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  • Book: Carl M. Bender, Steven A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I Asymptotic Methods and Perturbation Theory, 1999


Pade technique for summing a series. Asymptotic series. Fuchs' theorem. Frobenius series.


Poles are "mocking up" the cut.

展示了 $\frac{\ln (1 + x)}{x}$ Pade 近似的结果. Mathematican 中可以用函数 PadeApproximant 来实现.


Stirling's Series

$$\begin{align} \Gamma(x) \sim x^{x-1/2} e^{-x}\sqrt{2\pi} \left(1 + \frac{1}{12 x} + \frac{1}{288 z^2} - \cdots \right) \end{align}$$

它收敛很慢. 但是变成 Pade 后, 就收敛很快. 但是 no idea why it works.

What is Asymptotic Series?

$$\begin{align} f(x) - \sum_{n = 0}^N a_n x^n \sim a_{N + 1} x^{N + 1} \quad \mathrm{as} \quad x\to 0 \end{align}$$

means that

$$\begin{align} f(x) - \sum_{n=0}^N a_n x^n \sim a_{N + 1} x^{N + 1} \quad \mathrm{as} \quad x\to 0 \quad \mathrm{for} \quad \mathrm{all} \quad N \end{align}$$


$$\begin{align} \lim_{x \to \infty} \frac{f(x) - \sum_{n=0}^N a_n x^n}{x^{N + 1}} = a_{N + 1} \quad \mathrm{for} \quad \mathrm{all} \quad N \end{align}$$

只有说一个 series asymptotic to 某个函数时, 才有意义. 直接说某个 series 是不是 asymptotic series 是没有意义的.

对于一个 series, 总有一个 function 是此级数 asymptotic to 的.

但对于一个 function 只有一个 series 是 asymptotic to 它的.

A function has an unique asymptotic series. But a series could be asymptotic to many different functions.

asymptotic series 也可以有其它的形式, 如

$$\begin{align} f(x) - \sum_{n = 0}^N (a_n - a) x^n \sim a_{N + 1} x^{N + 1} \quad \mathrm{as} \quad x\to a \end{align}$$

同样, $n$ 也不一定是整数. 可以是分数等等.

$$\begin{align} \frac{\Gamma(x)}{x^{x-1/2} e^{-x}\sqrt{2\pi}} \sim 1 + \frac{1}{12 x} + \frac{1}{288 z^2} - \cdots \quad \mathrm{as} \quad x \to \infty \end{align}$$

An Example about "How do asymptotic series arise?"

$$\begin{align} x^3 y'' = y \end{align}$$

在 near $x=0$ 处进行 local analysis.

偏题: 如何分析一个 second order linear DE?


$$\begin{align} a(x) y'' + b(x) y' + c(x)y = 0 \end{align}$$

的方程, 首先化成 normal form. 两边同时除以 $a(x)$ , 变成了如下形式

$$\begin{align} y'' + A(x) y' + B(x)y = 0 \end{align}$$

要分析它在 $x\to x_0$ 处的行为. How does $y(x)$ behave near $x = x_0$ ?

There are exactly three possibilities there.

case 1

$A(x)$ , $B(x)$ analytic near $x = x_0$ ( $x_0$ is a regular point)

此时有定理, Fuchs' theorem. (Lazarus Immanuel Fuchs (5 May 1833 – 26 April 1902) was a Jewish-German[1] mathematician who contributed important research in the field of linear differential equations)

All solutions have a Taylor series about $x = x_0$

$$\begin{align} y(x) = \sum_0^{\infty} (x - x_0)^n a_n \end{align}$$

case 2

Not both $A$ and $B$ are analytical. But $(x - x_0) A(x)$ and $(x - x_0) B(x)$ are analytical. ( $x_0$ is a regular singular point)

At least one solution has the form (Frobenius series)

$$\begin{align} y(x) = \sum_0^{\infty} (x - x_0)^{n + \alpha} a_n \end{align}$$


$$\begin{align} y(x) = \sum_0^{\infty} (x - x_0)^{n + \alpha} a_n \ln(x - x_0) \end{align}$$

case 1 and case 2 solutions' converge radius is often the distance from $x_0$ to the nearest singularity.

case 3

$x_0$ is an irregular point.

local analysis


$$\begin{align} x^3 y'' = y \end{align}$$


$$\begin{align} A = 0, \quad B = \frac{1}{x^3} \end{align}$$

它属于 case 3. 如果强行写一个 Frobenius series 的解

$$\begin{align} y(x) = \sum_{n = 0}^{\infty} a_n x^{n + \alpha} \quad(a_n \neq 0) \end{align}$$


$$\begin{align} \sum_{n = 1}^{\infty} (n - 1 + \alpha)(n - 2 + \alpha) a_{n - 1} x^{n + \alpha} = \sum_{n = 0}^{\infty} a_n x^{n + \alpha} \end{align}$$

结果是 $a_0 = 0$ , $a_n = 0$ . 也就是它没有 Frobenius series 的解.

