McSinyx

# Infinite Sequences: A Case Study in Functional Python

In this article, we will only consider sequences defined by a function whose domain is a subset of the set of all integers. Such sequences will be visualized, i.e. we will try to evaluate the first few (thousand) elements, using functional programming paradigm, where functions are more similar to the ones in math (in contrast to imperative style with side effects confusing to inexperenced coders). The idea is taken from subsection 3.5.2 of SICP and adapted to Python, which, compare to Scheme, is significantly more popular: Python is pre-installed on almost every modern Unix-like system, namely macOS, GNU/Linux and the *BSDs; and even at MIT, the new 6.01 in Python has recently replaced the legendary 6.001 (SICP).

One notable advantage of using Python is its huge standard library. For example the identity sequence (sequence defined by the identity function) can be imported directly from itertools:

>>> from itertools import count
>>> positive_integers = count(start=1)
>>> next(positive_integers)
1
>>> next(positive_integers)
2
>>> for _ in range(4): next(positive_integers)
...
3
4
5
6

To open a Python emulator, simply lauch your terminal and run python. If that is somehow still too struggling, navigate to the interactive shell on Python.org.

Let's get it started with somethings everyone hates: recursively defined sequences, e.g. the famous Fibonacci ($F_n = F_{n-1} + F_{n-2}$, $F_1 = 1$ and $F_0 = 0$). Since Python does not support tail recursion, it's generally not a good idea to define anything recursively (which is, ironically, the only trivial functional solution in this case) but since we will only evaluate the first few terms (use the Tab key to indent the line when needed):

>>> def fibonacci(n, a=0, b=1):
...     # To avoid making the code look complicated,
...     # n < 0 is not handled here.
...     return a if n == 0 else fibonacci(n - 1, b, a + b)
...
>>> fibo_seq = (fibonacci(n) for n in count(start=0))
>>> for _ in range(7): next(fibo_seq)
...
0
1
1
2
3
5
8

The fibo_seq above is just to demonstrate how itertools.count can be use to create an infinite sequence defined by a function. For better performance, this should be used instead:

def fibonacci_sequence(a=0, b=1):
yield a
yield from fibonacci_sequence(b, a + b)

It is noticable that the elements having been iterated through (using next) will disappear forever in the void (oh no!), but that is the cost we are willing to pay to save some memory, especially when we need to evaluate a member of (arbitrarily) large index to estimate the sequence's limit. One case in point is estimating a definite integral using left Riemann sum.

def integral(f, a, b):
def left_riemann_sum(n):
dx = (b-a) / n
def x(i): return a + i*dx
return sum(f(x(i)) for i in range(n)) * dx
return left_riemann_sum

The function integral(f, a, b) as defined above returns a function taking $n$ as an argument. As $n\to\infty$, its result approaches $\int_a^b f(x)\mathrm d x$. For example, we are going to estimate $\pi$ as the area of a semicircle whose radius is $\sqrt 2$:

>>> from math import sqrt
>>> def semicircle(x): return sqrt(abs(2 - x*x))
...
>>> pi = integral(semicircle, -sqrt(2), sqrt(2))
>>> pi_seq = (pi(n) for n in count(start=2))
>>> for _ in range(3): next(pi_seq)
...
2.000000029802323
2.514157464087051
2.7320508224700384

Whilst the first few aren't quite close, at index around 1000, the result is somewhat acceptable:

3.1414873191059525
3.1414874770617427
3.1414876346231577

Since we are comfortable with sequence of sums, let's move on to sums of a sequence, which are called series. For estimation, again, we are going to make use of infinite sequences of partial sums, which are implemented as itertools.accumulate by thoughtful Python developers. Geometric and p-series can be defined as follow:

from itertools import accumulate as partial_sums

def geometric_series(r, a=1):
return partial_sums(a*r**n for n in count(0))

def p_series(p):
return partial_sums(1 / n**p for n in count(1))

