PDE: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -64xy, \quad 0 \le x \le 1,\ 0 \le y \le 1$$ So $f(x,y) = -64xy$. Boundary Conditions: - $u(0,y) = 0$ (left) - $u(x,0) = 0$ (bottom) - $u(1,y) = 150$ (right) ...

$$A = \begin{bmatrix} 2 & 1 & 1 \ 3 & 2 & 3 \ 1 & 4 & 9 \end{bmatrix}$$ Cholesky's method requires the matrix to be symmetric (and positive definite), factorising it as: $$A = L L^T$$ where $L$ is lower triangular. The given matrix is ...

Newton's Backward Difference Interpolation to Estimate f(0.6)

STEP 1 - EXTRACT: Given Data

• Step size $h = 0.1$ (equally spaced)
• Target: $f(0.6)$

Since $x = 0.6$ lies beyond the last point $x = 0.5$, use Newton's Backward Difference Formula with base $x_n = 0.5$.

STEP 2 - SOLVE

Backward Difference Table

Verification of differences:

• $\nabla f$: $0.36,\ 0.34,\ 0.31,\ 0.28$
• $\nabla^2 f$: $-0.02,\ -0.03,\ -0.03$
• $\nabla^3 f$: $-0.01,\ 0.00$
• $\nabla^4 f$: $0.01$

Compute s

$$s = \frac{x - x_n}{h} = \frac{0.6 - 0.5}{0.1} = 1$$

Apply the Formula

$$P_n(x) = f(x_n) + s,\nabla f + \frac{s(s+1)}{2!}\nabla^2 f + \frac{s(s+1)(s+2)}{3!}\nabla^3 f + \frac{s(s+1)(s+2)(s+3)}{4!}\nabla^4 f$$

With $s = 1$ (using the last row):

Term 1: $f(x_n) = 3.97$

Term 2: $1 \times 0.28 = 0.28$

Term 3: $\dfrac{1 \cdot 2}{2}\times(-0.03) = 1 \times (-0.03) = -0.03$

Term 4: $\dfrac{1 \cdot 2 \cdot 3}{6}\times 0.00 = 0.00$

Term 5: $\dfrac{1 \cdot 2 \cdot 3 \cdot 4}{24}\times 0.01 = 1 \times 0.01 = 0.01$

Sum

$$f(0.6) \approx 3.97 + 0.28 - 0.03 + 0.00 + 0.01$$

$$\boxed{f(0.6) \approx 4.23}$$

• Equation: $f(x) = \cos x + 2\sin x - x^2 = 0$ - Error precision (tolerance): $E = 0.01$ - Initial guesses: not specified in question. I choose $x0 = 1$, $x1 = 2$ (bracketing the root, angles in radians). Feature Newton-Raphson Secant -...

--- Approximate Error is the difference between the current approximation and the previous approximation of a root. It gives an estimate of how much the solution has changed between two successive iterations. $$Ea = x{new} - x{old}$$ Sin...

PDE: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -64xy, \quad 0 \le x \le 1,\ 0 \le y \le 1$$ So $f(x,y) = -64xy$. Boundary Conditions: - $u(0,y) = 0$ (left) - $u(x,0) = 0$ (bottom) - $u(1,y) = 150$ (right) ...

Newton's Backward Difference Interpolation to Estimate f(0.6)

STEP 1 - EXTRACT: Given Data

• Step size $h = 0.1$ (equally spaced)
• Target: $f(0.6)$

Since $x = 0.6$ lies beyond the last point $x = 0.5$, use Newton's Backward Difference Formula with base $x_n = 0.5$.

STEP 2 - SOLVE

Backward Difference Table

Verification of differences:

• $\nabla f$: $0.36,\ 0.34,\ 0.31,\ 0.28$
• $\nabla^2 f$: $-0.02,\ -0.03,\ -0.03$
• $\nabla^3 f$: $-0.01,\ 0.00$
• $\nabla^4 f$: $0.01$

Compute s

$$s = \frac{x - x_n}{h} = \frac{0.6 - 0.5}{0.1} = 1$$

Apply the Formula

$$P_n(x) = f(x_n) + s,\nabla f + \frac{s(s+1)}{2!}\nabla^2 f + \frac{s(s+1)(s+2)}{3!}\nabla^3 f + \frac{s(s+1)(s+2)(s+3)}{4!}\nabla^4 f$$

