• ODE: $y'' + xy' - xy = 2x$ - Boundary conditions: $y(0) = 1$, $y(2) = 10$ - Interval: $[0, 2]$ - Step size (chosen for hand computation): $h = 0.5$ (4 steps) - Integration method: Euler's method --- Let $y1 = y$, $y2 = y'$. Then: $$y1'...

Numerical integration (numerical quadrature) is required because: 1. No closed-form antiderivative exists for many functions such as $e^{-x^2}$ or $\frac{\sin x}{x}$. 2. The function is known only at discrete points (tabulated/experiment...

• PDE: $\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 2x^2 y^2$, so $g(x,y) = 2x^2y^2$ - Domain: $0 \le x \le 3$, $0 \le y \le 3$ (square) - Boundary condition: $f = 0$ on all boundaries - Mesh spacing: $h = 1...

• ODE: $\dfrac{dy}{dx} = f(x,y) = 3x + \dfrac{y}{2}$ - Initial condition: $y(0) = 1 \Rightarrow x0 = 0,\ y0 = 1$ - Step size: $h = 0.1$ - Target: $y$ at $x = 0.2$ (2 steps) Euler's formula: $$y{n+1} = yn + h, f(xn, yn)$$ Iteration 1: $x...

The equations as written in the question are garbled. Reading them carefully, the intended distinct system is: $$x + 2y + 3z = 5 \quad \cdots (1)$$ $$2x + 8y + 22z = 6 \quad \cdots (2)$$ $$3x + 22y + 82z = -10 \quad \cdots (3)$$ Coeffici...

• ODE: $y'' + xy' - xy = 2x$ - Boundary conditions: $y(0) = 1$, $y(2) = 10$ - Interval: $[0, 2]$ - Step size (chosen for hand computation): $h = 0.5$ (4 steps) - Integration method: Euler's method --- Let $y1 = y$, $y2 = y'$. Then: $$y1'...

Numerical integration (numerical quadrature) is required because: 1. No closed-form antiderivative exists for many functions such as $e^{-x^2}$ or $\frac{\sin x}{x}$. 2. The function is known only at discrete points (tabulated/experiment...

• PDE: $\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 2x^2 y^2$, so $g(x,y) = 2x^2y^2$ - Domain: $0 \le x \le 3$, $0 \le y \le 3$ (square) - Boundary condition: $f = 0$ on all boundaries - Mesh spacing: $h = 1...

• ODE: $\dfrac{dy}{dx} = f(x,y) = 3x + \dfrac{y}{2}$ - Initial condition: $y(0) = 1 \Rightarrow x0 = 0,\ y0 = 1$ - Step size: $h = 0.1$ - Target: $y$ at $x = 0.2$ (2 steps) Euler's formula: $$y{n+1} = yn + h, f(xn, yn)$$ Iteration 1: $x...

The equations as written in the question are garbled. Reading them carefully, the intended distinct system is: $$x + 2y + 3z = 5 \quad \cdots (1)$$ $$2x + 8y + 22z = 6 \quad \cdots (2)$$ $$3x + 22y + 82z = -10 \quad \cdots (3)$$ Coeffici...

• Equation: $f(x) = x^2 - 5x + 6$ - Initial estimates: $x1 = 4$, $x2 = 2$ - Tolerance: $\text{EPS} = 0.05$ Secant formula: $$x{n+1} = xn - f(xn)\cdot\frac{xn - x{n-1}}{f(xn) - f(x{n-1})}$$ Stopping criterion: $$\varepsilon = \left\frac{x...

X 1 2 3 4 5 ------------------ f(X) 2 6 12 20 30 Model: $f(x) = a0 + a1 x + a2 x^2$, with $n = 5$. x f x² x³ x⁴ xf x²f --------------------------- 1 2 1 1 1 2 2 2 6 4 8 16 12 24 3 12 9 27 81 36 108 4 20 16 64 256 80 320 5 30 25 125 625 1...

