By 苏剑林 | March 07, 2015
Why is the Second Part Long Overdue?
Actually, before writing this series, I had already planned out the content for the next several posts. I originally intended to confidently introduce some integral expansion techniques I had thought of; moreover, since I am quite familiar with the perturbation method itself, normally it wouldn't have taken this long for the second part to appear. However, while I was finishing the first part and preparing to write the second, I came across this response on Zhihu: http://www.zhihu.com/question/24735673
That article greatly expanded my understanding of series. It mentioned that the expansion of an integral is an asymptotic series. This made me hesitate, wondering if this series had any value, because an asymptotic series implies that no matter what expansion technique is used, the radius of convergence of the resulting series is zero.
Later, upon re-thinking, I realized that even for asymptotic series, there is room for improvement and methods to accelerate convergence. So I think these few articles of mine should still have a little bit of significance, and can also serve as an opportunity to introduce theories related to asymptotic series and singularities. Well, let's proceed with that in mind.
Step-by-Step Expansion in the Exponent
The technique mentioned in the previous article was to vary the constant coefficients and set the first-order term to 0 by adjusting variables. The main problem with that method is that it cannot solve for higher-order approximations step-by-step. However, its underlying idea is quite good: introducing adjustable variables to make the terms of the expansion zero, rather than performing a pure power series expansion. Below, we first try to place the expansion within the exponent to obtain a method that can find approximations step-by-step, serving as a prototype for the methods we want to develop further.
In general, we want to evaluate the integral
$$\int_{-\infty}^{+\infty} e^{-a x^2-\varepsilon x^4} dx$$
In the previous article, we hoped to use the integral $\int_{-\infty}^{+\infty} e^{-A x^2} dx$ to approximate it. The idea was to adjust $A$ to make the first-order term zero, but this only allowed us to go that far because there was only one adjustable variable, and generally, only one term can be set to zero. To achieve a step-by-step expansion, i.e., introducing multiple adjustable quantities, we set
$$A=a+a_1 \varepsilon + a_2\varepsilon^2 + a_3 \varepsilon^3 + \dots$$
And then consider
$$\int_{-\infty}^{+\infty} \left[e^{-a x^2-\varepsilon x^4}-e^{-(a+a_1 \varepsilon + a_2\varepsilon^2 + a_3 \varepsilon^3 + \dots)x^2}\right] dx$$
Expanding the integrand inside the brackets in terms of the order of $\varepsilon$, we get
$$\begin{aligned}x^2 e^{-a x^2} \left(a_1-x^2\right)\varepsilon &+ \frac{1}{2} x^2 e^{-a x^2} \left(-a_1^2 x^2+2 a_2+x^6\right)\varepsilon^2 \\&- \frac{1}{6} x^2 e^{-a x^2} \left(-a_1^3 x^4+6 a_1 a_2 x^2-6 a_3+x^{10}\right)\varepsilon^3+\dots\end{aligned}$$
By integrating term by term, we have
$$\begin{aligned}\sqrt{\frac{\pi}{a} }&\left[\frac{1}{4}(2 a a_1 -3)\left(\frac{\varepsilon}{a^2} \right)+\frac{1}{32}\left(4 a^2 \left(4 a a_2-3 a_1^2\right)+105\right)\left(\frac{\varepsilon}{a^2} \right)^2\right.\\
&\left.+\frac{1}{128}\left(3465-8 a^3 \left(8 a^2 a_3-12 a a_1 a_2+5 a_1^3\right)\right)\left(\frac{\varepsilon}{a^2} \right)^3+\dots\right]\end{aligned}$$
Setting each term to 0, we obtain a system of equations:
$$\left\{\begin{aligned}&2 a a_1 -3=0\\
&4 a^2 \left(4 a a_2-3 a_1^2\right)+105=0\\
&3465-8 a^3 \left(8 a^2 a_3-12 a a_1 a_2+5 a_1^3\right)=0\\
&\dots\end{aligned}\right.$$
Solving gives:
$$a_1=\frac{3}{2 a},\,a_2=-\frac{39}{8 a^3},\,a_3=\frac{657}{16 a^5},\,\dots$$
Therefore, we have the series:
\begin{equation}
\begin{aligned}
&\int_{-\infty}^{+\infty} e^{-a x^2-\varepsilon x^4} dx \\
=&\int_{-\infty}^{+\infty} e^{-\left(a+\frac{3}{2}\frac{\varepsilon}{a}-\frac{39}{8}\frac{\varepsilon^2}{a^3}+\frac{657}{16}\frac{\varepsilon^3}{a^5}\dots\right)x^2} dx\\
=&\sqrt{\frac{\pi}{a+\frac{3}{2}\frac{\varepsilon}{a}-\frac{39}{8}\frac{\varepsilon^2}{a^3}+\frac{657}{16}\frac{\varepsilon^3}{a^5}\dots}}
\end{aligned}
\tag{5}
\end{equation}
This is another asymptotic expression for the original integral. Equation $(5)$ is slightly more effective than equation $(3)$; compared to equation $(4)$, when $\varepsilon/a^2$ is very small, truncating the first few terms of $(5)$ for calculation can yield more significant digits than equation $(4)$. Of course, it fails when $a\to 0$ or $\varepsilon\to \infty$—but even in failure, it merely yields a finite 0 rather than other indeterminate results. This is one of the advantages of this expansion method.
The Significance of Asymptotic Series
As already mentioned, when taking infinitely many terms, the radius of convergence of an asymptotic series is actually 0. So what is the point of an asymptotic series? In fact, first of all, for numerical approximation, one can truncate the first few terms of the asymptotic series to obtain a quite good approximation for a certain range of the independent variable. Secondly, for some asymptotic series, we can use re-summation techniques to reconstruct its original series and obtain an exact solution—there are many examples of this in quantum field theory!
Therefore, we need asymptotic series, and we can also consider ways to improve them (i.e., reducing the rate of divergence). There are various techniques to do such things, as long as you have enough creativity. In the next article, we will demonstrate this once again.