# Padé approximants again

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An $(n, m)$th Padé approximant for a generating function $f(x)$ is a pair of polynomials $(P_{n, m}(x), Q_{n, m}(x))$ such that $P_{n, m}(x)$ and $Q_{n, m}(x)$ have degrees not exceeding $n$ and $m$, respectively, and

$\begin{equation} \label{pade} \tag{Padé} f(x) Q_{n, m}(x) - P_{n, m}(x) = O(x^{n + m + 1}), \end{equation}$

where $O(x^k)$ stands for any generating function which is zero before the $k$th term (and possibly later).

Given $n$ and $m$, we write the polynomials $P_{n, m}(x)$ and $Q_{n, m}(x)$ explicitly as

\begin{align*} P_{n, m}(x) &= \sum_{k = 0}^n a(n, m, k) x^k \\ Q_{n, m}(x) &= \sum_{k = 0}^m b(n, m, k) x^k. \end{align*}

We assume that $a(n, m, k) = 0$ if $k > n$ or $k < 0$, and that $b(n, m, k) = 0$ if $k < 0$ or $k > m$. Such coefficients are guaranteed to exist and, for a fixed $f(x)$, are unique up to a constant multiple1. It is common to take $b(n, m, 0) = 1$ just so there is a fixed starting point.

In a previous post I derived a closed form for Padé approximants of the function $(1 - 4x)^{-1/2}$. Here, I want to show that we can generalize this by choosing an arbitrary exponent.

Highlights:

• The Padé approximants of the function $f(x) = (1 + tx)^c$ are uniquely determined up to a constant multiple2. Choosing the constant to be $1$, the coefficients are as follows:
\begin{align*} a(n, m, k) &= t^k {n \choose k} {m + c \choose k} {n + m \choose k}^{-1} \\ b(n, m, k) &= t^k {m \choose k} {n - c \choose k} {n + m \choose k}^{-1}. \end{align*}
• This is proven by establishing (automatically!) the identity
$\sum_{j = 0}^k {m \choose j} {n - c \choose j} {c \choose k - j} {n + m \choose j}^{-1} = {n \choose k} {m + c \choose k} {n + m \choose k}^{-1}.$

Differences between previous post and this one:

• In my last post, I considered only diagonal approximants. Also, I defined $a(n, k)$ to be the coefficient on $x^{n - k}$ rather than $x^k$. This seemed useful at the time, but Dr. Z has shown me the error of my ways.

• My last post tried to simplify the coefficients by taking $a(n, 0) = 1$. The smart thing is to realize that it doesn’t really matter.

More on these two points later. For now, let’s get to the good stuff.

# Padé approximants in detail

From an analytic perspective, the point of a Padé approximant as defined in \eqref{pade} is that the power series of the rational function $P_{n, m}(x) / Q_{n, m}(x)$ and $f(x)$ ought to agree for the first $n + m$ terms. The form in \eqref{pade} is slightly more general, and suggests how to actual compute the coefficients.

Looking at the coefficient on $x^k$ for $0 \leq k \leq n + m$ in \eqref{pade} shows that necessary and sufficient conditions to be the coefficients of a Padé approximant are

$\begin{equation} \label{convolution} \sum_j b(n, m, j) c(k - j) = a(n, m, k), \qquad 0 \leq k \leq n + m, \end{equation}$

and the boundary conditions on $a(n, m, k)$ and $b(n, m, k)$. (These are implicitly assumed in the sum since it ranges over all integers $j$.)

The boundary conditions ensure that the coefficients are unique and that we don’t need to go beyond the degrees $n$ and $m$ for our polynomials. For example, lots of sequences satisfy \eqref{convolution}. In fact, for any fixed $b(n, m, k)$ and $c(k)$, we could just define $a(n, m, k)$ by \eqref{convolution}. (Equivalently, just pick a generating function $Q(x)$ and define another one called $P(x) = f(x) Q(x)$.) This gets you the formal identity, but with possibly infinite generating functions rather than proper polynomials.

As an example, take $f(x) = (1 + x)^{-1/2}$ and $b(n, m, k) = (n + m) / 2^k$. Computing the “$(4, 4)$th Padé” approximant by defining $a(n, m, k)$ by \eqref{convolution} yields the rational function

$\frac{27x^4 - 16x^3 + 48x^2 + 128}{8x^4 + 16x^3 + 32x^2 + 64x + 128}.$

However, the series expansion of this rational function disagrees with that of $(1 + x)^{-1/2}$ at the fifth term, when it ought to agree until at least the eighth!