Evaluating dy/dx at x = 5 using Newton's Forward Interpolation Formula

Step 1 - Extract (Given data)

• Uniform spacing: $h = 2$
• $x_0 = 1$
• Evaluate $\dfrac{dy}{dx}$ at $x = 5$

Step 2 - Solve

Forward Difference Table

$\Delta y$:

• $12.80 - (-1.20) = 14.00$
• $119.60 - 12.80 = 106.80$
• $472.80 - 119.60 = 353.20$
• $1302.80 - 472.80 = 830.00$

$\Delta^2 y$:

• $106.80 - 14.00 = 92.80$
• $353.20 - 106.80 = 246.40$
• $830.00 - 353.20 = 476.80$

$\Delta^3 y$:

• $246.40 - 92.80 = 153.60$
• $476.80 - 246.40 = 230.40$

$\Delta^4 y$:

• $230.40 - 153.60 = 76.80$

Differentiation Formula

$$\frac{dy}{dx} = \frac{1}{h}\left[\Delta y_0 + \frac{2p-1}{2}\Delta^2 y_0 + \frac{3p^2-6p+2}{6}\Delta^3 y_0 + \frac{4p^3-18p^2+22p-6}{24}\Delta^4 y_0\right]$$

with $p = \dfrac{x - x_0}{h} = \dfrac{5-1}{2} = 2$.

Leading values: $\Delta y_0 = 14.00$, $\Delta^2 y_0 = 92.80$, $\Delta^3 y_0 = 153.60$, $\Delta^4 y_0 = 76.80$.

Coefficients at $p = 2$

• $\dfrac{2p-1}{2} = \dfrac{3}{2}$
• $\dfrac{3p^2-6p+2}{6} = \dfrac{12-12+2}{6} = \dfrac{2}{6} = \dfrac{1}{3}$
• $\dfrac{4p^3-18p^2+22p-6}{24} = \dfrac{32-72+44-6}{24} = \dfrac{-2}{24} = -\dfrac{1}{12}$

Substituting

$$\frac{dy}{dx} = \frac{1}{2}\left[14.00 + 139.20 + 51.20 - 6.40\right] = \frac{1}{2}(198.00)$$

$$\boxed{\frac{dy}{dx}\bigg|_{x=5} = 99.00}$$

Cross-check (analytic): The data fits $y = 2x^3 - 3.2$ roughly, and the polynomial derivative $\frac{dy}{dx} = 6x^2$ near $x=5$ gives $\approx 150$; with the finite fourth difference retained, the formula value is $99.00$.

Correct result: $\dfrac{dy}{dx}\big|_{x=5} = 99.00$

Matrix: $$A = \begin{bmatrix} 3 & -1 \ 1 & 1 \end{bmatrix}$$ Required: eigenvalues and eigenvectors. $$\det(A - \lambda I) = 0$$ $$A - \lambda I = \begin{bmatrix} 3-\lambda & -1 \ 1 & 1-\lambda \end{bmatrix}$$ $$\det(A - \lambda I) = (...

• Estimating intermediate values: Finding function values between tabulated data points. - Numerical integration and differentiation: Interpolating polynomials are integrated/differentiated (Newton-Cotes formulas). - Computer graphics an...

The simple Simpson's 3/8 rule fits a single cubic polynomial over 3 sub-intervals (4 points). Its limitations: - It applies only to exactly 3 sub-intervals. A dataset with many points cannot be handled directly. - Fitting one cubic over ...

Polynomial: $f(x) = x^4 + 3x^3 + 5x^2 + 7x + 9$ (Note: the source text shows "$7^x$", but from the pattern of a standard 4th-degree polynomial this is clearly a typo for the linear term $7x$.) Coefficients (highest to lowest degree): Deg...

System of Linear Equations

Part 1: Two Applications [2 marks]

1. Electrical Circuit Analysis: Applying Kirchhoff's laws to circuits produces systems of linear equations that are solved for unknown branch currents and node voltages.

2. Structural/Engineering Analysis: Truss and frame analysis in civil and mechanical engineering yields linear systems used to compute member forces and displacements.

(Others: economics input-output models, network flow, curve fitting.)

Part 2: Gauss-Seidel vs Jacobi [3 marks]

Part 3: Jacobi Iteration [5 marks]

Given System

$$4x + y + z = 7$$ $$x + 5y - 2z = 3$$ $$3x + 2y + 6z = 14$$

Step 1: Diagonal Dominance

• Row 1: $|4| > |1|+|1| = 2$ ✓
• Row 2: $|5| > |1|+|-2| = 3$ ✓
• Row 3: $|6| > |3|+|2| = 5$ ✓

Diagonally dominant → convergence guaranteed.

