Padé approximations and EKHAD: A lab journal
Published:
Inspired by a section in Zeilberger’s “COMPUTERIZED DECONSTRUCTION” (Z), I have written a small Padé approximator in Python. Here I will attempt to experimentally discover the Padé approximation of $(1  4x)^{1/2}$.
An $(m, n)$th Padé approximant is a rational function $P_{n, m}(x) / Q_{n, m}(x)$ such that
\[f(x)  \frac{P_{n, m}(x)}{Q_{n, m}(x)} = O(x^{n + m + 1}).\]These are like “rational Taylor polynomials.”
Here, we will consider the diagonal $(n, n)$ approximants, denoted
\[\begin{align*} P_n(x) &= \sum_{k = 0}^n a(n, n  k) x^k \\ Q_n(x) &= \sum_{k = 0}^n b(n, n  k) x^k. \end{align*}\]We will take $a(n, 0) = 1$.
Highlights:

For $f(x) = (1  4x)^{1/2}$, the sequences $a(n, k)$ and $b(n, k)$ are uniquely determined.

$a(n, k) = (1)^k {n + k \choose 2k}$.

$b(n, k) = (1)^k {n + k \choose 2k} \frac{2n + 1}{2k + 1}$.

This is rigorously established by (automatically!) proving the identity
I have written the rest of this like an informal “lab journal” with little ex post facto cleaning.
First steps
Here’s the full listing:
import sympy as sp
def pade(f, x, n, m, leading_numer=1):
a = sp.symbols("a:{}".format(n + 1))
b = sp.symbols("b:{}".format(m + 1))
P = sum(a[k] * x**k for k in range(n + 1)).subs(a[1], leading_numer)
Q = sum(b[k] * x**k for k in range(m + 1))
expansion = sp.series(f * Q  P, x, n=n + m + 1)
eqns = [expansion.coeff(x, k) for k in range(n + m + 1)]
soln = sp.solve(eqns)
return (P / Q).subs(soln)
def diagonal_pade(f, x, n, *args, **kwargs):
return pade(f, x, n, n, *args, **kwargs)
And here’s an example run:
In [1]: run pade
In [2]: diagonal_pade(exp(x), x, 1)
Out[2]:
x + 2
─────
2  x
In [3]: diagonal_pade(exp(x), x, 2)
Out[3]:
2
x + 6⋅x + 12
─────────────
2
x  6⋅x + 12
In [4]: diagonal_pade(exp(x), x, 3)
Out[4]:
3 2
x + 12⋅x + 60⋅x + 120
─────────────────────────
3 2
 x + 12⋅x  60⋅x + 120
In [5]: diagonal_pade(exp(x), x, 4)
Out[5]:
4 3 2
x + 20⋅x + 180⋅x + 840⋅x + 1680
──────────────────────────────────
4 3 2
x  20⋅x + 180⋅x  840⋅x + 1680
These results agree with Z.
Let’s look at some diagonal Padé approximants for $(14x)^{1 / 2}$:
In [2]: diagonal_pade(1 / sqrt(1  4 * x), x, 1)
Out[2]:
x  1
───────
3⋅x  1
In [3]: diagonal_pade(1 / sqrt(1  4 * x), x, 2)
Out[3]:
2
x  3⋅x + 1
──────────────
2
5⋅x  5⋅x + 1
In [4]: diagonal_pade(1 / sqrt(1  4 * x), x, 3)
Out[4]:
3 2
x  6⋅x + 5⋅x  1
──────────────────────
3 2
7⋅x  14⋅x + 7⋅x  1
In [5]: diagonal_pade(1 / sqrt(1  4 * x), x, 4)
Out[5]:
4 3 2
x  10⋅x + 15⋅x  7⋅x + 1
──────────────────────────────
4 3 2
9⋅x  30⋅x + 27⋅x  9⋅x + 1
The involved polynomials look like they have nice integer coefficients. Let’s investigate them.
In [6]: a = lambda n, m: numer(diagonal_pade(1 / sqrt(1  4 * x), x, n)).coeff(x, n  m)
In [7]: # a(n, m) is the coefficient on x^{n  m} in the numerator of the nth diagonal Pade approximant.
# ...
In [13]: [a(k, 1) for k in range(1, 10)]
Out[13]: [1, 3, 6, 10, 15, 21, 28, 36, 45]
In [14]: [abs(a(k, 1)) for k in range(1, 10)]
Out[14]: [1, 3, 6, 10, 15, 21, 28, 36, 45]
This last sequence has a linear difference. It looks like $\Delta a(n, 1) = 2 + n$, which means that $a(n, 1)$ is probably some quadratic function. In fact, it’s easy to see that these are just triangular numbers. These can be written ${n \choose 2}$, and we should remember that for the next sequence.