To learn more about Padé approximants than you would ever care to know, see Chapter 20 and on of Wall’s Analytic Theory of Continued Fractions.

# Recurrences

Suppose that $f(x) = (1 + x)^c$ for some real $c$. Using the Maple packages PADE and FindRec, I have conjectured the following recurrences3 for the coefficients $a(n, m, k)$ and $b(n, m, k)$ of the $(n, m)$th Padé approximant for $f$:

\begin{align} \begin{split} a(n, m, k + 1) &= \frac{(n - k)(m + c - k)}{(k + 1) (n + m - k)} a(n, m, k) \\ b(n, m, k + 1) &= \frac{(m - k)(n - c - k)}{(k + 1)(n + m - k)} b(n, m, k) \end{split} \label{rec} \tag{Recurrence} \end{align}

Unrolling this and shifting $k$ back by $1$ gives the following solutions:

\begin{align*} a(n, m, k + 1) &= a(n, m, 0) \frac{n^{\underline{k}} (m + c)^{\underline{k}}}{k! (n + m)^{\underline{k}}} \\ b(n, m, k + 1) &= b(n, m, 0) \frac{m^{\underline{k}} (n - c)^{\underline{k}}}{k! (n + m)^{\underline{k}}}. \end{align*}

But if we just look at these long enough, we see how we should write them:

\begin{align} \begin{split} a(n, m, k) &= a(n, m, 0) {n \choose k} {m + c \choose k} {n + m \choose k}^{-1} \\ b(n, m, k) &= b(n, m, 0) {m \choose k} {n - c \choose k} {n + m \choose k}^{-1}. \end{split} \label{solution} \tag{Solution} \end{align}

All that’s left are the two constant terms. Setting $x = 0$ in \eqref{pade} tells us that $a(n, m, 0) = f(0) b(n, m, 0)$, and $f(x) = (1 + x)^c$ gives $f(0) = 1$, so the constant terms are actually equal. Padé approximants are only equal up to a constant multiple anyway, so from now on let’s just assume that $a(n, m, 0) = b(n, m, 0) = 1$. We’ll come back to this later.

# Verification

If we define $a(n, m, k)$ and $b(n, m, k)$ to vanish when they should, then all we need to check to prove our conjecture is that \eqref{convolution} holds with $c(k) = [x^k] (1 + x)^c = {c \choose k}$. Writing this out, it is our challenge to prove the following:

$\sum_{j = 0}^k {m \choose j} {n - c \choose j} {c \choose k - j} {n + m \choose j}^{-1} = {n \choose k} {m + c \choose k} {n + m \choose k}^{-1}.$

This looks quite burly, but it is completely routine to prove these days with WZ theory. The proof goes like this in Maple:

with(SumTools[Hypergeometric]):

a := (n, m, k, c) -> binomial(n, k) * binomial(m + c, k) / binomial(n + m, k):
b := (n, m, k, c) -> binomial(m, k) * binomial(n - c, k) / binomial(n + m, k):
d := (k, c) -> binomial(c, k):
T := b(n, m, j, c) * d(k - j, c) / a(n, m, k, c):

ZeilbergerRecurrence(T, k, j, f, 0..k);


This code outputs -f(k) + f(k + 1) = 0, where $f(k)$ is our sum in $j$. This means, essentially, that our identity is true so long as it is true for $k = 0$. Plugging in $k = 0$ gives the trivially true statement $1 = 1$, so we’re good!

# The constant factor

I may have been slightly misleading when I said that Padé approximants are unique up to a constant multiple. This is true, but in practice packages will choose different constants for each $(n, m)$th approximant, to clear denominators or something like that. Put another way, the Padé approximates are some function of $n$ and $m$ times the sequences in \eqref{solution}.

The PADE package assumes that $b(n, m, 0) = 1$, but then normalizes the approximant by clearing denominators and making everything look nice. I believe that, for PADE, the correct “constant” is $(n + m)! / \min(n, m)$, so

$a(n, m, 0) = b(n, m, 0) = \frac{(n + m)!}{\min(n, m)!}.$

Multiplying through by $(n + m)!$ gives integer coefficients, and I guess the $\min(n, m)!$ term is there to get rid of any common integer factors after that.

