Newton‑Raphson Method | Root‑finding Tutorial with Examples (GATE / Engineering Math)

What is the Newton‑Raphson Method?
Derivation of the Algorithm
Step-by-Step Example
Convergence and Limitations
Application in GATE / Engineering Maths
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Newton-Raphson Method:

In this article, we discuss the formula of the Newton-Raphson method, its limitations, and its advantages. Also, we provide a few solved examples and a few unsolved questions for practice.

We discuss Newton iterative formula and then solve a few questions using these iterative formulae. For practice unsolved questions are also provided. This method is generally used to improve the results obtained by one of the previous methods. This method can be derived from Taylor's series.

The formula used as follows: $x_{n+1}= x_n - \frac{f(x_n)}{f'(x_n)}$

NOTE:

(1)] This method is useful in cases of large values of $f'(x)$ that is , when the graph of $f(x)$ while crossing the x-axis is nearly vertical.

(2)] If $f'(x)$ is zero or nearly zero, the method fails.

(3)] Newton's formula converges, provided that the initial approximation $x_0$ is chosen sufficiently close to the root. This method is also called the method of tangents.

(4)] This method is also used to obtain complex roots.

(5)] The order of convergence of Newton-Rphson method is 2, that is, Newton-Rphson method is quadratic convergent.

(6)] Method converges rapidly.

(7)] The formula is also termed as second order formula due to its quadratic convergence.

Finding the Root of $f(x)=x\log_{10}x-1.2$ with Newton’s Method

Step 1: Locate an interval $[a,b]$ that brackets the root by evaluating successive integers $x=0,1,2,3,\dots$. Stop when $f(x)$ changes sign.

$ f(0) = -1.2 $ (negative)
$ f(1) = -1.2 $ (negative)
$ f(2) = -0.5979 $ (negative)
$ f(3) = 0.23136 $ (positive)

Because the sign changes between $x=2$ and $x=3$, the root lies in the interval $[2,3]$. Choose the initial approximation $x_{0}=2$.

Step 2: Differentiate:

$f'(x)=\log_{10}x+\log_{10}e=\log_{10}x+0.43429.$

Step 3: Write Newton’s iteration formula

$ x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})} =x_{n}-\frac{x_{n}\log_{10}x_{n}-1.2}{\log_{10}x_{n}+0.43429}=\frac{0.43429\,x_{n}+1.2}{\log_{10}x_{n}+0.43429}. $

Step 4: Iterate until successive values stabilize.

Iteration	Value
$x_{1}$	$\displaystyle \frac{0.43429(2)+1.2}{\log_{10}2+0.43429}=2.81$
$x_{2}$	$\displaystyle \frac{0.43429(2.81)+1.2}{\log_{10}2.81+0.43429}=2.741$
$x_{3}$	$\displaystyle \frac{0.43429(2.741)+1.2}{\log_{10}2.741+0.43429}=2.74064$
$x_{4}$	$\displaystyle \frac{0.43429(2.74064)+1.2}{\log_{10}2.74064+0.43429}=2.74065$

Hence the approximate root is

$ \boxed{2.7406}\quad\text{(correct to four decimal places).} $

Newton–Raphson Method — Root of $x^{3}-3x-5=0$

Step 1: Evaluate $f(x)=x^{3}-3x-5$ at successive integers to locate a sign change.

$ f(0) = -5 $ (negative)
$ f(1) = 1 - 3 - 5 = -7 $ (negative)
$ f(2) = 8 - 6 - 5 = -3 $ (negative)
$ f(3) = 27 - 9 - 5 = 13 $ (positive)

The sign changes between $x=2$ and $x=3$, so the root lies in $[2,3]$. Choose $x_{0}=2.4$ as the initial guess.

Step 2: The derivative is

$f'(x) = 3x^{2}-3. $

Step 3: Newton’s iteration formula is

$ x_{n+1}= x_{n}-\frac{f\!\left(x_{n}\right)}{f'\!\left(x_{n}\right)} = x_{n}-\frac{x_{n}^{3}-3x_{n}-5}{3x_{n}^{2}-3} = \frac{2x_{n}^{3}+5}{3x_{n}^{2}-3}. $

Step 4: Iterate until the values stabilize.

Iteration	Value ( $x_{n}$ )
$x_{1}$	$\displaystyle \frac{2(2.4)^{3}+5}{3(2.4)^{2}-3}=2.28627$
$x_{2}$	$\displaystyle \frac{2(2.28627)^{3}+5}{3(2.28627)^{2}-3}=2.279047$
$x_{3}$	$\displaystyle \frac{2(2.279047)^{3}+5}{3(2.279047)^{2}-3}=2.2790188$

The approximate root is

$ \boxed{2.2790}\quad\text{(correct to four decimal places).} $

Newton–Raphson Method for $3x = \cos x + 1$

Step 1: Define the function:

$ f(x) = 3x - \cos x - 1 $

Evaluate $f(x)$ at values of $x = 0, 1, 2, \ldots$:

$f(0) = -2$ (negative)
$f(1) = 1.4597$ (positive)

The sign changes between $x = 0$ and $x = 1$, so the root lies in $[0, 1]$. Choose $x_0 = 0.6$ as the initial approximation.

