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

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:

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

Starting with $\frac{(a_i+\frac{n}{a_i})}{2}$ we can derive the Joy expression to compute it using abstract dummy variables to stand in for actual values. First undivide by two:

$(a_i+\frac{n}{a_i})$ 2 /

Then unadd terms:

$a_i$ $\frac{n}{a_i}$ + 2 /

Undivide again:

$a_i$ $n$ $a_i$ / + 2 /

Finally deduplicate the $a_i$ term:

$a_i$ $n$ over / + 2 /

Let's try out this function over / + 2 / on an example:

F == over / + 2 /

In order to use this function F we have to provide an initial estimate for the value of the square root, and we want to keep the input value n handy for iterations (we don't want the user to have to keep reentering it.)

The initial estimate can be 2, and we can cons the input value onto a quote with F:

broken due to no float div

clear

7 [] true [i [1000 F] [first] [pop sqr] fork - abs 10 >] loop pop

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

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

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)

Base-Case¶

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

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 "tail-recursive" 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

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 ...

Try it out...

Check it.