BE Civil Engineering (IOE, TU) Numerical Methods (IOE, SH 553) Question Paper 2076 Nepal

(a) Bracketing the root

Evaluate $f(x)=x^3-6x^2+11x-6.1$ at the interval ends:

$$f(1.0) = 1 - 6 + 11 - 6.1 = -0.1 < 0$$ $$f(1.5) = 3.375 - 13.5 + 16.5 - 6.1 = 0.275 > 0$$

Since $f(1.0)\cdot f(1.5) < 0$ and $f$ is continuous, a real root lies in $[1.0,1.5]$.

(b) Newton–Raphson iteration

The derivative is $f'(x) = 3x^2 - 12x + 11$. The iteration formula is

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$

Start with $x_0 = 1.0$, where $f'(1.0) = 3 - 12 + 11 = 2$.

Working for $n=0$: $x_1 = 1.0 - \dfrac{-0.1}{2} = 1.05$. Working for $n=1$: $f(1.05)=1.157625-6.615+11.55-6.1=0.001625$, $f'(1.05)=3.3075-12.6+11=1.7075$, so $x_2 = 1.05 - \dfrac{0.001625}{1.7075} = 1.049048$. At $n=2$ the value repeats to six decimals, so convergence is achieved.

Root (4 d.p.): $\boxed{x \approx 1.0490\ \text{m}}$

(c) Failure cases of Newton–Raphson

1. When $f'(x_n)$ is zero or very near zero, the correction term blows up (the tangent is nearly horizontal).
2. When the initial guess is far from the root or near a turning point, successive iterates may oscillate or diverge.

(a) Divided-difference table

First differences:

$$f[1,2]=\frac{7-3}{2-1}=4,\quad f[2,4]=\frac{21-7}{4-2}=7,\quad f[4,7]=\frac{60-21}{7-4}=13.$$

Second differences:

$$f[1,2,4]=\frac{7-4}{4-1}=1,\quad f[2,4,7]=\frac{13-7}{7-2}=1.2.$$

Third difference:

$$f[1,2,4,7]=\frac{1.2-1}{7-1}=\frac{0.2}{6}=0.033333.$$

(b) Newton's divided-difference polynomial

$$P(x)=3 + 4(x-1) + 1(x-1)(x-2) + 0.033333(x-1)(x-2)(x-4).$$

Estimate at $x=5$:

• $(x-1)=4,\ (x-2)=3,\ (x-4)=1$
• Term 0: $3$
• Term 1: $4\times4 = 16$
• Term 2: $1\times 4\times 3 = 12$
• Term 3: $0.033333\times 4\times 3\times 1 = 0.4$

$$P(5) = 3 + 16 + 12 + 0.4 = 31.4\ \text{mm}.$$

Estimated settlement at $x=5$ months: $\boxed{31.4\ \text{mm}}$

(c) Advantage: When a new data point is added, Newton's divided-difference form only requires computing one additional bottom-edge difference and term; the existing polynomial is reused, whereas the Lagrange form must be rebuilt entirely.

(a) Diagonal dominance

For each row, $|a_{ii}| \ge \sum_{j\ne i}|a_{ij}|$:

• Row 1: $|10| = 10 > |2|+|1| = 3$ ✓
• Row 2: $|8| = 8 > |1|+|2| = 3$ ✓
• Row 3: $|10| = 10 > |2|+|1| = 3$ ✓

The coefficient matrix is strictly diagonally dominant, so Gauss–Seidel converges for any starting vector.

(b) Gauss–Seidel iteration

Rearranged update equations:

$$x_1 = \frac{9 - 2x_2 - x_3}{10},\quad x_2 = \frac{19 - x_1 - 2x_3}{8},\quad x_3 = \frac{34 - 2x_1 - x_2}{10}.$$

Iteration 1 (start $0,0,0$):

• $x_1 = (9-0-0)/10 = 0.9000$
• $x_2 = (19-0.9-0)/8 = 18.1/8 = 2.2625$
• $x_3 = (34-2(0.9)-2.2625)/10 = (34-1.8-2.2625)/10 = 29.9375/10 = 2.9938$

Iteration 2:

