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 log_{10}10 = 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

x_{1} = (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

x_{2} = (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

x_{3} = (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

X_{4} = (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 + x^{3} 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

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