BE Civil Engineering (IOE, TU) Numerical Methods (IOE, SH 553) Question Paper 2079 Nepal

Derivation. Let $x_n$ be an approximation to a root $\alpha$ of $f(x)=0$. Expand $f$ in a Taylor series about $x_n$ and retain terms up to first order:

$$f(x_{n+1}) \approx f(x_n) + (x_{n+1}-x_n)\,f'(x_n).$$

Setting $f(x_{n+1})=0$ and solving for $x_{n+1}$ gives the Newton-Raphson formula

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

Order of convergence. If $\alpha$ is a simple root, the error $e_{n+1} \approx \dfrac{f''(\alpha)}{2f'(\alpha)}e_n^{2}$, so convergence is quadratic (order 2).

Application. Here

$$f(x) = x^3 + 2x^2 + 10x - 20, \qquad f'(x) = 3x^2 + 4x + 10.$$

Iteration table (working to 6 d.p.):

Since $x_3$ and $x_4$ agree to four decimal places ($1.3688$), the iteration is stopped.

Root: $\;\boxed{x \approx 1.3688\ \text{m}}$ (check: $f(1.3688) \approx 2\times10^{-6} \approx 0$).

Divided-difference table. With $f[x_i,\dots]$ denoting divided differences:

Newton's divided-difference polynomial (using the leading entries):

$$P(x) = 1 + 4(x-1) + 0.66667(x-1)(x-2) + 0.05556(x-1)(x-2)(x-4).$$

Estimate at $x=3$:

$$P(3) = 1 + 4(2) + 0.66667(2)(1) + 0.05556(2)(1)(-1)$$ $$= 1 + 8 + 1.33333 - 0.11111 = \boxed{10.2222\ \text{m}}.$$

Estimate at $x=5$:

$$P(5) = 1 + 4(4) + 0.66667(4)(3) + 0.05556(4)(3)(1)$$ $$= 1 + 16 + 8.00000 + 0.66667 = \boxed{25.6667\ \text{m}}.$$

Thus the interpolated reduced levels are approximately 10.22 m at $x=3$ m and 25.67 m at $x=5$ m.

Function values with $f(x)=\dfrac{1}{1+x^2}$, $h=1$:

(a) Trapezoidal rule.

$$I_T = \frac{h}{2}\Big[(f_0+f_6) + 2(f_1+f_2+f_3+f_4+f_5)\Big]$$ $$= \frac{1}{2}\Big[1.027027 + 2(0.897286)\Big] = \frac{1}{2}(2.821599) = \boxed{1.410799}.$$

(b) Simpson's $1/3$ rule ($n=6$, even):

$$I_S = \frac{h}{3}\Big[(f_0+f_6) + 4(f_1+f_3+f_5) + 2(f_2+f_4)\Big]$$ $$= \frac{1}{3}\Big[1.027027 + 4(0.638462) + 2(0.258824)\Big]$$ $$= \frac{1}{3}\Big[1.027027 + 2.553846 + 0.517647\Big] = \frac{1}{3}(4.098520) = \boxed{1.366173}.$$

Comparison. Exact value $= \tan^{-1}(6) = 1.405648$.

Comment. Simpson's rule is normally more accurate (error $O(h^4)$ versus $O(h^2)$). Here the integrand is sharply peaked near $x=0$ where the curvature is large, so with the coarse spacing $h=1$ the trapezoidal estimate happens to be numerically closer to the exact value. Reducing $h$ (more sub-intervals) makes Simpson's rule converge faster and clearly outperform the trapezoidal rule.

Convergence condition. Gauss-Seidel converges (for any starting vector) if the coefficient matrix is strictly diagonally dominant, i.e. for each row $i$

$$|a_{ii}| > \sum_{j\neq i} |a_{ij}|.$$

Check: Row 1: $20 > 1+2=3$ ✓; Row 2: $20 > 3+1=4$ ✓; Row 3: $20 > 2+3=5$ ✓. The system is diagonally dominant, so convergence is guaranteed.

Iteration formulae (solve each equation for its diagonal unknown):

$$x = \frac{17 - y + 2z}{20},\quad y = \frac{-18 - 3x + z}{20},\quad z = \frac{25 - 2x + 3y}{20}.$$

Using the latest available values within each sweep:

From iteration 3 onwards the values are stable to three decimal places.

Solution: $\;\boxed{x = 1.000,\; y = -1.000,\; z = 1.000}$ (verifies all three original equations exactly).

