BE Civil Engineering (IOE, TU) Numerical Methods (IOE, SH 553) Question Paper 2080 Nepal

(a) Newton-Raphson iteration

The iteration formula is

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \qquad f(x)=x^3-2x-5,\quad f'(x)=3x^2-2.$$

Iteration 1 ($x_0=2$): $f(2)=8-4-5=-1$, $\;f'(2)=12-2=10$.

$$x_1 = 2 - \frac{-1}{10} = 2.1.$$

Iteration 2 ($x_1=2.1$): $f(2.1)=9.261-4.2-5=0.061$, $\;f'(2.1)=3(4.41)-2=11.23$.

$$x_2 = 2.1 - \frac{0.061}{11.23} = 2.1 - 0.005432 = 2.094568.$$

Iteration 3 ($x_2=2.094568$): $x_2^3 = 9.188566$, so $f=9.188566-4.189136-5=0.000430$. $f'(2.094568)=3(4.387215)-2=11.161645$.

$$x_3 = 2.094568 - \frac{0.000430}{11.161645}=2.094568-0.0000385=2.0945295.$$

At iteration 3 the relative error $|(x_3-x_2)/x_3|\times100 = 0.00184\% < 0.01\%$, so we stop.

Root $\boldsymbol{x \approx 2.0945}$ (to 4 decimal places).

(b) Order of convergence and failure

Newton-Raphson has quadratic convergence (order $p=2$) near a simple root: the error satisfies

$$|e_{n+1}| \approx \left|\frac{f''(\xi)}{2f'(\xi)}\right|\,|e_n|^2 .$$

The method fails or diverges when $f'(x_n)=0$ (or is very small) at an iterate — for example near a turning point or an inflection — because the tangent is nearly horizontal and the next iterate is thrown far away. It can also oscillate if the initial guess is poor.

(a) Divided-difference table

Let $x=P$, $y=S$.

First divided differences:

$$f[50,100]=\frac{9-4}{100-50}=\frac{5}{50}=0.10,$$ $$f[100,150]=\frac{16-9}{150-100}=\frac{7}{50}=0.14,$$ $$f[150,200]=\frac{27-16}{200-150}=\frac{11}{50}=0.22.$$

Second divided differences:

$$f[50,100,150]=\frac{0.14-0.10}{150-50}=\frac{0.04}{100}=0.0004,$$ $$f[100,150,200]=\frac{0.22-0.14}{200-100}=\frac{0.08}{100}=0.0008.$$

Third divided difference:

$$f[50,100,150,200]=\frac{0.0008-0.0004}{200-50}=\frac{0.0004}{150}=2.6667\times10^{-6}.$$

Newton forward divided-difference polynomial:

$$S(P) = 4 + 0.10(P-50) + 0.0004(P-50)(P-100) + 2.6667\times10^{-6}(P-50)(P-100)(P-150).$$

(b) Estimate at $P=125$ kN

$(P-50)=75,\;(P-100)=25,\;(P-150)=-25.$

• Term 1: $4$
• Term 2: $0.10\times75 = 7.5$
• Term 3: $0.0004\times75\times25 = 0.0004\times1875 = 0.75$
• Term 4: $2.6667\times10^{-6}\times75\times25\times(-25) = 2.6667\times10^{-6}\times(-46875) = -0.125$

$$S(125) = 4 + 7.5 + 0.75 - 0.125 = 12.125\ \text{mm}.$$

Estimated settlement at 125 kN $\boldsymbol{\approx 12.125}$ mm.

Diagonal dominance check

• Row 1: $|10| \ge |2|+|1| = 3$ ✓
• Row 2: $|20| \ge |2|+|-2| = 4$ ✓
• Row 3: $|10| \ge |-2|+|3| = 5$ ✓

The matrix is strictly diagonally dominant, so Gauss-Seidel converges.

Rearranged update equations:

$$x = \frac{9 - 2y - z}{10},\quad y = \frac{-44 - 2x + 2z}{20},\quad z = \frac{22 + 2x - 3y}{10}.$$

Iteration 1 (start $x=y=z=0$):

• $x = (9-0-0)/10 = 0.9$
• $y = (-44 - 2(0.9) + 0)/20 = (-44-1.8)/20 = -2.29$
• $z = (22 + 2(0.9) - 3(-2.29))/10 = (22+1.8+6.87)/10 = 3.067$

Iteration 2:

• $x = (9 - 2(-2.29) - 3.067)/10 = (9+4.58-3.067)/10 = 1.0513$
• $y = (-44 - 2(1.0513) + 2(3.067))/20 = (-44-2.1026+6.134)/20 = -1.998430$
• $z = (22 + 2(1.0513) - 3(-1.998430))/10 = (22+2.1026+5.99529)/10 = 3.009789$

Iteration 3:

