Pade Approximation -- unfortunately missed in most Caclulus courses

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[welcomeblack]

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

[mulletronuk]

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

[miraj]

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 + ... -

[demenion]

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?

[Calcprof]

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

[MooImABunny]

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)

[jaafarmejri]

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

[GeoffryGifari]

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?

[ffggddss]

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

[divisix]

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.

[JacobHa]

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.

[TimMaddux]

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

[tomoki-vo]

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

[MyOneFiftiethOfADollar]

Thx for this.

Learned about Pade Approximants in a numerical analysis textbook by Burden an Faires

[Bodyknock]

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.

[xinpingdonohoe]

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

[LouisEmery]

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

[AndrewJonkers]

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

[fizikchy]

Another, simpler, approximation to sin x in the same interval [0, \pi] can be constructed just by a quadratic polynomial y=ax^2+bx+c. Finding the coefficients a, b, c is left as an exercise to the reader haha! Hint: Use three points (0, 0), (\pi/2, 1), (\pi, 0). This is a really "perfect" approximation, believe me.

[Kapomafioso]

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.