We can then use these to determine whether a series is convergent or divergent. For instance, one can easily verify that the $p$-series with $p = 2$ converges to $\pi^2 / 6 \approx 1.6449340668482264$ via

>>> s = p_series(p=2)
>>> for _ in range(11): next(s)
...
1.0
1.25
1.3611111111111112
1.4236111111111112
1.4636111111111112
1.4913888888888889
1.511797052154195
1.527422052154195
1.5397677311665408
1.5497677311665408
1.558032193976458

We can observe that it takes quite a lot of steps to get the precision we would generally expect ($s_{11}$ is only precise to the first decimal place; second decimal places: $s_{101}$; third: $s_{2304}$). Luckily, many techniques for series acceleration are available. Shanks transformation for instance, can be implemented as follow:

from itertools import islice, tee

def shanks(seq):
return map(lambda x, y, z: (x*z - y*y) / (x + z - y*2),
*(islice(t, i, None) for i, t in enumerate(tee(seq, 3))))

In the code above, lambda x, y, z: (x*z - y*y) / (x + z - y*2) denotes the anonymous function $(x, y, z) \mapsto \frac{xz - y^2}{x + z - 2y}$ and map is a higher order function applying that function to respective elements of subsequences starting from index 1, 2 and 3 of seq. On Python 2, one should import imap from itertools to get the same lazy behavior of map on Python 3.

>>> s = shanks(p_series(2))
>>> for _ in range(10): next(s)
...
1.4500000000000002
1.503968253968257
1.53472222222223
1.5545202020202133
1.5683119658120213
1.57846371882088
1.5862455815659202
1.5923993101138652
1.5973867787856946
1.6015104548459742

The result was quite satisfying, yet we can do one step futher by continuously applying the transformation to the sequence:

>>> def compose(transform, seq):
... 	yield next(seq)
... 	yield from compose(transform, transform(seq))
...
>>> s = compose(shanks, p_series(2))
>>> for _ in range(10): next(s)
...
1.0
1.503968253968257
1.5999812811165188
1.6284732442271674
1.6384666832276524
1.642311342667821
1.6425249569252578
1.640277484549416
1.6415443295058203
1.642038043478661

Shanks transformation works on every sequence (not just sequences of partial sums). Back to previous example of using left Riemann sum to compute definite integral:

>>> pi_seq = compose(shanks, map(pi, count(2)))
>>> for _ in range(10): next(pi_seq)
...
2.000000029802323
2.978391111182236
3.105916845397819
3.1323116570377185
3.1389379264270736
3.140788413965646
3.140921512857936
3.1400282163913436
3.1400874774021816
3.1407097229603256
>>> next(islice(pi_seq, 300, None))
3.1415061302492413

Now having series defined, let's see if we can learn anything about power series. Sequence of partial sums of power series $\sum c_n (x - a)^n$ can be defined as

from operator import mul

def power_series(c, start=0, a=0):
return lambda x: partial_sums(map(mul, c, (x**n for n in count(start))))

We can use this to compute functions that can be written as Taylor series:

from math import factorial
def exp(x):
return power_series(1/factorial(n) for n in count(0))(x)

def cos(x):
c = ((1 - n%2) * (1 - n%4) / factorial(n) for n in count(0))
return power_series(c)(x)

def sin(x):
c = (n%2 * (2 - n%4) / factorial(n) for n in count(1))
return power_series(c, start=1)(x)

Amazing! Let's test 'em!

>>> e = compose(shanks, exp(1)) # this should converges to 2.718281828459045
>>> for _ in range(4): next(e)
...
1.0
2.749999999999996
2.718276515152136
2.718281825486623

Impressive, huh? For sine and cosine, series acceleration is not even necessary:

>>> from math import pi as PI
>>> s = sin(PI/6)
>>> for _ in range(5): next(s)
...
0.5235987755982988
0.5235987755982988
0.49967417939436376
0.49967417939436376
0.5000021325887924
>>> next(islice(cos(PI/3), 8, None))
0.500000433432915