NUMERICAL METHODS

The following will be discussed in this Topic

1. Iteration
2. linear interpolation
3. Newton-Raphson method
4. Bisection method
5. Secant method
6. Regular falsi method
7. Numerical integration

1. Trapezium rule
2. Simpson rule

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1. ITERATION

Also known as successive approximation which is the process of finding the roots of the number.

Suppose we need to find the square root of 18 i.e. √ (18) but 18 lies between 4 and 5 it means 4² = 16, less 2 from18 and 5² = 25, 7 more than 18 that’s why it’s said to lie between 4 and 5.

Then x₁₌ first approximation and x₂ = second approximation then:

1. X₁= 4 and X₂ = 5
2. Take the average say X₃ = ^{(x1 + x2)} / 2 that is (4 + 5)/2 = 4.5   then X₃ = 4.5.
3. Take the average again say X₄ = ^{(x1 + x3)} /2 that is (4 + 4.5)/2 = 4.25 then X₄ = 4.25

Therefore by two iteration the square root of 18 is 4.25.

In general the square root of positive number N can be calculated from the iterative formula:

where X_r is the first approximation square root of the given positive number N. e.g. if given to find √(27) then X_r is 5.

Qn: find the square roots of the following using iterative formula.

a). 145,   b). 47   c).65   d).82

HOW TO FIND THE FORMULAR FOR FINDING ROOTS OF EQUATION

Example:

Establish the formula to solve the equation x³ – 5x – 3 = 0 for the roots which lie between x = 2 and x = 3.

Solution:

1. Make x the subject of the formula using x³ that will be x³ = 5x + 3, and by dividing by x² the formula will be x = ⁵/_x + ³/x_²……….. (I)
2. Also make x the subject of the formula using 5x that will 5x = x³ – 3 and dividing by 5 both sides the formula will become x = x³/₅ – ³/₅………. (II)
3. Then we test for converges and diverges in each equation above, if f ́(x) < 1 the formula will be converges and if f ́(x) > 1 the formula will be diverges, but we need converges.

How to test for converges and diverges.

1. Find the average of the two roots given above i.e. x = 2 and x = 3 then the average is 2.5
2. find the derivative of the equations above i.e. eqn. (I) will be f ́(x) = -5x^-2 – 6x^-3 then substitute the value of x by average above say x = 2.5 then by substitution f’(2.5) = 0.835, that is less than 1 then the formula is converges and there is no need for testing another because the converges equation is already obtained.

There fore the iterative formula is X_{n + 1} = ⁵/X_n + ³/X_2n from the first equation.

To find the roots we start by x₁ = 2.5 in order to get X₂ i.e. X₂ = ⁵/_2.5 + ³/ _(2.5)² then continue up to 3 or 4 iteration.

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