Numerical solutions of linear and nonlinear algebraic equations | Engineering Mathematics for Mechanical Engineering PDF Download

Solved Numericals

Q1. If a and (a + h) are two consecutive approximate roots of the equation f(x) = 0 obtained by Newton's method, then h is equal to.
Solution: 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,

∴ The value of h is .

Q2. While solving the equation x² – 3x + 1 = 0 using Newton – Raphson method the initial guess of the root is as 1, then the value of the root will be?
Solution: Given, f(x) = x² – 3x + 1 and x₀ = 1

∴ The value of root will be x = 0.

Q3. The iterative formula to find the root of the equation f(x) = x³ - 5x + 7 = 0 by the Newton Raphson method is ______. 
Solution: Given,

Q4. If f(0) = 3, f(1) = 5, f(3) = 21, then the unique polynomial of degree 2 or less using Newton divided difference interpolation will be:
Solution: Given,
f(0) = 3, f(1) = 5, f(3) = 21;

⇒ (0,3), (1,5), (3,21) are the data points;

The polynomial will be f(x) = b₀ + b₁(x) + b₂ (x)(x – 1);

⇒ b₀ = f(0) = 3;

Substituting the constant b₀, b₁, b₂ in the quadratic interpolant,
⇒ f(x) = 3 + 2x + 2 (x)(x – 1) = 3 + 2x + 2x² – 2x = 3 + 2x²;
The unique polynomial of degree 2 will be f(x) = 3 + 2x².

Q5. The 2^nd approximation to a root of the equation x² - x - 1 = 0 in the interval (1, 2) by Bisection method will be: 
Solution: Given:
f(x) = x2 - x - 1 = 0  , a = 1 , b = 2
f(1) = 1 - 1 -1 = -1 < 0 , f(2) = 22 - 2 - 1 = 1 > 0 , Hence root lies between 1 and 2

By Bi-section method,

Which is positive. Hence, the root lies between 1.5 and 2

By Bi-section method,

The 2^nd approximation to a root of the equation x2 - x - 1 = 0 in the interval (1, 2) by Bisection method is 1.75

Q6. The approximate value of a root of x³ – 13 = 0, then 3.5 as initial value, after one iteration using Newton-Raphson method, is
Solution: 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

∴ x₁ = 2.6871

Q7. The iterative formula to find the root of the equation f(x) = x³ - 5x + 7 = 0 by the Newton Raphson method is ______.
Solution: Given,

=

Q8. Find the positive real root of x³ - x - 3 = 0 using Newton-Raphson method. If the starting guess (x₀) is 2, the numerical value of the root after two iterations (x₂) is __________ (round off to two decimal places).
Ans: 1.66 - 1.68
Solution:
f(x) = x³ - x - 3

Starting guess (x₀ = 2)
Now first iterations

Second iterations

x₁ = 1.7273

x₂ = 1.67 

Q9. The function f(x) = e^x – 1 is to be solved using Newton-Raphson method. If the initial value of x₀ is taken as 1.0, then the absolute error Ans: 0.05 - 0.07
Solution:
f(x) = e^x – 1
f’(x) = e^x
x₀ = 1
Iteration 1:
At x₀ = 1, f(x) = 1.718
f'(x) = 2.718

Iteration 2:
At x₂ = 0.3678, f(x) = 0.4445
f'(x) = 1.4445

By inspection, the actual solution for the given equation is, x = 0
Therefore, the error = 0.06005.

Q10. The iteration formula to find the reciprocal of a given number N by Newton’s method is
Solution: Given, X = 1/N
Let, 

Q11. If the equation sin (x) = x² is solved by Newton Raphson’s method with the initial guess of x = 1, then the value of x after 2 iterations would be 
Ans: 0.86 - 0.89
Solution: Given:
Initial guess (x0) = 1; sin(x) = x2 ⇒ f(x) = x2 - sin(x);
f'(x) = 2x - cos(x)
First iteration (n = 0)

x₁ = 0.891
Second Iteration (n = 1)

x² = 0.87