• $x_1 = (9-2(2.2625)-2.9938)/10 = (9-4.525-2.9938)/10 = 1.4812/10 = 0.1481$
• $x_2 = (19-0.1481-2(2.9938)/1)/8 = (19-0.1481-5.9875)/8 = 12.8644/8 = 1.6080$
• $x_3 = (34-2(0.1481)-1.6080)/10 = (34-0.2962-1.6080)/10 = 32.0957/10 = 3.2096$

Iteration 3:

• $x_1 = (9-2(1.6080)-3.2096)/10 = (9-3.2161-3.2096)/10 = 2.5743/10 = 0.2574$
• $x_2 = (19-0.2574-2(3.2096)/1)/8 = (19-0.2574-6.4191)/8 = 12.3235/8 = 1.5404$
• $x_3 = (34-2(0.2574)-1.5404)/10 = (34-0.5149-1.5404)/10 = 31.9447/10 = 3.1945$

The iterates are approaching the true solution $x_1\approx0.25,\ x_2\approx1.55,\ x_3\approx3.19$ mm.

Approximate displacements after 3 iterations: $\boxed{x_1\approx0.2574,\ x_2\approx1.5404,\ x_3\approx3.1945\ \text{mm}}$

(c) Gauss–Seidel uses the most recently updated values within the same iteration, so it generally converges faster than the Jacobi method, which uses only previous-iteration values.

(a) Simpson's 1/3 rule

There are $n=8$ intervals (even number — applicable), with step $h=1$ m. Label ordinates $y_0,\dots,y_8$.

$$A = \frac{h}{3}\Big[ (y_0+y_8) + 4(y_1+y_3+y_5+y_7) + 2(y_2+y_4+y_6) \Big].$$

• Ends: $y_0+y_8 = 0+0 = 0$
• Odd ordinates: $y_1+y_3+y_5+y_7 = 4+9+11+5 = 29$, times 4 $= 116$
• Even interior: $y_2+y_4+y_6 = 7+12+8 = 27$, times 2 $= 54$

$$A = \frac{1}{3}\big[0 + 116 + 54\big] = \frac{170}{3} = 56.6667\ \text{m}^2.$$

Area (Simpson's 1/3): $\boxed{56.667\ \text{m}^2}$

(b) Trapezoidal rule

$$A = \frac{h}{2}\Big[(y_0+y_8) + 2(y_1+y_2+\dots+y_7)\Big].$$

Interior sum: $4+7+9+12+11+8+5 = 56$.

$$A = \frac{1}{2}\big[0 + 2(56)\big] = \frac{1}{2}(112) = 56.000\ \text{m}^2.$$

Area (Trapezoidal): $\boxed{56.000\ \text{m}^2}$

Comparison: The two estimates differ by $0.667\ \text{m}^2$. Simpson's rule fits parabolic arcs through triples of points and is more accurate for smoothly curving boundaries, while the trapezoidal rule uses straight chords and tends to underestimate area on a convex (bulging) boundary such as this.

(c) Condition: Simpson's 1/3 rule requires the number of intervals $n$ to be even (equivalently, an odd number of ordinates), with equal spacing $h$. Here $n=8$ is even, so the rule applies directly.

(a) RK4 method

Let $f(t,T) = -0.5(T-25)$, $h=0.5$. The RK4 update is

$$T_{n+1} = T_n + \tfrac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4),$$

where

$$k_1=h f(t_n,T_n),\ k_2=h f(t_n+\tfrac{h}{2},T_n+\tfrac{k_1}{2}),\ k_3=h f(t_n+\tfrac{h}{2},T_n+\tfrac{k_2}{2}),\ k_4=h f(t_n+h,T_n+k_3).$$

Step 1: $t_0=0,\ T_0=85$.

• $k_1 = 0.5\cdot(-0.5)(85-25) = 0.5\cdot(-30) = -15.0000$
• $k_2 = 0.5\cdot(-0.5)(85 + (-15/2) - 25) = 0.5\cdot(-0.5)(52.5) = 0.5\cdot(-26.25) = -13.1250$
• $k_3 = 0.5\cdot(-0.5)(85 + (-13.125/2) - 25) = 0.5\cdot(-0.5)(53.4375) = 0.5\cdot(-26.71875) = -13.3594$
• $k_4 = 0.5\cdot(-0.5)(85 + (-13.3594) - 25) = 0.5\cdot(-0.5)(46.6406) = 0.5\cdot(-23.3203) = -11.6602$

$$T_1 = 85 + \tfrac{1}{6}\big[-15 + 2(-13.125) + 2(-13.3594) + (-11.6602)\big]$$ $$= 85 + \tfrac{1}{6}(-79.7539) = 85 - 13.2923 = 71.7077\,^{\circ}\text{C}\ \text{at}\ t=0.5.$$

Step 2: $t_1=0.5,\ T_1=71.7077$.

