Thun/docs/Newton-Raphson.ipynb

Newton's method

Let's use the Newton-Raphson method for finding the root of an equation to write a function that can compute the square root of a number.

Cf. "Why Functional Programming Matters" by John Hughes

In [1]:

from notebook_preamble import J, V, define

A Generator for Approximations

To make a generator that generates successive approximations let’s start by assuming an initial approximation and then derive the function that computes the next approximation:

   a F
---------
    a'

A Function to Compute the Next Approximation

This is the equation for computing the next approximate value of the square root:

a_{i+1} = \frac{(a_i+\frac{n}{a_i})}{2}

a n over / + 2 /
a n a    / + 2 /
a n/a      + 2 /
a+n/a        2 /
(a+n/a)/2

The function we want has the argument n in it:

F == n over / + 2 /

Make it into a Generator

Our generator would be created by:

a [dup F] make_generator

With n as part of the function F, but n is the input to the sqrt function we’re writing. If we let 1 be the initial approximation:

1 n 1 / + 2 /
1 n/1   + 2 /
1 n     + 2 /
n+1       2 /
(n+1)/2

The generator can be written as:

23 1 swap  [over / + 2 /] cons [dup] swoncat make_generator
1 23       [over / + 2 /] cons [dup] swoncat make_generator
1       [23 over / + 2 /]      [dup] swoncat make_generator
1   [dup 23 over / + 2 /]                    make_generator

In [2]:

define('gsra == 1 swap [over / + 2 /] cons [dup] swoncat make_generator')

In [3]:

J('23 gsra')

[1 [dup 23 over / + 2 /] codireco]

Let's drive the generator a few time (with the x combinator) and square the approximation to see how well it works...

In [4]:

J('23 gsra 6 [x popd] times first sqr')

23.0000000001585

Finding Consecutive Approximations within a Tolerance

From "Why Functional Programming Matters" by John Hughes:

The remainder of a square root finder is a function within, which takes a tolerance and a list of approximations and looks down the list for two successive approximations that differ by no more than the given tolerance.

(And note that by “list” he means a lazily-evaluated list.)

Using the output [a G] of the above generator for square root approximations, and further assuming that the first term a has been generated already and epsilon ε is handy on the stack...

   a [b G] ε within
---------------------- a b - abs ε <=
      b


   a [b G] ε within
---------------------- a b - abs ε >
   b [c G] ε within

Predicate

a [b G]             ε [first - abs] dip <=
a [b G] first - abs ε                   <=
a b           - abs ε                   <=
a-b             abs ε                   <=
abs(a-b)            ε                   <=
(abs(a-b)<=ε)

In [5]:

define('_within_P == [first - abs] dip <=')

Base-Case

a [b G] ε roll< popop first
  [b G] ε a     popop first
  [b G]               first
   b

In [6]:

define('_within_B == roll< popop first')

Recur

a [b G] ε R0 [within] R1

1. Discard a.
2. Use x combinator to generate next term from G.
3. Run within with i (it is a primrec function.)

Pretty straightforward:

a [b G]        ε R0           [within] R1
a [b G]        ε [popd x] dip [within] i
a [b G] popd x ε              [within] i
  [b G]      x ε              [within] i
b [c G]        ε              [within] i
b [c G]        ε               within

b [c G] ε within

In [7]:

define('_within_R == [popd x] dip')

Setting up

The recursive function we have defined so far needs a slight preamble: x to prime the generator and the epsilon value to use:

[a G] x ε ...
a [b G] ε ...

In [8]:

define('within == x 0.000000001 [_within_P] [_within_B] [_within_R] primrec')
define('sqrt == gsra within')

Try it out...

In [9]:

J('36 sqrt')

6.0

In [10]:

J('23 sqrt')

4.795831523312719

Check it.

In [11]:

4.795831523312719**2

Out [11]:

22.999999999999996

In [12]:

from math import sqrt

sqrt(23)

Out [12]:

4.795831523312719