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1. Using Newton's Method of Approximation Approximate the Cube Root of 4, Assume the Root Is Close to 2, Using Only 2 Iterations. 2. I.

Question

1. Using Newton's method of approximation approximate the cube root of 4, assume the root is close to 2, using only 2 iterations. 2. i. What is a (a).linear spline (b)a cubic spline. (6 marks) iii. ii. Construct a linear spline that interpolates the points (1,2),(2,1),(4,4) and (5,3) Study the function below and check if it satisfies conditions for a cubic spline f(x)= ) S_(1)(x)=5+4x^2+x^2+2x^3 on [0,1] S_(2)(x)=12+12(x-1)+7(x-1)^2+(x-1)^3 on [1,2] . Let f(x) be a function with a continuous second derivative on the interval [x_(0),x_(1)] Denote the corresponding function values by y_(0)=f(x_(0)) and y_(1)=f(x_(1)) . Consider the degree 1 interpolating polynomial p(x) through (x_(0),y_(0)) and (x_(1),y_(1)) a. Using the Lagrange formulation, find the interpolating polynomial p(x) at the points (x_(0),y_(0)) and (x_(1),y_(1)) b. State the Trapezoid Formula for numerical integration for the approximation of I=int _(x_(0))^x_(1)f(x)dx and use your result in a) above to show how it is derived. c. The interval [x_(0),x_(1)] can be partitioned into n intervals to give a more accurate results , show how this formula can be extend to cater for this situation. (3+7+5=15 marks)

Solution

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4.3 (267 Votos)
Julia Maria Avançado · Tutor por 1 anos

Resposta

1. Using Newton's method of approximation to approximate the cube root of 4, we start with an initial guess of 2. We will perform two iterations.Let and .Iteration 1: Iteration 2: Therefore, the cube root of 4 is approximately 1.62.2. i. A linear spline is a piecewise linear function that interpolates a set of points. It consists of a set of line segments connected end-to-end, where each line segment is defined by two points.A cubic spline is a piecewise cubic function that interpolates a set of points. It consists of a set of cubic curves connected end-to-end, where each cubic curve is defined by four points.ii. To construct a linear spline that interpolates the points , , , and , we can use the following steps:1. Arrange the points in order of their x-coordinates: , , , .2. Choose two points, say and , and find the equation of the line passing through them. The equation of the line is .3. Choose another two points, say and , and find the equation of the line passing through them. The equation of the line is .4. The linear spline is then given by the piecewise function: \begin{cases}y = -x + 3, & \text{if } x \leq 2 \\y = -\frac{1}{3}x + \frac{16}{3}, & \text{if } x > 2\end{cases} iii. To check if the function satisfies the conditions for a cubic spline, we need to check if it is continuous and has a continuous second derivative on the interval and .The function is defined as: f(x) = \begin{cases}S_1(x) = 5 + 4x^2 + x^2 + 2x^3, & \text{on } [0,1] \\S_2(x) = 12 + 12(x-1) + 7(x-1)^2 + (x-1)^3, & \text{on } [1,2]\end{cases} To check if is continuous, we need to check if the left-hand limit and right-hand limit of are equal at . We can do this by evaluating the left-hand limit and right-hand limit of as approaches 1.Left-hand limit: \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} S_1(x) = \lim_{x \to 1^-} (5 + 4x^2 + x^2 + 2x^3) = 5 + 4(1)^2 + (1)^2 + 2(1)^3 = 12 Right-hand limit: \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} S_2(x) = \lim_{x \to 1^+} (12 + 12(x-1) +