It's an extremely useful technique in the field of asymptotics (see Carl Bender's lecture series). For example, imagine that you have some function f(x), and you can calculate the derivatives of this function near x=0. However, the Taylor series of this function has zero radius of convergence. For example, its terms might be (-x)^n n!. How do you evaluate this function at x=1? Taylor says that such a limit doesn't exist, but Pade says (by seeing that the main sequence of approximants converges) that the limit is actually 0.596... See "Padé approximation of Stieltjes series" by Allen et al (1975) for more details

At the start of the video, the 4th taylor series displayed is for ln(1+x) rather than arctan(x). arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ... -

18:12 body once told me the world is gonna roll me

Another higher order approximation that is ordinarily skipped in Calculus is Gaussian quadrature. A fun application of fitting polynomials at variable points

a good place to stop should maybe include a word about what happens for different values of n and m. is a (1,5) approximation just as good as a (3,3) since they agree with the function up to the same degree?

In EE it is used quite ofren in control theory to approximate the transfer function of a delay exp(-sT)

One specifically good property of the Podé approximation for exp is that taking e^ix, while its Taylor polynomials don't satisfy |Pn(x)| = 1 for real x, the Podé rationals do. An example of its usefulness is in physics numerical computation. You'd sometimes have to multiply a bunch of e^ix's, and if their magnitudes are not 1, things diverge quickly. e^(-iHt) ≈ (1-iHt/2)/(1+iHt/2) is pretty great already. It's also used in creating digital filters from analog filters with the Z-transform z = e^(s*T) ≈ (1+sT/2)/(1+sT/2)

Yep. I didn't come across Padé approximants until well after (at least 10 years) I finished graduate school.

This is a really good tool to have in one's toolbox for approximating functions. Taylor/Maclaurin series have some well-known pitfalls for numerical approximations. For sin & cos, those Taylor series include a mess of large terms of alternating sign, which makes computation useless, despite their exact convergence everywhere. [Although, to be fair, the reflection & translation symmetries of those functions allow us to limit our interval of approximation to where the series is more manageable.] Fred

I have seen in 2 or 3 Chinese math books about the higher math background of the public exam questions, the authers have introduced Pade approximation in their books.

Some things that come to mind after watching the video: 1. Does it make sense to take the limit of m or n goes to infinity of the Pade ratio? 2. Does the difference between the largest powers m and n matter? are there classes of functions for which m and n are separated by a fixed amount in the approximation? 3. How often do we get the largest power of the denominator (n in this case) being larger than that of the numerator (m)? I imagine in this case the ratio is "more different" to Taylor series compared to when m is larger than n 4. How to make sure we get the right powers m and n for the ansatz at the beginning? What would happen if initially I didn't know what the Pade ratio for sin(x) look like and I chose a4 x^4 + a3 x^3 + .... / a2 x^2 + a1 x +... ? 5. What are the advantages of using Pade approximation compared to Taylor series seeing that the coefficients are more cumbersome to calculate?

Neat video. 👍 You should consider doing a short follow-up that compares the Pade vs the Taylor approximations for some functions to explain when you might prefer using Pade over Taylor.

I used least squares rational aproximation for sin(x) . impressed by the precision and genralization

15:24 A bit of correction is needed here: A function having derivatives of all orders doesn’t guarantee it to have a Taylor approximation/expansion. That is, the Taylor series might not converge to the original function, and even possibly fail to do so on any interval, however small. The standard example is the function defined as exp(-1/x) on the positive reals and 0 everywhere else.

Calculus course without Pade = Calclueless. I see what you did there.😂

15:00 note that numerator has odd powers, and denominator has even powers, required for the symmetry of sin (x).

outstanding!

You could also force the rational function through a mesh of known points for a more global approach. 🙂

I first encountered this in numerical methods course

This method is also useful for analytic continuation. It's insane. But it works because "nice" complex functions are really limited in what they can possibly look like if you already know some values. In other words, you can't just arbitrarily joint the known points by whatever you want, it won't work. Let's say I have a mysterious function f(z), for which I know the following values: f(i), f(2i), f(3i), ..., f(n*i). You can do the Pade approximation of this (for well-chosen numerator and denominator, often informed by known asymptotics) and then simply plug in any complex or real value of z. This is actually used in physics. There are also recursive formulas for calculating the coefficients if you write Pade in a different form.

At 12:00 rhe contradicting terms for b2 are not in the x^3 coefficient, but in the x^5 instead

What rules do we want for m and n to make a good approximation? m≥n, m≈n, etc.

In analysis, all functions are approximated using Taylor's series. What would be nice if one can replace the Taylor series with a Pade approximate. Interesting exercise. Rational functions seem to work better than polynomials.

Always glad to catch up on a bit of "Caclulus". I wonder how you pronounce that?

I had it on calculus Ii

What happens if you take n and m tending towards infinity. Is there any closed form for that like we have for Taylor series?

at 3:00 isn’t that the series expansion of the logarithm?

At 7:03 it should be a1 - a0*b1 in the numerator. Since a0 = 1 it doesn't change the result. But still

Can the degree of denominator be larger using this approximation? Also, what makes a function “nice” for us to be able to do this?

3:01 Well, the series for arctgx is actually wrong (needs to be x-x³/3+x⁵/5-x⁷/7+ο(x⁷)) By the way, it's the series for log(1+x)

Can they be used analogusly to powersieries to solve to ODEs?

I don't get why in the general case we truncate the Taylor series to the order m+n and not m-n... By truncating to the order m-n, we know that the maximum degree of the polynomial resulting from the multiplication of this truncated series with the denominator will be m (the same order as the numerator) ; isn't it what we want in order to identify coefficients corresponding to the same power of the left and right hand side polynomials ?

Nice! But, it seems like this technique yields a Pade approximation equal to the truncated Taylor series which is only an approximation of the original function. Can the Pade coefficients be refined, perhaps by a Nelder-Meade algorithm (or maybe something better?) to yield a better approximation of the original function?

At least in physics this and gaussian quadrature are offen used . They are not skiped at all

I was wondering if there are other series with different function building blocks besides polynomials (Taylor), rational functions (Pade and Laurent) and sin or cos (Fourier) series that can approximate given functions? Such as a sum of exponentials, logs, abs, square roots or others? Or maybe rather than a sum of these, a product of the building block functions? Thanks!

There's a typo in the title "Caclulus" :)

3:05 I don't think that arctan x series is correct. Arctan(x) is antisymmetric around x = 0, so there should be no even terms present in its series.

I recently stumbled upon this sum A(x) = x^L\sum_{t=0}^T (T-t)x^{t}\binom{t + L}{t}. It would be cool if you could cover its solutions for x1 in a video. Keep up with the good work!

Setting b_0=1 works well untill you try to approximate 1/x

The Taylor series for arctangent is wrong.

Please give it a try : N=12345678910111213......2024.(here 1 to 2024 are written in a row).Now the question is what is the remainder if N is divided by 2025?

I guess that the genious of Padé was to put b0 = 1 in order to make this work!

Nice

you made a mistake in arctan. instead of (-1,1] its [-1,1]

WHY does this even work..why are m and n different..when does m equal n..isnt everypne else wondering this??

with x^2+1 i garanti that the denominator is not zero

How has no one noticed "Caclulus"

Am I the only one who kept thinking _Padmé_ approximation, à la Star Wars? 😂

Divisions are not nice approximations for computer.

Audio is so low.

I'm glad this is not taught in courses. Seems pretty difficult to get and not at all as useful as Taylor series.

Its seen in most caclulus classes, i think you switched caclulus and calculus up.