mmm crunchy symbols edit: holy shit the idea of creating an arbitrary probability distribution was something I messed with just a few days ago this is creepy

There is a prime tick missing, in the upper formula at 4:19

urf... 9:05 you got me with that beautiful track, love it

There will never be a video about the most overlooked subject of a branch of mathematics; for the moment a video is made about it it is now by definition no longer the most overlooked as there will always be a subject without a video.

If you ever make a series on slightly more advanced calculus, ill definitely watch it

Thank you, this was a very clear and concise explanation of these ideas.I hope that high school calculus teachers see this and adopt it in their pedagogy. This view of integration by parts and implicit differentiation were deemphasized when I was going through BC Calculus due to our focus on teaching to the test.

This channel is so underated! Having geometric interpretations and visualizations make the concepts so much easier to understand, precisely the way math should be learned. Unfortunately, most teachers don't provide this kind of explaination as most of them don't even understand these interpretations themselves, only with memorization. Keep up the good work!

YES LOOK FORWARD TO NEXT VIDEO

Excellent video!

This concept is core to Spivak's trig and calculus on trig function. Using Inverse functions he bypasses having to prove Lim x goes to 0 of sin(x)/x is 1 that most other calculus books need to prove derivatives of trig functions.

Amazing explanation. I never properly understood this concept before I watched this video. Thank you for sharing.

Nice! Really hope you cover the probability transform.

omg new upload :DDDdddd

very cool video I will be following the rest of your videos ^_^

0:08 There are 3 types of functions: 1. One-to-one function: These functions are STRICTLY increasing or decreasing throughout its domain and has no first-order critical points besides the endpoints if they exist. The inverse function is defined throughout its entire domain. 2. Piecewise inverse function: These functions are not one-to-one, but an inverse exists for a PORTION of the function. These functions have at least one first-order critical point. 3. Singular function: The term "singular" in mathematics refers to anything that are not invertible. Because of that, a singular function is a function that doesn't have an inverse. These functions exist as annihilation functions, where all values in its domain get "annihilated" to a single value as its range. Because of that, a function is singular if and only if (iff) its derivative is 0 throughout its domain, similar to how a singular matrix has a determinant of 0. Let's say you have this function f(x)=1. This means that for every value of x inserted, the function returns 1 every time. Therefore, this graph takes the form of a horizontal line at y=1. If we say we want to invert f(x), we need to switch the domains and ranges, and since this function annihilates all values of x and returns 1, meaning the domain of the function is all numbers, but the range of the function is 1, if an inverse exists, it will then map 1 for the domain and, uh-oh, we have a problem here. A function operates like a time-distance relation: You can be in the same position at 2 different Planck Times, but you, at the same Planck Time, cannot exist in 2 different locations. If f(x)=1, f^-1(x) will have f(n!=1) not making sense, as f(x) cannot not equal 1. However, f^-1(1) maps out all real numbers, meaning that it wouldn't be a function, as at the same time the function simultaneously exists everywhere. This means there doesn't exist any continuous piece of this function that makes sense; it only makes sense for that one point, and even for that point, it violates the basic rules for functions. Graphically, to invert a function, you need to reflect it over the line f(x)=x, and reflecting a horizontal line gives you a vertical line, which violates the vertical line test. Finally, using singular functions, we are able to introduce a concept called indeterminate forms. As you can see, for the function f(x)=1, f^-1(1) gives an indeterminate form due to it being a vertical line at x=1. Given a singular function f(x), f^-1(f^(x)) is indeterminate. Even though applying the inverse function after applying the function should result in the identity function f(x), for annihilation functions, it all has lots of paradoxes. An indeterminate form comes from basic algebra, where a declared variable has no value assigned to it, as variables are MEANT to be indeterminate until you assign a certain value of it or a transformation of it. For example, x is indeterminate. With n variables, you need n equations to make the values determinate. Some equations are degenerate and continue to evaluate an indeterminate form, such as x=x, as it is true for all x. Using that, we can prove the 11 indeterminate forms are indeterminate using these formulas: 0x = 0 for all Aleph-Null x, so 0 / 0 is indeterminate. 0 * infinity and infinity / infinity are variants of 0 / 0, as 0^-1 = infinity. infinity + x = infinity for all Aleph-Null x, so infinity - infinity is indeterminate. The infinitieth root of x is 1 for all Aleph-Null x, so 1^infinity is indeterminate. 1^x = 1 for all Aleph-Null x, so log_1(1) is indeterminate. 0^x is either 0, 1, or infinity, so log_0(0), log)_0(infinity), log_infinity(0), and log_infinity(infinity) are all indeterminate. Using these equations, we can prove which forms are NOT indeterminate, to see how many indeterminate forms are there. Calculi operate on indeterminate forms, as a derivative is equal to 0 / 0, and integrals are 0*infinity, so we can see how many calculi are there. For example, the product/geometric calculus consists of the product derivative, which is limit as h goes to 1 of log_h(xh/x), which is log_1(1), and the product integral, which is 1^infinity. Infinity is a fixed-point of X = X+1, so when dealing arithmetic with infinity, we use a variable x to denote the fixed-point, which is useful on checking if forms are indeterminate. Case 1: 0*infinity, x=0*infinity=0+0+0+0+0+...+0+0+0+0=0+(0+0+0+0+...+0+0+0+0), so x = x+0, and x = x. Indeterminate. Case 2: 1^infinity, x = 1^infinity = 1*1*1*1*1*...1*1*1*1*1=1*(1*1*1*1*...1*1*1*1*1), so x = 1*x, and x = x. Indeterminate. Case 3: 1^^infinity, x = 1^^infinity=1^1^1^1^1...1^1^1^1^1=1^(1^1^1^1^1...1^1^1^1^1), so x = 1^x. This only is true if x=1, so 1^^infinity is NOT indeterminate. However, if x=infinity, it equals an indeterminate form, so it is considered indeterminate in the complex world. Case 4: 0^infinity, x = 0^infinity = 0*0*0*0*0*...0*0*0*0*0=0*(0*0*0*0*...0*0*0*0*0), so x = 0*x, and x = 0. Not indeterminate in the real world. Case 5: 0^^infinity, x = 0^^infinity = 0^0^0^0^0...0^0^0^0^0=0^(0^0^0^0^0...0^0^0^0^0), so x = 0^x, but NO SOLUTIONS EXIST. This is because 0^^x = 1 if x is even, and 0 if x is odd. Same thing goes with 0^^^x, 0^^^^x, and 0{n}x. In Boolean algebra, if a function has no solutions, it evaluates to false. If it is an indeterminate form, it evaluates to true. If it is only true for a value or a finite set of values, it returns that value or set of values. Therefore, 0/0=0*infinity=infinity/infinity=infinity-infinity=infinity^0=1^infinity=log_1(1)=log_0(0)=log_infinity(0)=log_0(infinity)=log_infinity(infinity)=True

