TU BCA Numerical Methods Master Guide: Bisection, Newton-Raphson, Gauss Elimination & Simpson’s 1/3 Rule (Solved Numericals)

Author: Bhuban Subedi | Subject: Numerical Methods (CACS251) | Semester: Fourth Semester

In scientific computing, game physics engines, machine learning optimization, and financial modeling, analytical solutions for high-degree polynomial or transcendental equations are often impossible. In the Tribhuvan University BCA fourth semester, Numerical Methods (CACS251) provides algorithmic techniques for finding approximate roots, solving large linear systems, and performing numerical integration with controlled error tolerances.

In the final 60-mark TU board examination, Numerical Methods is a 100% computational paper. Group B and Group C questions demand rigorous step-by-step iteration tables, calculation of relative error tolerances ($|x_{n+1} – x_n| < \epsilon$), and integration formulas.

In this guide, I will solve standard TU board numericals step-by-step.

1. Non-Linear Equation: Bisection Method

The Bisection Method uses the Intermediate Value Theorem. If a continuous function $f(x)$ satisfies $f(a) \cdot f(b) < 0$, at least one real root exists in the interval $(a, b)$.
$$c = \frac{a + b}{2}$$

Solved TU Board Problem 1: Bisection Method

Problem: Find a real root of the equation $f(x) = x^3 – x – 1 = 0$ correct to 3 decimal places using the Bisection Method.

Step 1: Bracket the Root:
– $f(1) = 1^3 – 1 – 1 = -1 < 0$
– $f(2) = 2^3 – 2 – 1 = 8 – 3 = 5 > 0$
– Since $f(1) \cdot f(2) < 0$, a root lies between $a = 1$ and $b = 2$.

Iteration Table:

+-----+----------+----------+----------+------------+--------------------+
| Iter| a (-)    | b (+)    | c = (a+b)/2| f(c)      | Interval Update    |
+-----+----------+----------+----------+------------+--------------------+
| 1   | 1.0000   | 2.0000   | 1.5000   | +0.8750    | b = 1.5000         |
| 2   | 1.0000   | 1.5000   | 1.2500   | -0.2969    | a = 1.2500         |
| 3   | 1.2500   | 1.5000   | 1.3750   | +0.2246    | b = 1.3750         |
| 4   | 1.2500   | 1.3750   | 1.3125   | -0.0515    | a = 1.3125         |
| 5   | 1.3125   | 1.3750   | 1.3438   | +0.0826    | b = 1.3438         |
| 6   | 1.3125   | 1.3438   | 1.3281   | +0.0145    | b = 1.3281         |
| 7   | 1.3125   | 1.3281   | 1.3203   | -0.0187    | a = 1.3203         |
| 8   | 1.3203   | 1.3281   | 1.3242   | -0.0021    | a = 1.3242         |
| 9   | 1.3242   | 1.3281   | 1.3262   | +0.0062    | b = 1.3262         |
| 10  | 1.3242   | 1.3262   | 1.3252   | +0.0020    | Root Found         |
+-----+----------+----------+----------+------------+--------------------+

Approximate Root: $\mathbf{x \approx 1.325}$

2. Fast Convergence: Newton-Raphson Method

The Newton-Raphson Method uses the tangent line slope to converge quadratically (order of convergence $p = 2$):
$$x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)}$$

Solved TU Board Problem 2: Newton-Raphson Method

Problem: Find a real root of $f(x) = x \sin x + \cos x = 0$ near $x_0 = \pi \approx 3.1416$.

Step 1: Compute Derivative $f'(x)$:
$$f'(x) = \frac{d}{dx}(x \sin x + \cos x) = (1 \cdot \sin x + x \cos x) – \sin x = x \cos x$$

Step 2: Apply Iteration Formula:
$$x_{n+1} = x_n – \frac{x_n \sin x_n + \cos x_n}{x_n \cos x_n}$$

Iterations (Calculations in Radians):
– $n = 0$: $x_0 = 3.1416$
– $f(x_0) = 3.1416 \sin(3.1416) + \cos(3.1416) \approx 0 – 1 = -1$
– $f'(x_0) = 3.1416 \cos(3.1416) \approx -3.1416$
– $x_1 = 3.1416 – \frac{-1}{-3.1416} = 3.1416 – 0.3183 = \mathbf{2.8233}$

• $n = 1$: $x_1 = 2.8233$
• $f(x_1) = 2.8233 \sin(2.8233) + \cos(2.8233) = 2.8233(0.3129) – 0.9498 = -0.0654$
• $f'(x_1) = 2.8233(-0.9498) = -2.6816$

• $x_2 = 2.8233 – \frac{-0.0654}{-2.6816} = 2.8233 – 0.0244 = \mathbf{2.7989}$

• $n = 2$: $x_2 = 2.7989 \implies x_3 = \mathbf{2.7984}$

Approximate Root: $\mathbf{x \approx 2.798}$

3. Numerical Integration: Simpson’s 1/3 Rule

For an even number of intervals $n$ with step size $h = \frac{b – a}{n}$:
$$\int_a^b f(x) dx \approx \frac{h}{3} \left[ (y_0 + y_n) + 4(y_1 + y_3 + \dots + y_{n-1}) + 2(y_2 + y_4 + \dots + y_{n-2}) \right]$$

+-------------------------------------------------------------------------------+
|                       SIMPSON'S 1/3 RULE MNEMONIC:                            |
|       Area = (h / 3) * [ (First + Last) + 4 * (Odds) + 2 * (Evens) ]          |
+-------------------------------------------------------------------------------+

Solved TU Board Problem 3: Simpson’s 1/3 Rule

Problem: Evaluate $\int_0^6 \frac{1}{1 + x^2} dx$ using Simpson’s $1/3$ Rule with $n = 6$ sub-intervals.

Step 1: Compute Step Size $h$:
$$h = \frac{b – a}{n} = \frac{6 – 0}{6} = 1$$

Step 2: Table of Values ($y = \frac{1}{1 + x^2}$):

+---+---+--------------------+
| i | x | y_i = 1 / (1 + x^2)|
+---+---+--------------------+
| 0 | 0 | y_0 = 1.0000       |
| 1 | 1 | y_1 = 0.5000       |
| 2 | 2 | y_2 = 0.2000       |
| 3 | 3 | y_3 = 0.1000       |
| 4 | 4 | y_4 = 0.0588       |
| 5 | 5 | y_5 = 0.0385       |
| 6 | 6 | y_6 = 0.0270       |
+---+---+--------------------+

Step 3: Calculate Integral:
$$\text{Sum} = (y_0 + y_6) + 4(y_1 + y_3 + y_5) + 2(y_2 + y_4)$$
$$\text{Sum} = (1.0000 + 0.0270) + 4(0.5000 + 0.1000 + 0.0385) + 2(0.2000 + 0.0588)$$
$$\text{Sum} = 1.0270 + 4(0.6385) + 2(0.2588) = 1.0270 + 2.5540 + 0.5176 = 4.0986$$
$$\int_0^6 \frac{1}{1 + x^2} dx \approx \frac{1}{3} \times 4.0986 = \mathbf{1.3662}$$

Frequently Asked Questions (FAQ)

Q1: Why does the Newton-Raphson method fail when $f'(x) = 0$?

Because the formula divides by $f'(x_n)$. When $f'(x_n) = 0$, the tangent is horizontal and does not intersect the x-axis, causing a division by zero error.

Q2: What is the restriction on interval count $n$ for Simpson’s 1/3 Rule?

The number of sub-intervals $n$ must always be an even number ($n = 2, 4, 6, 8, \dots$).