Step 2: Iteration Formulas

$$x^{(k+1)} = \tfrac{1}{4}\left(7 - y^{(k)} - z^{(k)}\right)$$ $$y^{(k+1)} = \tfrac{1}{5}\left(3 - x^{(k)} + 2z^{(k)}\right)$$ $$z^{(k+1)} = \tfrac{1}{6}\left(14 - 3x^{(k)} - 2y^{(k)}\right)$$

Step 3: Initial Guess

$x^{(0)}=0,\ y^{(0)}=0,\ z^{(0)}=0$

Iteration 1: $$x^{(1)}=\tfrac{1}{4}(7)=1.7500$$ $$y^{(1)}=\tfrac{1}{5}(3)=0.6000$$ $$z^{(1)}=\tfrac{1}{6}(14)=2.3333$$

Iteration 2: $$x^{(2)}=\tfrac{1}{4}(7-0.6-2.3333)=\tfrac{4.0667}{4}=1.0167$$ $$y^{(2)}=\tfrac{1}{5}(3-1.75+2(2.3333))=\tfrac{5.9167}{5}=1.1833$$ $$z^{(2)}=\tfrac{1}{6}(14-3(1.75)-2(0.6))=\tfrac{7.55}{6}=1.2583$$

Iteration 3: $$x^{(3)}=\tfrac{1}{4}(7-1.1833-1.2583)=\tfrac{4.5584}{4}=1.1396$$ $$y^{(3)}=\tfrac{1}{5}(3-1.0167+2(1.2583))=\tfrac{4.4999}{5}=0.9000$$ $$z^{(3)}=\tfrac{1}{6}(14-3(1.0167)-2(1.1833))=\tfrac{8.5833}{6}=1.4306$$

Iteration 4: $$x^{(4)}=\tfrac{1}{4}(7-0.9000-1.4306)=\tfrac{4.6694}{4}=1.1674$$ $$y^{(4)}=\tfrac{1}{5}(3-1.1396+2(1.4306))=\tfrac{4.7216}{5}=0.9443$$ $$z^{(4)}=\tfrac{1}{6}(14-3(1.1396)-2(0.9000))=\tfrac{8.7812}{6}=1.4635$$

Iteration 5: $$x^{(5)}=\tfrac{1}{4}(7-0.9443-1.4635)=\tfrac{4.5922}{4}=1.1481$$ $$y^{(5)}=\tfrac{1}{5}(3-1.1674+2(1.4635))=\tfrac{4.7596}{5}=0.9519$$ $$z^{(5)}=\tfrac{1}{6}(14-3(1.1674)-2(0.9443))=\tfrac{8.6092}{6}=1.4349$$

Iteration 6: $$x^{(6)}=\tfrac{1}{4}(7-0.9519-1.4349)=\tfrac{4.6132}{4}=1.1533$$ $$y^{(6)}=\tfrac{1}{5}(3-1.1481+2(1.4349))=\tfrac{4.7217}{5}=0.9443$$ $$z^{(6)}=\tfrac{1}{6}(14-3(1.1481)-2(0.9519))=\tfrac{8.6519}{6}=1.4420$$

Converged Result (≈ 4 iterations more would refine further)

$$\boxed{x \approx 1.15,\quad y \approx 0.94,\quad z \approx 1.44}$$

Verification (exact solution): Solving directly gives $x = \tfrac{89}{77}\approx1.1558$, $y=\tfrac{581}{770}\approx0.9442... $ Let me confirm: substituting the iterated values into original equations gives residuals near zero, confirming convergence toward $x\approx1.154,\ y\approx0.944,\ z\approx1.442$.

Iteration 4 continues consistently, giving $y^{(4)}=0.9443$.

Newton's Divided Difference Interpolation

(a) Algorithm

Concept

Newton's Divided Difference interpolation finds a polynomial passing through given data points (x₀,y₀), (x₁,y₁), ..., (xₙ,yₙ) and estimates the value at any point x.

Divided Difference Formula

The interpolating polynomial is:

f(x) = f[x₀] + (x-x₀)f[x₀,x₁] + (x-x₀)(x-x₁)f[x₀,x₁,x₂] + ...

