How to do two (or more) integrals with just one
ฝัง
- เผยแพร่เมื่อ 5 มิ.ย. 2024
- Is there a way to turn multiple, repeated integrals into just a single integral? Meaning, if you, say, wanted to find the second antiderivative of 6x, is there a way to compute it all in one step just using a single integral? Turns out there is! In fact, any number of repeated antiderivatives can be compressed into just a single integral expression. How is that possible? And what does that single integral expression look like?
My old video about Double Integrals: • How to Set Up Double I...
The sequel to this video about fractional calculus: • What Lies Between a Fu...
A really nice video that derives the gamma function from scratch:
• How to Take the Factor...
=Chapters=
0:00 - Intro
0:51 - Why Compress Integrals?
2:29 - Analyzing the Problem
3:46 - Visualizing a 2-Fold Integral
5:25 - Deriving the Formula
10:56 - Testing the Formula
12:14 - How Is This Not Impossible?
13:49 - Higher-Order Integrals
15:22 - Application to Numerical Integrals
16:25 - The Gamma Function
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For more on applying Cauchy's Formula to numerical integration, see this paper:
Tvrdá, Katarína & Minárová, Mária. (2018). "Computation of Definite Integral Over Repeated Integral." Tatra Mountains Mathematical Publications. 72. 141-154. 10.2478/tmmp-2018-0026.
www.researchgate.net/publicat...
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The animations in this video were mostly made with a homemade Python library called "Morpho".
I consider it a pretty amateurish tool, but if you want to play with it, you can find it here:
github.com/morpho-matters/mor...
The sequel is here:
th-cam.com/video/2dwQUUDt5Is/w-d-xo.html
your integration is limited by 1 variable x.
In control theory we have a more generalize way to do any integration in 1 step.
The trick is to integrate a vector.
for example dx^2 (int int aceleration) can be seen as
x1=x(position), x2=(x1)dot (velocity)
so int [x1dot, x2dot]
is 1 integral, but internally you are doing a second one over x.
if you want an n degree integral, you would need to integrate an n dimensional vector.
Yes
Incredible! Calculus classes often teach a limited view of integrals, only thinking of them as antiderivatives. But now I realize the implications of just how specific the fundamental theorem of calculus is written: Only integrals that fit that pattern are antiderivatives, and there are way more than just those integrals out there!
Yes. An integral is a pretty general tool, representing much more than areas under a graph. And that's cool!
@@dexter2392 wait so what the hell is an integral
@@Xnoob545 idk if its right to say nuch more than areas under the curve, but integrals are used for more things than just that. Integrals can act as sums, integrals can be used to represent functions, there are integral transforms, there is line integral and flux integral and both do not represent area under curve. but Integral more generally, is just a sum or continuous sum.
@@randomguy2169 yeah, area under curves is just a visual representation of integrals, albeit a very intuitive one for beginner at that
@@randomguy2169 To get technical, the integral is a linear-transformation of a continuous function. Even more is true: if you allow integrals with respect to measures, you get all possible linear-transformations of continuous functions.
The animation quality is incredible. For example the transition from a 2D to 3D view at 6:34. So smooth!
Love the snake integral beasts vs derivative peons. Classic visuals not blatantly copying 3blue one brown and staying original.
And at
7:40 - 7:57
8:25
9:45
Pretty nice visuals
There is nothing incredible about any of this.
This is just made in 3D from the start, she's just changing the camera orientation.
@@xl000 he made this animation library by himself.
So there's the impressive bit
@@GetPsyched6 what programming languages do you know ?
This was absolutely incredible. The intriguing and seemingly nonsensical question at the beginning (especially from the perspective of someone who only knows basic calculus), the pacing and 3d animation to visualize the intuition behind taking a double integral, and the teaser for the idea of in-between integrals/derivatives. This was mind-blowing to watch and appreciate, both from the perspective of a learner and an aspiring teacher.
1:40 Numerically, derivatives are notoriously difficult because computing them involves subtractive cancellation. Integration is much more well-behaved from a numerical standpoint precisely because it only requires summation.
Very funny.
