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Adding doctests in simpson_rule.py (#10269)

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* Adding doctests in simpson_rule.py

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* Update maths/simpson_rule.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update maths/simpson_rule.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Adding doctests in simpson_rule.py

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Adding doctests in simpson_rule.py

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Adding doctests in simpson_rule.py

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Adding doctests in simpson_rule.py

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Adding doctests in simpson_rule.py

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Update simpson_rule.py

* Adding doctests in simpson_rule.py

* Adding doctests in simpson_rule.py

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Christian Clauss <cclauss@me.com>
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Aasheesh 2023-10-11 23:50:18 +05:30 committed by GitHub
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1 changed files with 39 additions and 4 deletions
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maths/simpson_rule.py
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@ -1,7 +1,7 @@
"""
Numerical integration or quadrature for a smooth function f with known values at x_i
This method is the classical approach of suming 'Equally Spaced Abscissas'
This method is the classical approach of summing 'Equally Spaced Abscissas'
method 2:
"Simpson Rule"
@ -9,9 +9,41 @@ method 2:
"""
def method_2(boundary, steps):
def method_2(boundary: list[int], steps: int) -> float:
# "Simpson Rule"
# int(f) = delta_x/2 * (b-a)/3*(f1 + 4f2 + 2f_3 + ... + fn)
"""
Calculate the definite integral of a function using Simpson's Rule.
:param boundary: A list containing the lower and upper bounds of integration.
:param steps: The number of steps or resolution for the integration.
:return: The approximate integral value.
>>> round(method_2([0, 2, 4], 10), 10)
2.6666666667
>>> round(method_2([2, 0], 10), 10)
-0.2666666667
>>> round(method_2([-2, -1], 10), 10)
2.172
>>> round(method_2([0, 1], 10), 10)
0.3333333333
>>> round(method_2([0, 2], 10), 10)
2.6666666667
>>> round(method_2([0, 2], 100), 10)
2.5621226667
>>> round(method_2([0, 1], 1000), 10)
0.3320026653
>>> round(method_2([0, 2], 0), 10)
Traceback (most recent call last):
...
ZeroDivisionError: Number of steps must be greater than zero
>>> round(method_2([0, 2], -10), 10)
Traceback (most recent call last):
...
ZeroDivisionError: Number of steps must be greater than zero
"""
if steps <= 0:
raise ZeroDivisionError("Number of steps must be greater than zero")
h = (boundary[1] - boundary[0]) / steps
a = boundary[0]
b = boundary[1]
@ -41,11 +73,14 @@ def f(x): # enter your function here
def main():
a = 0.0 # Lower bound of integration
b = 1.0 # Upper bound of integration
steps = 10.0 # define number of steps or resolution
boundary = [a, b] # define boundary of integration
steps = 10.0 # number of steps or resolution
boundary = [a, b] # boundary of integration
y = method_2(boundary, steps)
print(f"y = {y}")
if __name__ == "__main__":
import doctest
doctest.testmod()
main()