Newton-Raphson Method MCQ Quiz - Objective Question with Answer for Newton-Raphson Method - Download Free PDF

The correct answer is option 1: 1.909

Key Points

We are given the equation:

$f(x)=x^3-x-5$

Its derivative is:

$f^{\prime }(x)=3x^2-1$

Initial approximation: $x_0=2$

Newton-Raphson iteration formula is:

$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$

Step-by-step Calculation:

• $f(2)=2^3-2-5=8-2-5=1$
• $f^{\prime }(2)=3(2{)}^2-1=12-1=11$
• $x_1=2-\frac{1}{11}\approx 2-0.0909=1.909$

But this gives 1.909, which matches option 1.

However, options provided show 1.904 as correct (option 2), but mathematically, the correct value after first iteration is:

$\boxed{{\displaystyle x_1=1.909}}$

Additional Information

• The Newton-Raphson method converges quadratically near the root.
• This method is highly efficient for equations with simple real roots.
• Only one iteration gives an approximate root: 1.909

Hence, the correct answer is: option 1: 1.909

The correct answer is 1.909.

Key Points

• The Newton-Raphson method is an iterative numerical method used to find the roots of a real-valued function.
• Given a function f(x)" id="MathJax-Element-159-Frame" role="presentation" style="position: relative;" tabindex="0">f(x)$f(x)$ $f(x)$ and its derivative f′(x)" id="MathJax-Element-160-Frame" role="presentation" style="position: relative;" tabindex="0">f′(x)$f^{\prime }(x)$ $f'(x)$ , the Newton-Raphson iteration formula is:
  xn+1=xn−f(xn)f′(xn)" id="MathJax-Element-161-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">xn+1=xn−f(xn)f′(xn)$$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$$
  $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
• For the given equation f(x)=x3−x−5" id="MathJax-Element-162-Frame" role="presentation" style="position: relative;" tabindex="0">f(x)=x3−x−5$f(x)=x^3-x-5$ $f(x) = x^3 - x - 5$ , we first need to compute its derivative:
  f′(x)=3x2−1" id="MathJax-Element-163-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">f′(x)=3x2−1$$f^{\prime }(x)=3x^2-1$$
  $$f'(x) = 3x^2 - 1$$
• Starting with the initial approximation x0=2" id="MathJax-Element-164-Frame" role="presentation" style="position: relative;" tabindex="0">x0=2$x_0=2$ $x_0 = 2$ , we apply the Newton-Raphson iteration formula:
  x1=x0−f(x0)f′(x0)" id="MathJax-Element-165-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">x1=x0−f(x0)f′(x0)$$x_1=x_0-\frac{f(x_0)}{f^{\prime }(x_0)}$$
  $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$$
• Substituting x0=2" id="MathJax-Element-166-Frame" role="presentation" style="position: relative;" tabindex="0">x0=2$x_0=2$ $x_0 = 2$ :
  f(2)=23−2−5=8−2−5=1" id="MathJax-Element-167-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">f(2)=23−2−5=8−2−5=1$$f(2)=2^3-2-5=8-2-5=1$$
  $$f(2) = 2^3 - 2 - 5 = 8 - 2 - 5 = 1$$
  f′(2)=3(22)−1=3(4)−1=12−1=11" id="MathJax-Element-168-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">f′(2)=3(22)−1=3(4)−1=12−1=11$$f^{\prime }(2)=3(2^2)-1=3(4)-1=12-1=11$$
  $$f'(2) = 3(2^2) - 1 = 3(4) - 1 = 12 - 1 = 11$$
  x1=2−111=2−0.0909=1.909" id="MathJax-Element-169-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">x1=2−111=2−0.0909=1.909$$x_1=2-\frac{1}{11}=2-0.0909=1.909$$
  $$x_1 = 2 - \frac{1}{11} = 2 - 0.0909 = 1.909$$

Additional Information

• The Newton-Raphson method is highly efficient and converges quickly if the initial approximation is close to the actual root.
• It is important to ensure that the derivative f′(x)" id="MathJax-Element-170-Frame" role="presentation" style="position: relative;" tabindex="0">f′(x)$f^{\prime }(x)$ $f'(x)$ is not zero at the initial approximation to avoid division by zero.
• The method may fail to converge or converge to a different root if the initial approximation is not chosen carefully.

Explanation:

f(x) = x4 + 2x3 - 11x2 - 12x + 36 for x ∈ ℝ and we are finding the root α = 2using

the Newton-Raphson method $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

The Newton-Raphson method typically has quadratic convergence, i.e., the error in the approximation

$\mid x_{n+1}-\alpha |$ satisfies: $\mid x_{n+1}-\alpha |$ = $c|x_n-\alpha {|}^2$

for some constant c > 0. provided $f^{\prime }(\alpha )\ne 0$

$f^{\prime }(x)=4x^3+6x^2-22x-12$

we have f(2) = 0 and f'(2) = 0

$f^{″}(x)=12x^2+12x$ -22

Thus f''(2) This indicates that the Newton-Raphson method will converge with linear order for this root.