• $x = (9 - 2(-1.998430) - 3.009789)/10 = (9+3.99686-3.009789)/10 = 0.998707$
• $y = (-44 - 2(0.998707) + 2(3.009789))/20 = (-44-1.997414+6.019578)/20 = -1.998892$
• $z = (22 + 2(0.998707) - 3(-1.998892))/10 = (22+1.997414+5.996676)/10 = 2.999409$

Iteration 4:

• $x = (9 - 2(-1.998892) - 2.999409)/10 = (9+3.997784-2.999409)/10 = 0.999838$
• $y = (-44 - 2(0.999838) + 2(2.999409))/20 = (-44-1.999676+5.998818)/20 = -2.000043$
• $z = (22 + 2(0.999838) - 3(-2.000043))/10 = (22+1.999676+6.000129)/10 = 2.999981$

Changes at iteration 4 are all $<0.001$ except $x$ ($\approx0.00113$); one more pass gives $x\to1.00000$. The solution converges to

$$\boxed{x \approx 1.000,\quad y \approx -2.000,\quad z \approx 3.000.}$$

RK4 formulae for $\dfrac{dT}{dt}=f(t,T)=-0.5(T-25)$:

$$k_1=hf(t_n,T_n),\;k_2=hf\!\left(t_n+\tfrac h2,T_n+\tfrac{k_1}{2}\right),\;k_3=hf\!\left(t_n+\tfrac h2,T_n+\tfrac{k_2}{2}\right),\;k_4=hf(t_n+h,T_n+k_3),$$ $$T_{n+1}=T_n+\tfrac16(k_1+2k_2+2k_3+k_4).$$

Here $h=0.5$, and $f$ does not depend on $t$ explicitly.

Step 1: $t_0=0,\ T_0=100$.

• $k_1 = 0.5\times[-0.5(100-25)] = 0.5\times(-37.5) = -18.75$
• $T_0+k_1/2 = 100-9.375 = 90.625$; $\;k_2 = 0.5\times[-0.5(90.625-25)] = 0.5\times(-32.8125) = -16.40625$
• $T_0+k_2/2 = 100-8.203125 = 91.796875$; $\;k_3 = 0.5\times[-0.5(91.796875-25)] = 0.5\times(-33.3984375) = -16.69921875$
• $T_0+k_3 = 100-16.69921875 = 83.30078125$; $\;k_4 = 0.5\times[-0.5(83.30078125-25)] = 0.5\times(-29.150390625) = -14.5751953125$

$$T_1 = 100 + \tfrac16\big(-18.75 + 2(-16.40625) + 2(-16.69921875) - 14.5751953125\big).$$

Sum inside: $-18.75 -32.8125 -33.3984375 -14.5751953125 = -99.5361328125$.

$$T_1 = 100 - 16.5893554688 = 83.4106445\ ^\circ\text{C}.$$

Step 2: $t_1=0.5,\ T_1=83.4106445$.

• $k_1 = 0.5\times[-0.5(83.4106445-25)] = 0.5\times(-29.20532225) = -14.602661125$
• $T_1+k_1/2 = 83.4106445-7.301330563 = 76.10931394$; $\;k_2 = 0.5\times[-0.5(51.10931394)] = -12.77732849$
• $T_1+k_2/2 = 83.4106445-6.388664243 = 77.02198026$; $\;k_3 = 0.5\times[-0.5(52.02198026)] = -13.00549506$
• $T_1+k_3 = 83.4106445-13.00549506 = 70.40514944$; $\;k_4 = 0.5\times[-0.5(45.40514944)] = -11.35128736$

$$T_2 = 83.4106445 + \tfrac16\big(-14.602661125 + 2(-12.77732849) + 2(-13.00549506) - 11.35128736\big).$$

Sum: $-14.602661125 -25.55465698 -26.01099012 -11.35128736 = -77.51959559$.

$$T_2 = 83.4106445 - 12.91993260 = 70.4907119\ ^\circ\text{C}.$$

Exact solution: $T(t) = 25 + 75e^{-0.5t}$.

$$T(1.0) = 25 + 75e^{-0.5} = 25 + 75(0.6065307) = 25 + 45.4898 = 70.4898\ ^\circ\text{C}.$$

Comparison: RK4 gives $\boldsymbol{T(1.0)\approx 70.491\ ^\circ}$C versus exact $70.490\ ^\circ$C — error $\approx 0.001\ ^\circ$C, demonstrating RK4's high accuracy.

Grid layout ($h=1$). Interior nodes at $(1,1),(2,1),(1,2),(2,2)$. Label:

  y=3:  100   100   100   100
  y=2:   0    u3    u4    0
  y=1:   0    u1    u2    0
  y=0:   0     0     0    0

where $u_1=u(1,1),\,u_2=u(2,1),\,u_3=u(1,2),\,u_4=u(2,2)$.