With $s = 1$ (using the last row):

Term 1: $f(x_n) = 3.97$

Term 2: $1 \times 0.28 = 0.28$

Term 3: $\dfrac{1 \cdot 2}{2}\times(-0.03) = 1 \times (-0.03) = -0.03$

Term 4: $\dfrac{1 \cdot 2 \cdot 3}{6}\times 0.00 = 0.00$

Term 5: $\dfrac{1 \cdot 2 \cdot 3 \cdot 4}{24}\times 0.01 = 1 \times 0.01 = 0.01$

Sum

$$f(0.6) \approx 3.97 + 0.28 - 0.03 + 0.00 + 0.01$$

$$\boxed{f(0.6) \approx 4.23}$$

$$A = \begin{bmatrix} 2 & 1 & 1 \ 3 & 2 & 3 \ 1 & 4 & 9 \end{bmatrix}$$ Cholesky's method requires the matrix to be symmetric (and positive definite), factorising it as: $$A = L L^T$$ where $L$ is lower triangular. The given matrix is ...

Regression Analysis and Second Order Polynomial Fitting

What is Regression Analysis?

Regression analysis is a statistical technique used to determine the best-fit mathematical relationship between a dependent variable and one or more independent variables. Unlike interpolation (which forces the curve through every data point), regression minimizes the total squared error between the observed data and the fitted curve (least squares principle). It is widely used for prediction and trend estimation.

Given Data

$$ \begin{array}{c|ccccc} x & 2 & 4 & 6 & 8 & 10\ \hline y & 1.4 & 2.0 & 2.4 & 2.6 & 2.8 \end{array} $$

$n = 5$. Fit $y = a_0 + a_1 x + a_2 x^2$.

Step 1: Summation Table

$\sum x = 30$, $\sum y = 11.2$, $\sum x^2 = 220$, $\sum x^3 = 1800$, $\sum x^4 = 15664$, $\sum xy = 74.0$, $\sum x^2y = 570.4$

(Check $\sum x^4$: $16+256+1296+4096+10000 = 15664$ ✓)

Step 2: Normal Equations

$$5a_0 + 30a_1 + 220a_2 = 11.2 \quad (1)$$ $$30a_0 + 220a_1 + 1800a_2 = 74.0 \quad (2)$$ $$220a_0 + 1800a_1 + 15664a_2 = 570.4 \quad (3)$$

Step 3: Solve

$(2) - 6\times(1)$: $$40a_1 + 480a_2 = 6.8 \quad (4)$$

$(3) - 44\times(1)$: $$480a_1 + 5984a_2 = 77.6 \quad (5)$$

$(5) - 12\times(4)$: $12\times(4)$ gives $480a_1 + 5760a_2 = 81.6$. $$(5984 - 5760)a_2 = 77.6 - 81.6$$ $$224a_2 = -4.0 \implies a_2 = -0.017857$$

From (4): $$40a_1 = 6.8 - 480(-0.017857) = 6.8 + 8.5714 = 15.3714$$ $$a_1 = 0.38429$$

From (1): $$5a_0 = 11.2 - 30(0.38429) - 220(-0.017857)$$ $$= 11.2 - 11.5286 + 3.9286 = 3.6000$$ $$a_0 = 0.7200$$

(Using the more precise $a_1=0.384286$ and $a_2=-0.0178571$ gives $a_0 = 0.72$; rounding earlier would give $0.7188$.)

Result

$$\boxed{y = 0.72 + 0.3843,x - 0.01786,x^2}$$

Rounding during the intermediate steps would give $a_0 \approx 0.7188$, $a_1 \approx 0.3848$ and $a_2 \approx -0.0179$; the more accurate value is $a_0 \approx 0.72$, and the fit is essentially the same.

• Equation: $f(x) = \cos x + 2\sin x - x^2 = 0$ - Error precision (tolerance): $E = 0.01$ - Initial guesses: not specified in question. I choose $x0 = 1$, $x1 = 2$ (bracketing the root, angles in radians). Feature Newton-Raphson Secant -...

--- Approximate Error is the difference between the current approximation and the previous approximation of a root. It gives an estimate of how much the solution has changed between two successive iterations. $$Ea = x{new} - x{old}$$ Sin...

