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1 -- | The Roots.Simple module contains root-finding algorithms. That
2 -- is, procedures to (numerically) find solutions to the equation,
3 --
4 -- > f(x) = 0
5 --
6 -- where f is assumed to be continuous on the interval of interest.
7 --
8
9 module Roots.Simple
10 where
11
12 import Data.List (find)
13
14 import qualified Roots.Fast as F
15
16 -- | Does the (continuous) function @f@ have a root on the interval
17 -- [a,b]? If f(a) <] 0 and f(b) ]> 0, we know that there's a root in
18 -- [a,b] by the intermediate value theorem. Likewise when f(a) >= 0
19 -- and f(b) <= 0.
20 --
21 -- Examples:
22 --
23 -- >>> let f x = x**3
24 -- >>> has_root f (-1) 1 Nothing
25 -- True
26 --
27 -- This fails if we don't specify an @epsilon@, because cos(-2) ==
28 -- cos(2) doesn't imply that there's a root on [-2,2].
29 --
30 -- >>> has_root cos (-2) 2 Nothing
31 -- False
32 -- >>> has_root cos (-2) 2 (Just 0.001)
33 -- True
34 --
35 has_root :: (Fractional a, Ord a, Ord b, Num b)
36 => (a -> b) -- ^ The function @f@
37 -> a -- ^ The \"left\" endpoint, @a@
38 -> a -- ^ The \"right\" endpoint, @b@
39 -> Maybe a -- ^ The size of the smallest subinterval
40 -- we'll examine, @epsilon@
41 -> Bool
42 has_root f a b epsilon =
43 F.has_root f a b epsilon Nothing Nothing
44
45
46
47
48 -- | We are given a function @f@ and an interval [a,b]. The bisection
49 -- method checks finds a root by splitting [a,b] in half repeatedly.
50 --
51 -- If one is found within some prescribed tolerance @epsilon@, it is
52 -- returned. Otherwise, the interval [a,b] is split into two
53 -- subintervals [a,c] and [c,b] of equal length which are then both
54 -- checked via the same process.
55 --
56 -- Returns 'Just' the value x for which f(x) == 0 if one is found,
57 -- or Nothing if one of the preconditions is violated.
58 --
59 -- Examples:
60 --
61 -- >>> bisect cos 1 2 0.001
62 -- Just 1.5712890625
63 --
64 -- >>> bisect sin (-1) 1 0.001
65 -- Just 0.0
66 --
67 bisect :: (Fractional a, Ord a, Num b, Ord b)
68 => (a -> b) -- ^ The function @f@ whose root we seek
69 -> a -- ^ The \"left\" endpoint of the interval, @a@
70 -> a -- ^ The \"right\" endpoint of the interval, @b@
71 -> a -- ^ The tolerance, @epsilon@
72 -> Maybe a
73 bisect f a b epsilon =
74 F.bisect f a b epsilon Nothing Nothing
75
76
77
78 -- | The sequence x_{n} of values obtained by applying Newton's method
79 -- on the function @f@ and initial guess @x0@.
80 --
81 -- Examples:
82 --
83 -- Atkinson, p. 60.
84 -- >>> let f x = x^6 - x - 1
85 -- >>> let f' x = 6*x^5 - 1
86 -- >>> tail $ take 4 $ newton_iterations f f' 2
87 -- [1.6806282722513088,1.4307389882390624,1.2549709561094362]
88 --
89 newton_iterations :: (Fractional a, Ord a)
90 => (a -> a) -- ^ The function @f@ whose root we seek
91 -> (a -> a) -- ^ The derivative of @f@
92 -> a -- ^ Initial guess, x-naught
93 -> [a]
94 newton_iterations f f' x0 =
95 iterate next x0
96 where
97 next xn =
98 xn - ( (f xn) / (f' xn) )
99
100
101
102 -- | Use Newton's method to find a root of @f@ near the initial guess
103 -- @x0@. If your guess is bad, this will recurse forever!
104 --
105 -- Examples:
106 --
107 -- Atkinson, p. 60.
108 --
109 -- >>> let f x = x^6 - x - 1
110 -- >>> let f' x = 6*x^5 - 1
111 -- >>> let Just root = newtons_method f f' (1/1000000) 2
112 -- >>> root
113 -- 1.1347241385002211
114 -- >>> abs (f root) < 1/100000
115 -- True
116 --
117 newtons_method :: (Fractional a, Ord a)
118 => (a -> a) -- ^ The function @f@ whose root we seek
119 -> (a -> a) -- ^ The derivative of @f@
120 -> a -- ^ The tolerance epsilon
121 -> a -- ^ Initial guess, x-naught
122 -> Maybe a
123 newtons_method f f' epsilon x0
124 = find (\x -> abs (f x) < epsilon) x_n
125 where
126 x_n = newton_iterations f f' x0
127
128
129
130 -- | Takes a function @f@ of two arguments and repeatedly applies @f@
131 -- to the previous two values. Returns a list containing all
132 -- generated values, f(x0, x1), f(x1, x2), f(x2, x3)...
