Padé Approximants

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Dr. Will Wood

Dr. Will Wood

Күн бұрын

In this video we'll talk about Padé approximants: What they are, How to calculate them and why they're useful.
Chapters:
0:00 Introduction
0:33 The Problem with Taylor Series
2:11 Constructing Padé Approximants
4:50 Why Padé Approximants are useful
5:45 Summary
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Thank you so much, I hope you find the content useful.

Пікірлер: 625
@3blue1brown
@3blue1brown 2 жыл бұрын
I had never known about Padé approximations, and you did such a good job motivating and explaining them. Also, the way you mentioned how it seems almost unreasonably effective, like getting something for nothing, definitely felt like it was speaking directly to the thought passing through my mind at that moment.
@DrWillWood
@DrWillWood 2 жыл бұрын
Thanks so much! Padé approximants still feel kind of magic to me and it's definitely that they aren't well known that made me want to make a video on them. Also, does this mean I have bragging rights that "I taught 3b1b some mathematics" :-D
@T3sl4
@T3sl4 2 жыл бұрын
I'm excited!
@remicornwall754
@remicornwall754 2 жыл бұрын
Qualitatively and intuitively, is it because when you truncate a Taylor series at some order O(n) you are doing precisely that but when you have a rational polynomial approximation O(m)/O(n) it really could be order greater than m or n if it is divided out? I.e. if you do 1/(1-x), the series is infinite, so you must be accruing all that accuracy by doing O(m)/O(n) because it is not really truncated at order O(m) or O(n) but is potentially an infinite series, even though you write it compactly as a rational fraction.
@curtiswfranks
@curtiswfranks 2 жыл бұрын
It is not quite for nothing. You are calculating twice as many interrelated coefficients (minus 1), and the overall function is more complicated (which makes computation for any given x harder and further dependent on precision). And that is AFTER having to calculate the Taylor expansion, so it is triple the amount of work, and is 'refining' an approximation which was already rather nice and effective in some respects. It still seems unreasonably effective to me, a bit, but it is not entirely for free. There probably are also some theoretical costs associated with going from polynomials (extremely nice, highly-constrained functions) to rational functions (which are just slightly less so).
@phiefer3
@phiefer3 2 жыл бұрын
@@curtiswfranks I don't think it's fair to say that you calculate twice as many coefficients, in general the number of coefficients would be roughly equal. In the Pade aproximation you have m coefficients in the numerator and n coefficients in the denominator, and you construct it from a m+n order taylor series, so they literally should be equal. In the example the taylor series only had fewer because some of the coefficients were 0, but you still have to calculate those coefficients anyways. As far as the "something from nothing", I think what is meant is that you get this extra precision without taking any more information from the original function. I could literally give you a taylor series without telling you what it is supposed to be approximating, and you could build a better approximation even though you don't even know what it is you are approximating.
@jamesblank2024
@jamesblank2024 2 жыл бұрын
Padé approximations shine in analog filter design where you have poles and zeros. They are particularly effective in analog delay lines.
@5ty717
@5ty717 11 ай бұрын
Beautiful
@Eizengoldt
@Eizengoldt 11 ай бұрын
Nerd
@pbs1516
@pbs1516 11 ай бұрын
We also use them in control theory, for the same reasons (a finite-order modeling of a time delay). It's becoming an old man trick though, now that most of what we're doing practically is in discrete time (were delays can be treated as a linear system with the good multiplicity!)
@alexanderlea2293
@alexanderlea2293 11 ай бұрын
As soon as he mentioned the fractional polynomial, my mind went to laplace and filters. It's such a natural fit.
@yash1152
@yash1152 10 ай бұрын
thanks i was looking for some comment mentioning its use case areas.
@carl8703
@carl8703 2 жыл бұрын
In general, expressions of the form at 0:23 are interesting since they have a few nice properties: 1.) Much like how polynomials are capable of expressing any function composed from addition, subtraction, and multiplication (for a finite number of operations, at the very least), expressions of the form at 0:23 do the same thing, but for addition, subtraction, multiplication, *and division*. Another way of putting it is that they are capable of describing any function defined on a field. This might help towards explaining why Padé approximates can be so much more effective than Taylor series. 2.) Approximates of that form are used all the time in highly realistic graphics shaders. This is because they can be used to create fast approximates of functions whose real values could not be calculated in the time it takes to render a frame. Unlike polynomials, they can behave very well over their entire domain, and they avoid large exponents that could introduce floating point precision issues, both of which are important when you need to guarantee that a shader will not create graphical artifacts in a limited environment where all you have to work with is 32 bit floating point precision. They also avoid calls to advanced functions like sin() or exp(), which again makes their execution especially fast. 3.) You don't always need the derivatives of a function to find such an approximate. For instance, if you know that a function has an asymptote, or that it assumes a certain value at 0, or that it's symmetric, or that it tends towards a number at ±∞, then that automatically tells you something about the coefficients within the approximate. It then becomes much easier for you to run an optimization algorithm on a dataset to find good values for the remaining coefficients. Christophe Schlick gives an excellent example of this approach in "An Inexpensive BRDF Model for Physically-based Rendering" (1994). 4.) Multivariate versions of the approximate are a thing, too. To see how such a thing can be done, simply start from the proof for the statement in 1.) but now instead of working with real values and a variable "x" as elements, you'll also be working with another variable "y". As an example, for 3rd order bivariate approximates you'll wind up with polynomials in your numerators and denominators that have the form p₁xxx +p₂xxy + p₃xyy + p₄yyy + p₅xx + p₆xy + p₇yy + p₈x + p₉y + p₁₀
@DrWillWood
@DrWillWood 2 жыл бұрын
Wow, that's so awesome! All news to me as well! Thanks for taking the time to write this post I appreciate it, as will people watching this video I'm sure.
