# Newton Raphson Method

Newton Raphson Method or Newton Method is a powerful technique for solving equations numerically. It is most commonly used for approximation of the roots of the real-valued functions. Newton Rapson Method was developed by Isaac Newton and Joseph Raphson, hence the name Newton Rapson Method.

Newton Raphson Method involves iteratively refining an initial guess to converge it toward the desired root. However, the method is not efficient to calculate the roots of the polynomials or equations with higher degrees but in the case of small-degree equations, this method yields very quick results. In this article, we will learn about Newton Raphson Method and the steps to calculate the roots using this method as well.

## What is Newton Raphson Method?

The Newton-Raphson method which is also known as Newton’s method, is an iterative numerical method used to find the roots of a real-valued function. This formula is named after Sir Isaac Newton and Joseph Raphson, as they independently contributed to its development. Newton Raphson Method or Newton’s Method is an algorithm to approximate the roots of zeros of the real-valued functions, using guess for the first iteration (x0) and then approximating the next iteration(x1) which is close to roots, using the following formula.

x1 = x0 – f(x0)/f'(x0)

where,

• x0 is the initial value of x,
• f(x0) is the value of the equation at initial value, and
• f'(x0) is the value of the first order derivative of the equation or function at the initial value x0.

Note: f'(x0) should not be zero else the fraction part of the formula will change to infinity which means f(x) should not be a constant function.

## Newton Raphson Method Formula

In the general form, the Newton-Raphson method formula is written as follows:

xn = xn-1 – f(xn-1)/f'(xn-1)

Where,

• xn-1 is the estimated (n-1)th root of the function,
• f(xn-1) is the value of the equation at (n-1)th estimated root, and
• f'(xn-1) is the value of the first order derivative of the equation or function at xn-1.

## Newton Raphson Method Calculation

Assume the equation or functions whose roots are to be calculated as f(x) = 0.

In order to prove the validity of Newton Raphson method following steps are followed:

Step 1: Draw a graph of f(x) for different values of x as shown below:

Step 2: A tangent is drawn to f(x) at x0. This is the initial value.

Step 3:This tangent will intersect the X- axis at some fixed point (x1,0) if the first derivative of f(x) is not zero i.e. f'(x0) ≠ 0.

Step 4: As this method assumes iteration of roots, this x1 is considered to be the next approximation of the root.

Step 5: Now steps 2 to 4 are repeated until we reach the actual root x*.

Now we know that the slope-intercept equation of any line is represented as y = mx + c,

Where m is the slope of the line and c is the x-intercept of the line.

Using the same formula we, get

y = f(x0) + f'(x0) (x − x0)

Here f(x0) represents the c and f'(x0) represents the slope of the tangent m. As this equation holds true for every value of x, it must hold true for x1. Thus, substituting x with x1, and equating the equation to zero as we need to calculate the roots, we get:

0 = f(x0) + f'(x0) (x1 − x0)

x1 = x0 – f(x0)/f'(x0)

Which is the Newton Raphson method formula.

Thus, Newton Raphson’s method was mathematically proved and accepted to be valid.

### Convergence of Newton Raphson Method

The Newton-Raphson method tends to converge if the following condition holds true:

|f(x).f”(x)| < |f'(x)|2

It means that the method converges when the modulus of the product of the value of the function at x and the second derivative of a function at x is lesser than the square of the modulo of the first derivative of the function at x. The Newton-Raphson Method has a convergence of order 2 which means it has a quadratic convergence.

Note:

Newton Raphson’s method is not valid if the first derivative of the function is 0 which means f'(x) = 0. It is only possible when the given function is a constant function.

## Newton Raphson Method Example

Let’s consider the following example to learn more about the process of finding the root of a real-valued function.

Example: For the initial value x0 = 3, approximate the root of f(x)=x3+3x+1.

Solution:

Given, x0 = 3 and f(x) = x3+3x+1

f'(x) = 3x2+3

f'(x0) = 3(9) + 3 = 30

f(x0) = f(3) = 27 + 3(3) + 1 = 37

Using Newton Raphson method:

x1 = x0 – f(x0)/f'(x0)

= 3 – 37/30

= 1.767

## Solved Problems of Newton Raphson Method

Problem 1: For the initial value x= 1, approximate the root of f(x)=x2−5x+1.

Solution:

Given, x= 1 and f(x) = x2-5x+1

f'(x) = 2x-5

f'(x0) = 2 – 5 = -3

f(x0) = f(1) = 1 – 5 + 1 = -3

Using Newton Raphson method:

x1 = x0 – f(x0)/f'(x0)

⇒ x1 = 1 – (-3)/-3

⇒ x1 = 1 -1

⇒ x1 = 0

Problem 2: For the initial value x= 2, approximate the root of f(x)=x3−6x+1.

Solution:

Given, x= 2 and f(x) = x3-6x+1

f'(x) = 3x2 – 6

f'(x0) = 3(4) – 6 = 6

f(x0) = f(2) = 8 – 12 + 1 = -3

Using Newton Raphson method:

x1 = x0 – f(x0)/f'(x0)

⇒ x1 = 2 – (-3)/6

⇒ x1 = 2 + 1/2

⇒ x1 = 5/2 = 2.5

Problem 3: For the initial value x= 3, approximate the root of f(x)=x2−3.

Solution:

Given, x= 3 and f(x) = x2-3

f'(x) = 2x

f'(x0) = 6

f(x0) = f(3) = 9 – 3 = 6

Using Newton Raphson method:

x1 = x0 – f(x0)/f'(x0)

⇒ x1 = 3 – 6/6

⇒ x1 = 2

Problem 4: Find the root of the equation f(x) = x3 – 3 = 0, if the initial value is 2.

Solution:

Given x0 = 2 and f(x) = x3 – 3

f'(x) = 3x2

f'(x0 = 2) = 3 × 4 = 12

f(x0) = 8 – 3 = 5

Using Newton Raphson method:

x1 = x0 – f(x0)/f'(x0)

⇒ x1 = 2 – 5/12

⇒ x1 = 1.583

Using Newton Raphson method again:

x2 = 1.4544

x3 = 1.4424

x4 = 1.4422

Therefore, the root of the equation is approximately x = 1.442.

Problem 5: Find the root of the equation f(x) = x3 – 5x + 3 = 0, if the initial value is 3.

Solution:

Given x0 = 3 and f(x) = x3 – 5x + 3 = 0

f'(x) = 3x2 – 5

f'(x0 = 3) = 3 × 9 – 5 = 22

f(x0 = 3) = 27 – 15 + 3 = 15

Using Newton Raphson method:

x1 = x0 – f(x0)/f'(x0)

⇒ x1 = 3 – 15/22

⇒ x1 = 2.3181

Using Newton Raphson method again:

x2 = 1.9705

x3 = 1.8504

x4 = 1.8345

x5 = 1.8342

Therefore, the root of the equation is approximately x = 1.834.

## FAQs of Newton Raphson Method

### Q1: Define Newton Raphson Method.

Newton Raphson Method is a numerical method to approximate the roots of any given real-valued function. In this method, we used various iterations to approximate the roots, and the higher the number of iterations the less error in the value of the calculated root.

### Q2: What is the Advantage of Newton Raphson Method?

Newton Raphson method has an advantage that it allows us to guess the roots of an equation with a small degree very efficiently and quickly.

### Q3: What is the Disadvantage of Newton Raphson Method?

The disadvantage of Newton Raphson method is that it tends to become very complex when the degree of the polynomial becomes very large.