--- title: Simpson's Methods published: true description: Article about Simpson's Methods tags: #NumericalMethods #SimpsonsMethods #Rust --- Before explaining what the Simpson's methods are used for and give an example, it is necessary to give the definition of numerical method. A numerical method is a mathematical technique used for solving mathematical problems that cannot be solved or are difficult to solve analytically. To solve a problem analytically is to give an exact answer in the form of a mathematical expression. In other words, a numerical method is an algorithm that converges to a solution that approximates to the exact answer. This solution is called _numerical solution_. So, let's start with the main topic. Simpson's methods are used for approximating the value of an integral I( _f_ ) of a function _f_(_x_) over an interval from _a_ to _b_ using quadratic (Simpson's 1/3 method) and cubic (Simpson's 3/8 method) polynomials. These methods are used when analytical integration is difficult or not possible, and when the integrand is given as a set of discrete points. > Simpson's 1/3 method uses a quadratic polynomial to approximate the integrand. We need three points to determine the coefficients of this polynomial. These points are _x_₁ = _a_, _x_₃ = _b_ and _x_₂ = (_a_+_b_)/2 > ![1/3](https://thepracticaldev.s3.amazonaws.com/i/milyzy904t8vhkj8v0u3.png) >> The name 1/3 in the method comes from the factor in the expression. > If you want a more accurate evaluation of the integral with this method, you can use the composite Simpson's 1/3 method in which you must divide the whole interval into _n_ subintervals using an even number, because Simpson's 1/3 method needs three points for defining a quadratic polynomial, that means that this method applies two adjacent subintervals at a time. > ![1/3C](https://thepracticaldev.s3.amazonaws.com/i/4a0qsfv9wczvdvxl1g6e.png) >> Where, the subintervals _n_ must be equally space. And _h_ = (_b_-_a_)/_n_ --- > Simpson's 3/8 method uses a cubic polynomial to approximate the integrand. We need four points to determine the coefficients of this polynomial. These points are _x_₁ = _a_, _x_₂ = _a_+_h_, _x_₃ = _a_+2 _h_ and _x_₄ = _b_ > ![3/8](https://thepracticaldev.s3.amazonaws.com/i/pin7tseoulymiru9eonm.png) >> The name 3/8 in the method comes from the factor in the expression. > If you want a more accurate evaluation of the integral with this method, you can use the composite Simpson's 3/8 method in which you have to divide the whole interval into a number _n_ of subintervals that is divisible by 3, because Simpson's 3/8 method need four points for constructing a cubic polynomial, that mean that this method applies three adjacent subintervals at a time. > ![3/8C](https://thepracticaldev.s3.amazonaws.com/i/b3rsjarh9ku9ekovbyxr.png) >> Where, the subintervals _n_ must be equally space. And _h_ = (_b_-_a_)/_n_ These methods are applied in the real world to calculate areas, volumes, curve lengths and other problems related to integrals. For example: the company ECO wants to drain and fill a polluted marsh (see the image below) that has a depth of 5 feet. The CEO of ECO wants to know how many cubic feet of land are needed to fill the area after draining the marsh. ![Marsh](https://thepracticaldev.s3.amazonaws.com/i/r89dpf5gxg0853j18syq.png) To solve this problem I used the composite Simpson's 1/3 method. ```rust fn simpson(a: f64, b: f64, n: i32) -> f64 { let mut y: f64 = funcion(a) + funcion(b); let mut x: f64 = a; let h: f64 = (b-a)/(n as f64); for i in 1..n { x = x + h; if i % 2 == 0{ y= y + 2.0*funcion(x); }else{ y= y + 4.0*funcion(x); } } return (b-a) * y / (3.0*(n as f64)); } ``` Solution: To calculate the volume of the marsh, we must first estimate the surface area using the composite Simpson's 1/3 method. ```rust let mut resultado: i32 = simpson(vec![146, 122, 76, 54, 40, 30, 13], 20); fn simpson(v: Vec, h: i32) -> i32 { let mut y: i32 = 0; let mut con: i32 = 0; let size: usize = v.len()-1; for i in v { if con == 0{ y = y + i; }else if con == (size as i32) { y = y + i; }else{ if con % 2 == 0{ y= y + 2*i; }else{ y= y + 4*i; } } con = con +1; } return h*y/3; } ``` And finally multiply by 5. ```rust resultado = resultado * 5; ``` Obtaining that the approximate volume is `40 500 cubic feet`. In conclusion, integration with numerical methods is a useful technique when we try to integrate a complicated function or if we only have tabulated data. With the Simpson's methods we can approximate a complex integral to the integral of a polynomial and obtain a solution that approximates the exact answer, even in some cases we can obtain the exact answer. ### References * Gilat A., Subramaniam, V. (2013). Numerical methods for engineers and scientists. Wiley. Third Edition. * Heath M., (2002). Scientific Computing: An Introductory Survet. McGraw Hill. Second Edition. * Chapra, S., Canale, R. (2011). Métodos Numéricos para ingenieros. McGraw Hill. Sexta edición. ### Additional resources * [Numerical Methods](https://github.com/LindaEliza/NumericalMethods) repository in GitHub, by LindaEliza.

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Simpson's Methods

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