@Peter-bg1ku
@Peter-bg1ku 2 жыл бұрын
Thank you @carl. You almost made me shed a tear with your post.
@diegohcsantos
@diegohcsantos 2 жыл бұрын
@@DrWillWood is there any result about the error estimation with a certain Pade aproximante? Something like Lagrange's form for the remainder fo a Taylor polynomial?
@dlevi67
@dlevi67 2 жыл бұрын
This comment should be pinned - it adds so much extra information to an already excellent video!
@TheBasikShow
@TheBasikShow 2 жыл бұрын
I don’t see how (1) could be true. The Weierstrass function is defined on a commutative field (ℝ) but doesn’t have a Taylor approximation, and therefore (if I understand correctly) doesn’t have a Padé approximation. Maybe you meant that Padé approximations can express any functions which are derived from adding, subtracting, multiplying and dividing the identity f(x) = x?
@Elies313E
@Elies313E 2 жыл бұрын
The algorithm recommended this video to me, I'm so thankful because this is beatiful and very useful.
@Elies313E
@Elies313E 9 ай бұрын
@@chonchjohnch Which part?
@Sarsanoa
@Sarsanoa 2 жыл бұрын
Oh nice. e^-x is a very common function to Padé approximate in linear control theory because it's the Laplace transform of a uniform time delay. Notably, x in this context is a complex number, yet it still works. I've never understood how it was computed until now. I think the aha moment is realizing we are discarding all higher order terms when we perform the equation balance. This is the key reason why the Padé approximation isn't just equal to the Taylor approximation.
@mstarsup
@mstarsup 2 жыл бұрын
The case you're talking about can be computed much more easily actually. Just write e^(-x) = e^(-x/2)/e^(x/2), and use Taylor expansion for e^(-x/2) and for e^(x/2) :)
@romajimamulo
@romajimamulo 2 жыл бұрын
I've never heard of this before, and after judging so many bad entries, this is a breath of fresh air
@DrWillWood
@DrWillWood 2 жыл бұрын
Thanks a lot! Its one of my favourite things about maths KZfaq, coming across concepts you wouldn't have otherwise! Also, good luck with SoME1 :-)
@romajimamulo
@romajimamulo 2 жыл бұрын
@@DrWillWood thank you for the good luck. Check out mine if you have time (it's a 5 part series, but for judging, only the first part is required
@ciarfah
@ciarfah 2 жыл бұрын
We used these in Control Theory to approximate time delays (exponentials) in the transfer function
2 жыл бұрын
This is an excellent explanation of something I didn't know existed. Yet it's so simple and elegant. I'm working on a Machine Learning playlist on linear regression and kernel methods and I wish I had seen this video earlier! I'll play around with Padé approximants for a while and see where this leads me. Thank you for this interesting new perspective!
@helpicantgetoffofyoutube
@helpicantgetoffofyoutube 11 ай бұрын
Hello there! It's been a year. I just watched the video, and now I wonder what you managed to do since then
@eulefranz944
@eulefranz944 2 жыл бұрын
Finally! I encountered these so often in physics papers. Finally I get it!
@MrKatana333
@MrKatana333 2 жыл бұрын
What physics papers? I was just wondering how this could be applied in physics. Can you give some reference please? Thanks!
@felosrg1266
@felosrg1266 11 ай бұрын
In which fields did you find those pate approximations in use?
@tbucker2247
@tbucker2247 2 жыл бұрын
Mech Eng grad student here. This is my "did you know?!" flex for the next couple weeks. Amazing video, thanks!!
@isaackay5887
@isaackay5887 2 жыл бұрын
Let's be realistic though...you and I both know we're still gonna regard *sin(x)≈x* _for small x_ lol
@janami-dharmam
@janami-dharmam 2 жыл бұрын
Basically you are using M+N terms in the taylor series but using only M and N terms for evaluation. This is computationally very efficient.
