BE Civil Engineering (IOE, TU) Numerical Methods (IOE, SH 553) Question Paper 2077 Nepal

(a) Existence of a root and bisection (3 marks)

$f(x)=x^3-x-11$.

$f(2)=8-2-11=-5<0$, and $f(3)=27-3-11=13>0$.

Since $f$ is continuous and $f(2)\cdot f(3)<0$, by the Intermediate Value Theorem a real root lies in $[2,3]$.

Iteration 1: $x_m=\tfrac{2+3}{2}=2.5$, $f(2.5)=15.625-2.5-11=2.125>0$. Root lies in $[2,\,2.5]$.

Iteration 2: $x_m=\tfrac{2+2.5}{2}=2.25$, $f(2.25)=11.39063-2.25-11=-1.85937<0$. Root lies in $[2.25,\,2.5]$.

(b) Newton-Raphson (5 marks)

$f'(x)=3x^2-1$, so $x_{n+1}=x_n-\dfrac{x_n^3-x_n-11}{3x_n^2-1}$.

Iterates $x_2=2.3737$ and $x_3=2.3737$ agree to four decimals.

Root $\approx \mathbf{2.3737}$.

(c) Convergence (2 marks)

Newton-Raphson has quadratic (second-order) convergence near a simple root: the error satisfies $e_{n+1}\approx C\,e_n^2$. It may fail when $f'(x_n)=0$ (or is very small), giving division by zero / divergence; it can also overshoot or cycle if the initial guess is far from the root.

(a) Forward difference table (3 marks)

(b) Newton forward interpolation at $x=1.5$ (3 marks)

Here $x_0=0$, $h=1$, $p=\dfrac{x-x_0}{h}=\dfrac{1.5-0}{1}=1.5$.

$$y = y_0 + p\,\Delta y_0 + \frac{p(p-1)}{2!}\Delta^2 y_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 y_0$$ $$y = 1 + (1.5)(1) + \frac{(1.5)(0.5)}{2}(8) + \frac{(1.5)(0.5)(-0.5)}{6}(6)$$ $$y = 1 + 1.5 + 3.0 - 0.375 = 5.125$$

Settlement at $x=1.5$ is $\mathbf{5.125\ \text{mm}}$.

(c) Interpolating polynomial (2 marks)

With general $p=x$ (since $h=1$, $x_0=0$):

$$y = 1 + x + \frac{x(x-1)}{2}(8) + \frac{x(x-1)(x-2)}{6}(6)$$ $$= 1 + x + 4x(x-1) + x(x-1)(x-2)$$ $$= 1 + x + 4x^2 - 4x + (x^3 - 3x^2 + 2x)$$ $$\boxed{y(x) = x^3 + x^2 - x + 1}$$

Check: $y(2)=8+4-2+1=11$ ✓, $y(3)=27+9-3+1=34$ ✓.

(a) Diagonal dominance and iteration formulae (2 marks)

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

The matrix is diagonally dominant, so Gauss-Seidel converges. Rearranging each equation for its diagonal variable:

$$x = \frac{8 + y - z}{4},\quad y = \frac{3 - 2x - 2z}{5},\quad z = \frac{11 - x - 2y}{4}.$$

(b) Iterations (5 marks) — each new value uses the most recent estimates.

From iteration 6 to 7 the changes are $|{\Delta x}|=0.00062$, $|{\Delta y}|=0.00026$, $|{\Delta z}|=0.00029$, all $<0.01$. Stop.

(c) Solution (1 mark)

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

Exact solution $(1,-1,3)$ confirms the result.

RK4 formulae: with $f(x,y)=x+y$,

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

Step 1: $x_0=0,\;y_0=1,\;h=0.1$ (3 marks)

• $k_1 = 0.1\,(0+1) = 0.100000$
• $k_2 = 0.1\,(0.05 + 1.05) = 0.110000$
• $k_3 = 0.1\,(0.05 + 1.055) = 0.110500$
• $k_4 = 0.1\,(0.1 + 1.1105) = 0.121050$

$$y_1 = 1 + \tfrac{1}{6}(0.1 + 2(0.11) + 2(0.1105) + 0.12105) = 1 + 0.110342 = 1.110342$$

$y(0.1) \approx \mathbf{1.110342}$.

