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Function bisection

maths/numerical_analysis/bisection.py:4–43  ·  view source on GitHub ↗

finds where function becomes 0 in [a,b] using bolzano >>> bisection(lambda x: x ** 3 - 1, -5, 5) 1.0000000149011612 >>> bisection(lambda x: x ** 3 - 1, 2, 1000) Traceback (most recent call last): ... ValueError: could not find root in given interval. >>> bisectio

(function: Callable[[float], float], a: float, b: float)

Source from the content-addressed store, hash-verified

2
3
4def bisection(function: Callable[[float], float], a: float, b: float) -> float:
5 """
6 finds where function becomes 0 in [a,b] using bolzano
7 >>> bisection(lambda x: x ** 3 - 1, -5, 5)
8 1.0000000149011612
9 >>> bisection(lambda x: x ** 3 - 1, 2, 1000)
10 Traceback (most recent call last):
11 ...
12 ValueError: could not find root in given interval.
13 >>> bisection(lambda x: x ** 2 - 4 * x + 3, 0, 2)
14 1.0
15 >>> bisection(lambda x: x ** 2 - 4 * x + 3, 2, 4)
16 3.0
17 >>> bisection(lambda x: x ** 2 - 4 * x + 3, 4, 1000)
18 Traceback (most recent call last):
19 ...
20 ValueError: could not find root in given interval.
21 """
22 start: float = a
23 end: float = b
24 if function(a) == 0: # one of the a or b is a root for the function
25 return a
26 elif function(b) == 0:
27 return b
28 elif (
29 function(a) * function(b) > 0
30 ): # if none of these are root and they are both positive or negative,
31 # then this algorithm can't find the root
32 raise ValueError("could not find root in given interval.")
33 else:
34 mid: float = start + (end - start) / 2.0
35 while abs(start - mid) > 10**-7: # until precisely equals to 10^-7
36 if function(mid) == 0:
37 return mid
38 elif function(mid) * function(start) < 0:
39 end = mid
40 else:
41 start = mid
42 mid = start + (end - start) / 2.0
43 return mid
44
45
46def f(x: float) -> float:

Callers 1

bisection.pyFile · 0.70

Calls 1

functionFunction · 0.85

Tested by

no test coverage detected