@oniflrog4487
@oniflrog4487 2 жыл бұрын
@@isaackay5887 MechEs that work in Rotordynamics: *small* x? what you mean!? 🤣
@cdenn016
@cdenn016 11 ай бұрын
If you want to be a summation god then check out the book by Carl Bender. Elite 💪
@henrikd.8818
@henrikd.8818 2 жыл бұрын
I really like how fast you managed to explain it! Only few math videos get a topic like this explained in under 7 minutes
@inigolarraza5599
@inigolarraza5599 11 ай бұрын
Yeah, if you skip the mostly unnecessarily long proofs and theorems, and use a more accesible language, most 1st-2nd year college mathematics cousld be explained this way. Concepts like eigenvalues and eigenvectors, Taylor polynomials and series, or Lagrange multipliers could easily be taught in 10-20 minutes (of course, if you already know matrices, determinants, derivatives and some multivariable calculus) but easily take up entire lectures, because of the excessive attention on proofs or "preliminary definitions" that are not necessary to understand the concepts in the first place. (only to rigorously define them). The sad reality is that most students get lost or mentally exhausted on the theoretical chip-chat and they end up NOT learning the methods or even understanding what they're doing.
@ZakaiOlsen
@ZakaiOlsen 2 жыл бұрын
Having spent a great deal of time reading up on Pade approximants and struggling to find easy to understand introductory examples it is extremely exciting to see content such as this being put out there for people to learn. Fantastic job motivating the need and demonstrating the utility for these rational approximations. In my personal explorations, I have found multipoint Pade approximations to be very cool, being able to capture asymptotic behaviors for both large and small x, or around poles / points of interest is very cool. Keep up the awesome work!
@DonMeaker
@DonMeaker 2 жыл бұрын
Ratios of polynomials are used in aircraft flight controls. Normally, flight test attempts to measure aircraft handling qualities in a ratio of two second order polynomials, even though modern digital flight controls may be much higher order functions.
@mjmlvp
@mjmlvp 11 ай бұрын
For a detailed description you got to also see the video series on youtube of lectures by Carl Bender, "Mathematical Physics"
@rolfexner9557
@rolfexner9557 2 жыл бұрын
It seems natural to choose M = N. What are the situations where there is an advantage in choosing M > N or N > M, where I have a "budget" of M+N coefficients that I want to work with?
@ishimarubreizh3726
@ishimarubreizh3726 2 жыл бұрын
Having M=N means you can choose a finite non zero limit to the padé approximant according to the function you want to capture the behavior. In this case the denominator had a larger order and therefore made the fraction goes to 0 at inf. If it was the other way around (N>M) it would blow up and you would lose one of the motivation to use padé, but having polynomials with roots at the denominator allows to better describe poles located at finite x I would say, where Taylor expansion fail to even exist Edit : it seems this reply was liked a few times so I would like to add that going to inf is not an actual loss of motivation. It is great to have an expression that also captures the asymptotic behavior even when there is no finite limit, whereas Taylor blows up without following the divergence of the function
@pierrecurie
@pierrecurie 2 жыл бұрын
M=0 is just an ordinary Taylor series, so that's at least 1 use for N>M. From that perspective, it can also be seen that the Taylor series is a special case for pade approx.
@viliml2763
@viliml2763 2 жыл бұрын
Your function will behave like x^(N-M) asymptotically. For example, with a function like sine that tends to stay constant, you might want M=N. With a function e^x that tends to zero, you might want M>N. And so on.
@cornevanzyl5880
@cornevanzyl5880 2 жыл бұрын
I really didn't like calculus in University but I find this very interesting. I can appreciate the beauty much more now that I'm not suffering through it
@yan.weather
@yan.weather 11 ай бұрын
Suffering indeed 😂🎉
@dodokgp
@dodokgp 11 ай бұрын
The best and yet the simplest explanation for Padé approximation I have seen! We use it a lot in finite element simulation software in engineering, but I was always in search for a more intuitive explanation for its merits over default Taylor series! I am happy today.
@ichigonixsun
@ichigonixsun 11 ай бұрын
The Padé Approximant might be closer to the actual function in the long run, but it actually has a larger relative error compared to the Taylor series around x=0. Since we only care about approximating sin(x) from x=0 to x=pi/4, because we can then use reflection and other properties to get the value for other angles, the benefits are overcome by the disadvantages (i.e. you have to do more arithmetic operations, including a division).
@ere4t4t4rrrrr4
@ere4t4t4rrrrr4 11 ай бұрын
That's interesting. Are there any approximation that is locally better than the Taylor series, around some specific point?
@ichigonixsun
@ichigonixsun 11 ай бұрын
​@@ere4t4t4rrrrr4 I know there are algorithms that can compute the value of a function at a given point with (arbitrarily) better precision, but i don't know about any other closed algebraic formula which locally approximates a given function better than the Taylor Series.
