Josh Crotty

The Newton Raphson Method

What is the Newton Raphson Method and how do you use it?

Numerical Methods
The Newton Raphson Method

Newton-Raphson Method: Finding Roots with Calculus

The Newton-Raphson Method is one of the most powerful and widely used numerical techniques for finding the roots of equations.

A root (or zero) of a function is a value of (x) such that:

f(x)=0f(x) = 0

When equations become too difficult to solve algebraically, the Newton-Raphson method provides a fast and elegant way to approximate the solution using calculus.

What Is the Newton-Raphson Method?

The Newton-Raphson method starts with an initial guess (x_0) and repeatedly improves that guess by using the tangent line to the function.

This creates a sequence of increasingly accurate approximations:

x0, x1, x2, x3, …x_0,\ x_1,\ x_2,\ x_3,\ \dots

Under the right conditions, the method converges extremely quickly to the true root.

The Newton-Raphson Formula

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Where:

  • (x_n) is the current approximation.
  • (x_{n+1}) is the next approximation.
  • (f(x_n)) is the value of the function at (x_n).
  • (f’(x_n)) is the derivative (the slope of the tangent line).

Why This Formula Works

At the current estimate (x_n), we draw the tangent line to the graph of the function.

The tangent line equation is:

y=f(xn)+f′(xn)(x−xn)y = f(x_n) + f'(x_n)(x - x_n)

To estimate the root, we find where this tangent line crosses the x-axis by setting (y = 0):

0=f(xn)+f′(xn)(x−xn)0 = f(x_n) + f'(x_n)(x - x_n)

Solving for (x) gives:

x=xn−f(xn)f′(xn)x = x_n - \frac{f(x_n)}{f'(x_n)}

This becomes the update formula used in the Newton-Raphson method.

Step by Step Algorithm

  1. Define the function (f(x)).
  2. Compute its derivative (f’(x)).
  3. Choose an initial guess (x_0).
  4. Calculate the next approximation using the Newton-Raphson formula.
  5. Repeat until the estimates stop changing significantly.

A common stopping condition is:

∣xn+1−xn∣<10−6|x_{n+1} - x_n| < 10^{-6}

Example: Finding (\sqrt{2})

To compute (\sqrt{2}), we solve:

x2−2=0x^2 - 2 = 0

Step 1: Define the Function

f(x)=x2−2f(x) = x^2 - 2

Step 2: Compute the Derivative

f′(x)=2xf'(x) = 2x

Step 3: Substitute into the Formula

xn+1=xn−xn2−22xnx_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n}

Step 4: Choose an Initial Guess

Let:

x0=1.5x_0 = 1.5

Successive tangent lines converging to the root of x² - 2

Iteration 1

x1=1.5−1.52−22(1.5)x_1 = 1.5 - \frac{1.5^2 - 2}{2(1.5)} x1=1.4166666667x_1 = 1.4166666667

Iteration 2

x2=1.4166666667−1.41666666672−22(1.4166666667)x_2 = 1.4166666667 - \frac{1.4166666667^2 - 2}{2(1.4166666667)} x2≈1.4142156863x_2 \approx 1.4142156863

Iteration 3

x3≈1.4142135624x_3 \approx 1.4142135624

Final Answer

2≈1.4142135624\sqrt{2} \approx 1.4142135624

The actual value is:

2=1.414213562373095…\sqrt{2} = 1.414213562373095\ldots

After only three iterations, the approximation is accurate to many decimal places.

Iteration Table

IterationApproximation
(x_0)1.5000000000
(x_1)1.4166666667
(x_2)1.4142156863
(x_3)1.4142135624

Advantages of the Newton-Raphson Method

  • Extremely fast convergence near the root.
  • Simple formula to implement.
  • Widely used in mathematics, engineering, and computer science.
  • Forms the basis of many scientific computing algorithms.

Limitations

The method can fail if:

  • The derivative is zero or very close to zero.
  • The initial guess is too far from the root.
  • The function is not smooth.
  • The iterations diverge instead of converging.

Because of this, choosing a reasonable starting point is important.

Python Implementation

def newton_raphson(f, df, x0, tol=1e-6, max_iter=100):
    x = x0

    for i in range(max_iter):
        x_new = x - f(x) / df(x)

        if abs(x_new - x) < tol:
            return x_new

        x = x_new

    return x

# Example: sqrt(2)
f = lambda x: x**2 - 2
df = lambda x: 2*x

root = newton_raphson(f, df, 1.5)
print(root)  # 1.4142135623730951
© 2026 Josh Crotty