Numerical Methods Vedamurthy Solution 203

Numerical Methods Vedamurthy Solution 203 >>> __https://shoxet.com/2tuXHg__

Numerical Methods by Vedamurthy: Solution to Exercise 1.3, Question 3

In this article, we will show how to solve the following question from the book Numerical Methods by Vedamurthy [^1^]:

Find a negative root of the following equations using Newtonâs method:

(a) x - 2x - 5 = 0

(b) x + 3x - 9x - 10 = 0

(c) x - x - 10 = 0

Newton's method is an iterative technique for finding the roots of a function f(x) by using the tangent line approximation. The formula for the nth iteration is:

xn+1 = xn - f(xn)/f'(xn)

where f'(x) is the derivative of f(x). The method converges to a root if the initial guess x0 is sufficiently close to it and f(x) is continuous and differentiable near the root.

We will use four decimal places of accuracy and stop the iteration when xn+1 - xn < 0.0001.

(a) x - 2x - 5 = 0

The derivative of f(x) = x - 2x - 5 is f'(x) = 3x - 2. We choose x0 = -2 as an initial guess. The iterations are:

nxnf(xn)f'(xn)xn+1xn+1 - xn

0-2.0000-1.0000-10.0000-1.90000.1000

1-1.9000-0.4590-8.3700-1.84510.0549

2-1.8451-0.1128-7.5189-1.83020.0149

3-1.8302-0.0286-7.3287-1.82640.0038

4-1.8264-0.0073-7.2796-1.82550.0009

5-1.8255-0.0019-7.2730-1.82540.0001

Therefore, the negative root of f(x) = x - 2x - 5 is approximately -1.8254.

(b) x + 3x - 9x - 10 = 0

The derivative of f(x) = x + 3x - 9x - 10 is f'(x) = 3x + 6x - 9. We choose x0 = -3 as an initial guess. The iterations are:

nxnf(xn)f'(xn)xn+1xn+1 - xn

0-3.0000-16.0000-36.0000-2.55560.4444

1-2.5556-7.4815-25.0370-2.25480.3008

2-2.2548-3.3149-18.0570-2.07730.1775

3-2.0773-1.3627-13.8538-1.97560.1017

4-1.9756-0.5459-11.6074-1.92250.0531

5-1.9225-0.1739-11.0579-1.90840.0141

6-1.9084-0.0557-10.9428-1.90390.0045 ec8f644aee