• $k_1 = 0.5\cdot(-0.5)(71.7077-25) = 0.5\cdot(-0.5)(46.7077) = -11.6769$
• $k_2 = 0.5\cdot(-0.5)(71.7077 + (-11.6769/2) - 25) = 0.5\cdot(-0.5)(40.8692) = -10.2173$
• $k_3 = 0.5\cdot(-0.5)(71.7077 + (-10.2173/2) - 25) = 0.5\cdot(-0.5)(41.5990) = -10.3998$
• $k_4 = 0.5\cdot(-0.5)(71.7077 + (-10.3998) - 25) = 0.5\cdot(-0.5)(36.3079) = -9.0770$

$$T_2 = 71.7077 + \tfrac{1}{6}\big[-11.6769 + 2(-10.2173) + 2(-10.3998) + (-9.0770)\big]$$ $$= 71.7077 + \tfrac{1}{6}(-61.9882) = 71.7077 - 10.3314 = 61.3763\,^{\circ}\text{C}.$$

RK4 estimate at $t=1.0$ h: $\boxed{T \approx 61.376\,^{\circ}\text{C}}$

(b) Exact value and error

$$T_{\text{exact}}(1.0) = 25 + 60e^{-0.5} = 25 + 60(0.606531) = 25 + 36.3919 = 61.3919\,^{\circ}\text{C}.$$

Absolute error:

$$|61.3919 - 61.3763| = 0.0156\,^{\circ}\text{C}.$$

Absolute error: $\boxed{\approx 0.016\,^{\circ}\text{C}}$ — RK4 is highly accurate even with a relatively large step.

Bisection method on $f(x)=x^2-5$, interval $[2,3]$

Check: $f(2)=4-5=-1<0$, $f(3)=9-5=4>0$ — root bracketed.

Working:

• Iter 1: $f(2.5)=6.25-5=1.25>0$, so root in $[2,2.5]$.
• Iter 2: $f(2.25)=5.0625-5=0.0625>0$, so root in $[2,2.25]$.
• Iter 3: $f(2.125)=4.515625-5=-0.484375<0$, so root in $[2.125,2.25]$.

Approximate root (next midpoint): $\dfrac{2.125+2.25}{2} = 2.1875$.

Error bound after 3 iterations: $\dfrac{b-a}{2^{3}} = \dfrac{3-2}{8} = 0.125$.

$$\boxed{x \approx 2.1875,\quad |\text{error}| \le 0.125}$$

(true root $\sqrt5 = 2.2361$).

Forward-difference table

First differences: $18-10=8,\ 30-18=12,\ 46-30=16$. Second differences: $12-8=4,\ 16-12=4$. Third difference: $4-4=0$.

Newton's forward formula with $h=1$, $p=\dfrac{t-t_0}{h}=\dfrac{0.5-0}{1}=0.5$:

$$Q(t)=Q_0 + p\,\Delta Q_0 + \frac{p(p-1)}{2!}\Delta^2 Q_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 Q_0.$$

Substitute $Q_0=10,\ \Delta Q_0=8,\ \Delta^2 Q_0=4,\ \Delta^3 Q_0=0$:

• Term 1: $10$
• Term 2: $0.5\times 8 = 4$
• Term 3: $\dfrac{0.5(0.5-1)}{2}\times 4 = \dfrac{0.5(-0.5)}{2}\times 4 = (-0.125)\times 4 = -0.5$
• Term 4: $0$

$$Q(0.5) = 10 + 4 - 0.5 + 0 = 13.5\ \text{m}^3/\text{s}.$$

Estimated discharge at $t=0.5$ h: $\boxed{13.5\ \text{m}^3/\text{s}}$

Velocity at $t=0.1$ s (central difference)

$$v \approx \frac{s(0.2)-s(0.0)}{2\,\Delta t} = \frac{0.28 - 0.00}{2(0.1)} = \frac{0.28}{0.2} = 1.40\ \text{m/s}.$$

