Information

  • 原出处: https://www.perimeterinstitute.ca/video-library/collection/11/12-psi-mathematical-physics
  • 可以看自动生成的字幕: [[https://www.youtube.com/playlist?list=PLzcd6SoIscwjHuWRE38UXWG92uq0Sy4UF]]
  • Bilibili: https://www.bilibili.com/video/BV1w4411q7x6?fromsearch&seid7852838902448285010
  • Book: Carl M. Bender, Steven A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I Asympotic Methods and Perturbation Theory, 1999

Keywords

Continued fractions. Pade sequence. Shear magic. Stieltjes series

Continued Factions

有一个 $HP(\epsilon)$ 微扰得到一个 asympotic series 形式的结果 $ANS(\epsilon) \sim \sum a_n \epsilon^n$ , 但是一般只有前几项, 而且收敛半径是 $0$ . 思路上将其转化成 continued functions, a sequence of "approximants" , 而这个 sequence 的收敛性比那个 asympotic series 的收敛性好的多. 之前讲 $e^{ze^{ze^{z\cdots}}}$ 的例子只因为 Bender love 这个例子, 因为它太 weird, 它 glorious, 并且鲜为人知. 之后并不会再讨论 continued exponential.

下面考虑 continued fractions

$$\begin{align} \label{eq:cf} \sum a_n x^n \to \frac{b_0}{1 - \frac{b_1x}{1 - \frac{ b_2x}{1 - b_3 x \cdots}}} \end{align}$$

idea 是, 我们求得了 (\ref{eq:cf}) 的左边, 但它是发散的. 如果把它转换成 (\ref{eq:cf}) 右边的 continued fractions 的形式, the sequence of approximants you get from the continued fractions might well converge. the sequence of approximants 指的是(假设收敛到 $L$ )

$$\begin{align} b_0, \frac{b_0}{1 - b_1x}, \frac{b_0}{1 - \frac{b_1x}{ 1 - b_2x}}, \cdots, \to L \end{align}$$

也就是说, 我们从一个发散的, 但是包含我们所需要的信息的级数, 转化成一个收敛的级数, 因此我们从其中提取所需要的信息.

IQ Test

用一个特别的方式来说明这个.

像 IQ 测试一样, 下面数例的下一个数是什么

$$\begin{align} 1, 5, 61, ?? \end{align}$$

这太难了, 你之前知道就会, 之前不知道就不会. 它们是 Euler numbers. 下面讲一种方法 来计算它. 将上面的数列记为(手动加上 $a_0 = 1$ )

$$\begin{align} a_0 = 1, a_1= 1, a_2 = 5, a_3 = 61, \cdots \end{align}$$

我们想将它 transform 成一个更明显, 更好猜的数例 $b_n$

$$\begin{align} b_0, b_1, b_2, b_3, \cdots \end{align}$$

下面分三个步骤进行

step1

Given $\{a_n\}$, imagine that (moment problems) \begin{align} a_n = \int_{-L}^L \mathrm{d}x\cdot w(x) x^{2n} \end{align}

step2

If you have $\{b_n\}$ \begin{align} P_0 (x) =& 1 \\ P_1(x) \equiv& x \\ P_{n + 1}(x) =& x P_n - b_n P_{n - 1}(x) \end{align} so (monic) \begin{align} P_2(x) =& x^2 - b_1 \\ P_3(x) =& x^{3} - (b_1 + b_2) x \\ P_4(x) =& x^4 - (b_1 + b_2 + b_3) x^2 + b_3 b_1 \end{align}

step3

Let's require these polynomials are orthogonal with respect to $w$ \begin{align} \int_{-L} ^L \mathrm{d}x \cdot w(x) P_n(x) P_m(x) = 0 , \quad n\neq m \end{align} so \begin{align} P_0 \bot P_2 :& a_1 - b_1 = 0 \Rightarrow a_1 = b_1 \\ P_1 \bot P_3 :& a_2 - (b_1 + b_2 ) b_1 = 0 \Rightarrow a_2 = b_1(b_1 + b_2) \\ P_4 \bot P_o :& a_2 = b_1(b_1 + b_2) \\ P_4 \bot P_2 :& a_3 = b_1(b_1 + b_2)^2 + b_1 b_2 b_3 \end{align} $\mathscr{P}$

发现结果恰好是之前的数例

$$\begin{align} a_1 = 1 \to& b_1 = 1 \\ a_2 = 5 \to& b_2 = 4 \\ a_3 = 61\to& b_3 = 9 \\ a_4 = 1385 \to& b_4 = 16 \end{align}$$

If $a_n\sim n!$ , then $b_n \sim n$ . If $a_n \sim (2n)!$ , then $b_n\sim n^2$ . 这里的 $a_n, b_n$ 的关系, 就是前面的级数和 continued fractions 的关系. 比如

$$\begin{align} E_n(1, 5, 61, 1385, \cdots) \sim (2n)! C^n,\quad \mathrm{as}\, n\to \infty \end{align}$$

