Newton's Method Example

In 17 th Century Newton discovered a method for solving algebraic equations by defining a sequences of numbers that become closer to the root sought. This method is also called Newton Raphon Method.

Newton's Method Example:
Compute the real root of x log₁₀10 = 1.2 which lie between 2 and 3 and correct the result to three decimal places.
f(x) = x log(x) -1.2
f(2) = 2 log(2) -1.2 = - 0.5980 (negative)
f(3) = 3 log(3) - 1.2 = 0.231 (positive)
Since f(2) and f(3) have opposite signs, so the root of f(x) will be lies between 2 and 3 let a= 2 and b =3
Hence the approximation to the root is
x₁ = (a+b)/2 = 5/2 = 2.5
Now f(2.5) = 2.5 log(2.5) - 1.2= -0.2051

Hence the root will lies between 2.5 and 3
x₂ = (2.5 + 3)/2 = 2.75
And f(2.75) = 2.75 8 log(2.75) - 1.2 = 0.0081 (positive)

Hence the root is lies between 2.5 and 2.75

The third approximation to the root is
x₃ = (2.5 + 2.75)/2 = 2.625
f(2.625) = -0.0997 (negative)
Hence the root lie between 2.75 and 2.625

The fourth approximation to the root is
X₄ = (2.75+ 2.625)/2 = 2. 6875
So the approximation to the root is 2.6875( after fourth iteration)

Newton's method Practice problem:
1. Compute the real root of 3x - cos x -1 = 0 by newton's Raphson method
2. Find the root of the equation sin x = 1 + x³ between (-2,-1) to 3 decimal places by using newton's Raphson method
3. If an approximation root of the equation x (1- log_e x ) = 0.5 lies between 01 and 0.2 find the value of the root and correct to three decimal places