Test: Numerical solutions of linear & non-linear algebraic equations

Concept: 
Let x₀ be an approximation root of f(x) = 0 and x₁ = x₀^th be the correct root so that f(x₁) = 0 expanding f(x₀^th) by Taylor's series we obtain

Neglecting the second and higher order derivatives we have f(x₀) + hf'(x₀) = 0

A better approximation than x0 is therefore given by x₁, where x₁ = x₀ + h =  successive approximations are given by x₂, x₃ ....x_n+1 where 
For p^th root of a given number N, is not of equation f(x) = x^p - N = 0
Iteration equation:

for cube root, put p = 3

Therefore, option (2) is correct one.

Given
Now,
f(x) = x² – N = 0
Differentiating,
f’(x) = 2x
Now,
Using Newton-Raphson Method,

x_i+1 = (x_i + N/x_i)/2

Concept:
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
• It is sensitive to the starting value. Convergence fails if the starting point is not near the root.
• It is not preferred when the graph of f(x) is nearly horizontal where it crosses the x-axis as the values of f’(x) have negative values in this case.

Concept:
Newton’s divided difference polynomial method:
Second order polynomial interpolation using Newton’s divided difference polynomial method is as follows,
Given (x₀,y₀), (x₁,y₁), (x₂,y₂) be the data points and f(x) be the quadratic interpolant, then f(x) is given by
f(x) = b₀ + b₁(x – x₀) + b₂ (x – x₀)(x – x₁);
Where
b₀ = f(x₀);

Calculation:
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;
Easy method:
To save time, simply substitute the data points in the polynomials given in options and find the polynomial that is satisfying all data points.

Concept:
Newton-Raphson method: It has order of convergence 2 and number of guesses required is 1.
Iteration formula, 
Calculation:

Concept:
Newton-Raphson Method:
The iteration formula is given by

Where x₀ is the initial value/root of the equation f(x) = 0
Given,
f(x) = x³ + x² + 3x + 4 = 0
f'(x) = 3x² + 2x + 3
∴ f(-1) = 1 > 0 and f(-2) = -6 < 0
∴ f(-1).f(-2) < 0
⇒ ∃ a root lies in [-1, -2]
Let, x₀ = -1
By Newton Raphson method
First approximation


x₁ = -1.25

Concept:
The iterative formula for Newton Raphson method is given as, 

[NOTE: Take the trignometric terms in Radian while using scitific calculator for this type of numericals]
Calculation:
Given:
f(x) = 2 sin x - x
∴ f'(x) = 2 cos x - 1
Initial guess is x₀ = 2.0
The first iteration by Newton Raphson method is given by,

⇒ x₁ = 1.901

Order of convergence of the Newton Raphson method is two
Important Point
Order of convergence of various numerical methods

Concept:
Bisection method:
Used to find the root for a function. Root of a function f(x) = a such that f(a)= 0
Property: if a function f(x) is continuous on the interval [a…b] and sign of f(a) ≠  sign of f(b). There is a value c belongs to [a…b] such that f(c) = 0, means c is a root in between [a….b]
Note:
Bisection method cut the interval into 2 halves and check which half contains a root of the equation.
1) Suppose interval [a…b] .
2) Cut interval in the middle to find m : m = (a+b)/2
3) sign of f(m) not matches with f(a) proceed the search in the new interval.
Calculation:
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.

Concept:
Newton-Raphson Method:
The iteration formula is given by

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

∴ x₁ = 2.6871