Heun's method is a predictor-corrector method (also called the improved Euler's method) for solving ODEs of the form: $$\frac{dy}{dx} = f(x, y), \quad y(x0) = y0$$ The core formula is: $$y{n+1} = yn + \frac{h}{2}(m1 + m2)$$ Where: - $m1 ...

Inherent Errors and Newton-Raphson Method

Part 1: Inherent Errors (2 marks)

Inherent errors are errors that already exist in a problem before any numerical computation is carried out. They are caused by the data or by the mathematical model, not by the numerical technique used.

Main sources:

• Data errors: Values obtained from physical measurements or observation carry uncertainty.
• Modelling / simplifying assumptions: Approximating a real system by a simplified mathematical model introduces error.
• Errors in empirical constants: Constants determined experimentally are not exact.

Such errors cannot be removed by using a better numerical method; they can only be controlled so that they are not amplified.

Part 2: Derivation of Newton-Raphson Method (4 marks)

Let $f(x)=0$ be the equation whose root is required, and let $x_n$ be an approximate root.

Expand $f(x_n+h)$ by Taylor's series:

$$f(x_n+h)=f(x_n)+h,f'(x_n)+\frac{h^2}{2!}f''(x_n)+\cdots$$

Since $h$ is small, neglect second and higher order terms:

$$f(x_n+h)\approx f(x_n)+h,f'(x_n)$$

If $x_n+h$ is the exact root, then $f(x_n+h)=0$:

$$0=f(x_n)+h,f'(x_n)\quad\Rightarrow\quad h=-\frac{f(x_n)}{f'(x_n)}$$

Hence the improved approximation $x_{n+1}=x_n+h$ gives:

$$\boxed{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},\qquad f'(x_n)\neq 0}$$

Part 3: Solve $x^2-5x+6=0$ to 3 Decimal Places (4 marks)

$$f(x)=x^2-5x+6,\qquad f'(x)=2x-5$$

Iteration formula:

$$x_{n+1}=x_n-\frac{x_n^2-5x_n+6}{2x_n-5}$$

(Exact roots: $(x-2)(x-3)=0 \Rightarrow x=2,,3$.)

Root near $x=3$, take $x_0=4$

Iteration 1: $$f(4)=16-20+6=2,\quad f'(4)=3$$ $$x_1=4-\tfrac{2}{3}=3.3333$$

Iteration 2: $$f(3.3333)=11.1109-16.6665+6=0.4444,\quad f'(3.3333)=1.6666$$ $$x_2=3.3333-\tfrac{0.4444}{1.6666}=3.0667$$

Iteration 3: $$f(3.0667)=9.4046-15.3335+6=0.0711,\quad f'(3.0667)=1.1334$$ $$x_3=3.0667-\tfrac{0.0711}{1.1334}=3.0040$$

Iteration 4: $$f(3.0040)=9.0240-15.0200+6=0.0040,\quad f'(3.0040)=1.0080$$ $$x_4=3.0040-\tfrac{0.0040}{1.0080}=3.0000$$

$$\boxed{x=3.000}$$

Root near $x=2$, take $x_0=1$

Iteration 1: $$f(1)=1-5+6=2,\quad f'(1)=-3$$ $$x_1=1-\tfrac{2}{-3}=1.6667$$

Iteration 2: $$f(1.6667)=2.7779-8.3335+6=0.4444,\quad f'(1.6667)=-1.6666$$ $$x_2=1.6667-\tfrac{0.4444}{-1.6666}=1.9334$$

Iteration 3: $$f(1.9334)=3.7380-9.6670+6=0.0710,\quad f'(1.9334)=-1.1332$$ $$x_3=1.9334-\tfrac{0.0710}{-1.1332}=1.9961$$

Iteration 4: $$f(1.9961)=3.9844-9.9805+6=0.0039,\quad f'(1.9961)=-1.0078$$ $$x_4=1.9961-\tfrac{0.0039}{-1.0078}=2.0000$$

$$\boxed{x=2.000}$$

Final Result

The roots of $x^2-5x+6=0$ correct to 3 decimal places are:

$$x=2.000 \quad\text{and}\quad x=3.000$$

Gauss-Jordan Elimination Method

Given Data

System of equations: $$x + 2y - 3z = 4 \quad \cdots (1)$$ $$2x + 4y - 6z = 8 \quad \cdots (2)$$ $$x - 2y + 5z = 4 \quad \cdots (3)$$