The order of convergence of the Newton-Raphson method for finding the root $$
\alpha = 2 is 1.

$$$$
\textbf{1 (Linear Convergence)}.

Concept:

Newton’s method or Newton-Raphson method:

It helps finds an approximate root of f(x) = 0 from a initial guess x_n by approximating f(x) as its tangent line.

Successively, we get x_n+1 = x_n - $\frac{f(x_n)}{f^{\prime }(x_n)}$

Calculation:

Given, a and (a + h) are two consecutive approximate roots of the equation f(x) = 0 

Let x_n = a and x_n+1 = a + h 

∴ Using Newton's method, xn+1 = xn - $\frac{f(x_n)}{f^{\prime }(x_n)}$

⇒ a + h = a - $\frac{f(a)}{f^{\prime }(a)}$

⇒ h = $-\frac{f(a)}{f^{\prime }(a)}$

∴ The value of h is $-\frac{f(a)}{f^{\prime }(a)}$.

The correct answer is Option 4.

Concept:

Newton’s method or Newton-Raphson method:

It helps finds an approximate root of f(x) = 0 from a initial guess xn by approximating f(x) as its tangent line.

Successively, we get xn+1 = xn - $\frac{f(x_n)}{f^{\prime }(x_n)}$

Calculation:

Given, f(x) = x² – 3x + 1 and x₀ = 1

∴ x₁ = x₀ - $\frac{f(x_0)}{f^{\prime }(x_0)}$

= 1 - $\frac{(1{)}^2-3(1)+1}{2(1)-3}$

= 1 - 1 

= 0

∴ The value of root will be x = 0.

The correct answer is Option 4.

Order of convergence of the Newton Raphson method is two

Important Point

Order of convergence of various numerical methods

Concept:

Newton-Raphson Method:

The iteration formula is given by

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

Where x0 is the initial value/root of the equation f(x) = 0

Calculation:

Given:

f(x) = x³ - 13, x₀ = 3.5

f^'(x) = 3x²

f(x₀) = f(3.5) = 3.5³ - 13 = 29.875

f^'(x₀) = f^'(3.5) = 3 × 3.5² = 36.75

We know that

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

$x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}$

_{$x_1=3.5-\frac{29.875}{36.75}$}

∴ x₁ = 2.6871

Concept:

According to Newton-Raphson Method

$X_{n+1}=X_n-\frac{f\left(X_n\right)}{f^{\prime }\left(X_n\right)}$

Calculation:

f(x) = e^x – 1

f’(x) = e^x

x₀ = 1

Iteration 1:

At x₀ = 1, f(x) = 1.718

f'(x) = 2.718

$x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}$

$=1-\frac{1.718}{2.718}=0.3678$

Iteration 2:

At x₂ = 0.3678, f(x) = 0.4445

f'(x) = 1.4445

$x_2=x_1-\frac{f\left(x_1\right)}{f^{\prime }\left(x_1\right)}$

$=0.3678-\frac{0.4447}{1.4447}=0.06005$

By inspection, the actual solution for the given equation is, x = 0

Therefore, the error = 0.06005

Concept:

Newton Raphson method:

The iteration formula is  $x_{n+1}=x_n-\frac{f(x)}{f^{\prime }(x)}$

Calculation:

Given

$f(x)=x^3-5x+7$

$x_{n+1}=x_n-\frac{f(x)}{f^{\prime }(x)}$

$f^{\prime }(x)=3x^2-5$

$x_{k+1}=x_k-(\frac{x_k^3-5x+7}{3x_k^2-5})$ ⇒$x_{k+1}=\frac{3x_k^3-5x_k-x_k^3+5x_k-7}{3x_k^2-5}$ ⇒ $x_{k+1}=\frac{2x_k^3-7}{3x_k^2-5}$

Concept:

According to the Newton-Raphson method:

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

where x_n is starting guess.

Calculation:

Given:

f(x) = x3 - x - 3 

$f^{\prime }\left(x\right)=3x^2-1$

Starting guess (x0 = 2)

Now first iterations

$x_1=x_0-\frac{(x_0^3-x_0-3)}{3x_0^2-1}$

$x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}=2-\frac{{\left(2\right)}^3-2-3}{3{\left(2\right)}^2-1}=1.7273$

Second iterations

$x_2=x_1-\frac{(x_1^3-x_1-3)}{3x_1^2-1}$

x₁ = 1.7273

$x_2=x_1-\frac{f\left(x_1\right)}{f^{\prime }\left(x_1\right)}=1.7273-\frac{{\left(1.7273\right)}^3-1.7273-3}{3{\left(1.7273\right)}^2-1}=1.67369$

x₂ = 1.67 

Concept:

Newton-Raphson method: It has order of convergence 2 and number of guesses required is 1.