@beeble2003
@beeble2003 11 ай бұрын
Is that true in general for periodic functions? What about non-periodic functions?
@Solution4uTx
@Solution4uTx 3 ай бұрын
@@ichigonixsun that's interesting could you please share the name of those algorithm i want to test
@ichigonixsun
@ichigonixsun 3 ай бұрын
@@Solution4uTx CORDIC, for example. There are many others with different use cases.
@DeathStocker
@DeathStocker 2 жыл бұрын
A hidden gem of a channel! Never really considered other approximations because the Taylor ones are so commonly used in computation. I remember reading about polynomial approximations of trigonometric functions for low-end hardware but maybe those were less general than the Padé approximation.
@DrWillWood
@DrWillWood 2 жыл бұрын
Thanks a lot! Ah yeah that's a really nice application for this sort of stuff (how can we pack the most powerful approximation into the smallest memory/compute). I don't know much about it but definitely cool! I remember a lecturer saying splines were important in this area but I'll be honest I can't remember the details!
@DeathStocker
@DeathStocker 2 жыл бұрын
@@DrWillWood Found the book that I read! It is "The Art of Designing Embedded Systems (2nd ed)" by Jack Ganssle. Chapter 4.4 has floating point approximations for common functions like exponent, log, and trigonometric.
@DrWillWood
@DrWillWood 2 жыл бұрын
@@DeathStocker Awesome!! thanks for that :-)
@HELLO-mx6pt
@HELLO-mx6pt 2 жыл бұрын
Pretty cool video! The general idea for the approximation reminds me a lot of the process one uses to expand a fraction into a p-adic sum, to get its p-adic representation. After years of doing math, this whole idea of actually using rational functions, a thing that we commonly ignore due to them being "ugly" is shinning a new light. Keep up the good work!
@paris_mars
@paris_mars 11 ай бұрын
This is a great video. Well made, simple, clearly explained, genuinely interesting. Awesome.
@nikolaimikuszeit3204
@nikolaimikuszeit3204 2 жыл бұрын
Definitively interesting, but if I get it right: if you decide you need a higher order you cannot re-use the low order coefficients that you already have. That I'd consider a disadvantage.
@beeble2003
@beeble2003 11 ай бұрын
Doesn't seem like much of a disadvantage: you calculate the approximant only a few times compared to how many times you use that approximation function.
@nikolaimikuszeit3204
@nikolaimikuszeit3204 11 ай бұрын
@@beeble2003 Well, I'd assume that this is true for many---definitively not all---applications using approximations and I agree. Then it is "not much" of a disadvantage but it is one. ;) Cheers.
@9WEAVER9
@9WEAVER9 Ай бұрын
Well it's not like those new coefficients are impossible to find. ​You'll have a recurrence relation, at least, for the coefficients of the next higher order Padé. Otherwise you wouldn't even have a Pade' to begin with. See it as problematic as you'd like, Padé gives you information in proportion to the work you give the Padé. That's just a commonality of Asymptotic Analysis.@@nikolaimikuszeit3204
@vector8310
@vector8310 6 ай бұрын
Superb introduction. I was browsing thru Hall's book on continued fractions and happened upon a section on Padé approximants, which peaked my curiosity and led me to this video. I can't wait to study these further. Thank you.
@dank.
@dank. Сағат бұрын
I found this video a year ago, before I knew calculus, so I wasn't able to follow through the part about "Taylor series(s)". Having just finished calc 1 and 2, I'm so glad I found this video again! Very insightful, thank you for sharing.
@user-de1td7jh9y
@user-de1td7jh9y Жыл бұрын
Thank you very much! All the explanations I found on the Internet were quite difficult for me to understand. You've done a really cool work!
@knpark2025
@knpark2025 11 ай бұрын
I once needed to look into a dataset that seemed to have an asymptotic line. I remembered of rational equations from high school algebra, did a regression analysis with that instead of polynomials, and it worked wonders. I never expected this to be also a thing for approximating existing functions and I am so happy to learn about it here.
@Peter-bg1ku
@Peter-bg1ku 2 жыл бұрын
This video is way better than the book I read which skipped a lot of the information relating to the Pade approximation. Thank you! This is brilliant!
@DrWillWood
@DrWillWood 2 жыл бұрын
Thanks so much!
@philipoakley5498
@philipoakley5498 2 жыл бұрын
Excellent. Hadn't known about that method. I like (typically) that the estimate tends to zero in the long term (no mention of NM Effects)
@MuradBeybalaev
@MuradBeybalaev 11 ай бұрын
I appreciate that you switched to stressing the correct syllable midway through the video. Not only is the man French, but there's even an explicit diacritic in the last syllable of his name to make stress extra clear.