Five-point Laplace formula: $u_{i,j}=\tfrac14(u_{i+1,j}+u_{i-1,j}+u_{i,j+1}+u_{i,j-1})$.

Equations (using boundary values):

$$u_1 = \tfrac14(u_2 + 0 + u_3 + 0),$$ $$u_2 = \tfrac14(0 + u_1 + u_4 + 0),$$ $$u_3 = \tfrac14(u_4 + 0 + 100 + u_1),$$ $$u_4 = \tfrac14(0 + u_3 + 100 + u_2).$$

Symmetry about the vertical centre line gives $u_1=u_2$ and $u_3=u_4$, but we solve the general iteration to demonstrate.

Liebmann (Gauss-Seidel) iterations, start all $=0$:

Iteration 1:

• $u_1=\tfrac14(0+0+0+0)=0$
• $u_2=\tfrac14(0+0+0+0)=0$
• $u_3=\tfrac14(0+0+100+0)=25$
• $u_4=\tfrac14(0+25+100+0)=31.25$

Iteration 2:

• $u_1=\tfrac14(u_2+u_3)=\tfrac14(0+25)=6.25$
• $u_2=\tfrac14(u_1+u_4)=\tfrac14(6.25+31.25)=9.375$
• $u_3=\tfrac14(u_4+100+u_1)=\tfrac14(31.25+100+6.25)=34.375$
• $u_4=\tfrac14(u_3+100+u_2)=\tfrac14(34.375+100+9.375)=35.9375$

Iteration 3:

• $u_1=\tfrac14(u_2+u_3)=\tfrac14(9.375+34.375)=10.9375$
• $u_2=\tfrac14(u_1+u_4)=\tfrac14(10.9375+35.9375)=11.71875$
• $u_3=\tfrac14(u_4+100+u_1)=\tfrac14(35.9375+100+10.9375)=36.71875$
• $u_4=\tfrac14(u_3+100+u_2)=\tfrac14(36.71875+100+11.71875)=37.109375$

Continued iteration converges to the exact values $u_1=u_2=12.5$ and $u_3=u_4=37.5$.

After 3 iterations: $\boldsymbol{u_1\approx10.94,\ u_2\approx11.72,\ u_3\approx36.72,\ u_4\approx37.11}$, trending toward the symmetric solution $(12.5,12.5,37.5,37.5)$.

Bisection method. $f(x)=x^2-\cos x-1$.

Check the bracket: $f(1)=1-\cos 1-1 = 1-0.540302-1 = -0.540302<0$. $f(2)=4-\cos 2-1 = 4-(-0.416147)-1 = 3.416147>0$. Root lies in $[1,2]$.

Iteration 1: $c_1=1.5$. $f(1.5)=2.25-\cos1.5-1=2.25-0.070737-1=1.179263>0$. Root in $[1,1.5]$.

Iteration 2: $c_2=1.25$. $f(1.25)=1.5625-\cos1.25-1=1.5625-0.315322-1=0.247178>0$. Root in $[1,1.25]$.

Iteration 3: $c_3=1.125$. $f(1.125)=1.265625-\cos1.125-1=1.265625-0.428913-1=-0.163288<0$. Root in $[1.125,1.25]$.

Iteration 4: $c_4=1.1875$. $f(1.1875)=1.410156-\cos1.1875-1=1.410156-0.373123-1=0.037033>0$. Root in $[1.125,1.1875]$.

Estimated root $\approx \dfrac{1.125+1.1875}{2}=1.15625$.

Error bound: after $n$ iterations the error is at most $\dfrac{b-a}{2^n}=\dfrac{2-1}{2^4}=\dfrac{1}{16}=\boldsymbol{0.0625}$.

Simpson's 1/3 rule. With $n=6$ intervals (even) and $h=2$ m:

$$A = \frac{h}{3}\Big[(y_0+y_6) + 4(y_1+y_3+y_5) + 2(y_2+y_4)\Big].$$

Values:

• End ordinates: $y_0+y_6 = 0 + 0 = 0$.
• Odd ordinates: $y_1+y_3+y_5 = 1.5+2.8+1.6 = 5.9$, so $4\times5.9 = 23.6$.
• Even (interior) ordinates: $y_2+y_4 = 2.4+2.5 = 4.9$, so $2\times4.9 = 9.8$.

Sum $= 0 + 23.6 + 9.8 = 33.4$.

$$A = \frac{2}{3}\times 33.4 = \frac{66.8}{3} = 22.2\overline{6}\ \text{m}^2.$$

Cross-sectional area of flow $\boldsymbol{\approx 22.27\ \text{m}^2}$.