Pade Sequence

There is very deep connetction between continued fractions and orthogonal polynomials. 当然, 最简单的方法是直接 Taylor 展开, 如

$$\begin{align} \frac{1}{1 - x} =& 1 + x + \mathcal{O}(x^2) \\ \frac{1}{1 - \frac{x}{1 - 4x}} =& 1 + x + 5 x^2 + \mathcal{O}(x^2) \\ \frac{1}{1 - \frac{x}{1 - \frac{4x}{1 - 9 x}}} =& 1 + x + 5 x^2 + 61 x^3 + \cdots \\ \vdots & \end{align}$$

如果把 continued fractions 有理化, 那么

$$\begin{align} b_0 \to& \frac{\mathscr{P}_0}{\mathscr{P}_0} \label{eq:padeStart}\\ \frac{b_0}{1 - b_1x}\to& \frac{\mathscr{P}_0}{\mathscr{P}_{1}} \\ \frac{b_0}{1 - \frac{b_1 x}{ 1 - b_2x}} \to& \frac{\mathscr{P}_1}{\mathscr{P}_1}\\ \frac{b_0}{1 - \frac{b_1 x}{ 1 - \frac{b_2x}{1 - b_3x}}} \to& \frac{\mathscr{P}_1}{\mathscr{P}_2} \\ \vdots \to& \frac{\mathscr{P}_2}{\mathscr{P}_2} \\ \vdots \to& \frac{\mathscr{P}_2}{\mathscr{P}_3} \label{eq:padeEnd}\\ \vdots & \end{align}$$

其中 $\mathscr{P}_n$ 代表 $n$ 阶多项式. 右边叫做 Pade sequence. 更 general 的

$$\begin{align} \sum_0^{p+q} a_n x^n = \frac{\mathscr{P}_p}{\mathscr{P}_q} = \mathcal{P}^p_q \end{align}$$

两边的系数相等. 被称作 Pade approximates. 对于一个级数, 可以写一个 Pade table (Bender 在黑板上笔误了上下标标反了) 而前面的 (\ref{eq:padeStart}) ~ (\ref{eq:padeEnd}) 是 Pade table 中(副)对角的部

$q=1$ $q=2$ $q=3$ $\cdots$
$p=1$ $\mathcal{P}^1_1$ $\mathcal{P}^1_2$ $\mathcal{P}^1_3$ $\cdots$
$p=2$ $\mathcal{P}^2_1$ $\mathcal{P}^2_2$ $\mathcal{P}^2_3$ $\cdots$
$p=3$ $\mathcal{P}^3_1$ $\mathcal{P}^3_2$ $\mathcal{P}^3_3$ $\cdots$
$\vdots$ $\vdots$ $\vdots$ $\vdots$ $\ddots$

分, 称为 main/diagonal Pade sequence.

总的来说, 我们将一个不收敛的级数, 写成了 Pade sequence 的形式, 而这通常收敛非常 快, really really rapidly.

接下来是精辟的评论

It's the antidote to what they teach you in calsses on quantum field theory and on quantum mechanics. Because in a class on quantum mechanics or quantum field theory they teach you perturbation theory and without teaching you this. It's junk! Because you can't do anything with what you've learned. You calculate a bunch of Feynman diagrams, big deal, but you convert that to Pade, and you can sum it up.

Pade 不仅收敛地快, 而且 $\mathcal{P}^n_n$ 和 $\mathcal{P}^n_{n + 1}$ 分别从 $L$ 的 上边和下边逼近, 因此还可以做 Shanks transform, 会收敛地更快.

之后举了几个例子, 来说明 Pade sequence 比 Taylor series 收敛地更快.

An Intuitive Explanation

Pade 是多项式, 它的分母可以有零点. 比如 $\Gamma$ function 有许多 pole, Taylor series 没有 pole, 但 Pade 可以有. Pade 可以 mock up (模拟) poles. 因此 Pade 是一个 very flexible 的表示.

同时, Taylor series 的收敛区域是一个圆, 它要求圆内所有的点都收敛. 这就限制了它的 收敛的区域. 而 Pade 就不必, 它的收敛区域可以是整个复平切去割线. Pade 所要求的更 少, 它的收敛区域不必是一个圆, 因此, 它可以给我们的更多.

Taylor series hate you, because you force them to converge in a full circle. It just say 'go away, I'm not working for you today.' Whereas Pade, you don't force it to converge here, you're not requiring it and it doesn't converge, and therefore where it does converge it converges much better and much faster than a Taylor series. It's wonderful!

Summary

讲了如何将发散的级数求和, 给出的方法是将原来发散的级数转化成 Pade sequence.

Reference

  • https://en.wikipedia.org/wiki/Pad%C3%A9_table