这个问题太难了, 让我们看一下一阶的问题, 能不能猜一下它的解.

$y' = \frac{y}{2}$

它的 case 1, 在 $x = 0$ 处是 regular point. 那么它有 Taylor series 形式的解. 当 然的, 它的解是

$$\begin{align} y(x) = C e^{\frac{x}{2}} = C \sum_{n = 0}^{\infty} \frac{(x/2)^n}{n!} \end{align}$$

$y' = \frac{y}{2x}$

它是 case 2, 在 $x = 0$ 处是 regular singular point. 它的解为

$$\begin{align} y(x) = C \sqrt{x} \end{align}$$

这是 Frobenius series, $\alpha = 1/2$ , 只有一项. Frobenius was right!

$y' = \frac{y}{2x^2}$

它是 case 3, irregular point. 但是它的可解的, 结果是

$$\begin{align} y = C e^{- \frac{1}{2x}} \end{align}$$


$$\begin{align} y = e^{s(x)} \end{align}$$

How do asymptotic series arise?

因此, Green (是 George Green?)给出了这样的 suggested, 一个 $y = e^{s(x)}$ 形式的 试探解( $s(x)$ is singular, blowing up at $x = 0$ )

$$\begin{align} y(x) = e^{a x^b}, \quad b<0 \end{align}$$

如果直接将 $y = e^{s(x)}$ 代回 $x^3 y'' = y$ , 会得到 Riccati

$$\begin{align} x^3(s'' + s'^2) &= 1 \\ \Downarrow& \\ x^3(T' + T^2) &= 1 \end{align}$$

如果 $s = a x^b$ 那么会得到

$$\begin{align} s' =& b a x^{b - 1}\\ s'' =& b(b - 1) a x^{b-2}\\ s'^2 =& b^2a^2 x^{2b - 2} \end{align}$$

Mr. Green noticed that if $b < 0$ which of $s'', s'^2$ is more important as $x\to 0$ ? (Its an issue of 2b or not 2b. That's the point! -_- !!! Bender 讲了这么多就 只是为了玩 shakespeare 的梗...当然不是...) 当然是

$$\begin{align} s'' \ll (s')^2 \quad \mathrm{as} \quad x \to 0 \end{align}$$

所以用了 the method of dominant balance( 第一节课讲的 ), 将等号的换成 $\sim$ , 问题 有了进展. 这也再次说明 equal sign make no progress! Replace this equation by an asymptotic relations

$$\begin{align} x^3(s'' + s'^2) &= 1 \\ \Downarrow& \\ x^3 s'^2 &\sim 1 \quad \mathrm{as} \quad x\to 0\\ \end{align}$$

ha! I can solve that equation. It's only a first order euqation.


$$\begin{align} s \sim \mp C\frac{2}{\sqrt{x}}\quad \mathrm{as} \quad x\to 0 \end{align}$$

Just like that! We cracked into the equation!

We know what $s$ is asymptotic to. 但是我们想要的是 $y(x) = e^{s(x)}$ . 那么问题 来了. If $f(x)\sim g(x)$ as $x\to x_0$ , is it true $e^{f(x)}\sim e^{g(x)}$ as $x\to x_0$ ? (因为 asymptotic 并不意味着 $f(x) - g(x)$ 很小, 它可以很大, 比如 $x^2 + x \sim x^{2}$ as $x\to \infty$ , 但它们相差无穷大!)

Only true when $f(x) - g(x) \ll 1$ as $x \to x_0$ . The one thing that you cannot do both sides of an asymptotic approximation is to exponent it.

$\mp \frac{2}{\sqrt{x}}\quad \mathrm{as}$ 叫做 controlling factor. 因此我们只能 roughly speaking

$$\begin{align} y(x) \sim e^{\frac{2}{\sqrt{x}}} \quad \mathrm{or} \quad e^{-\frac{2}{\sqrt{x}}} \quad \mathrm{as} \quad x\to 0 \end{align}$$

有一个解在 $x \to 0$ 时很快地 blow up. 另一个趋于 $0$ . 这两个解都没有 Frobenius 或者 Taylor series.

Bender 最后总结, 新的概念就是这样被发明的. 这个方程的解一定是一个函数, 但是通常 的方法都没法表示, 所以 we're forced to a new way to represent the function.


$$\begin{align} y(x) \sim e^{\mp \frac{2}{\sqrt{x}}} x^{3/4} \sum_{n = 0}^{\infty} a_n x^{n/2} \quad \mathrm{as} \quad x \to 0 \end{align}$$

之后会发现这是个 divergent series. 所以说, divergent series are forced upon us. This is the only way to represent the solution. We don't have a convergent series representation.

之后会讲如何 sum that series and actually write down the answer.


介绍了 asymptotic series 的定义, 之后重点用一个例子说明我们是如何有了 asymptotic series 这个概念的: 我们发现了一些方程, 它的解的形式只能用 asymptotic series 来表 示. 这就像 $x^2 = -1$ 的解让我们不得不拥有了复数的概念一样.