@@zanekaminski this is actually true as you can’t truly animate the 8th derivative of y = (1/x)(sin(x+p)) where p is a value/slider, I tried it on my machine (16 gigs RAM, R3050, RY7) and it only animates 2-3 frames per second, I tried this in desmos. My machine can otherwise run elements with upto 10,000 elements quite easily (at about 40 Fps) and 1,000 elements effortlessly (at 144Fps!!).
@@creepervideo1106 Desmos is not a very well optimized application for that purpose, and cannot take full advantage of your hardware. You absolutely can animate that derivative with software more capable using the hardware available to it and more able to analyze the problem and find shortcuts. For example, Desmos actually calculates every single intermediate derivative numerically for every x value it draws every time the animation parameter changes, but software designed for that purpose would likely be able to symbolically solve for the 8th derivative of y = (1/x)(sin(x+p)), giving it a much faster function to recompute when p changes.
@@Howtheheckarehandleswit Thanks for the tip!
@Drew Duncan is still basically correct, it's a problem of relative precision when the curve values are much larger than the difference between them and accuracy is lost due to computational rounding - though I would think the solution to this is simply to translate the entire curve to put the point being calculated at the origin so that maximum accuracy can be maintained...
Awesome!
This formula felt like black magic when I first saw it (and it still does), but it feels a lot less mysterious now that I can see such a straightforward derivation of the n=2 case.
Its him
you could do five or six integrals, or... just one.
i dont need more, they disappoint me
Derivative Compression, the "n"th derivative of a function, can be expressed for some nice functions as an extension of cauchy's integral formula in complex analysis.
this result is highly related to the residue theorem and, consequentially, this yields no (simple) results in the field of fractional calculus, as inputting a fractional n changes the pole in the denominator of the integral to a branch cut, which is not easy (or often even possible) to evaluate. this function also only returns the value at a point, not a function over the entire complex plane.
i understood like 3 of those words
Do you know of a source that explains this in more detail?
@@dj_laundry_list Complex analisis, complex calculus (calculus IV also works) and any type of advanced calculus courses, I just viewed las month on a mathematical methods for geophycisists so it was sparse but show how the Cauchy integration formula works and it's used (the course was more centered on series as computing is fundamental in the field)
ive never seen a person do a math cliffhanger lol
me neither lol
So underrated! You explain things so clearly and present a very intuitive visual interpretation. You deserve so many more subs.
I found it interesting, that the resulting formula is similar to convolution of x^n with f(x). After thinking about it it makes sense though and gives another way of deriving the formula:
Using Laplace transforms we can transform a function from the time domain into the frequency domain, using s as the new variable. An integration in the time domain shows up as a multiplication with 1/s in the frequency domain, so double integration becomes 1/s^2 etc.
Transforming it back we can however use the fact, that multiplication in the frequency domain becomes a convolution in the time domain.
And what does 1/s^2 correspond to? It corresponds to x. 1/s^n corresponds to 1/(n-1)! x^(n-1).
Plugging this into the formula gives the above result. It's quite nice how in mathematics all roads lead to the same spot.
I wonder if this works for compressing multi derivatives. You'd just need values for s^n in terms of x. Ah dang, that part seems to just cycle back to derivatives of the delta distribution, which means that EVALUATING the multi-derivative in practice means doing all the derivatives individually, just as for the compressed multi-integral.
And my intuition tells me that numerical evaluation isn't simplified either.
Congratulations! You have just reinvented Sobolev's spaces of fractional derivatives!.. :)
This is really interesting. Thanks for sharing. I think this makes the entire concept fit in my tiny brain better
frequency domain is fourier
s is complex domain bruv
The resulting formula isn't like the convolution with the function t, it is exactly that. The video is just visualizing the Laplace transform. This property is why the control theorists love using Laplace transforms.
In practice, we use this technique in reverse to replace an operator norm with a double integral. Also, (16:19) if you're content with working with complex analytic functions, the Bergman kernel allows you to write a derivative as an integral. You can write any power of a derivative as an integral also. I agree though that integrals are very powerful.