In [15]: [abs(a(k, 2)) for k in range(1, 10)]
Out[15]: [0, 1, 5, 15, 35, 70, 126, 210, 330]
In [16]: import numpy as np
In [17]: seq = [abs(a(k, 2)) for k in range(1, 10)]
In [18]: np.diff(seq)
Out[18]: array([1, 4, 10, 20, 35, 56, 84, 120], dtype=object)
In [19]: np.diff(seq, 2)
Out[19]: array([3, 6, 10, 15, 21, 28, 36], dtype=object)
In [20]: np.diff(seq, 3)
Out[20]: array([3, 4, 5, 6, 7, 8], dtype=object)
In [21]: np.diff(seq, 4)
Out[21]: array([1, 1, 1, 1, 1], dtype=object)
These don’t look as straightforward, but iterated differences tell us that these coefficients probably come from a quartic polynomial. Based on the last guess, I would say it’s ${n + 3 \choose 4}$, and that looks right.
There seems to be a metapattern developing. Let’s see one more to get the feel for what it should be.
In [48]: [abs(a(k, 3)) for k in range(1, 10)]
Out[48]: [0, 0, 1, 7, 28, 84, 210, 462, 924]
# ...
In [55]: [binomial(k + 4, 6) for k in range(9)]
Out[55]: [0, 0, 1, 7, 28, 84, 210, 462, 924, 1716]
We have enough to hazard a first guess. The function $a(n, k)$ is something like ${n + k + 1 \choose 2k}$, but we need to get the details right.
Getting the details right
Getting the details right is an important skill. Until you can teach a computer to do it automatically, it pays to practice catching all the nasty mistakes you can make while staring at sequences.
I started with guess(n, k) = binomial(n + k + 1, 2k)
. This is definitely
wrong. First of all, $a(n, k)$ is sometimes negative. Oops. I forgot that
I took absolute values at some point. I suspect that the signs alternate in
$k$, so I can just multiply by $(1)^k$ at the end to fix that.
The formula is off some other way as well. For example, guess(4, 2) = 35
while a(4, 2) = 15
. But guess(3, 2) = 15
, so maybe we’re just off by one in
$n$. Looking back at the listcomprehension I used to create the binomial
coefficients, I think I am. So let’s say guess(n, k) = binomial(n + k, 2k)
.
Picking a value at random, we have a(10, 3) = 1716
and guess(10, 3)
= 1716
. Good! And it looks like the signs do alternate in $k$, so maybe we
actually have guess(n, k) = (1)^k * binomial(n + k, 2k)
.
Here’s a quick verification that we might be right:
In [117]: guess = lambda n, k: (1)**k * binomial(n + k, 2 * k)
In [118]: from itertools import product
# ...
In [123]: [guess(n, k)  a(n, k) for n, k in product(range(10), repeat=2) if n >= k]
Out[123]:
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Looks good! So we now conjecture that
\[a(n, k) = (1)^k {n + k \choose 2k}.\]The denominator
It feels like the denominator should be easier now that we know the numerator, but that isn’t exactly the case. The coefficients here aren’t just binomial coefficients.
Here is a helper functions to make metaguessing easier:
# I am using the `fit_poly()` functions described here:
# https://groups.google.com/forum/m/#!msg/sympy/twjvFft6i3w/tkYMq6oM20QJ
def guess_denom_poly(k, deg, n_terms=10):
def b(n, k):
return sp.denom(diagonal_pade(1 / sqrt(1  4 * x), x, n)).coeff(x, n  k)
bs = [b(j, k) for j in range(k, n_terms + k)]
print([abs(t) for t in bs])
return fit_poly(list(range(k, n_terms + k)), bs, n, degree=deg)[0].factor()
I suspect that $b(n, k)$ is again a polynomial in $n$ for each fixed $k$, hence all the stuff above about fitting and polynomials.
Let’s see some examples:
# This is the coefficient on x^{n  3} in the denominator of order n.
In [214]: guess_denom_poly(3, 7)
[1, 9, 44, 156, 450, 1122, 2508, 5148, 9867, 17875]
Out[214]:
(n + 1)⋅(n + 2)⋅(n + 3)⋅(n + 4)⋅(n + 5)⋅(n + 6)⋅(2⋅n + 7)
───────────────────────────────────────────────────────────
5040
# This is the coefficient on x^{n  2} in the denominator of order n.
In [215]: guess_denom_poly(2, 5)
[1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289]
Out[215]:
(n + 1)⋅(n + 2)⋅(n + 3)⋅(n + 4)⋅(2⋅n + 5)
─────────────────────────────────────────
120
# This is the coefficient on x^{n  1} in the denominator of order n.
In [216]: guess_denom_poly(1, 3)
[1, 5, 14, 30, 55, 91, 140, 204, 285, 385]
Out[216]:
(n + 1)⋅(n + 2)⋅(2⋅n + 3)
───────────────────────────
6
These might look intimidating, but closer inspection reveals that we probably
have b(n, k) = binomial(n + 2k, 2k) * (2n + 2k + 1)
.
