`Newton's method `__ ===================================================================== Newton-Raphson for finding the root of an equation. .. code:: ipython2 from notebook_preamble import J, V, define Cf. `"Why Functional Programming Matters" by John Hughes `__ Finding the Square-Root of a Number ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Let's define a function that computes this equation: :math:`a_{i+1} = \frac{(a_i+\frac{n}{a_i})}{2}` :: 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') Compute the Error ^^^^^^^^^^^^^^^^^ 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') ``square-root`` ^^^^^^^^^^^^^^^ 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 a recursive function ``K`` from here. :: n a' e ε K K == [P] [E] [R0] [R1] genrec Base-case ~~~~~~~~~ The predicate and the base case are obvious: :: K == [<] [popop popd] [R0] [R1] genrec :: n a' e ε popop popd n a' popd a' Recur ~~~~~~~~~~ The recursive branch is pretty easy. Discard the error and recur. :: K == [<] [popop popd] [R0] [R1] genrec K == [<] [popop popd] [R0 [K] R1] ifte :: n a' e ε R0 [K] R1 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. (``R1`` is ``i`` so this is a ``primrec`` "primitive recursive" function.) .. 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 Initial Approximation and Epsilon ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 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') Examples ~~~~~~~~~~ .. 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