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Created composite Simpson's integration method. Tests included. #819

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Merged
raklaptudirm merged 8 commits into TheAlgorithms:master from ggkogkou:simpson_branch
Oct 28, 2021
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1 change: 1 addition & 0 deletions DIRECTORY.md
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* [ReverseNumber](https://github.com/TheAlgorithms/Javascript/blob/master/Maths/ReverseNumber.js)
* [ReversePolishNotation](https://github.com/TheAlgorithms/Javascript/blob/master/Maths/ReversePolishNotation.js)
* [SieveOfEratosthenes](https://github.com/TheAlgorithms/Javascript/blob/master/Maths/SieveOfEratosthenes.js)
* [SimpsonIntegration](https://github.com/TheAlgorithms/Javascript/blob/master/Maths/SimpsonIntegration.js)
* [Softmax](https://github.com/TheAlgorithms/Javascript/blob/master/Maths/Softmax.js)
* [SquareRoot](https://github.com/TheAlgorithms/Javascript/blob/master/Maths/SquareRoot.js)
* [SumOfDigits](https://github.com/TheAlgorithms/Javascript/blob/master/Maths/SumOfDigits.js)
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67 changes: 67 additions & 0 deletions Maths/SimpsonIntegration.js
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/*
*
* @file
* @title Composite Simpson's rule for definite integral evaluation
* @author: [ggkogkou](https://github.com/ggkogkou)
* @brief Calculate definite integrals using composite Simpson's numerical method
*
* @details The idea is to split the interval in an EVEN number N of intervals and use as interpolation points the xi
* for which it applies that xi = x0 + i*h, where h is a step defined as h = (b-a)/N where a and b are the
* first and last points of the interval of the integration [a, b].
*
* We create a table of the xi and their corresponding f(xi) values and we evaluate the integral by the formula:
* I = h/3 * {f(x0) + 4*f(x1) + 2*f(x2) + ... + 2*f(xN-2) + 4*f(xN-1) + f(xN)}
*
* That means that the first and last indexed i f(xi) are multiplied by 1,
* the odd indexed f(xi) by 4 and the even by 2.
*
* N must be even number and a<b. By increasing N, we also increase precision
*
* More info: [Wikipedia link](https://en.wikipedia.org/wiki/Simpson%27s_rule#Composite_Simpson's_rule)
*
*/

function integralEvaluation (N, a, b, func) {
// Check if N is an even integer
let isNEven = true
if (N % 2 !== 0) isNEven = false

if (!Number.isInteger(N) || Number.isNaN(a) || Number.isNaN(b)) { throw new TypeError('Expected integer N and finite a, b') }
if (!isNEven) { throw Error('N is not an even number') }
if (N <= 0) { throw Error('N has to be >= 2') }

// Check if a < b
if (a > b) { throw Error('a must be less or equal than b') }
if (a === b) return 0

// Calculate the step h
const h = (b - a) / N

// Find interpolation points
let xi = a // initialize xi = x0
const pointsArray = []

// Find the sum {f(x0) + 4*f(x1) + 2*f(x2) + ... + 2*f(xN-2) + 4*f(xN-1) + f(xN)}
let temp
for (let i = 0; i < N + 1; i++) {
if (i === 0 || i === N) temp = func(xi)
else if (i % 2 === 0) temp = 2 * func(xi)
else temp = 4 * func(xi)

pointsArray.push(temp)
xi += h
}

// Calculate the integral
let result = h / 3
temp = 0
for (let i = 0; i < pointsArray.length; i++) temp += pointsArray[i]

result *= temp

if (Number.isNaN(result)) { throw Error('Result is NaN. The input interval doesnt belong to the functions domain') }

return result
}

export { integralEvaluation }
16 changes: 16 additions & 0 deletions Maths/test/SimpsonIntegration.test.js
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import { integralEvaluation } from '../SimpsonIntegration'

test('Should return the integral of f(x) = sqrt(x) in [1, 3] to be equal 2.797434', () => {
const result = integralEvaluation(16, 1, 3, (x) => { return Math.sqrt(x) })
expect(Number(result.toPrecision(7))).toBe(2.797434)
})

test('Should return the integral of f(x) = sqrt(x) + x^2 in [1, 3] to be equal 11.46410161', () => {
const result = integralEvaluation(64, 1, 3, (x) => { return Math.sqrt(x) + Math.pow(x, 2) })
expect(Number(result.toPrecision(10))).toBe(11.46410161)
})

test('Should return the integral of f(x) = log(x) + Pi*x^3 in [5, 12] to be equal 15809.9141543', () => {
const result = integralEvaluation(128, 5, 12, (x) => { return Math.log(x) + Math.PI * Math.pow(x, 3) })
expect(Number(result.toPrecision(12))).toBe(15809.9141543)
})

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