Step 1: Augmented Matrix

$$[A|b] = \begin{bmatrix} 1 & 2 & -3 & | & 4 \ 2 & 4 & -6 & | & 8 \ 1 & -2 & 5 & | & 4 \end{bmatrix}$$

Step 2: Eliminate column 1 below pivot

$R_2 \to R_2 - 2R_1$: $[0,\ 0,\ 0,\ |\ 0]$

$R_3 \to R_3 - R_1$: $[0,\ -4,\ 8,\ |\ 0]$

$$\begin{bmatrix} 1 & 2 & -3 & | & 4 \ 0 & 0 & 0 & | & 0 \ 0 & -4 & 8 & | & 0 \end{bmatrix}$$

Step 3: Swap $R_2 \leftrightarrow R_3$

$$\begin{bmatrix} 1 & 2 & -3 & | & 4 \ 0 & -4 & 8 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix}$$

Step 4: Normalize Row 2 ($R_2 \to R_2 / (-4)$)

$$\begin{bmatrix} 1 & 2 & -3 & | & 4 \ 0 & 1 & -2 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix}$$

Step 5: Eliminate column 2 above pivot ($R_1 \to R_1 - 2R_2$)

$R_1: [1,\ 0,\ 1,\ |\ 4]$

$$\begin{bmatrix} 1 & 0 & 1 & | & 4 \ 0 & 1 & -2 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix}$$

Step 6: Interpretation

Row 3 gives $0 = 0$, which is consistent. Rank of coefficient matrix $= 2 <$ number of unknowns $= 3$, so there are infinitely many solutions.

Let $z = t$ (free parameter, $t \in \mathbb{R}$).

From Row 2: $y - 2t = 0 \implies y = 2t$

From Row 1: $x + t = 4 \implies x = 4 - t$

Verification with equation (3): $x - 2y + 5z = (4-t) - 2(2t) + 5t = 4 - t - 4t + 5t = 4$ ✓

Final Answer

$$x = 4 - t, \quad y = 2t, \quad z = t, \quad t \in \mathbb{R}$$

The system has infinitely many solutions since equation (2) $= 2 \times$ equation (1), reducing to two independent equations in three unknowns.

Cubic Spline Interpolation on [1, 3], Estimate f(2)

STEP 1 - Given Data

Number of intervals $n = 2$. We use natural cubic spline (standard assumption when no end conditions given).

STEP 2 - Solve

Interval widths

$$h_0 = x_1 - x_0 = 2, \qquad h_1 = x_2 - x_1 = 1$$

Moment (second-derivative) equations

Natural boundary conditions: $M_0 = 0,\ M_2 = 0$.

Interior equation at $i=1$:

$$h_0 M_0 + 2(h_0+h_1)M_1 + h_1 M_2 = 6\left[\frac{f_2-f_1}{h_1} - \frac{f_1-f_0}{h_0}\right]$$

RHS:

$$\frac{f_2-f_1}{h_1} = \frac{9.0-4.5}{1} = 4.5, \qquad \frac{f_1-f_0}{h_0} = \frac{4.5-1.5}{2} = 1.5$$

$$\text{RHS} = 6(4.5-1.5) = 18$$

Substituting $M_0 = M_2 = 0$:

$$2(2+1)M_1 = 18 \Rightarrow 6M_1 = 18 \Rightarrow M_1 = 3$$

So $M_0 = 0,\ M_1 = 3,\ M_2 = 0$.

Spline on [1, 3]

$$S_0(x) = \frac{M_0}{6h_0}(x_1-x)^3 + \frac{M_1}{6h_0}(x-x_0)^3 + \left(\frac{f_0}{h_0}-\frac{M_0 h_0}{6}\right)(x_1-x) + \left(\frac{f_1}{h_0}-\frac{M_1 h_0}{6}\right)(x-x_0)$$

With $x_0=1,\ x_1=3,\ h_0=2$:

$$S_0(x) = \frac{3}{12}(x-1)^3 + \left(\frac{1.5}{2}\right)(3-x) + \left(\frac{4.5}{2}-\frac{3\cdot 2}{6}\right)(x-1)$$

$$S_0(x) = 0.25(x-1)^3 + 0.75(3-x) + (2.25-1)(x-1)$$

Simplify linear terms:

$$0.75(3-x) + 1.25(x-1) = 2.25 - 0.75x + 1.25x - 1.25 = 1 + 0.5x$$

$$\boxed{S_0(x) = 0.25(x-1)^3 + 0.5x + 1.0}$$

Estimate f(2)