@VLSrinivas
@VLSrinivas 11 ай бұрын
I have read about Padé schemes to discretize a partial differential operator in Computational fluid dynamics a while ago but thanks for making the advantage over Taylor series more visible. This could be of great use in computational engineering simulations to avoid divergence.
@abdulkadiryilmaz4085
@abdulkadiryilmaz4085 2 жыл бұрын
Immediately subbed. Definitely a great content, keep up the good work
@TheQxY
@TheQxY 2 жыл бұрын
After seeing this video I remembered that I actually did learn about this in an advanced mathematics class during my Master's, although then with with much higher order terms. However, they never explained very well how usefull it actually is, for simple approximations as well, so I quickly forgot about this. Thank you for the refesher, very well explained, and I will definitely keep PAdé approximation in mind as a useful tool in the future! EDIT: I remember now that the assignment was to find a singularity in a polynomial of order 130. By using Mathematica and gradually decreasing the orders of N and M in the Padé polynomial allowed the finding of a function with the same symmetry as the original giant polynomial, but with much reduced terms. This derivative of the approximated polynomial could then be used to find the singularity, which did not change location during approximation. Just a cool example of a more complicated application of the Padé approximation method for those who are interested.
@curiousaboutscience
@curiousaboutscience 11 ай бұрын
This was fun to watch! Definitely good points on the typical Taylor series divergence too. How cool!
@fizzygrapedrink4835
@fizzygrapedrink4835 Ай бұрын
Such a great video! I'm cramming for exams in my uni right now, and this was super useful and pleasant to listen to! Way more understandable than our professor's notes lol
@charlessmyth
@charlessmyth 2 жыл бұрын
This stuff is like the authors you never heard of, but were notable for their time :-)
@xyzct
@xyzct 2 жыл бұрын
I'm stunned I've never heard of this before, given how important Taylor series approximations are.
@diegocrnkowise3102
@diegocrnkowise3102 9 ай бұрын
Very interesting explanation, I'm currently studying electrical engineering and I've just been slightly introduced to a Padé approximant during a Control Systems practice, but never learned the algebra behind it in previous calculus classes.
@frentz7
@frentz7 2 жыл бұрын
You have such a nice style, in the communicating (both voice and visual) but also / maybe even more so, a nice "pace" of thinking and a swell difficulty level.
@geoffrygifari3377
@geoffrygifari3377 2 жыл бұрын
Physics often use power series expansion because its so easy to just cut off terms higher than some order we want for small values of x, saying that those are "negligible". I imagine picking a polynomial order "just high enough" would be tougher if its in a ratio like the Padé approximant
@fatcatzero
@fatcatzero 2 жыл бұрын
Is the Padé Approximant ever less accurate than the same-order Taylor series?
@brandongroth4569
@brandongroth4569 2 жыл бұрын
​@@fatcatzero By construction, a Pade Approximant is at least as accurate as a Taylor series because we force it to match the coefficients of the Taylor series. It is just a lot more work to find them via a N+M linear system of equations. If the approximant reduces to something nice like sin, they can be very useful, but that is probably a rare case. In numerical analysis, you often trade simplicity for accuracy, which is what is happening here with Taylor vs Pade.
@fatcatzero
@fatcatzero 2 жыл бұрын
@@brandongroth4569 ah, I misinterpreted the original comment (mistook the point about "just high enough" to mean "bounding it to smaller than O(x^n) from the actual solution", not "do the easiest thing that will get me within O(x^n)").
@geoffrygifari3377
@geoffrygifari3377 2 жыл бұрын
@@fatcatzero oh yeah, my point was close to that. what i'm saying is because taylor series results in a *sum* , you can just terminate the sum at nth power to get high-enough accuracy, using polynomial of order n. Now how can you do something like that if the approximant is a ratio? do we terminate the powers at both numerator and denominator? by no means obvious to me
@fatcatzero
@fatcatzero 2 жыл бұрын
@@geoffrygifari3377 if we start with our n'th degree Taylor series and set N+M=n for the Padé Approximant, it seems like it's always going to be at least as good of an approximation as O(x^n) since it's equal-to-or-better than our O(x^n) Taylor approximation. I literally just learned about this concept by watching this video so I by no means know the specifics, but if it is true that the M+N approximant is always at least as good the associated n-degree Taylor series, yes it's by definition more work and it could be hard to determine how much better it is, but the upper bound on deviation from the actual function seems to be the Taylor series where we know very well how to determine the size of our error. Do you think any of that is incorrect and/or am I still missing something about your concern with applying this method?
@connorfrankston5548
@connorfrankston5548 11 ай бұрын
Very nice and very simple, makes perfect sense. I feel that these Padé approximants can greatly improve approximate series solutions to difference and differential equations in general.
@StratosFair
@StratosFair 11 ай бұрын
Short, clear, and instructive. Congratulations for the great work 👍🏾👍🏾
@DrWillWood
@DrWillWood 11 ай бұрын
Thank you very much!