Step 2: Compute the derivative:

$ f'(x) = 3 + \sin x $

Step 3: Apply Newton’s iteration formula:

$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{3x_n - \cos x_n - 1}{3 + \sin x_n} = \frac{x_n \sin x_n + \cos x_n + 1}{3 + \sin x_n} $

Step 4: Perform iterations:

Iteration	Value ( $x_n$ )
$x_1$	$\displaystyle \frac{0.6 \cdot \sin(0.6) + \cos(0.6) + 1}{3 + \sin(0.6)} = 0.6071$
$x_2$	$\displaystyle \frac{0.6071 \cdot \sin(0.6071) + \cos(0.6071) + 1}{3 + \sin(0.6071)} = 0.6071$

The iterations converge. Hence, the approximate root is:

$ \boxed{0.6071} \quad \text{(correct to four decimal places).} $

Newton–Raphson Method for $x^{4}-x-10=0$

Step 1 — Locate an interval bracketing the root.

$ f(x) = x^{4} - x - 10 $
$f(0) = -10$ (negative),
$f(1) = 1-1-10 = -10 $ (negative),
$f(2) = 16-2-10 = 4$ (positive).

The sign change between $x=1$ and $x=2$ shows a root in the interval $[1,2]$. Choose the initial guess $x_{0}=1.8$.

Step 2 — Derivative.

$ f'(x)=4x^{3}-1. $

Step 3 — Newton iteration.

$ x_{n+1} = x_{n}-\frac{f(x_{n})}{f'(x_{n})} = x_{n}-\frac{x_{n}^{4}-x_{n}-10}{4x_{n}^{3}-1} = \frac{3x_{n}^{4}+10}{4x_{n}^{3}-1}. $

Step 4 — Compute successive approximations.

Iteration	Value ( $x_{n}$ )
$x_{1}$	$\displaystyle \frac{3(1.8)^{4}+10}{4(1.8)^{3}-1}=1.85833$
$x_{2}$	$\displaystyle \frac{3(1.85833)^{4}+10}{4(1.85833)^{3}-1}=1.85559$
$x_{3}$	$\displaystyle \frac{3(1.85559)^{4}+10}{4(1.85559)^{3}-1}=1.85558$

After convergence, the real root (to four decimal places) is

$ \boxed{1.8556}. $

EXERCISE

Use Newton–Raphson to find the positive root of $x^{3}-x-1=0$ (correct to four decimal places).
Apply Newton–Raphson to approximate the root of $2\sin x - x = 0$ between $(1,2)$.
Determine the real root of $e^{x} + x = 5$ using Newton–Raphson.
Find a root of the equation $\sin x = 1 - x$ using the Newton–Raphson method, correct to four decimal places.
Find the real root of the equation $x^3 - 5x + 3 = 0$ correct to three decimal places using Newton–Raphson method.
Find the real root of the equation $\cos x - x^2 - x = 0$ correct to five decimal places using Newton–Raphson method.

Iteration	Value
\(x_{1}\)	\(\displaystyle \frac{0.43429(2)+1.2}{\log_{10}2+0.43429}=2.81\)
\(x_{2}\)	\(\displaystyle \frac{0.43429(2.81)+1.2}{\log_{10}2.81+0.43429}=2.741\)
\(x_{3}\)	\(\displaystyle \frac{0.43429(2.741)+1.2}{\log_{10}2.741+0.43429}=2.74064\)
\(x_{4}\)	\(\displaystyle \frac{0.43429(2.74064)+1.2}{\log_{10}2.74064+0.43429}=2.74065\)

Iteration	Value ( \(x_{n}\) )
\(x_{1}\)	\(\displaystyle \frac{2(2.4)^{3}+5}{3(2.4)^{2}-3}=2.28627\)
\(x_{2}\)	\(\displaystyle \frac{2(2.28627)^{3}+5}{3(2.28627)^{2}-3}=2.279047\)
\(x_{3}\)	\(\displaystyle \frac{2(2.279047)^{3}+5}{3(2.279047)^{2}-3}=2.2790188\)

Iteration	Value ( \(x_n\) )
\(x_1\)	\(\displaystyle \frac{0.6 \cdot \sin(0.6) + \cos(0.6) + 1}{3 + \sin(0.6)} = 0.6071\)
\(x_2\)	\(\displaystyle \frac{0.6071 \cdot \sin(0.6071) + \cos(0.6071) + 1}{3 + \sin(0.6071)} = 0.6071\)

Iteration	Value ( \(x_{n}\) )
\(x_{1}\)	\(\displaystyle \frac{3(1.8)^{4}+10}{4(1.8)^{3}-1}=1.85833\)
\(x_{2}\)	\(\displaystyle \frac{3(1.85833)^{4}+10}{4(1.85833)^{3}-1}=1.85559\)
\(x_{3}\)	\(\displaystyle \frac{3(1.85559)^{4}+10}{4(1.85559)^{3}-1}=1.85558\)

Ganitagya Gyanam

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