Step 2: $x_1=0.1,\;y_1=1.110342$ (3 marks)

• $k_1 = 0.1\,(0.1 + 1.110342) = 0.121034$
• $k_2 = 0.1\,(0.15 + 1.170859) = 0.132086$
• $k_3 = 0.1\,(0.15 + 1.176385) = 0.132639$
• $k_4 = 0.1\,(0.2 + 1.242981) = 0.144298$

$$y_2 = 1.110342 + \tfrac{1}{6}(0.121034 + 2(0.132086) + 2(0.132639) + 0.144298) = 1.110342 + 0.132464 = 1.242806$$

$y(0.2) \approx \mathbf{1.242806}$.

Comparison with exact (2 marks)

Exact: $y(0.2) = 2e^{0.2} - 0.2 - 1 = 2(1.221403) - 1.2 = 1.242806$.

Absolute error $= |1.242806 - 1.242806| \approx \mathbf{1\times10^{-6}}$, showing RK4's high accuracy.

Five-point formula (2 marks): each interior node equals one-quarter of the sum of its four neighbours.

Neighbour bookkeeping (left/right/top/bottom):

• $u_1$: left $=0$, right $=u_2$, top $=100$, bottom $=u_3$
• $u_2$: left $=u_1$, right $=0$, top $=100$, bottom $=u_4$
• $u_3$: left $=0$, right $=u_4$, top $=u_1$, bottom $=0$
• $u_4$: left $=u_3$, right $=0$, top $=u_2$, bottom $=0$

Equations (2 marks):

$$4u_1 = 100 + u_2 + u_3 \quad(1)$$ $$4u_2 = 100 + u_1 + u_4 \quad(2)$$ $$4u_3 = u_1 + u_4 \quad(3)$$ $$4u_4 = u_2 + u_3 \quad(4)$$

Exact solution (4 marks): By symmetry of the boundary (left-right mirror), $u_1=u_2$ and $u_3=u_4$. Let $u_1=u_2=a$, $u_3=u_4=b$.

From (1): $4a = 100 + a + b \Rightarrow 3a - b = 100.$ From (3): $4b = a + b \Rightarrow 3b = a \Rightarrow a = 3b.$

Substitute: $3(3b) - b = 100 \Rightarrow 8b = 100 \Rightarrow b = 12.5,\; a = 37.5.$

$$\boxed{u_1 = u_2 = 37.5,\qquad u_3 = u_4 = 12.5}$$

Check (2): $4(37.5)=150$ and $100+37.5+12.5=150$ ✓. The nodes nearer the hot top edge are hotter, as physically expected.

Setup (2 marks): $a=0$, $b=1$, $n=4$, $h=\dfrac{1-0}{4}=0.25$, $f(x)=\dfrac{1}{1+x^2}$.

Simpson's 1/3 rule (3 marks):

$$I = \frac{h}{3}\Big[f_0 + f_4 + 4(f_1+f_3) + 2f_2\Big]$$ $$= \frac{0.25}{3}\Big[1.0 + 0.5 + 4(0.941176 + 0.640000) + 2(0.8)\Big]$$ $$= \frac{0.25}{3}\big[1.5 + 4(1.581176) + 1.6\big] = \frac{0.25}{3}(9.424706) = 0.785392$$

$I \approx \mathbf{0.785392}$.

Comparison (1 mark): Exact value $=\arctan(1)=\dfrac{\pi}{4}=0.785398$.

Absolute error $= |0.785398 - 0.785392| = \mathbf{6\times10^{-6}}$.

Trapezoidal rule (2 marks): distance $= \displaystyle\int_0^{10} v\,dt \approx \frac{h}{2}\Big[v_0 + v_n + 2\sum_{i=1}^{n-1} v_i\Big]$, with $h=2$ s, $n=5$.

Computation (3 marks):

$$\text{Distance} = \frac{2}{2}\Big[(0 + 40) + 2(12 + 25 + 33 + 38)\Big]$$ $$= 1\times\big[40 + 2(108)\big] = 40 + 216 = 256$$

Total distance $\approx \mathbf{256\ \text{m}}$.

Source of error (1 mark): The trapezoidal rule replaces the true velocity curve by straight-line segments between data points. Where the curve is concave (as during acceleration here), the chords lie below the curve, so the estimate is slightly lower than the true distance; the error is $O(h^2)$ and shrinks as the sampling interval $h$ decreases.