@ianprado1488
@ianprado1488 2 жыл бұрын
As a calc 2 TA, I absolutely loved this
@easymathematik
@easymathematik 2 жыл бұрын
I´ve met Padé-Approximation at university in 5th semester. The name of the course was - as you can guess - "Approximation". :D There are another very interesting methods as well. Nice video from you. :)
@algorev8679
@algorev8679 11 ай бұрын
What other interesting methods did you see?
@seanleith5312
@seanleith5312 11 ай бұрын
We know Spanish peple have no contribution to Math or science in General, is this the first one?
@ere4t4t4rrrrr4
@ere4t4t4rrrrr4 11 ай бұрын
@@algorev8679 one are Chebyshev polynomials, another are Lagrange polynomials (they are used if you want to approximate minimizing the error around a large interval of the function and not just approximate around a point like Taylor series or Padé approximants). Check out the approximation theory article on Wikipedia
@jacquardscootch8939
@jacquardscootch8939 2 жыл бұрын
This was really interesting. A professor at my college did a lot of research with approximants known as “Chromatic Derivatives”, and these share a similar motivation.
@tedburke525
@tedburke525 2 жыл бұрын
So clear and concise! Thank you.
@TypicalAlec
@TypicalAlec 2 жыл бұрын
This is fantastic, genuinely might use these at work 👍🏻
@zaccandels6695
@zaccandels6695 2 жыл бұрын
Very cool. I'm somewhat surprised I was never introduced to this in numerical analysis
@tadtastic
@tadtastic 11 ай бұрын
this is a really neat method of approximation explained very clearly and practically! nice vid
@algorithminc.8850
@algorithminc.8850 2 жыл бұрын
Great explanation. These are the kinds of topics you want to share with everyone, as a scientist, but want to keep quiet, as a company, in order to have an edge. Thank you much.
@feraudyh
@feraudyh 11 ай бұрын
I have friends who live in a house in rue Paul Padé near Paris. I'm going to ring them and explain what the video has done so well. I'm sure it will make their day.
@MW-vg9dn
@MW-vg9dn 2 жыл бұрын
Very cool channel, thanks for making these videos!
@karambiout9737
@karambiout9737 2 жыл бұрын
Oh thank you for the explanation, I will try to apply the padé approximants
@1997CWR
@1997CWR 2 жыл бұрын
Excellent Presentation. Concise, clear, and thoroughly enjoyable.
@DrWillWood
@DrWillWood 2 жыл бұрын
Thanks a lot!
@robertdavie1221
@robertdavie1221 11 ай бұрын
Very well explained. Thank you!
@kongolandwalker
@kongolandwalker 11 ай бұрын
Rewatching because the thing has a lot of potential and i might apply it in my projects.
@EconJohnTutor
@EconJohnTutor 2 жыл бұрын
This is incredible. I never knew about this, thank you!
@speedsterh
@speedsterh 2 жыл бұрын
Very interesting and very well explained ! Thank you
@shubhamg9495
@shubhamg9495 2 жыл бұрын
Such an informative video. Thank you so much!
@editvega803
@editvega803 2 жыл бұрын
Thank you very much!! I didn't know about that. Very interesting 😀🤔
@MeanSoybean
@MeanSoybean 11 ай бұрын
This is absolutely brilliant.
@abelferquiza1627
@abelferquiza1627 2 жыл бұрын
I didnt know and i liked, so simple and useful.thanks
@EngMostafaEssam
@EngMostafaEssam 2 жыл бұрын
Thanks a lot, you deserve more than million subscribers ❤
@ericthecyclist
@ericthecyclist 11 ай бұрын
I always regretted not taking a splines class when i was a grad student because I didn't understand NURBS (Non-uniform rational B-splines) and how to compute their coefficients. In very few minutes, you made it apparent.
@the_math_behind
@the_math_behind 11 ай бұрын
Great video with a concise and effective explanation!
@pjplaysdoom
@pjplaysdoom 11 ай бұрын
Remarkable that in my university Maths degree, I didn't meet Pade Approximants. You explain the topic very clearly. I believe the B_0 = 1 assumption is fine unless the denominator polynomial vanishes at x = 0, which would imply a vertical asymptote in the overall function.
@energyeve2152
@energyeve2152 11 ай бұрын
Cool! Thanks for sharing!
@AJ-et3vf
@AJ-et3vf 2 жыл бұрын
Awesome video! Thank you!
@hrissan
@hrissan 2 жыл бұрын
Thanks, never heard of this approximation before.
@modolief
@modolief 3 ай бұрын
The first time I saw Padé approximants was in a paper proving the transcendence of e. Thanks for the useful discussion!
@pepsithebunny2404
@pepsithebunny2404 2 жыл бұрын
Never heard of it, may enhance the results i am currently having with Taylor series alone. Tankyou.