Nice!

Thanks. Great video. I look forward to checking out your channel. Subscribed. Cheers ...

Youngs inequality; One of my favorites.

Stopped too soon! I want to see how you go from your desired pdf to the prng algo to sample it! Great video

good job

Finally!

I’m not sure if chi distribution is in the standard libraries, but chi^2 definitely is. Generating Boltzman variables would simply involve taking the root if chi^2 variables. (Intuitively, that is how one derives the Boltzman distirubtion anyways as the components of the velocities are i.i.d. normal.)

Very nice, though I'm kinda confused on the steps at/before 8:14. Are we just rearranging the above equation to get an expression for the red term? And are we also setting alpha to 0 so the alpha coefficient disappears? It's a very nice presentation, though i Thin using maybe a couple more colors even might help keep things straight for me. Personally i once saw a video that utilized a different color for every term (like 6) and it was a revelation. What would have been very difficult to follow became very easy, i knew exactly what was going where. With so much green and red in equations, in bounds, and as areas in a graph all on the same screen i think my brain is getting kinda lost. That said, i love you for making this video. Please keep making more as long as it makes you happy. I can tell you've put a LOT of work into this and it shows. Thank you so much

Very good. You should first test if the function (which you want to integrate the inverse of) indeed has an inverse (within its stated domain)! How can you do this? e.g 1 Consider f(x) = x^3 - 3x, does it have an inverse function between x = 2 to x = 3? Differentiate, f'(x) = 3x^2 - 3 = 3(x^2 -1), Solve for any f'(x) = 0, stationary points at x= 1, -1, so SPs here are outwith stated domain (also no other critical points) so inverse function exists. Should the domain be chosen so that say, from x=0 to x=2 then the inverse function would not exist. e.g 2 Find def. integral of inverse function of (x^3+x+1) between x=1 to x=3? Before you go ahead and try to do this you should test if indeed the inverse f^(-1)(x) exists? f'(x) = 3x^2 +1 which is irreducible and has no stationary points, so yes it exists everywhere on ℝ. Hope that helps! 🥺