Lagrange formula (2 marks): for three points,

$$f(x) = \sum_{i} f(x_i)\, L_i(x),\qquad L_i(x)=\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}.$$

Data: $(x_0,f_0)=(1,1)$, $(x_1,f_1)=(3,27)$, $(x_2,f_2)=(4,64)$. Evaluate at $x=2$.

Basis polynomials at $x=2$ (3 marks):

$$L_0 = \frac{(2-3)(2-4)}{(1-3)(1-4)} = \frac{(-1)(-2)}{(-2)(-3)} = \frac{2}{6} = 0.33333$$ $$L_1 = \frac{(2-1)(2-4)}{(3-1)(3-4)} = \frac{(1)(-2)}{(2)(-1)} = \frac{-2}{-2} = 1.00000$$ $$L_2 = \frac{(2-1)(2-3)}{(4-1)(4-3)} = \frac{(1)(-1)}{(3)(1)} = \frac{-1}{3} = -0.33333$$

Interpolated value (1 mark):

$$f(2) = 1(0.33333) + 27(1.00000) + 64(-0.33333)$$ $$= 0.33333 + 27 - 21.33333 = 6.0$$

$f(2) \approx \mathbf{6}$. (Consistent with the cubic trend $x^3$ giving $8$; the quadratic Lagrange fit through three points yields $6$.)

Central difference for first derivative (2 marks):

$$f'(x) \approx \frac{f(x+h)-f(x-h)}{2h}$$ $$f'(1.1) \approx \frac{f(1.2)-f(1.0)}{2(0.1)} = \frac{3.3201 - 2.7183}{0.2} = \frac{0.6018}{0.2} = 3.0090$$

$f'(1.1) \approx \mathbf{3.0090}$.

Central difference for second derivative (2 marks):

$$f''(x) \approx \frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$ $$f''(1.1) \approx \frac{3.3201 - 2(3.0042) + 2.7183}{(0.1)^2} = \frac{3.3201 - 6.0084 + 2.7183}{0.01} = \frac{0.0300}{0.01} = 3.0000$$

$f''(1.1) \approx \mathbf{3.0000}$.

Comparison (1 mark): Exact $f'(1.1)=e^{1.1}=3.0042$. Absolute error $=|3.0090-3.0042| = 0.0048$, the central-difference truncation error of order $O(h^2)$. (The small discrepancy is consistent with rounded four-decimal table data.)

Euler's formula (1 mark): $y_{n+1} = y_n + h\,f(x_n,y_n)$ with $f(x,y)=-2y$ and $h=0.25$, so $y_{n+1}=y_n + 0.25(-2y_n) = y_n(1-0.5) = 0.5\,y_n$.

Iterations (3 marks):

$y(1.0) \approx \mathbf{0.06250}$.

Comparison (1 mark): Exact $y(1.0)=e^{-2}=0.13534$. Absolute error $=|0.13534-0.06250|=0.07284$, a large error: with this big step Euler's first-order method substantially underestimates the decay. Accuracy would improve markedly with a smaller $h$ or a higher-order method (e.g. RK4).

(a) Secant method (5 marks)

Formula: $x_{n+1} = x_n - f(x_n)\dfrac{x_n - x_{n-1}}{f(x_n)-f(x_{n-1})}$, with $f(x)=\cos x - x$.

$f(0.5)=\cos 0.5 - 0.5 = 0.877583 - 0.5 = 0.377583$ $f(1.0)=\cos 1.0 - 1.0 = 0.540302 - 1.0 = -0.459698$

Iteration: $x_2=0.725482$; $x_3=0.738399$; $x_4=0.739087$. Successive values $0.7384$ and $0.7391$ agree to three decimals.

Root $\approx \mathbf{0.739}$ rad.

(b) Least-squares straight line (5 marks)

With $n=5$, build the sums:

Normal equations:

$$\Sigma y = na + b\,\Sigma x \;\Rightarrow\; 29 = 5a + 15b$$ $$\Sigma xy = a\,\Sigma x + b\,\Sigma x^2 \;\Rightarrow\; 102 = 15a + 55b$$

Solve: multiply the first by 3: $87 = 15a + 45b$. Subtract from the second: $102-87 = (55-45)b \Rightarrow 15 = 10b \Rightarrow b = 1.5$.

Then $5a = 29 - 15(1.5) = 29 - 22.5 = 6.5 \Rightarrow a = 1.3$.

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

At $x=6$: $y = 1.3 + 1.5(6) = 1.3 + 9.0 = \mathbf{10.3}$.