This isn't a Circle - Why is Pi here? 

See this is why i read the comments.

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.

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.

Glad you liked it!

Nice.

BobTheRossGuy

XD

That's high praise

So damn right xD

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

Brutal

I would stuck at the very first step

That is a standart homework for multivariable calculus students.

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

Have a great day!

@Certyfikowany Przewracacz Hulajnóg Elektrycznych 355/113

​@@certyfikowanyprzewracaczhu33909/3

​@@certyfikowanyprzewracaczhu3390π

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.

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

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

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

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.

Very cool!

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.

Glad you enjoyed it!

Glad you liked it

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

Really glad you think so!

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

Glad you think so!

Yes :)

Glad you like it!

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.

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.

Thanks so much!

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)

@@tddupaid That makes a lot of sense, thanks !

Probability means desired result over all possible results.

No, that is irrelevant. The range of the function has nothing to do with the y-coordinate.

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

No. Technically speaking, you cannot "use degrees." That is unfortunately not how these functions work. I find it thoroughly problematic that degrees and radians are thought of as units of measurements you can convert between. It is just mathematically and metrologically incorrect.

This is because of the transformation sqrt(x) |-> x in the integrand.

What do you mean "how do you justify it"? You can just do it arbitrarily. That is how these things work. You can multiply a real number by a real number. That is still a real number. The integral is a real number. So it follow you can multiply the integral by itself. There is really nothing to justify.

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.

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.

Glad you think so!

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.

That's the Jacobian. Whenever we make a change in coordinates, we have to account for how little changes in the input affect little changes in the output. We compute this by taking the determinant of the Jacobian matrix, which in this case simplifies to r, therefore dx dy = r dr dθ

@@chielvooijs2689 thanks, dude

Video editing :)

Thank you! Cheers!

😂