Just a small tip, when you derive the formula for the derivative of an inverse function, you're using the chain rule so you're implicitly assuming that the inverse is differentiable, which is not immediately obvious

and what about mentioning the symmetry versus y = x ?

funnily enough the inverse of sin(x)+x when placed in the cosine fucntion leads to a cycloid curve

The definite integral by parts form is intuitive if you draw a rectangle on (a,f(a)), (b,f(b)) - the area of the rectangle is (b-a)(f(b)-f(a)), that is, the first term of integration by parts. The area above the curve in the rectangle is the 2nd term in integration by parts, and the formula is obvious if you flip the axes. The integral we are interested is then obviously the area of the rectangle, minus the area above the curve in the rectangle. The indefinite integral giving rise to this is then obvious as well. There's no need to deal with (0, f(0) at all, and it just clutters up and confuses the intuition.

Yo the pokemon background music 🎶

explainin why finding inverses is hard while my brain is just thinking of toothless dancing because of the music lmao

What about Donald in Mathmagic Land? Seriously, it influenced my eventual decision to study mathematics.

4:15 Formula is missing a derivative symbol on f, viz RHS denominator should be f'(f^{-1}(x)), not f(f^{-1}(x)). Otherwise, of course, the function and inverse would simply cancel out.

soon

Wonderful video! Is there any chance you can use colours other than red and green to distinguish areas and algebra? Those of us with the most common type of colour blindness can't see the different (blue and yellow is good, or one solid and one hatched). When you're talking about manipulating the red and green parts of the algebra, I don't know what you mean until you do it and then I'm trying to play catch up with what I missed

4:12 that's a mistake in the derivative of inverse. There should be a ' next to the first f

Is there a term for a function that does not include an integral in its definition? In other words, you have an integral where there is no simpler way to express the term?

so if there's an ode that's like f''=-1/(2f) then it's related to urf

I thought a function has to be one-to-one *and* surjective to have an inverse, not just one-to-one?

The red equation at 4: 28 appears to have an error. Differentiating the right side yields. x/ [f’(f inv (x))] + f inv (x) - x which does not equal integrand f inv(x) on left hand side of equation.

Slide at 4:20 seems to be incorrect. Denominator should be f'(...), instead of f(...), no? (If it's hard to read, it's f-prime

Amazing video! I actually have been looking for someone to collaborate with when it comes to explaining mathematical topics like these. Would you perhaps be interested?

4:19 I know I haven’t taken calculus yet, but why does the derivative of an inverse seem to just be 1/x? I know that most of the time, f(f^-1(x)) is just x, but I know it’s not that simple, did I miss something obvious?

It’s fun to say urf

At 2:52, you explicitly said "x to a rational power", but then showed x^x^log(x), which isn't exactly a rational composition? I assume you meant real power? Or am I missing something?

what do you mean the inverse of e^x is a function?

I used to watch Urf.

The easiest way to get a standard distribution with an RNG is just to generate two or more and add them together.

4:48 the derivative of logx is 1/(x*ln10)

Lagrange put above Fourier?

4:08 Surely the derivative of the factorial function is defined. We'd need a continuous version of the factorial function to even have a derivative, and thus make that symbol meaningful. And the continuos factorial function is defined using an integral of elementary functions. So surely, by the Fundamental Theorem of Calculus, the derivative would be composed of elementary functions. Where did I go wrong?

the way I see it, it's totally fine to not teach quotient and inverse method for integrals because they boil down to special cases of product and u sub.

3:12 you mean X^5-X-1

You can rotate it 90 degrees

Literally just got a question where the integral asked was of the form f(x) + f^-1(x)

wdym e^x is hiding in the derivative of lnx

You said *most* functions are not one to one. How do you prove that? Why is there no surjection from family of one to one functions; to the family of all functions?

0:11 isn’t it just multivalued function?

I don't understand shit but shit looks cool maybe one day I will understand of this

Oh, you missed the apostrophe, got it

is that just the same as f(y)dy

erf urf

Go and watch Herbert gross lecture of mit

hi

ENGINEER MENTIONED

That music really isn't helping me to concentrate 😯

Rather arrogant of you to conclude this is the most overlooked concept in Calculus. The result is an immediate consequence of d/dx(f(f^-1(x)) = x) and the chain rule. I don't know of a colleague who did not use this result to find the derivatives of the inverse trig functions. Oh well, its the norm to exaggerate/lie to get page views.

sneaky jojo refrence

Stop copying, I had found the demo of the derivative of log(x) before you

what's most difficult to get is how much you are just showing off or really intending to convey a useful insight...