This has practical ramifications to the experimental side of this problem. There are, I think, three ways to guess the coefficients:

1. Guess a recurrence for $a(n, m, k)$ in $k$ (read fixed polynomials from left to right);

2. Guess a recurrence for $a(n, m, k)$ in $n$ (read fixed degrees from top to bottom); and

3. Guess a recurrence for $a(n, m, n - k)$ in $n$ (read fixed relative degrees from top to bottom).

If every $(n, m)$ pair has a different constant, then the last two options could be more difficult because the “normalizing constants” could be very complicated. In other words, reading across different values of $(n, m)$ forces you to guess the “base” sequences as well as the “normalizing” sequences. The first option, in comparison, only needs you to guess the “base” sequence! You know a priori that there is a normalizing sequence, and you can file that problem away for another day.

# Small generalizations and special cases

Given the Padé approximants of $f(x) = (1 + x)^c$, we can get the approximants of $(1 + tx)^c$ by scaling the old coefficients by $t^k$. Thus, to be completely general, we should write

\begin{align*} a(n, m, k) &= t^k {n \choose k} {m + c \choose k} {n + m \choose k}^{-1} \\ b(n, m, k) &= t^k {m \choose k} {n - c \choose k} {n + m \choose k}^{-1}. \end{align*}

(If you don’t believe me, just modify the above Maple code to prove it yourself.)

The function $f(x) = (1 - 4x)^{-1/2}$ is of special interest since it generates the central binomial coefficients. Let’s find the diagonal approximants (i.e., $n = m$) for this $f$. Our work here tells us that

\begin{align*} a(n, n, k) &= (-4)^k {n \choose k} {n - \frac{1}{2} \choose k} {2n \choose k}^{-1} \\ b(n, n, k) &= (-4)^k {n \choose k} {n + \frac{1}{2} \choose k} {2n \choose k}^{-1}. \end{align*}

By some elementary binomial coefficient identities,

\begin{align*} {n - \frac{1}{2} \choose k} &= \frac{ {2n \choose 2k} {2k \choose k}}{4^k {n \choose k}} \\ {n + \frac{1}{2} \choose k} &= \frac{ {2(n + 1) \choose 2k} {2k \choose k}}{4^k {n + 1 \choose k}}. \end{align*}

Therefore, after some simplifying,

\begin{align*} a(n, n, k) &= (-1)^k {2n - k \choose k} \\ b(n, n, k) &= (-1)^k \frac{2n + 1}{2(n - k) + 1} {2n - k \choose k}. \end{align*}

This tells us that, say, the leading coefficient in the denominator is $b(n, n, n) = (-1)^n (2n + 1)$. This checks out:

> Pade1((1 - 4 * x)^(-1/2), x, 2, 2);
2
x  - 3 x + 1
--------------
2
5 x  - 5 x + 1

> Pade1((1 - 4 * x)^(-1/2), x, 3, 3);
3      2
-x  + 6 x  - 5 x + 1
-----------------------
3       2
-7 x  + 14 x  - 7 x + 1

> Pade1((1 - 4 * x)^(-1/2), x, 4, 4);
4       3       2
x  - 10 x  + 15 x  - 7 x + 1
------------------------------
4       3       2
9 x  - 30 x  + 27 x  - 9 x + 1


# Conclusion

This seems to conclude the story of Padé approximants for $(1 + tx)^c$. We have exhausted all meaningful generalizations. We could consider $(a + tx)^c$, but that’s really just the same function. No, I think that we’re done here.

Maybe we shouldn’t be surprised that these approximants turned out nice. The coefficients of $(1 + x)^c$ are just binomial coefficients. Of course they would spit out exactly the right kind of convolution identity.

Perhaps a more fruitful approach would be to look at known convolution identities and work backwards to discovering the involved functions. That way you at least know that you’ll have some nice Padé approximants, whatever the function is. A problem to be taken up soon!

1. See Chapter 20 of Wall’s Analytic Theory of Continued Fractions

2. If you rush off to check these with PADE or something similar, be aware that normalizing the computed Padé coefficients (e.g., cancelling denominators or something like that) will produce different constants for each $(n, m)$ pair. For PADE in particular, I believe that the constant is $(n + m)! / \min(n, m)!$.

3. Strictly speaking, these recurrences don’t make sense for $k = n$ or $k = m$. I should have cleared denominators, and doing that would show that $a(n, m, k)$ vanishes if $k > n$, and $b(n, m, k)$ vanishes if $k > m$, as we would expect.