Make numbers == 3000.1.* required.

[numerical-analysis.git] / src / Roots / Simple.hs

-- | The Roots.Simple module contains root-finding algorithms. That
--   is, procedures to (numerically) find solutions to the equation,
--
--   > f(x) = 0
--
--   where f is assumed to be continuous on the interval of interest.
--

module Roots.Simple
where

import Data.List (find)

import qualified Roots.Fast as F

-- | Does the (continuous) function @f@ have a root on the interval
--   [a,b]? If f(a) <] 0 and f(b) ]> 0, we know that there's a root in
--   [a,b] by the intermediate value theorem. Likewise when f(a) >= 0
--   and f(b) <= 0.
--
--   Examples:
--
--   >>> let f x = x**3
--   >>> has_root f (-1) 1 Nothing
--   True
--
--   This fails if we don't specify an @epsilon@, because cos(-2) ==
--   cos(2) doesn't imply that there's a root on [-2,2].
--
--   >>> has_root cos (-2) 2 Nothing
--   False
--   >>> has_root cos (-2) 2 (Just 0.001)
--   True
--
has_root :: (Fractional a, Ord a, Ord b, Num b)
         => (a -> b) -- ^ The function @f@
         -> a       -- ^ The \"left\" endpoint, @a@
         -> a       -- ^ The \"right\" endpoint, @b@
         -> Maybe a -- ^ The size of the smallest subinterval
                   --   we'll examine, @epsilon@
         -> Bool
has_root f a b epsilon =
  F.has_root f a b epsilon Nothing Nothing




-- | We are given a function @f@ and an interval [a,b]. The bisection
--   method checks finds a root by splitting [a,b] in half repeatedly.
--
--   If one is found within some prescribed tolerance @epsilon@, it is
--   returned. Otherwise, the interval [a,b] is split into two
--   subintervals [a,c] and [c,b] of equal length which are then both
--   checked via the same process.
--
--   Returns 'Just' the value x for which f(x) == 0 if one is found,
--   or Nothing if one of the preconditions is violated.
--
--   Examples:
--
--   >>> bisect cos 1 2 0.001
--   Just 1.5712890625
--
--   >>> bisect sin (-1) 1 0.001
--   Just 0.0
--
bisect :: (Fractional a, Ord a, Num b, Ord b)
       => (a -> b) -- ^ The function @f@ whose root we seek
       -> a       -- ^ The \"left\" endpoint of the interval, @a@
       -> a       -- ^ The \"right\" endpoint of the interval, @b@
       -> a       -- ^ The tolerance, @epsilon@
       -> Maybe a
bisect f a b epsilon =
  F.bisect f a b epsilon Nothing Nothing



-- | The sequence x_{n} of values obtained by applying Newton's method
--   on the function @f@ and initial guess @x0@.
--
--   Examples:
--
--   Atkinson, p. 60.
--   >>> let f x = x^6 - x - 1
--   >>> let f' x = 6*x^5 - 1
--   >>> tail $ take 4 $ newton_iterations f f' 2
--   [1.6806282722513088,1.4307389882390624,1.2549709561094362]
--
newton_iterations :: (Fractional a, Ord a)
                    => (a -> a) -- ^ The function @f@ whose root we seek
                    -> (a -> a) -- ^ The derivative of @f@
                    -> a       -- ^ Initial guess, x-naught
                    -> [a]
newton_iterations f f' x0 =
  iterate next x0
  where
  next xn =
    xn - ( (f xn) / (f' xn) )



-- | Use Newton's method to find a root of @f@ near the initial guess
--   @x0@. If your guess is bad, this will recurse forever!
--
--   Examples:
--
--   Atkinson, p. 60.
--
--   >>> let f x = x^6 - x - 1
--   >>> let f' x = 6*x^5 - 1
--   >>> let Just root = newtons_method f f' (1/1000000) 2
--   >>> root
--   1.1347241385002211
--   >>> abs (f root) < 1/100000
--   True
--
--   >>> import Data.Number.BigFloat
--   >>> let eps = 1/(10^20) :: BigFloat Prec50
--   >>> let Just root = newtons_method f f' eps 2
--   >>> root
--   1.13472413840151949260544605450647284028100785303643e0
--   >>> abs (f root) < eps
--   True
--
newtons_method :: (Fractional a, Ord a)
                 => (a -> a) -- ^ The function @f@ whose root we seek
                 -> (a -> a) -- ^ The derivative of @f@
                 -> a       -- ^ The tolerance epsilon
                 -> a       -- ^ Initial guess, x-naught
                 -> Maybe a
newtons_method f f' epsilon x0
  = find (\x -> abs (f x) < epsilon) x_n
  where
    x_n = newton_iterations f f' x0