Lagrange interpolation for three points $(1,1),(2,8),(4,64)$:

$$y(x)=y_0 L_0 + y_1 L_1 + y_2 L_2,$$

where

$$L_0=\frac{(x-2)(x-4)}{(1-2)(1-4)},\quad L_1=\frac{(x-1)(x-4)}{(2-1)(2-4)},\quad L_2=\frac{(x-1)(x-2)}{(4-1)(4-2)}.$$

At $x=3$:

• $L_0=\dfrac{(3-2)(3-4)}{(-1)(-3)}=\dfrac{(1)(-1)}{3}=-\dfrac13$
• $L_1=\dfrac{(3-1)(3-4)}{(1)(-2)}=\dfrac{(2)(-1)}{-2}=1$
• $L_2=\dfrac{(3-1)(3-2)}{(3)(2)}=\dfrac{(2)(1)}{6}=\dfrac13$

$$y(3)=1\left(-\tfrac13\right) + 8(1) + 64\left(\tfrac13\right) = -\tfrac13 + 8 + \tfrac{64}{3} = 8 + \tfrac{63}{3} = 8 + 21 = 29.$$

$\boldsymbol{y(3) = 29}$. (Note: the underlying data fit $y=x^3$ would give $27$; the quadratic Lagrange fit through these three points yields $29$.)

Central-difference formulae with $h=0.1$ s, centred at $t=0.2$ s ($y_{i-1}=0.9,\ y_i=1.6,\ y_{i+1}=2.1$):

Velocity (first derivative):

$$\left.\frac{dy}{dt}\right|_{0.2} = \frac{y_{i+1}-y_{i-1}}{2h} = \frac{2.1 - 0.9}{2(0.1)} = \frac{1.2}{0.2} = 6.0\ \text{mm/s}.$$

Acceleration (second derivative):

$$\left.\frac{d^2y}{dt^2}\right|_{0.2} = \frac{y_{i+1}-2y_i+y_{i-1}}{h^2} = \frac{2.1 - 2(1.6) + 0.9}{(0.1)^2} = \frac{2.1 - 3.2 + 0.9}{0.01} = \frac{-0.2}{0.01} = -20\ \text{mm/s}^2.$$

Results: velocity $\boldsymbol{= 6.0\ \text{mm/s}}$, acceleration $\boldsymbol{= -20\ \text{mm/s}^2}$ (the negative sign indicates the motion is decelerating at $t=0.2$ s).

Modified Euler (Heun's) method: predictor then corrector

$$y^{p}_{n+1}=y_n+h\,f(x_n,y_n),\qquad y_{n+1}=y_n+\frac h2\big[f(x_n,y_n)+f(x_{n+1},y^{p}_{n+1})\big],$$

with $f(x,y)=x+y$, $h=0.1$.

Step 1: $x_0=0,\ y_0=1$.

• $f(x_0,y_0)=0+1=1$.
• Predictor: $y^p_1 = 1 + 0.1(1) = 1.1$.
• $f(x_1,y^p_1)=0.1+1.1=1.2$.
• Corrector: $y_1 = 1 + \dfrac{0.1}{2}(1 + 1.2) = 1 + 0.05(2.2) = 1 + 0.11 = 1.11$.

Step 2: $x_1=0.1,\ y_1=1.11$.

• $f(x_1,y_1)=0.1+1.11=1.21$.
• Predictor: $y^p_2 = 1.11 + 0.1(1.21) = 1.11 + 0.121 = 1.231$.
• $f(x_2,y^p_2)=0.2+1.231=1.431$.
• Corrector: $y_2 = 1.11 + \dfrac{0.1}{2}(1.21 + 1.431) = 1.11 + 0.05(2.641) = 1.11 + 0.13205 = 1.24205$.

$\boldsymbol{y(0.2) \approx 1.2421}$. (Exact $y=2e^x-x-1$ gives $y(0.2)=2e^{0.2}-1.2=2.44281-1.2=1.24281$; close agreement.)

Least-squares normal equations:

$$\sum y = na + b\sum x,\qquad \sum xy = a\sum x + b\sum x^2.$$

Compute the sums ($n=5$):

So $\sum x=15,\ \sum y=29,\ \sum x^2=55,\ \sum xy=102$.

Normal equations:

$$29 = 5a + 15b,\qquad 102 = 15a + 55b.$$

From the first: $a = \dfrac{29-15b}{5}$. Substitute into the second:

$$102 = 15\cdot\frac{29-15b}{5} + 55b = 3(29-15b) + 55b = 87 - 45b + 55b = 87 + 10b.$$ $$10b = 15 \Rightarrow b = 1.5.$$ $$a = \frac{29 - 15(1.5)}{5} = \frac{29 - 22.5}{5} = \frac{6.5}{5} = 1.3.$$

Best-fit line: $\boldsymbol{y = 1.3 + 1.5\,x}$.