The devil’s in the details, though, and this ain’t quite right. We fit these
polynomials using values which started from $k$, so we probably need to shift
$n$ back by $k$. Maybe guess2(n, k) = binomial(n + k, 2k) * (2n + 1)
?
I changed a line in the program to make sure that everything is starting at the same point, and now I get some more sensible results. For example:
In [281]: guess_denom_poly(1, 3)
[1, 5, 14, 30, 55, 91, 140, 204, 285, 385]
Out[281]:
n⋅(n + 1)⋅(2⋅n + 1)
─────────────────────
6
In [282]: guess_denom_poly(2, 5)
[1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289]
Out[282]:
n⋅(n  1)⋅(n + 1)⋅(n + 2)⋅(2⋅n + 1)
───────────────────────────────────
120
In [283]: guess_denom_poly(4, 9)
[1, 11, 65, 275, 935, 2717, 7007, 16445, 35750, 72930]
Out[283]:
n⋅(n  3)⋅(n  2)⋅(n  1)⋅(n + 1)⋅(n + 2)⋅(n + 3)⋅(n + 4)⋅(2⋅n + 1)
───────────────────────────────────────────────────────────────────
362880
Looking at the $k = 4$ case very hard, I notice that the denominator is
(2 * 4 + 1)!
, not (2 * 4)!
. That means that we should probably have
guess2(n, k) = binomial(n + k, 2k) * (2n + 1) / (2k + 1)
. This seems to work!
All that’s left is to note that b(n, k)
alternates in k
, so we probably
have
Putting it all together
We have some conjectures for the coefficients on our approximants. Let’s see if they work.
The equation that we’re using for Padé approximants is
\[f(x) Q_n(x)  P_n(x) = O(x^{2n + 1}),\]where, in our notation,
\[\begin{align*} P_n(x) &= \sum_{k = 0}^n a(n, n  k) x^k \\ Q_n(x) &= \sum_{k = 0}^n b(n, n  k) x^k. \end{align*}\]Thus the coefficients of $f(x) Q_n(x)$ with $f(x) = (1  4x)^{1/2}$ are (conjectured to be)
\[\begin{align*} [x^m] f(x) Q_n(x) &= \sum_{k = 0}^m b(n, n  k) {2(m  k) \choose m  k} \\ &= \sum_{0 \leq k \leq \min(n, m)} (1)^{n  k} {2n  k \choose 2(n  k)} \frac{2n + 1}{2(n  k) + 1} {2(m  k) \choose m  k}. \end{align*}\]I put bounds on the final sum since $b(n, n  k) = 0$ if $k > n$ or $k < 0$. But the conjectured expression does this as well, and the only part of the sum involving $m$ vanishes with $k > m$ and $k < 0$, so we can just drop the bounds. Thus we have (conjectured that)
\[[x^m] f(x) Q_n(x) = \sum_k (1)^{n  k} {2n  k \choose 2(n  k)} \frac{2n + 1}{2(n  k) + 1} {2(m  k) \choose m  k}.\]The coefficient on $x^m$ in $P_n(x)$ is just $a(n, n  m)$. Our conjectures hinge on the mighty identity
\[\sum_k {2n  k \choose 2(n  k)} {2(m  k) \choose m  k} \frac{(1)^k}{2(n  k) + 1} = \frac{(1)^m}{2n + 1} {2n  m \choose 2(n  m)}.\]If we clean this up, then it reads
\[\sum_k {n + k \choose 2k} {2(m  n + k) \choose m  n + k} \frac{(1)^k}{2k + 1} = \frac{(1)^{n  m}}{2n + 1} {2n  m \choose 2(n  m)}.\]After looking at some special cases and consulting Concrete Mathematics, I decided that this wasn’t an identity I had seen before. Luckily, it is entirely routine to prove thanks to Wilf–Zeilberger theory. The venerable package EKHAD can check this for us. The certificate function is
\[{\frac { \left( 2\,k+1 \right) \left( mn+k \right) k \left( 4\,mk6\,nk+4\,{m}^{2}12\,mn+ 10\,{n}^{2}5\,k10\,m+17\,n+7 \right) }{ \left( n1+k \right) \left( 2\,m2\,n1+2\,k \right) \left( 2\,n+1 \right) \left( 2\,n1+m \right) \left( 2\,n2+m \right) }}\]Marvelous.
So, we now have formulas for arbitrary diagonal Padé approximants of $(1  4x)^{1 / 2}$. I don’t think that this is of great practical importance, but it is pretty cool.