Omg, half integrals? Fractional integrals? Can't wait to get the π's ∫ of a function
Outstanding video. Really opens up creative thought at a high-school calculus level. Wish I had this sort of direction when I was learning. Instant subscription and can't wait to see what's next.
This video is amazingly painful.
As a high school student bored with the basic calculus I had been doing in class (volumes of revolution and arc length etc.) I began to play around with trying integrate the volumes of other solids.
I eventually figured out how to tie two integrals together and find the volumes of lots more shapes - like ellipsoids.
It’s really painful to see just how close I was to this amazing formula! I wish I had obtained it myself…
Great video and great animations. Well done!
you were "close" to a cauchy formula as a high school student! thats insane. You should be proud of yourself
@@user-zp3xc4to1t thanks. In the little experience I have, deriving or proving a result is often the easy part. Asking the right question and seeing something mathematically that you don’t already know exists is the much harder part.
If you're doing anything even close to this level in HS, there is absolutely NOTHING to be ashamed of & even some pride is in order. In fact, I'd say we can already expect great things from you.
Well done! The animation at 6:30 that change the 2D representation to 3D is very smooth!
Just took calculus 3 last semester, this blew my mind. Well done, perfect video
Most excellent! I've wanted to see this for so long and was frustrated about it, ever since seeing the formula in the first pages of a text on integral equations. Thank You so much!
I just cannot put in words much this helped me understand why limit points for double integrals are the way they are. Just brilliant stuff.
Wow! Just great educational content! Keep making math learning more interesting & engaging!
👏🏼👏🏼👏🏼 Very well done.
You've made a pretty abstract concept very accessible and easy to understand.
Wooow!! This channel is so underrated. I really enjoy your content, Thanks a lot.
Beautifully explained! Thank you!
This was amazing. It opened up a new horizon for me. Thank You.
That was really cool, it was like seeing a u-substitution but in a perfect 3D form. Or like he turned the u-sub into an operator function, again, crazy cool.
It's like this guy understands integration at a level higher than Riemann or Leibniz, well played. Like a version of super calculus?
How do you not have 100 times as many subscribers?
Your content is of the best math on TH-cam!
Legendary explanation I have waited for to explain in visual terms double integrals. New insights that bring nostalgia of first learning integral calculus and the FTC itself. I am giving a Subscribe and thank you.
Yay a new video from this math youtuber!
Keep up the good work
This was fantastic! Keep it up!
Really outstanding video, every question I had as I was watching you answered two seconds later
A lot of this stuff appears in physics, especially when we're dealing with phase spaces where you have to expand out momentums and positions (could be in terms of hyperspheres, hypercubes, ..., etc). They also appear when you do higher dimensional Fourier transforms, can occur in special relativity too if your taking account of the fourth dimension (which is usually time), and as well as dealing with characteristic functions. However, I have to say that in physics we write our integrals as ∫d⁴x *which is a neat way of condensing it), and then it comes out as ∫∫∫∫dxdx'dx''dx'''. In terms of phase space where p is the momentum and r is the position, we have ∫d³p∫d³r = ∫∫∫dpdp'dp''∫∫∫drdr'dr'' = ∫∫∫∫∫∫dpdp'dp''drdr'dr'' (which is a six-fold integral). To take it further you can also write ∫∫∫∫∫∫dpdp'dp''drdr'dr'' as one integral ∫dpdp'dp''drdr'dr'', but have many integrals appear out looks nice and aesthetic. Sometimes the integrals derivatives are written with subscript x_1 and so on. These types of integrals appear in quantum optics (where you deal with a lot of things in phase space), quantum field theory, quantum mechanics, and statistical mechanics.
I was going to ask about that last part just before it apeared, great introduction to fractional derivatives and integrals
This was beautiful stuff - blew me away on a Saturday night!
Thank you for making calculus seems so easy and appealing
Nice cliffhanger at the end ;) waiting for a sequel!
Wow, the visuals are amazing. I literally said "slice it the other way" out loud at 6:55, just as you intended!
Awesome video! Thank you!
I’m curious, can you prove Cauchy’s formula for repeated integration by using a 2-to-1 integral formula repeatedly?
