looks like he's writing with his left hand).
True ^^
@@BriTheMathGuy Do you use a mirror to invert the image optically? Or do you use software to invert the image digitally?
"It's not a circle" - he says before using the definition of a circle as a trick to integrate it.
09:59 - you should've put the square root under the integral, right?
i don't know what calculus is or how it works but this is cool
But how on earth can dx*dy equal to r*dr*dtheta? I mean, in polar coordinats x = r cos t and y = r sin t (t stands for theta). If we differentiate them, we get dx = dr cos t - r sin t dt and dy = dr sin t + r cos t dt. And from that dx dy = dr^2 cos t sin t - dt^2 r^2 cos t sin t + dr dt r (cos^2 t - sin^2 t). I know I am wrong somewhere, but please explain this for me!
r is the Jacobian of the rectangular -> polar conversion. This is determined though some gross matrix determinants and partial differentiation.
However, it's a lot easier to understand intuitively if you think about it visually. A dx*dy region is a rectangle with sides parallel to the axes of infinitesimal lengths dx and dy. A polar "rectangle" is curved, like a thick arc. Arc length is equal to radius * radians, so while the thickness is just dr, the length is actually r*dθ, radius * radians. So, you have r*dθ*dr as the infinitesimal region of area you integrate.
Also a quick note, your partial differentiation is incorrect (it seems like you were trying to sum implicit derivatives or something). If x = r*cosθ and y=r*sinθ then dx/dr = cosθ and dy/dr = sinθ. When differentiating with respect to r, θ is just a constant. On the other hand, dx/dθ = -r*sinθ and dy/dθ = r*cosθ. This is actually part of how the Jacobian is found mathematically. The Jacobian 𝔍 is equal to the determinant of a matrix with rows of the starting coordinate system differentiated by a variable in the target coordinate system according to column. So in this case, 𝔍 = (dx/dr)(dy/dθ) - (dy/dθ)(dx/dr) = r(cos²θ + sin²θ) = *r* which means that dx*dy = r*dr*dθ.
Ah great stuff!!!
Glad you like it!
Is this guy writing backwards or writing on a partially keyed surface and flipping it in post? Either way it’s cool
I came searching for copper and I found gold.
=2(1/2)! 😉
Got a sub from my side, respect!
Thanks so much!
bruh i just realised how people can do these type of math vids where they draw on a glass while facing the camera, but i think the trick is that they edit the video so that it is mirrored maybe?
we could also use gamma function to calculate the standard integral but alright