Newton Raphson Method (original) (raw)

Last Updated : 10 Sep, 2025

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.

• It is a numerical technique for approximating the roots of real-valued functions.
• It starts with initial guess of root and iteratively refines the result using a formula that involves derivative of the function.
• Compared to other root-finding methods like bisection and secant methods, the Newton-Raphson method stands out due to its significantly faster convergence rate (quadratic while other have linear).
• It requires computation of derivative and preferred over other methods when this computation easier and we can find good estimate of root.

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.

x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}

where,

• **x 0 is the initial value of x,
• **f(x 0 ) is the value of the equation at initial value, and
• **f'(x 0 ) 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:

x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}

Where,

• **x n-1 is the estimated (n-1)th root of the function
• **f(x n-1 ) is the value of the equation at (n-1)th estimated root
• **f'(x n-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'(x 0 ) ≠ 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)
x_1 = x_0 - \frac{ f(x_0)}{f'(x_0)}

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.

Applications of Newton Raphson Method

• **Root Finding in Mathematics: The primary use of the Newton-Raphson method is to find the roots (or zeros) of functions. Given an equation f(x)=0, the method iteratively approximates the solution by refining guesses.
• **Solving Non-linear Equations: In engineering and physics, many real-world problems are modeled by non-linear equations. The Newton-Raphson method is used to find solutions to these equations efficiently.
• **Optimization Problems: In optimization, the method is used to find the maximum or minimum of a function. By iterating towards a point where the derivative of the function equals zero, the method can identify critical points in functions.
• **Machine Learning: It is used in some optimization techniques, such as for training models by finding the minimum of a loss function using the method’s iterative approach to minimizing error.
• **Engineering (Structural Analysis): In structural engineering, the method is used to solve complex systems of equations, such as those that arise when analyzing stresses and strains in materials.

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 1: For the initial value x 0 **= 3, approximate the root of f(x)=x 3 +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:
x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}
= 3 - 37/30
= 1.767

**Example 2: For the initial value x 0 **= 1, approximate the root of f(x)=x 2 -5x+1.

**Solution:

Given, x0 = 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 = 1 - (-3)/-3
⇒ x1 = 1 -1
⇒ x1 = 0

**Problem 3: For the initial value x 0 **= 2, approximate the root of f(x)=x 3 -6x+1.

**Solution:

Given, x0 = 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 = 2 - (-3)/6
⇒ x1 = 2 + 1/2
⇒ x1 = 5/2 = 2.5

**Problem 4: For the initial value x 0 **= 3, approximate the root of f(x)=x 2 -3.

**Solution:

Given, x0 = 3 and f(x) = x2-3
f'(x) = 2x
f'(x0) = 6
f(x0) = f(3) = 9 - 3 = 6

Using Newton Raphson method:
⇒ x1 = 3 - 6/6
⇒ x1 = 2

**Problem 5: Find the root of the equation f(x) = x 3 - 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 = 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.

Articles related to Newton Raphson Method:

• Difference between Newton Raphson Method and Regular Falsi Method
• Difference between Bisection Method and Newton Raphson Method

Newton Raphson Method: Practice Problems

**Problem 1: Find the root of f(x) = x2-2 using the Newton-Raphson method starting with x0=1.

**Problem 2: Find the root of f(x) = x3-2x+1 using the Newton-Raphson method starting with x0=0.

**Problem 3: Find the root of f(x) = cos(x)-x using the Newton-Raphson method starting with x0=0.5.

**Problem 4: Find the root of f(x) = ex-3x using the Newton-Raphson method starting with x0=1.

**Problem 5: Find the root of f(x) = x3-4x2+6 using the Newton-Raphson method starting with x0=2.

**Problem 6: Find the root of f(x) = ln⁡(x)-1 using the Newton-Raphson method starting with x0=2.

**Problem 7: Find the root of f(x) = x4-8x2+16 using the Newton-Raphson method starting with x0=2.5.

**Problem 8: Find the root of f(x) = xsin⁡(x)-1 using the Newton-Raphson method starting with x0=1.

**Problem 9: Find the root of f(x)=x5-3x3+2 using the Newton-Raphson method starting with x0=1.

**Problem 10: Find a root of f(x) = x3-6x2+11x-6 using the Newton-Raphson method starting with x0=3.