Acceleration at $t=0.1$ s (central second difference)

$$a \approx \frac{s(0.2) - 2s(0.1) + s(0.0)}{(\Delta t)^2} = \frac{0.28 - 2(0.12) + 0.00}{(0.1)^2} = \frac{0.28 - 0.24}{0.01} = \frac{0.04}{0.01} = 4.0\ \text{m/s}^2.$$

Results: $\boxed{v(0.1) = 1.40\ \text{m/s},\quad a(0.1) = 4.0\ \text{m/s}^2}$

Modified Euler (Heun's) method

For $f(x,y)=x+y$, each step uses a predictor and one corrector:

$$y^{*}_{n+1} = y_n + h\,f(x_n,y_n)\quad(\text{predictor}),$$ $$y_{n+1} = y_n + \frac{h}{2}\big[f(x_n,y_n) + f(x_{n+1}, y^{*}_{n+1})\big]\quad(\text{corrector}).$$

Step 1: $x_0=0,\ y_0=1,\ h=0.1$.

• $f(x_0,y_0)=0+1=1$
• Predictor: $y^{*}_1 = 1 + 0.1(1) = 1.1$
• $f(x_1,y^{*}_1)=0.1+1.1=1.2$
• 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: $y^{*}_2 = 1.11 + 0.1(1.21) = 1.11 + 0.121 = 1.231$
• $f(x_2,y^{*}_2)=0.2+1.231=1.431$
• Corrector: $y_2 = 1.11 + \dfrac{0.1}{2}(1.21 + 1.431) = 1.11 + 0.05(2.641) = 1.11 + 0.13205 = 1.24205$

Estimate: $\boxed{y(0.2) \approx 1.2421}$ (Exact solution $y=2e^{x}-x-1$ gives $y(0.2)=2e^{0.2}-1.2=1.2428$, so the error is only $\approx 0.0008$.)

Five-point finite-difference (Laplace) formula

For a uniform grid with spacing $h$, the discretised Laplace equation at an interior node $(i,j)$ is

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

i.e. each interior value equals the average of its four neighbours.

Using symmetry

By the symmetry of the equations, $u_1 = u_2$ and $u_3 = u_4$. Let $u_1=u_2=A$ and $u_3=u_4=B$.

From equation 1: $A = \tfrac{1}{4}(100 + A + B)$ → $4A = 100 + A + B$ → $3A - B = 100.$ From equation 3: $B = \tfrac{1}{4}(A + B)$ → $4B = A + B$ → $3B - A = 0$ → $A = 3B.$

Substitute $A=3B$ into $3A - B = 100$:

$$3(3B) - B = 100 \Rightarrow 9B - B = 100 \Rightarrow 8B = 100 \Rightarrow B = 12.5.$$

Then $A = 3(12.5) = 37.5.$

Solution:

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

Check (equation 2): $\tfrac{1}{4}(100 + 37.5 + 12.5) = \tfrac{1}{4}(150) = 37.5$ ✓. (Equation 4): $\tfrac14(37.5+12.5)=\tfrac14(50)=12.5$ ✓.

Least-squares fit of $y = a + bx$

Normal equations:

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

Compute the sums ($n=5$):

Normal equations become:

$$82 = 5a + 15b \quad(1)$$ $$286 = 15a + 55b \quad(2)$$

From (1): $a = \dfrac{82 - 15b}{5}$. Multiply (1) by 3: $246 = 15a + 45b$. Subtract from (2):

$$286 - 246 = (55 - 45)b \Rightarrow 40 = 10b \Rightarrow b = 4.0.$$

Then $a = \dfrac{82 - 15(4)}{5} = \dfrac{82 - 60}{5} = \dfrac{22}{5} = 4.4.$

Best-fit line: $\boxed{y = 4.4 + 4.0x}$

Estimate at $x=7$ days:

$$y = 4.4 + 4.0(7) = 4.4 + 28.0 = 32.4\ \text{MPa}.$$

Predicted strength at 7 days: $\boxed{32.4\ \text{MPa}}$