@johanekekrantz7325
@johanekekrantz7325 11 ай бұрын
I really like this video. Very well explained. The concept could obviously (so im sure oneone did and that it has fancy name) be generalized to the idea of also thinking about how we want the approximation to behave for lim x -> inf by modelling things as a sum of functions F(X) (fractions of polynomials here) lim x -> F(x) where F defines how you want the approximation to behave for extrapolation. That way we could also think about the derivatives of the approximations at infinity.
@dhhan3100
@dhhan3100 10 ай бұрын
Very interesting. I heard a few times Pade approximaton, but i did not know why we use that.
@mikelezhnin8601
@mikelezhnin8601 2 жыл бұрын
I'm missing the point. It's cool and all, but there are two buts: 1) yes, taylor series do not extrapolate well, but it's not the point of taylor series, they are specifically used to approximate the function in some small area near to some point. 2) [N/0] padé approximant is the same as taylor series, and then you have the other N versions for padé approximants - [N-1/1], [N-2/2], etc. It seems unfair to say that padé approximants work better than taylor series, since padé approximants are a direct extension of taylor series, plus you can cheat by freely choosing how to split N into [N-M, M].
@olgittj1507
@olgittj1507 2 жыл бұрын
Not to mention that the Taylor expansion, if it exists, it is guaranteed to be analytic in the complex plane since it's a polynomial. N/M] Padé approximations will introduce poles if M > 0.
@theTweak0284
@theTweak0284 2 жыл бұрын
I agree. I would like to see a better example than for sin(x) because if your goal is to extrapolate an oscillating function, you want it to continue to oscillate as that behavior is completely lost and you are left with a much more troubling outcome: expecting some accuracy but getting none. The Pade approximation implies that at a certain point sin is basically 0, which is not true at all. Maybe it works better for some functions but unless there is some result I'm ignorant of that gives some kind of mention on how much "longer" it is accurate and what kind of accuracy you are given. I'm sure there's a use for it or this approximation would be buried in numerical analysis textbooks and never reach the light of KZfaq.
@kenansi1624
@kenansi1624 2 жыл бұрын
I guess the point is that it is a meaningful way to extend the Taylor expansion for a better approximation with the same amount of parameters. Often the limiting behavior tells something about the behavior near a point. Choosing among [N - M/M] also depends on the limiting behavior. In the example given here, e^-x and sin x, the limiting behavior is the order of constant. So, it’s natural to choose N=2M, and I have a feeling that it gives the best local approximation among all combinations including the Taylor approximation. Edit: the asymptotic behavior of e^-x should be approaching zero beyond all finite polynomials. So, the best one should be [0/N].
@JamesBlevins0
@JamesBlevins0 2 жыл бұрын
Worse-case analysis: There are functions for which Padé approximation is no better than Taylor series approximation. Most functions: Padé approximants are better, especially for functions with singularities.
@deathworld5253
@deathworld5253 2 жыл бұрын
Also, Taylor series is more suitable for both taking derivatives and integrating because it's just a sum of easy functions. It's much harder to to this with Pade approximations
@Aufenthalt
@Aufenthalt 2 жыл бұрын
I knew already the Pade approximants (see e.g. Bender-Orszag book) but I never understood why they give an advantage. Thanks for explaining.
@jim42078
@jim42078 2 жыл бұрын
I'm fairly sure this has givne me an idea that'll be useful, however indirectly, in a paper I'm working on. Thanks for the video!
@DrWillWood
@DrWillWood 2 жыл бұрын
Awesome! good luck with the paper!
@dvir-ross
@dvir-ross 2 жыл бұрын
Thanks for sharing! I learned something new today 🙂
@Cathal7707
@Cathal7707 2 жыл бұрын
Incredibly useful for curve fitting with functions that you know should tend to zero as x-> inf
@victorribera5796
@victorribera5796 2 жыл бұрын
It give you a lot of imporvement for only the extra step of rewritting the coefficients, quite impressive, indeed
@dr.rahulgupta7573
@dr.rahulgupta7573 2 жыл бұрын
Excellent presentation. vow !!
@amitozazad1584
@amitozazad1584 2 жыл бұрын
Simply amazing!
@drsamiaziz
@drsamiaziz 4 ай бұрын
A great explanation for Padé approximation. It is clear and simple. Allow me only to indicate a small error in Recap (at 6:27). In the last line the equation shows {0 = C_{2n} + B_1*C_{2n+1} + ...} and I think it should be written as {0 = C_{n+m} + B_1*C_{n+m+1} + ..} because the total number of equations is n+m; unless you were implying that n = m.
@spikypichu
@spikypichu 11 ай бұрын
Another great thing is that rational functions are often procedurally integratable. We can always factor the denominator into linears and quadratics (because of conjugate pairs) then we apply partial fractions. Terms of the form linear/quadratic can be dealt with using u-sub and arctan. If the quadratic is a perfect square then we only need u-sub. If there is an (ax+b)^2 - c form with positive c, we can further factor using difference of squares.
