Asymptotic Expansions

1. Asymptotic Sequences and Expansions
2. Uniqueness of Asymptotic Expansions
- 2.1. Non-Uniqueness
3. Transcendentally Small Terms
4. Arithmetic of Asymptotic Series
5. References

1. Asymptotic Sequences and Expansions

An Asymptotic Sequence (or "asymptotic scale") as \(x\to x_{0}\) consists of a sequence of functions \(\{\phi_{j}(x)\}_{j\in\NN}\) such that for all \(j\), \(\phi_{j}(x)\gg\phi_{j+1}(x)\).

Some definitions of an asymptotic sequence reverses the ordering, i.e., defines it as such that \(\phi_{j+1}(x)\gg\phi_{j}(x)\) as \(x\to x_{0}\), for any \(j\). This is useful for \(x_{0}=\infty\), for example.
The indexing set may be extended to be something like \(\{k\in\ZZ|k\geq n\}\). I'm not sure if we can extend it to the entire set of integers, we may run into difficulties.

An asymptotic sequence as \(x\to\infty\) which is frequently used are the inverse monomials \(\phi_{k}(x) = x^{-k}\). We see \(\phi_{k+1}(x)\ll\phi_{k}(x)\) as \(x\to\infty\).

Let \(f(x)\) be a function and \(\{\phi_{j}(x)\}_{j\in\NN}\) an asymptotic sequence as \(x\to x_{0}\). The Asymptotic Expansion of \(f(x)\) as \(x\to x_{0}\) is a function

\begin{equation} f(x)\sim a_{1}\phi_{1}(x) + \dots + a_{n}\phi_{n}(x) + \dots \end{equation}

such that for each \(n\in\NN\), as \(x\to x_{0}\) we have the partial sums

\begin{equation} f(x) - \sum^{n}_{j=1}a_{j}\phi_{j}(x)\ll\phi_{n}(x). \end{equation}

(Equivalently, the remainder from the n-th partial sum is dominated by the n-th element in the asymptotic sequence.)

Just take a note to compare an asymptotic series to a convergent series for a sequence of partial sums

\begin{equation} S_{n}(x) = \sum^{n}_{j=1}a_{j}\phi_{j}(x) \end{equation}

Convergence: \(S_{n}(x)\to f(x)\) as \(n\to\infty\) for each fixed \(x\)
Asymptotic: \(S_{n}(x)\sim f(x)\) as \(x\to x_{0}\), for each fixed \(n\).

2. Uniqueness of Asymptotic Expansions

Relative to an asymptotic sequence \(\{\phi_{j}(x)\}\) as \(x\to x_{0}\), the asymptotic expansion of \(f(x)\) is unique. But if we change asymptotic sequences, we get different coefficients, and so this uniqueness is with the asterisk "For a fixed asymptotic sequence".

More precisely, if we expand a given function as

\begin{equation} f(x)\sim a_{1}\phi_{1}(x) + \dots \end{equation}

then the claim is the coefficients \(a_{1}\), …, are uniquely determined for that sequence. Why?

Well, we determine the coefficients as

\begin{equation} a_{1} = \lim_{x\to x_{0}}\frac{f(x)}{\phi_{1}(x)} \end{equation}

since

\begin{equation} \lim_{x\to x_{0}}\frac{f(x)}{a_{1}\phi_{1}(x)} = 1. \end{equation}

(Otherwise we could not have \(f(x)\sim a_{1}\phi_{1}(x)\), contradicting the fact we are using an asymptotic expansion.)

Now, the next coefficient in the expansion is given by

\begin{equation} f(x)\sim a_{1}\phi_{1}(x) + a_{2}\phi_{2}(x) \end{equation}

which we find the coefficient by

\begin{equation} a_{2} = \lim_{x\to x_{0}}\frac{f(x) - a_{1}\phi_{1}(x)}{\phi_{2}(x)}. \end{equation}

Prove if \(f(x)\sim g_{1}(x) + g_{2}(x)\) as \(x\to x_{0}\), then \(f(x)-g_{1}(x)\sim g_{2}(x)\) as \(x\to x_{0}\). Do we need the additional hypothesis that \(g_{1}(x)\ll g_{2}(x)\) as \(x\to x_{0}\)?

The uniqueness of the coefficient depends on the uniqueness of these limits

\begin{equation} a_{n+1} = \lim_{x\to x_{0}}\frac{f(x) - S_{n}(x)}{\phi_{n+1}(x)}. \end{equation}

2.1. Non-Uniqueness

If we have two different asymptotic sequences \(\{\phi_{j}(x)\}\) and \(\{\psi_{j}(x)\}\), then the asymptotic expansions of \(f(x)\) as

\begin{equation} f(x)\sim\sum_{j=1}a_{j}\phi_{j}(x) \end{equation}

and

\begin{equation} f(x)\sim\sum_{j=1}b_{j}\psi_{j}(x) \end{equation}

does not necessarily imply \(a_{j}=b_{j}\) for all \(j\).

For a simple intuitive example, consider \(\phi_{j}(x) = x^{j}\) and \(\psi_{j}(x) = \exp(jx)\). We'd clearly get different asymptotic expansions as \(x\to\infty\).

