Thun/docs/sphinx_docs/notebooks/Newton-Raphson.rst

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`Newton's method <https://en.wikipedia.org/wiki/Newton%27s_method>`__
=====================================================================
.. code:: ipython2
from notebook_preamble import J, V, define
Cf. `"Why Functional Programming Matters" by John
Hughes <https://www.cs.kent.ac.uk/people/staff/dat/miranda/whyfp90.pdf>`__
:math:`a_{i+1} = \frac{(a_i+\frac{n}{a_i})}{2}`
Let's define a function that computes the above equation:
::
n a Q
---------------
(a+n/a)/2
n a tuck / + 2 /
a n a / + 2 /
a n/a + 2 /
a+n/a 2 /
(a+n/a)/2
We want it to leave n but replace a, so we execute it with ``unary``:
::
Q == [tuck / + 2 /] unary
.. code:: ipython2
define('Q == [tuck / + 2 /] unary')
And a function to compute the error:
::
n a sqr - abs
|n-a**2|
This should be ``nullary`` so as to leave both n and a on the stack
below the error.
::
err == [sqr - abs] nullary
.. code:: ipython2
define('err == [sqr - abs] nullary')
Now we can define a recursive program that expects a number ``n``, an
initial estimate ``a``, and an epsilon value ``ε``, and that leaves on
the stack the square root of ``n`` to within the precision of the
epsilon value. (Later on we'll refine it to generate the initial
estimate and hard-code an epsilon value.)
::
n a ε square-root
-----------------
√n
If we apply the two functions ``Q`` and ``err`` defined above we get the
next approximation and the error on the stack below the epsilon.
::
n a ε [Q err] dip
n a Q err ε
n a' err ε
n a' e ε
Let's define the recursive function from here. Start with ``ifte``; the
predicate and the base case behavior are obvious:
::
n a' e ε [<] [popop popd] [J] ifte
Base-case
::
n a' e ε popop popd
n a' popd
a'
The recursive branch is pretty easy. Discard the error and recur.
::
w/ K == [<] [popop popd] [J] ifte
n a' e ε J
n a' e ε popd [Q err] dip [K] i
n a' ε [Q err] dip [K] i
n a' Q err ε [K] i
n a'' e ε K
This fragment alone is pretty useful.
.. code:: ipython2
define('K == [<] [popop popd] [popd [Q err] dip] primrec')
.. code:: ipython2
J('25 10 0.001 dup K')
.. parsed-literal::
5.000000232305737
.. code:: ipython2
J('25 10 0.000001 dup K')
.. parsed-literal::
5.000000000000005
So now all we need is a way to generate an initial approximation and an
epsilon value:
::
square-root == dup 3 / 0.000001 dup K
.. code:: ipython2
define('square-root == dup 3 / 0.000001 dup K')
.. code:: ipython2
J('36 square-root')
.. parsed-literal::
6.000000000000007
.. code:: ipython2
J('4895048365636 square-root')
.. parsed-literal::
2212475.6192184356
.. code:: ipython2
2212475.6192184356 * 2212475.6192184356
.. parsed-literal::
4895048365636.0