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
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
define('gsra 1 swap [over / + 2 /] cons [dup] swoncat make_generator')
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…
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)
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
define('_within_B roll< popop first')
Recur¶
a [b G] ε R0 [within] R1
- Discard a.
- Use
x
combinator to generate next term fromG
. - Run
within
withi
(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
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 ...
define('within x 0.000000001 [_within_P] [_within_B] [_within_R] tailrec')
define('sqrt gsra within')
Try it out…
J('36 sqrt')
6.0
J('23 sqrt')
4.795831523312719
Check it.
4.795831523312719**2
22.999999999999996
from math import sqrt
sqrt(23)
4.795831523312719