133 --
134 -- Examples:
135 --
136 -- >>> let fibs = iterate2 (+) 0 1
137 -- >>> take 15 fibs
138 -- [0,1,1,2,3,5,8,13,21,34,55,89,144,233,377]
139 --
140 iterate2 :: (a -> a -> a) -- ^ The function @f@
141 -> a -- ^ The initial value @x0@
142 -> a -- ^ The second value, @x1@
143 -> [a] -- ^ The result list, [x0, x1, ...]
144 iterate2 f x0 x1 =
145 x0 : x1 : (go x0 x1)
146 where
147 go prev2 prev1 =
148 let next = f prev2 prev1 in
149 next : go prev1 next
150
151 -- | The sequence x_{n} of values obtained by applying the secant
152 -- method on the function @f@ and initial guesses @x0@, @x1@.
153 --
154 -- The recursion more or less implements a two-parameter 'iterate',
155 -- although one list is passed to the next iteration (as opposed to
156 -- one function argument, with iterate). At each step, we peel the
157 -- first two elements off the list and then compute/append elements
158 -- three, four... onto the end of the list.
159 --
160 -- Examples:
161 --
162 -- Atkinson, p. 67.
163 -- >>> let f x = x^6 - x - 1
164 -- >>> take 4 $ secant_iterations f 2 1
165 -- [2.0,1.0,1.0161290322580645,1.190577768676638]
166 --
167 secant_iterations :: (Fractional a, Ord a)
168 => (a -> a) -- ^ The function @f@ whose root we seek
169 -> a -- ^ Initial guess, x-naught
170 -> a -- ^ Second initial guess, x-one
171 -> [a]
172 secant_iterations f x0 x1 =
173 iterate2 g x0 x1
174 where
175 g prev2 prev1 =
176 let x_change = prev1 - prev2
177 y_change = (f prev1) - (f prev2)
178 in
179 (prev1 - (f prev1 * (x_change / y_change)))
180
181
182 -- | Use the secant method to find a root of @f@ near the initial guesses
183 -- @x0@ and @x1@. If your guesses are bad, this will recurse forever!
184 --
185 -- Examples:
186 --
187 -- Atkinson, p. 67.
188 -- >>> let f x = x^6 - x - 1
189 -- >>> let Just root = secant_method f (1/10^9) 2 1
190 -- >>> root
191 -- 1.1347241384015196
192 -- >>> abs (f root) < (1/10^9)
193 -- True
194 --
195 secant_method :: (Fractional a, Ord a)
196 => (a -> a) -- ^ The function @f@ whose root we seek
197 -> a -- ^ The tolerance epsilon
198 -> a -- ^ Initial guess, x-naught
199 -> a -- ^ Second initial guess, x-one
200 -> Maybe a
201 secant_method f epsilon x0 x1
202 = find (\x -> abs (f x) < epsilon) x_n
203 where
204 x_n = secant_iterations f x0 x1
205
206
207
208 fixed_point_iterations :: (a -> a) -- ^ The function @f@ to iterate.
209 -> a -- ^ The initial value @x0@.
210 -> [a] -- ^ The resulting sequence of x_{n}.
211 fixed_point_iterations f x0 =
212 iterate f x0
213
214
215 -- | Find a fixed point of the function @f@ with the search starting
216 -- at x0. This will find the first element in the chain f(x0),
217 -- f(f(x0)),... such that the magnitude of the difference between it
218 -- and the next element is less than epsilon.
219 --
220 fixed_point :: (Num a, Ord a)
221 => (a -> a) -- ^ The function @f@ to iterate.
222 -> a -- ^ The tolerance, @epsilon@.
223 -> a -- ^ The initial value @x0@.
224 -> a -- ^ The fixed point.
225 fixed_point f epsilon x0 =
226 fst winning_pair
227 where
228 xn = fixed_point_iterations f x0
229 xn_plus_one = tail $ fixed_point_iterations f x0
230
231 abs_diff v w =
232 abs (v - w)
233
234 -- The nth entry in this list is the absolute value of x_{n} -
235 -- x_{n+1}.
236 differences = zipWith abs_diff xn xn_plus_one
237
238 -- A list of pairs, (xn, |x_{n} - x_{n+1}|).
239 pairs = zip xn differences
240
241 -- The pair (xn, |x_{n} - x_{n+1}|) with
242 -- |x_{n} - x_{n+1}| < epsilon. The pattern match on 'Just' is
243 -- "safe" since the list is infinite. We'll succeed or loop
244 -- forever.
245 Just winning_pair = find (\(_, diff) -> diff < epsilon) pairs