Iteration formula, ${\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}-\frac{\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}}\right)}{{\mathrm{f}}^{\prime }\left({\mathrm{x}}_{\mathrm{n}}\right)}$

Calculation:

Given, $\mathrm{x}=\frac{1}{\mathrm{N}}$

Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}-\mathrm{N}$

${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\frac{1}{{\mathrm{x}}^2}$

${\mathrm{x}}_{\mathrm{i}+1}={\mathrm{x}}_{\mathrm{i}}-\left(\frac{\frac{1}{{\mathrm{x}}_{\mathrm{i}}}-\mathrm{N}}{-\frac{1}{{\mathrm{x}}_{\mathrm{i}}^2}}\right)={\mathrm{x}}_{\mathrm{i}}+\left(\frac{1-\mathrm{N}{\mathrm{x}}_{\mathrm{i}}}{{\mathrm{x}}_{\mathrm{i}}}\right)\left({\mathrm{x}}_{\mathrm{i}}^2\right)$

$\therefore {\mathrm{x}}_{\mathrm{i}+1}={\mathrm{x}}_{\mathrm{i}}\left(2-\mathrm{N}{\mathrm{x}}_{\mathrm{i}}\right)$

Concept:

Newton-Raphson method: It has an order of convergence 2 and the number of guesses required is 1.

Iterative formula is

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

Calculation:

Given:

Initial guess (x0) = 1; sin(x) = x2 ⇒ f(x) = x2 - sin(x);

f'(x) = 2x - cos(x)

First iteration (n = 0)

$x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}$

f(x) = x2 - sin(x) ⇒ f(x0) = x02 - sin(x0) = 1 - sin (1 × $\frac{180}{\pi }$) = 0.159

f'(x) = 2x - cos(x) ⇒ f'(x0) = 2x0 - cos(x0) = 2 - cos (1 × $\frac{180}{\pi }$) = 1.459

$x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}$ = $=1-\frac{0.159}{1.459}$

x1 = 0.891

Second Iteration (n = 1)

$x_2=x_1-\frac{f\left(x_1\right)}{f^{\prime }\left(x_1\right)}$

f(x) = x2 - sin(x) ⇒ f(x1) = x12 - sin(x1) = 0.891² - sin (0.891 × $\frac{180}{\pi }$) = 0.0161

f'(x) = 2x - cos(x) ⇒ f'(x₁) = 2x1 - cos(x1) = 1.782 - cos (0.891 × $\frac{180}{\pi }$) = 1.153

​$x_2=x_1-\frac{f\left(x_1\right)}{f^{\prime }\left(x_1\right)}$ $=0.891-\frac{0.0161}{1.153}$​ 

x₂ = 0.87

Mistake Points

All angles are in radians, convert them in degrees while calculating. 

Newton- Raphson method:

• The Newton - Raphson method is the type of open method (Extrapolation method).
• It is a powerful technique for solving algebraic and transcendental equations f( x ) = 0, numerically.
• It is an iteration method for solving a set of various nonlinear equations with an equal number of unknowns.

Advantages:

• It possesses quadratic convergence characteristics. Therefore, the convergence is very fast.
• The number of iterations is independent of the size of the system.
• The Newton-Raphson Method convergence is not sensitive to the choice of slack bus.
• Overall, there is a saving in computation time since a fewer number of iterations are required.

Disadvantages:

• It does not converge to a root when the second differential coefficient changes sign

Concept:

For Newton-Raphson’s

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

Analysis:

Given,

$x_n=\sqrt[3]{C}$

$\Rightarrow x_n^3-C=0$

Let,

$f\left(x_n\right)=x_n^3-C$

taking derivative:

$f\left(x_n^{\prime }\right)=3x_n^2$

∴ $x_{n+1}=x_n-\frac{x_n^3-C}{3x_n^2}$

= $\frac{3x_n^3-x_n^3+C}{3x_n^2}$

$x_{n+1}=\frac{2x_n^3+C}{3x_n^2}$

Concept:

According to Newton-Raphson Method

$X_{n+1}=X_n-\frac{f\left(X_n\right)}{f^{\prime }\left(X_n\right)}$

Calculation:

Given:

f(x) = x³ – x² + 4x – 4

f’(x) = 3x² – 2x + 4

Initial value x₀ = 2

f(x₀) = 8, f’(x₀) = 12

$x_1=2-\frac{8}{12}=\frac{4}{3}$