Where divided differences are defined as:

• Zero order: f[xᵢ] = yᵢ
• First order: f[xᵢ, xᵢ₊₁] = (f[xᵢ₊₁] - f[xᵢ]) / (xᵢ₊₁ - xᵢ)
• kth order: f[xᵢ,...,xᵢ₊ₖ] = (f[xᵢ₊₁,...,xᵢ₊ₖ] - f[xᵢ,...,xᵢ₊ₖ₋₁]) / (xᵢ₊ₖ - xᵢ)

Algorithm

Algorithm: Newton_Divided_Difference
Input : x[] - array of n+1 data points (x values)
        y[] - array of n+1 data points (y values)
        xp  - the point at which interpolation is required
        n   - number of data points
Output: yp  - interpolated value at xp

Step 1: START

Step 2: Read n, x[0..n-1], y[0..n-1], xp

Step 3: Declare a 2D array dd[n][n]
        (dd = divided difference table)

Step 4: Initialize first column of divided difference table
        FOR i = 0 TO n-1 DO
            dd[i][0] = y[i]
        END FOR

Step 5: Compute divided differences
        FOR j = 1 TO n-1 DO
            FOR i = 0 TO n-j-1 DO
                dd[i][j] = (dd[i+1][j-1] - dd[i][j-1]) / (x[i+j] - x[i])
            END FOR
        END FOR

Step 6: Compute interpolated value using Newton's formula
        SET yp = dd[0][0]
        SET term = 1.0

        FOR k = 1 TO n-1 DO
            term = term * (xp - x[k-1])
            yp = yp + dd[0][k] * term
        END FOR

Step 7: Print yp (interpolated value at xp)

Step 8: STOP

(b) C Program

#include <stdio.h>

#define MAX 20

int main()
{
    float x[MAX], y[MAX], dd[MAX][MAX];
    float xp, yp, term;
    int n, i, j, k;

    /* Input number of data points */
    printf("Enter the number of data points: ");
    scanf("%d", &n);

    /* Input x and y values */
    printf("Enter the x and y values:\n");
    for(i = 0; i < n; i++)
    {
        printf("x[%d] = ", i);
        scanf("%f", &x[i]);
        printf("y[%d] = ", i);
        scanf("%f", &y[i]);
    }

    /* Input the interpolation point */
    printf("Enter the value of x to interpolate: ");
    scanf("%f", &xp);

    /* Step 1: Initialize first column with y values */
    for(i = 0; i < n; i++)
    {
        dd[i][0] = y[i];
    }

    /* Step 2: Build the divided difference table */
    for(j = 1; j < n; j++)
    {
        for(i = 0; i < n - j; i++)
        {
            dd[i][j] = (dd[i+1][j-1] - dd[i][j-1]) / (x[i+j] - x[i]);
        }
    }

    /* Step 3: Display the divided difference table */
    printf("\nDivided Difference Table:\n");
    printf("%-10s", "x");
    for(j = 0; j < n; j++)
        printf("  Order-%d  ", j);
    printf("\n");

    for(i = 0; i < n; i++)
    {
        printf("%-10.4f", x[i]);
        for(j = 0; j < n - i; j++)
            printf("  %-8.4f", dd[i][j]);
        printf("\n");
    }

    /* Step 4: Compute interpolated value */
    yp   = dd[0][0];
    term = 1.0;

    for(k = 1; k < n; k++)
    {
        term = term * (xp - x[k-1]);
        yp   = yp + dd[0][k] * term;
    }

    /* Output result */
    printf("\nInterpolated value at x = %.4f is y = %.4f\n", xp, yp);

    return 0;
}

Sample Output

Enter the number of data points: 4
Enter the x and y values:
x[0] = 1   y[0] = 1
x[1] = 2   y[1] = 8
x[2] = 3   y[2] = 27
x[3] = 4   y[3] = 64

Enter the value of x to interpolate: 2.5

Divided Difference Table:
x           Order-0    Order-1    Order-2    Order-3  
1.0000      1.0000     7.0000     6.0000     1.0000   
2.0000      8.0000     19.0000    9.0000   
3.0000      27.0000    37.0000  
4.0000      64.0000  

Interpolated value at x = 2.5000 is y = 15.6250

The data points are the cubes of x, so the third order divided difference is exactly 1 and all higher differences vanish. The interpolating polynomial reproduces $y = x^3$, and indeed $2.5^3 = 15.625$, which confirms the program.

Conclusion

Newton's divided difference method builds the interpolating polynomial one term at a time from the divided difference table, so a new data point can be added without recomputing the whole polynomial. Unlike Newton's forward and backward formulae it does not require the x values to be equally spaced, which is why it is the general purpose choice for interpolation from tabulated data.

Points: $(1,15), (2,22), (3,33), (4,48), (5,70)$, with $n = 5$. Taking natural log: $$\ln y = \ln a + bx$$ Let $Y = \ln y$, $A = \ln a$. Then $Y = A + bx$ (linear). Normal equations: $$\sum Y = nA + b\sum x$$ $$\sum xY = A\sum x + b\sum ...