Since $1 \le 2 \le 3$, use $S_0$:

$$f(2) = 0.25(2-1)^3 + 0.5(2) + 1.0 = 0.25(1) + 1.0 + 1.0 = 2.25$$

$$\boxed{f(2) = 2.25}$$

Summary

Runge-Kutta 4th Order Method

Given Data

$$\frac{dy}{dx} = f(x,y) = \frac{x}{y}, \quad y_0 = 1 \text{ at } x_0 = 0, \quad \text{find } y(0.4)$$

Step size: $h = 0.4$ (single step).

RK4 Formulas

$$k_1 = h f(x_0, y_0)$$ $$k_2 = h f\left(x_0+\tfrac{h}{2}, y_0+\tfrac{k_1}{2}\right)$$ $$k_3 = h f\left(x_0+\tfrac{h}{2}, y_0+\tfrac{k_2}{2}\right)$$ $$k_4 = h f\left(x_0+h, y_0+k_3\right)$$ $$y_1 = y_0 + \tfrac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$$

Calculation

$k_1$: $$k_1 = 0.4 \cdot f(0, 1) = 0.4 \cdot \frac{0}{1} = 0$$

$k_2$: $$k_2 = 0.4 \cdot f(0.2, \ 1+0) = 0.4 \cdot \frac{0.2}{1} = 0.08$$

$k_3$: $$k_3 = 0.4 \cdot f(0.2, \ 1+0.04) = 0.4 \cdot \frac{0.2}{1.04} = 0.4 \cdot 0.192308 = 0.076923$$

$k_4$: $$k_4 = 0.4 \cdot f(0.4, \ 1+0.076923) = 0.4 \cdot \frac{0.4}{1.076923} = 0.4 \cdot 0.371429 = 0.148571$$

Final Result

$$y_1 = 1 + \frac{1}{6}\big(0 + 2(0.08) + 2(0.076923) + 0.148571\big)$$

$$= 1 + \frac{1}{6}\big(0 + 0.16 + 0.153846 + 0.148571\big)$$

$$= 1 + \frac{1}{6}(0.462418) = 1 + 0.077070$$

$$\boxed{y(0.4) \approx 1.0771}$$

Verification (Exact Solution)

Separating variables: $y,dy = x,dx \Rightarrow \tfrac{y^2}{2} = \tfrac{x^2}{2} + C$. With $y(0)=1$, $C = \tfrac12$, so $y = \sqrt{1+x^2}$.

At $x=0.4$: $y = \sqrt{1.16} = 1.07703$.

The RK4 result $1.0771$ matches the exact value very closely.

Pivoting is the process of rearranging (swapping) rows (or columns) of an augmented matrix during Gaussian elimination so that the element used as the divisor (pivot element) in each forward elimination step is the largest possible value...

In the Fixed Point Iteration Method, we rearrange the non-linear equation f(x) = 0 such that x is isolated on the left-hand side of the equation. The rearranged form is expressed as: $$x = g(x) \quad \cdots (1)$$ Since f(x) = 0 and x = g...

Gauss-Seidel Method

STEP 1 - EXTRACT: Given Data

The system of equations: $$x + 2y + 3z = 4 \quad \text{...(1)}$$ $$6x + 4y + 5z = 16 \quad \text{...(2)}$$ $$5x + 2y + 3z = 12 \quad \text{...(3)}$$

Initial guess (standard assumption): $x^{(0)} = y^{(0)} = z^{(0)} = 0$.

STEP 2 - SOLVE

Analysis of Diagonal Dominance

For Gauss-Seidel to converge reliably, we should try to arrange the largest coefficients on the diagonal. Let me examine:

• Coefficient of $x$: largest in (2) and (3), both equal to 6 and 5.
• Coefficient of $y$: largest is 4 (in eq 2).
• Coefficient of $z$: largest is 5 (in eq 2).

The largest coefficients cluster in equation (2), so a clean diagonally dominant arrangement is not possible. This is a poorly conditioned system for Gauss-Seidel. However, following the standard exam approach (partial pivoting), we rearrange to place the largest coefficient in each row on the diagonal.