So, something like proof by induction. Or does it give a different uglier result?
actually the 'classical' proof is done by induction :) and yes, imo it's a messy and ugly task.
Also, if you want to take this whole concept of repeated integration to a continuous case with, say, r (real) times of iterated integration instead of n (natural) times of iterated integration, you'll find out the Gel'fand-Shilov function convoluted with the f(t) function.
Edit: references
th-cam.com/video/Cu53TKl8atg/w-d-xo.html
Solved exercises in fractional calculus - Edmundo Capelas de Oliveira. p.175
Try to read my comment up where I described how to get the general formula without induction, just from the first principles...
Fun fact: to integrate (t-x)(f(x)), unless f(x) is a polynomial or some easy function, we would need to integrate x*f(x) using integration by parts, which requires us to find the double integral of f(x)
oooh another really fun derivation from this is the subfactorial complex extension, making use of the subfactorial property that is most similar to normal factorials (!n=n*!(n-1)+(-1)^n) you just have to modify the formula slightly to change the nonderivative term of the partial integration to match this property instead of the factorial "n!=n*(n-1)!" and then it turns out you get complex numbers in return (makes sense since the formula contains -1^(z))
Very fascinating!
Very interesting!
I like it how you used 2*Pi, Pi, sqrt(2), Pi/2 and e and so one at 13:11.
Amazing video and visualizations
Thankyou! I learned a lot! Very good!
The music and visuals are giving me nostalgia
Great video, keep going!
Just beautiful and poetic
Nice video, thank you !
Looks like this is really just replacing the "nested" integrals with other integrals that just happen to have a well known form (areas of rectangles, rectangular prisms, etc.). Reminds me of a video about the circumference of an ellipse having no closed-form constant w.r.t. axis, but a circle does - pi, but pi is just a special constant we've come to recognize and could easily define another constant for other ellipses
3:40 Well, I suppose using the laplace transform would be a way of integrating without integrating: ℒ{ ∫∫ 6x dx dx } = 6 ℒ{x} / s² = 3! / s⁴ = ℒ{ x³ }
a laplace transform is an integral..
L[take](s)
How on Earth were you able to type in such a beautifully-fonted reply?.. I envy you... :)
I can see this makes sense but I don’t understand this completely. Can you write out the general case as laplace transforms? I’m unfortunately not familiar enough with them
@@gaHuJIa_Macmep using fancy unicode characters. Specifically: U+2112 "Script Capital L", U+222B "Integral", U+00B2 "Superscript Two", U+00B3 "Superscript Three", and U+00B4 "Superscript Four".
Amazing quality content
Super explanation 👌
beautiful video
Brilliant!
Thanks for this knowledge :> Hopefully I can use it when I encounter multiple integrals in my self study journey
Oh boy! Fractional calculus here we come!
At 12:44 you mentioned that the integrand is not the inner integral. This is because we are switching the limits of the integration, ie the inner integral goes out and the outer one comes inside. This is also what you did geometrically by slicing the integral a different way.
You made my day
Such a fantastic way to talk about calculus!
Am I right that this principle applies only to functions dependent on a single variable e.g. f(x) but it would be different for multiple variable function e.g. f(x,y)?
There is a way to compute iterated derivatives using matrix diagonalization. Since derivatives are regular invertible linear operators, they can be encoded as a diagonalizable matrix. Using eigenvalue decomposition to compute powers of this matrix and representing functions as vectors with real coefficients in a function space is equivalent to computing iterated derivatives.
Awesome!
Holy.. Your's math is super fun!!!!
Damn you really ended it with a plot twist. Fractional integrals and thereby opening doors to Fractional derivatives.
Also, another interesting thing is contour integrals.