RK4 formula. For each step,

$$k_1 = h\,f(x_n,y_n),\quad k_2 = h\,f\!\left(x_n+\tfrac{h}{2},\,y_n+\tfrac{k_1}{2}\right),$$ $$k_3 = h\,f\!\left(x_n+\tfrac{h}{2},\,y_n+\tfrac{k_2}{2}\right),\quad k_4 = h\,f(x_n+h,\,y_n+k_3),$$ $$y_{n+1} = y_n + \tfrac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4\right).$$

Here $f(x,y)=x+y$, $h=0.1$. (Below the $k$ values are written as $f$-evaluations; multiply by $h$ in the weighted sum.)

Step 1: $x_0=0,\ y_0=1 \Rightarrow y(0.1)$

$$k_1=f(0,1)=1.000000$$ $$k_2=f(0.05,\,1+0.05\cdot1)=f(0.05,1.05)=1.100000$$ $$k_3=f(0.05,\,1+0.05\cdot1.1)=f(0.05,1.055)=1.105000$$ $$k_4=f(0.1,\,1+0.1\cdot1.105)=f(0.1,1.1105)=1.210500$$ $$y(0.1)=1+\tfrac{0.1}{6}\big(1+2(1.1)+2(1.105)+1.2105\big)=1+\tfrac{0.1}{6}(6.620500)=\boxed{1.110342}.$$

Step 2: $x_1=0.1,\ y_1=1.110342 \Rightarrow y(0.2)$

$$k_1=f(0.1,1.110342)=1.210342$$ $$k_2=f(0.15,\,1.110342+0.05\cdot1.210342)=f(0.15,1.170859)=1.320859$$ $$k_3=f(0.15,\,1.110342+0.05\cdot1.320859)=f(0.15,1.176385)=1.326385$$ $$k_4=f(0.2,\,1.110342+0.1\cdot1.326385)=f(0.2,1.242980)=1.442980$$ $$y(0.2)=1.110342+\tfrac{0.1}{6}\big(1.210342+2(1.320859)+2(1.326385)+1.442980\big)$$ $$=1.110342+\tfrac{0.1}{6}(7.947810)=\boxed{1.242805}.$$

Comparison. Exact: $y(0.2)=2e^{0.2}-0.2-1=1.242806$. The RK4 estimate $1.242805$ differs by only $\approx 1\times10^{-6}$, confirming the high (fourth-order) accuracy of the method.

Augmented matrix:

$$\left[\begin{array}{ccc|c} 2 & 1 & 1 & 10 \\ 3 & 2 & 3 & 18 \\ 1 & 4 & 9 & 16 \end{array}\right].$$

Step 1 — eliminate $x$ from rows 2 and 3 (pivot $a_{11}=2$): $R_2 \leftarrow R_2 - \tfrac{3}{2}R_1$ and $R_3 \leftarrow R_3 - \tfrac{1}{2}R_1$:

$$\left[\begin{array}{ccc|c} 2 & 1 & 1 & 10 \\ 0 & 0.5 & 1.5 & 3 \\ 0 & 3.5 & 8.5 & 11 \end{array}\right].$$

Step 2 — eliminate $y$ from row 3 (pivot $a_{22}=0.5$): $R_3 \leftarrow R_3 - \tfrac{3.5}{0.5}R_2 = R_3 - 7R_2$:

$$\left[\begin{array}{ccc|c} 2 & 1 & 1 & 10 \\ 0 & 0.5 & 1.5 & 3 \\ 0 & 0 & -2 & -10 \end{array}\right].$$

Back-substitution.

$$z = \frac{-10}{-2} = 5.$$ $$0.5y + 1.5(5) = 3 \;\Rightarrow\; 0.5y = 3 - 7.5 = -4.5 \;\Rightarrow\; y = -9.$$ $$2x + (-9) + 5 = 10 \;\Rightarrow\; 2x = 14 \;\Rightarrow\; x = 7.$$

Solution: $\;\boxed{x = 7,\; y = -9,\; z = 5}$ (check in eqn 3: $7 + 4(-9) + 9(5) = 7 - 36 + 45 = 16$ ✓).

Forward-difference table ($\Delta^k y$):

Leading differences: $y_0=1,\ \Delta y_0=1,\ \Delta^2 y_0=6,\ \Delta^3 y_0=6$.

Newton's forward formula with $p=\dfrac{x-x_0}{h}=\dfrac{1.5-0}{1}=1.5$:

$$y = y_0 + p\,\Delta y_0 + \frac{p(p-1)}{2!}\Delta^2 y_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 y_0.$$

Compute each term:

• $y_0 = 1$
• $p\,\Delta y_0 = 1.5(1) = 1.5$
• $\dfrac{1.5(0.5)}{2}(6) = \dfrac{0.75}{2}(6) = 0.375\times6 = 2.25$
• $\dfrac{1.5(0.5)(-0.5)}{6}(6) = \dfrac{-0.375}{6}(6) = -0.375$

$$y(1.5) = 1 + 1.5 + 2.25 - 0.375 = \boxed{4.375}.$$

(The data correspond to $y=x^3+1$; exact $1.5^3+1=4.375$, confirming the result.)