Rearranged system:

$$6x + 4y + 5z = 16 \Rightarrow x = \frac{16 - 4y - 5z}{6}$$ $$5x + 2y + 3z = 12 \Rightarrow y = \frac{12 - 5x - 3z}{2}$$ $$x + 2y + 3z = 4 \Rightarrow z = \frac{4 - x - 2y}{3}$$

Iteration 1 (from $x=y=z=0$)

$$x = \frac{16 - 0 - 0}{6} = 2.6667$$ $$y = \frac{12 - 5(2.6667) - 0}{2} = \frac{12 - 13.3333}{2} = -0.6667$$ $$z = \frac{4 - 2.6667 - 2(-0.6667)}{3} = \frac{4 - 2.6667 + 1.3333}{3} = \frac{2.6667}{3} = 0.8889$$

Iteration 2

$$x = \frac{16 - 4(-0.6667) - 5(0.8889)}{6} = \frac{16 + 2.6667 - 4.4444}{6} = \frac{14.2222}{6} = 2.3704$$ $$y = \frac{12 - 5(2.3704) - 3(0.8889)}{2} = \frac{12 - 11.8519 - 2.6667}{2} = \frac{-2.5185}{2} = -1.2593$$ $$z = \frac{4 - 2.3704 - 2(-1.2593)}{3} = \frac{4 - 2.3704 + 2.5185}{3} = \frac{4.1481}{3} = 1.3827$$

Iteration 3

$$x = \frac{16 - 4(-1.2593) - 5(1.3827)}{6} = \frac{16 + 5.0370 - 6.9136}{6} = \frac{14.1235}{6} = 2.3539$$ $$y = \frac{12 - 5(2.3539) - 3(1.3827)}{2} = \frac{12 - 11.7696 - 4.1481}{2} = \frac{-3.9177}{2} = -1.9588$$ $$z = \frac{4 - 2.3539 - 2(-1.9588)}{3} = \frac{4 - 2.3539 + 3.9177}{3} = \frac{5.5638}{3} = 1.8546$$

Iteration 4 (check on convergence)

$$x = \frac{16 - 4(-1.9588) - 5(1.8546)}{6} = \frac{16 + 7.8353 - 9.2730}{6} = \frac{14.5623}{6} = 2.4270$$ $$y = \frac{12 - 5(2.4270) - 3(1.8546)}{2} = \frac{12 - 12.1352 - 5.5638}{2} = \frac{-5.6990}{2} = -2.8495$$ $$z = \frac{4 - 2.4270 - 2(-2.8495)}{3} = \frac{4 - 2.4270 + 5.6990}{3} = \frac{7.2720}{3} = 2.4240$$

Summary Table

Important Observation on Convergence

The iterates are diverging, not converging. The values of $y$ move $-0.67 \to -1.26 \to -1.96 \to -2.85$ and $z$ moves $0.89 \to 1.38 \to 1.85 \to 2.42$; the gaps are growing, not shrinking. This confirms the earlier note that the system is not diagonally dominant and no rearrangement makes it so, so Gauss-Seidel does not converge here.

Exact solution (by direct elimination, subtracting (1) from (3): $4x = 8 \Rightarrow x = 2$; then $2y + 3z = 2$ and $4y + 5z = 4$, giving $y = -1,\ z = ...$). Solving:

• From $4x=8$: $x = 2$.
• $2y + 3z = 4 - 2 = 2$.
• $4y + 5z = 16 - 12 = 4$.
• Multiply first by 2: $4y + 6z = 4$; subtract: $z = 0$, then $2y = 2 \Rightarrow y = 1$.

So the true solution is $x = 2,\ y = 1,\ z = 0$. The Gauss-Seidel iterates ($x\approx 2.4,\ y\approx -2.8,\ z\approx 2.4$) are moving away from this, confirming divergence.

Final Result

The iterations up to iteration 3 and 4 are as tabulated. The method does not converge here: the iterates move away from the exact solution $x=2,\ y=1,\ z=0$, so any claim that it settles near $x=2,\ y=-2,\ z=2$ would be wrong.

Horner's rule is an efficient scheme for evaluating a polynomial and its derivative at a point by rewriting the polynomial in nested form. It is closely related to synthetic division and the Remainder Theorem. For a polynomial of degree ...