@lamediamond4172
@lamediamond4172 10 ай бұрын
Nice video, interesting stuff.
@Yxcell
@Yxcell 11 ай бұрын
Nice video, @DrWillWood!
@stefan11804
@stefan11804 11 ай бұрын
Never heard of Pade Approx. Thank you.
@zxborg9681
@zxborg9681 2 жыл бұрын
Very cool. This looks super useful in implementing DSP and FPGA based algorithms (my wheelhouse).
@nathanielvalla6142
@nathanielvalla6142 2 жыл бұрын
Yeah it very much reminds me of FIR versus IIRs
@tanmaysinha8138
@tanmaysinha8138 2 жыл бұрын
Very amazing video. You really explained the concepts very nicely and concisely. A few things: 1.) When we say that B_0 can be made 1 WLOG, are we not missing the cases when B_0=0? As such, we may be losing some generality 2.) Do we always know that the M+N dimensional linear system is non-singular ie has a solution? 3.) Since we are using an M+N dimensional Taylor expansion of f to get the [N/M] Padé approximation, would it not make more sense to compare against an M+N term Taylor expansion? For sin(x) the issue of diverging to infinity would still remain but atleast the expansion will agree with the curve for a bit longer.
11 ай бұрын
Such a greta video! So simple, elegant and effective. I would like to know however why the examples were given for the case where m=n, as in, when n=0, we just have a taylor polynomial. But you just explained how those are less efficient. I'm left wondering the question what is the best possible m, n pair for a given function, or perhaps more generally, a given degree.
@SuicideRedemption100
@SuicideRedemption100 11 ай бұрын
Also used in the scipy matrix exponential algorithm AND, it was used to solve the ising model at some point
@Geenimetsuri
@Geenimetsuri 2 жыл бұрын
I remember there was a Fortran course where we had to compute all sorts of fits using matrix algebra. Padé was one of them (as was Fourier, etc..). I had all but forgotten about it, thanks for sharing!
@nunyabeeswax7111
@nunyabeeswax7111 2 жыл бұрын
have you "nearly forgotten about it" or "not forgotten about it"? I'm dying to know, as I never get what "all but" is supposed to mean (I am not a native speaker). Thank you!
@spanishgame
@spanishgame 2 жыл бұрын
This is so cool??? How is this the first time I'm hearing of this after a decade of knowing about Taylor Series?????
@quintopia
@quintopia 2 жыл бұрын
Would you mind doing a followup about these? You could talk a lot more about end behavior, applications, and (the thing I wondered about the most) how to decide the best N and M.
@samieb4712
@samieb4712 2 жыл бұрын
Great video thanks !
@kaous5690
@kaous5690 11 ай бұрын
Saw this in complex analysis lessons. When a question asks for a gap between numbers and the tylor series gets infinite there, we had to use that so it goes to 0 instead. Didn't know it had a name and thought it was a part of the tylor series.
@Moe_Afkani
@Moe_Afkani 2 жыл бұрын
Helpful! Thanks.
@drgatsis
@drgatsis 11 ай бұрын
Nicely done!
@alessandro.calzavara
@alessandro.calzavara 2 жыл бұрын
Wow thanks! So interesting
@jimbrookhyser
@jimbrookhyser 2 жыл бұрын
Thanks! Will put this to good use! I'll have to try a few examples. The sine function worked, but both approximations seemed "bad" at similar points. Since we know sine is cyclic, maybe I need to compare the two approximations over 1, 1/2 and 1/4 cycle to see which might perform better (or even if there's a clear winner). Also, I would want to look more into e^-x, and I need to check the approximations for x
@antoinebrgt
@antoinebrgt 2 жыл бұрын
Very nice, thanks!
@vcubingx
@vcubingx 2 жыл бұрын
This was very well made. Great job!
@DrWillWood
@DrWillWood 2 жыл бұрын
I appreciate that, thank you!
@Formalec
@Formalec 11 ай бұрын
Very nice approximation to know
@luizhenriqueamaralcosta629
@luizhenriqueamaralcosta629 11 ай бұрын
Amazing job
@martinepstein9826
@martinepstein9826 2 жыл бұрын
It's really interesting that you choose the degree of the numerator and denominator, which sort of means you're deciding the asymptotic behavior. Then again, if a coefficient turns out to be 0 then the degree might not be what you wanted. On that note, you set this up as a 0th degree approximation but ended up with a -1st degree approximation. That's crazy to me. It's like the approximation knew that the function oscillates around 0 even though the first few terms of the Taylor series don't suggest this at all.
@beeble2003
@beeble2003 11 ай бұрын
"Then again, if a coefficient turns out to be 0 then the degree might not be what you wanted." Same is true for Taylor series, of course.
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