Forward-difference table:

Leading differences: $\Delta y_0=7,\ \Delta^2 y_0=12,\ \Delta^3 y_0=6,\ \Delta^4 y_0=0$.

First derivative at $x=1$ (start of table, forward formula):

$$\left.\frac{dy}{dx}\right|_{x_0} = \frac{1}{h}\left(\Delta y_0 - \frac{\Delta^2 y_0}{2} + \frac{\Delta^3 y_0}{3} - \frac{\Delta^4 y_0}{4}\right)$$ $$= \frac{1}{1}\left(7 - \frac{12}{2} + \frac{6}{3} - \frac{0}{4}\right) = 7 - 6 + 2 - 0 = \boxed{3}.$$

Second derivative at $x=1$:

$$\left.\frac{d^2y}{dx^2}\right|_{x_0} = \frac{1}{h^2}\left(\Delta^2 y_0 - \Delta^3 y_0 + \frac{11}{12}\Delta^4 y_0\right)$$ $$= \frac{1}{1}\left(12 - 6 + \frac{11}{12}\cdot 0\right) = \boxed{6}.$$

The data follow $y=x^3$, for which $y'(1)=3x^2=3$ and $y''(1)=6x=6$ — the numerical results match exactly.

Since $f(3) = 9-10 = -1 < 0$ and $f(4) = 16-10 = 6 > 0$, a root lies in $[3,4]$. At each step take $c=\frac{a+b}{2}$ and keep the half-interval where the sign of $f$ changes.

After five iterations the root is bracketed in $[3.1563, 3.1875]$.

Approximate root: $\;\boxed{x \approx 3.1563}$ (true value $\sqrt{10}=3.16228$; error $<0.007$).

Euler's formula: $y_{n+1} = y_n + h\,f(x_n, y_n)$, with $f(x,y)=x+y$, $h=0.1$.

Result: $\;\boxed{y(0.3) \approx 1.3620}$.

(The exact solution is $y=2e^{x}-x-1$, giving $y(0.3)=1.39972$; Euler's method underestimates because it uses only the slope at the start of each interval — accuracy improves with smaller $h$.)

Discretization. On a uniform square mesh of spacing $h$, replace each second derivative by a central difference:

$$\frac{\partial^2 u}{\partial x^2} \approx \frac{u_{i+1,j} - 2u_{i,j} + u_{i-1,j}}{h^2},\qquad \frac{\partial^2 u}{\partial y^2} \approx \frac{u_{i,j+1} - 2u_{i,j} + u_{i,j-1}}{h^2}.$$

Substituting into Laplace's equation, the $h^2$ cancels and the five-point formula becomes

$$u_{i+1,j} + u_{i-1,j} + u_{i,j+1} + u_{i,j-1} - 4u_{i,j} = 0,$$

which rearranges to the Liebmann (Gauss-Seidel) iteration formula

$$\boxed{\,u_{i,j} = \tfrac{1}{4}\big(u_{i+1,j} + u_{i-1,j} + u_{i,j+1} + u_{i,j-1}\big)\,}.$$

Each interior temperature equals the average of its four neighbours.

(a) Single interior node. Its four neighbours are the given boundary values $100,\ 80,\ 60,\ 40\ ^{\circ}\text{C}$:

$$u = \frac{100 + 80 + 60 + 40}{4} = \frac{280}{4} = \boxed{70^{\circ}\text{C}}.$$

With only one unknown the average is exact and no iteration is needed.

(b) Two interior nodes. Apply the five-point formula at each node:

$$u_1 = \frac{100 + 80 + 60 + u_2}{4} = \frac{240 + u_2}{4} \;\Rightarrow\; 4u_1 - u_2 = 240,$$ $$u_2 = \frac{u_1 + 80 + 60 + 40}{4} = \frac{180 + u_1}{4} \;\Rightarrow\; -u_1 + 4u_2 = 180.$$

Solve the $2\times2$ system. From the first, $u_2 = 4u_1 - 240$; substitute into the second:

$$-u_1 + 4(4u_1 - 240) = 180 \;\Rightarrow\; 15u_1 = 1140 \;\Rightarrow\; u_1 = 76^{\circ}\text{C}.$$ $$u_2 = 4(76) - 240 = 64^{\circ}\text{C}.$$

Result: $\;\boxed{u_1 = 76^{\circ}\text{C},\; u_2 = 64^{\circ}\text{C}}$ (a Liebmann iteration from a zero start also converges to these values).