-- | Takes a function @f@ of two arguments and repeatedly applies @f@
--   to the previous two values. Returns a list containing all
--   generated values, f(x0, x1), f(x1, x2), f(x2, x3)...
--
--   Examples:
--
--   >>> let fibs = iterate2 (+) 0 1
--   >>> take 15 fibs
--   [0,1,1,2,3,5,8,13,21,34,55,89,144,233,377]
--
iterate2 :: (a -> a -> a) -- ^ The function @f@
         -> a           -- ^ The initial value @x0@
         -> a           -- ^ The second value, @x1@
         -> [a]         -- ^ The result list, [x0, x1, ...]
iterate2 f x0 x1 =
  x0 : x1 : (go x0 x1)
  where
    go prev2 prev1 =
      let next = f prev2 prev1 in
        next : go prev1 next

-- | The sequence x_{n} of values obtained by applying the secant
--   method on the function @f@ and initial guesses @x0@, @x1@.
--
--   The recursion more or less implements a two-parameter 'iterate',
--   although one list is passed to the next iteration (as opposed to
--   one function argument, with iterate). At each step, we peel the
--   first two elements off the list and then compute/append elements
--   three, four... onto the end of the list.
--
--   Examples:
--
--   Atkinson, p. 67.
--   >>> let f x = x^6 - x - 1
--   >>> take 4 $ secant_iterations f 2 1
--   [2.0,1.0,1.0161290322580645,1.190577768676638]
--
secant_iterations :: (Fractional a, Ord a)
                    => (a -> a) -- ^ The function @f@ whose root we seek
                    -> a       -- ^ Initial guess, x-naught
                    -> a       -- ^ Second initial guess, x-one
                    -> [a]
secant_iterations f x0 x1 =
  iterate2 g x0 x1
  where
  g prev2 prev1 =
    let x_change = prev1 - prev2
        y_change = (f prev1) - (f prev2)
    in
      (prev1 - (f prev1 * (x_change / y_change)))


-- | Use the secant method to find a root of @f@ near the initial guesses
--   @x0@ and @x1@. If your guesses are bad, this will recurse forever!
--
--   Examples:
--
--   Atkinson, p. 67.
--   >>> let f x = x^6 - x - 1
--   >>> let Just root = secant_method f (1/10^9) 2 1
--   >>> root
--   1.1347241384015196
--   >>> abs (f root) < (1/10^9)
--   True
--
secant_method :: (Fractional a, Ord a)
                 => (a -> a) -- ^ The function @f@ whose root we seek
                 -> a       -- ^ The tolerance epsilon
                 -> a       -- ^ Initial guess, x-naught
                 -> a       -- ^ Second initial guess, x-one
                 -> Maybe a
secant_method f epsilon x0 x1
  = find (\x -> abs (f x) < epsilon) x_n
  where
    x_n = secant_iterations f x0 x1



fixed_point_iterations :: (a -> a) -- ^ The function @f@ to iterate.
                       -> a       -- ^ The initial value @x0@.
                       -> [a]     -- ^ The resulting sequence of x_{n}.
fixed_point_iterations f x0 =
  iterate f x0


-- | Find a fixed point of the function @f@ with the search starting
--   at x0. This will find the first element in the chain f(x0),
--   f(f(x0)),... such that the magnitude of the difference between it
--   and the next element is less than epsilon.
--
fixed_point :: (Num a, Ord a)
            => (a -> a) -- ^ The function @f@ to iterate.
            -> a       -- ^ The tolerance, @epsilon@.
            -> a       -- ^ The initial value @x0@.
            -> a       -- ^ The fixed point.
fixed_point f epsilon x0 =
  fst winning_pair
  where
    xn = fixed_point_iterations f x0
    xn_plus_one = tail $ fixed_point_iterations f x0

    abs_diff v w =
      abs (v - w)

    -- The nth entry in this list is the absolute value of x_{n} -
    -- x_{n+1}.
    differences = zipWith abs_diff xn xn_plus_one

    -- A list of pairs, (xn, |x_{n} - x_{n+1}|).
    pairs = zip xn differences

    -- The pair (xn, |x_{n} - x_{n+1}|) with
    -- |x_{n} - x_{n+1}| < epsilon. The pattern match on 'Just' is
    -- "safe" since the list is infinite. We'll succeed or loop
    -- forever.
    Just winning_pair = find (\(_, diff) -> diff < epsilon) pairs