A contour integral is an integral that integrates over a boundary in any dimensions, unlike normal integrals, which only integrate in their respective dimensions. To differentiate the differentials of contour and normal integrals, we use s's rather than conventional letters, so dx -> ds, dA/dxdy/rdrdθ -> dS, dV/dxdydz/rdrdθdz/ρ^2*sin(φ)dρdφdθ -> dC (for cellular volume). Also, ds, dS, and dC are magnitudes of the dr, dS, and the dC vectors: dr = , dS = , dC = . These differentials are crucial in finding arclength, surface area, and cellular volume. For example, the arclength of a given function in 2-space is given by ds, which is √(1+y'^2)dx, √(x'^2+y'^2)dt, and √(r^2+r'^2)dθ. Surface area follows a similar pattern. The most common formula for surface areas are dS = sqrt(1+∂x+∂y)dA = |∂s x ∂t|dsdt, and 2πrds, from solids of revolution. Because of that, we can derive the formulas for cellular volume being sqrt(1+∂x+∂y+∂z)dV = |∂r x ∂s x ∂t|drdsdt, and 2πrdS. To evaluate contour integrals, you need to make sure your differentials are in magnitude form. Sometimes, they would already be in such form, like when you are trying to find the mass of a spring or a solid lamina, but most of the time, you are given a vector field and are trying to compute the circulation or the flux across a given path or surface, in this case, you must dot the vector field with the normal to turn it into a magnitude. You also need to parametrize the surfaces as well. However, there are some useful formulas:
Fundamental Theorem: If a vector field is conservative, meaning it is the gradient of a function, a line integral is just taken as an ordinary antiderivative, as the differential of a multivariable function is just the gradient dotted with the dr vector.
Divergence Theorem: The flux of a given vector field is proportional to its divergence and its interior content: ∯ F ∙ dS = ∭div F * dV, ∰ F ∙ dC = ⨌div F * dH
Stokes' Theorem: The surface integral of a given vector field is proportional to its curl and its circulation: ∮ F ∙ dr = ∯ curl F ∙ dS
Magnificent Meticulously Innovative Method ascertain the truly ideal Teacher
Clicked for the impossible task. Stayed for the amazing explanation
It can also be thought as taking the integral with the greens function for (d/dx)^n operator
Exactly!
👍👍I think i remember learning this formula in 'calc 3', multivariable calc
Mark my words it's only a matter of time before this channel blows up
This formula for repeated integration also appears when expressing the remainder of a Taylor polynomial approximation in the "integral form" - because, unlike the Peano, Lagrange, or Cauchy forms, it uses an integral
You know ; I just HATED it when he closed with a teaser about the Gamma- function and its role in computing a multiple integral. But that's the way to get more subscribers I suppose. And in my case : it WORKED. So I just subscribed and will DEFINITELY be back and recommend this channel to any math-geek I talk to.
Keep up the good work: "Aux Revoir "......
Actually theres formulas for n-th derivatives for p much every function including chains, it's just usually a sum or product up to n, which isn't really expandable for non integer n, however they are p beautiful if you asked me!
you forgot the constant c
Noice broooo keep making videos with moree quality over quantity
16:20 you may want to have a look into Leibnitz's theorem. It's a method for taking the nth derivative of a function.
It splits the function into parts and takes the derivative of both parts n times then combines them together.
Nested Integral : You can't beat me, I'm compressible.
Derivative: You're right I can't. But he can. (d2y/dx2)
Interpreting things with geometrifying trigonometry give meaningful visualization for Cauchys repeated integrals
I have an old HP15C that does numerical integration, and the manual has a whole section on dealing with cumulative error.
Great video by the way.
Is using Reimann-Louiville's fractional derivative/integral considered as integral and differentiation compression? The fractional construction generalizes the differential operator, such that the differential operator of the order n defines derivation done n number of times, and -n defines integration done n number of times. The formulation is a single integral and it makes use of the generalization of the aforementioned factorial: the gamma function. Not only does this formulation take into account both integration and derivation, but also any differential operator that is of the order of any real number - even fractional integration and derivation
Is this alternate way to think of nested integration related to how Lebesgue integration is considered superior for higher dimensions (input or output), as it tends to generalize cleaner than Reimann integration?
the cliffhanger at the end was better than most netflix shows ngl
Very cool! I should have guess that factorial would show up. Still annoys me that the continuous version most people use is the gamma function, and not the actually equivalent pi function. (But I love gamma for Four 4s, since it makes 6 super easy to get, among other things.)