3. Transcendentally Small Terms

We call \(f(x)\) a Transcendentally Small Term if \(f(x)\ll x^{-n}\) as \(x\to\infty\) for any (fixed) \(n\geq0\).

Bender and Orszog calls such a term Subdominant to \(x^{-n}\).

The function \(\exp(-x)\) is a transcendentally small term as \(x\to\infty\). We'll need to show that \(\exp(-x)\ll x^{-n}\) as \(x\to\infty\) for any \(n\geq 0\). Check:

\begin{equation} \lim_{x\to\infty}\frac{\E^{-x}}{x^{-n}} = \lim_{x\to\infty}x^{n}\E^{-x} = 0. \end{equation}

This is true from basic calculus, so we have the desired result.

But while on the subject, note the asymptotic expansion

\begin{equation} \exp(-x)\sim a_{0} + \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} + \dots \end{equation}

has all zero coefficients \(a_{j}=0\) for \(j\geq0\). Hence \(f(x)\) and \(g(x)=f(x)+C\E^{-x}\) have the same asymptotic expansion, so transcendentally small terms are invisible to asymptotic power series and can cause numerical difficulties.

"Hyper-asymptotics" attempts to deal with transcendentally small terms (also called "asymptotics beyond all orders" or a perturbation problem beyond all orders). There's a note by Boyd, "The Devil's Invention" (2000) reviewing the field.
Trouble is because \(\exp(-z)\) is not analytic in \(\CC\), and has an essential singularity at \(\infty\).

4. Arithmetic of Asymptotic Series

We can add, subtract, multiply, divide, and integrate asymptotic series term-by-term. If, as \(x\to x_{0}\), we have

\begin{equation} \begin{split} f(x) &\sim f_{1}\phi_{1}(x) + f_{2}\phi_{2}(x) + \dots \\ g(x) &\sim g_{1}\phi_{1}(x) + g_{2}\phi_{2}(x) + \dots \end{split} \end{equation}

then we have

\begin{equation} f(x)g(x) \sim \left(f_{1}\phi_{1}(x) + f_{2}\phi_{2}(x) + \dots\right)\left(g_{1}\phi_{1}(x) + g_{2}\phi_{2}(x) + \dots\right) \end{equation}

and so on.

BUT we must be careful with the operations of substitution and differentiation.

Consider

\begin{equation} f(x) = \exp(x^{2}) \end{equation}

and

\begin{equation} x(\varepsilon) = \frac{1}{\varepsilon} + \varepsilon \end{equation}

as \(\varepsilon\to 0\) (so \(x\to\infty\)).

Correct way: we have exactly

\begin{align} f(x(\varepsilon)) &= \exp(x(\varepsilon)^{2}) \\ &= \E^{\varepsilon^{2} + 2 + (1/\varepsilon)^{2}} \\ &= \E^{2}\E^{\varepsilon^{2}}\E^{(1/\varepsilon)^{2}} \end{align}

But the Taylor expansion of the exponential function for small \(\varepsilon\) gives us

\begin{equation} \E^{\varepsilon^{2}} = 1 + \varepsilon^{2} + \frac{1}{2!}\varepsilon^{4}+\dots \end{equation}

which means

\begin{equation} \begin{split} f(x(\varepsilon)) &= \E^{\varepsilon^{2}}\E^{(1/\varepsilon)^{2}}\left(1 + \varepsilon^{2} + \frac{1}{2!}\varepsilon^{4}+\dots\right)\\ &\sim \E^{2}\E^{1/\varepsilon^{2}} \end{split} \end{equation}

as \(\varepsilon\to0\). We only used the asymptotic expansion of the "small part" after doing the substitution.

Wrong way: if we tried asymptotically expanding first, then substitution, we get the wrong results. We'd have \(x(\varepsilon)\sim1/\varepsilon\), and substitution gives us

\begin{equation} f(x(\varepsilon))\sim\E^{1/\varepsilon^{2}}\quad\mbox{(WRONG!)} \end{equation}

which is off numerically by a factor of \(\E^{2}\).

Differentiation also causes problems if done without care. Observe

\begin{equation} f(x) = \sin(x)+x\sim x \end{equation}

as \(x\to\infty\), since

\begin{equation} \lim_{x\to\infty}\frac{\sin(x)}{x}=0. \end{equation}

But naively differentiating after taking asymptotics predicts

\begin{equation} f'(x)\sim 1 \quad\mbox{(WRONG)}. \end{equation}

But

\begin{equation} f'(x) = 1 + \cos(x)\not\sim 1 \end{equation}

as \(x\to\infty\). The trouble is \(\sin(z)\) has an essential singularity at \(\infty\).

Tauberian theorems tell us when it is okay to differentiate an asymptotic expansion. For example, if \(f(x)\) and \(f'(x)\) have asymptotic expansions in \(\{\phi_{j}(x)\}\) as \(x\to x_{0}\), then it's OK to differentiate term-by-term the asymptotic expansion.

5. References

John P. Boyd,
"The Devil's Invention: Asymptotic, Superasymptotic and Hyperasymptotic Series".
Acta Applicandae Mathematicae 56, no.1 (2000) ResearchGate, CiteSeerX