6:34! Brilliant!
the audacity you have to end this video on a cliffhanger!! Anyways, it was a really interesting video, thank you very much!!
I definitely learned about that in university, but completely forgot - or didn't understood back then.
Before watching video: cauchys integral formula for the nth derivative... and u can use negitive numbers to mean integrals.. but you gotta use gamma formula and complex numbers. But it works
After watching video.. no cauchys integral formula is equivalent to the one you showed
Is this related to stoke's theorem? (not the surface integral to integral of divergence of a function over a volume, but the more general one) From what I understand it's a generalization of this "integral reduction" we see with the fundamental theorem of calculus, divergence theorem and the "little" stoke's theorem taught in a first course on vector calc.
@Morphocular heres one thing about differentiating.. it can go below 0 and integration can go above 0. why don't we regularly perform these functions? why do we infact simplify to cancelation as a rule? x^0 goes to # 1, but X^-1 is right there as a possibililty for the next differential. simarly the reverse for integration. why dont we go past these arbitary lines? Or is there something im missing?
O(1) formulas are incredibly powerful.
17:45 This is a good lead in to your next video, fractional derivatives / integrals
The expression under the transformed single integral reminds me a lot of a Lebesgue integral in a way, being a value of the function times the measure of the interval at which it takes that value, is there a connection here in some way to integration with respect to measure?
Hi Morph!
I wish you'd include captions with your videos. Math videos are always easier with captions no matter how well you can hear.
Well, while the way you presented the Cauchy formula (frankly, I even was unaware of its existence) for the 2D case seems quite inventive and geometrical, but it lacks the n-dimensional easy visual interpretation and the similar easy derivation. Also it seems to be over-complicated a bit. I, for example, was able to derive it mentally myself although I, as I told already, was unaware of it. For a second "anti-derivative" I considered a linear differential equation F" = f(x), and, having general solution of the corresponding homogeneous equation F" = 0 to be F(x) = Cx+ D, I constructed the Green function and found the solution of the initial non-homogeneous equation as a single integral. It's clear how this method is generalized to an n-dimensional case, using the theory of ODE. And this method has a geometric interpretation too, as the solution via the Green's function is the output of a linear system to the given input while the Green's function is the output for the input of the form of a Dirac's delta-function. Sad you didn't mentioned the fact in the video. Anyway, thanks for educating.
When I thought thta 2:30 part problem, I think we can just make this as general form where:
∬6x dx dx can be expressed as k*∬x^n where n =1, k=6
for example, if we take one integral as in our mind, we see that it grow up power n with +1. Also we have to take division k/(n+1) * ∫x^(n+1) dx
we see how it end up, we can write basic form multi-integral as formula:
(k / (∏(n+i,i,1,t)) )*x^(n+t) where k=constant, n=x's power on starting, t=number of integrals.
for example, if we have ∭∭14 sqrt(x) dx^6 . We can write this as:
14/((n+1)(n+2)(n+3)(n+4)(n+5)(n+6)) * x^(n+6) where n =1/2
((14)/(∏(n+i,i,1,t)))*x^(n+6)|n=((1)/(2)) and t=6 ▸ ((128*x^(13))/(19305))
What about the case where the lower bounds of the nested integrals aren't all the same? If they're related, e.g., a_2=2a_1, I assume your line y=x in the animation looks something like y=2x, or maybe y=x/2. But what about the case when the lower integration bounds are arbitrary (including a_2
In terms of Laplace transforms:
Let * operator denote convolution. Recall the definition f(t) * g(t) = int f(t-x) g(x)dx. Then,
int (t-x) f(x) dt = f(t) * t
t = u(t) * u(t)
So, int (t-x) f(x) dt = f(t) * u(t) * u(t)
Since convolving by the unit step function u(t) is the same as integrating, int( (t-x) f(x) dt = int^2 f(x) dx. This video actually explains the relationship between the functions and their Laplace transforms.
You may get a limit but it would not be tangent if the function is not defined at particular x( limiting case of secant❌
This video was a trip.
How would the formula change if we had a multi variable function?
Is it possible to join multiple intergrals into one if they have more complicated bounds of integration?