Answers

1. Graphical Method

Given

$$f(x)=x^3-2x-5$$

a) Equation is nonlinear because highest power of $x$ is $3$.

b) Sign-change intervals:

$$f(2) = 8-4-5 = -1$$ $$f(3) = 27-6-5 = 16$$

So root lies in $[2,3]$.

Again,

$$f(2.09) = 2.09^3-2(2.09)-5 \approx -0.050671$$ $$f(2.10) = 2.10^3-2(2.10)-5 \approx 0.061$$

So narrower interval is $[2.09,2.10]$.

c) Graph of cubic crosses $x$-axis near

$$x \approx 2.09$$

2. Bisection Method

Given

$$f(x)=x^3-7$$

a)

$$f(1)=1-7=-6,\qquad f(2)=8-7=1$$

Since $f(1)f(2)<0$, root exists in $[1,2]$.

b) Interval is suitable because $f(x)$ is continuous and signs at endpoints are opposite.

c) Bisection table with $\mathrm{TOL}=0.005$:

Final interval:

$$[1.90625,\ 1.9140625]$$

Approximate root:

$$x \approx \frac{1.90625+1.9140625}{2}=1.91015625 \approx 1.910$$

3. Fixed Point Iteration Method

Given

$$x^3+x-6=0$$

a) Two possible fixed-point forms:

$$x=\sqrt[3]{6-x}$$ $$x=6-x^3$$

Use

$$x_{n+1}=\sqrt[3]{6-x_n}$$

because it converges near root.

b) Suitable initial guess:

$$x_0=1.5$$

c) Iteration table:

Since error $<0.005$, stop.

Approximate root:

$$x \approx 1.6346$$

4. Secant Method

Given

$$f(x)=x^3-4x-9$$

a)

$$f(2)=8-8-9=-9$$ $$f(3)=27-12-9=6$$

So root lies in $(2,3)$.

b) Choose

$$x_0=2,\qquad x_1=3$$

c) Secant formula:

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

Five iterations:

Approximate root:

$$x \approx 2.706528$$

5. Newton-Raphson Method

Given

$$f(x)=x^3-2x^2+3$$

a) Stopping criterion:

$$|x_{n+1}-x_n|<\varepsilon \quad \text{or} \quad |f(x_{n+1})|<\varepsilon$$

b)

$$f'(x)=3x^2-4x$$

Newton-Raphson formula:

$$x_{n+1}=x_n-\frac{x_n^3-2x_n^2+3}{3x_n^2-4x_n}$$

c) Take $x_0=-1.5$.

Root after six iterations:

$$x \approx -1$$

6. Gaussian Elimination Method

System:

$$3x+y-z=4$$ $$2x-4y+5z=-3$$ $$x+2y+3z=10$$

a) Augmented matrix:

$$\left[\begin{array}{ccc|c} 3 & 1 & -1 & 4 \\ 2 & -4 & 5 & -3 \\ 1 & 2 & 3 & 10 \end{array}\right]$$

b) Upper triangular form:

$$R_2 \leftarrow R_2-\frac23R_1,\qquad R_3 \leftarrow R_3-\frac13R_1$$ $$\left[\begin{array}{ccc|c} 3 & 1 & -1 & 4 \\ 0 & -\frac{14}{3} & \frac{17}{3} & -\frac{17}{3} \\ 0 & \frac{5}{3} & \frac{10}{3} & \frac{26}{3} \end{array}\right]$$ $$R_3 \leftarrow R_3+\frac{5}{14}R_2$$ $$\left[\begin{array}{ccc|c} 3 & 1 & -1 & 4 \\ 0 & -\frac{14}{3} & \frac{17}{3} & -\frac{17}{3} \\ 0 & 0 & \frac{75}{14} & \frac{93}{14} \end{array}\right]$$

c) Back substitution:

$$z=\frac{93/14}{75/14}=\frac{31}{25}=1.24$$ $$y=\frac{68}{25}=2.72$$ $$x=\frac{21}{25}=0.84$$

So

$$x=0.84,\qquad y=2.72,\qquad z=1.24$$

7. Gaussian Elimination with Partial Pivoting

System:

$$0.2x+3y+z=8$$ $$4x+y-z=7$$ $$2x+5y+2z=10$$

a) Partial pivoting is required because first pivot $0.2$ is very small. Dividing by it may cause large round-off error.

b) Row interchange:

$$R_1 \leftrightarrow R_2$$

Largest absolute value in first column becomes pivot.

c) Start:

$$\left[\begin{array}{ccc|c} 0.2 & 3 & 1 & 8 \\ 4 & 1 & -1 & 7 \\ 2 & 5 & 2 & 10 \end{array}\right] \xrightarrow{R_1 \leftrightarrow R_2} \left[\begin{array}{ccc|c} 4 & 1 & -1 & 7 \\ 0.2 & 3 & 1 & 8 \\ 2 & 5 & 2 & 10 \end{array}\right]$$

Eliminate first column:

$$R_2 \leftarrow R_2-\frac{1}{20}R_1,\qquad R_3 \leftarrow R_3-\frac12R_1$$ $$\left[\begin{array}{ccc|c} 4 & 1 & -1 & 7 \\ 0 & \frac{59}{20} & \frac{21}{20} & \frac{153}{20} \\ 0 & \frac{9}{2} & \frac{5}{2} & \frac{13}{2} \end{array}\right]$$

Second pivot: $\left|\frac92\right|>\left|\frac{59}{20}\right|$, so swap

$$R_2 \leftrightarrow R_3$$

Then

$$R_3 \leftarrow R_3-\frac{59}{90}R_2$$ $$\left[\begin{array}{ccc|c} 4 & 1 & -1 & 7 \\ 0 & \frac{9}{2} & \frac{5}{2} & \frac{13}{2} \\ 0 & 0 & -\frac{53}{90} & \frac{61}{18} \end{array}\right]$$

Back substitution gives

$$z=-\frac{305}{53}\approx -5.7547$$ $$y=\frac{246}{53}\approx 4.6415$$ $$x=-\frac{45}{53}\approx -0.8491$$

8. Gauss-Seidel Method

Given

$$10x+y-z=11$$ $$2x+10y+z=12$$ $$-2x+3y+10z=14$$

a) Iterative form:

$$x=\frac{11-y+z}{10}$$ $$y=\frac{12-2x-z}{10}$$ $$z=\frac{14+2x-3y}{10}$$

b) Diagonal dominance:

$$|10|>|1|+|{-1}|=2$$ $$|10|>|2|+|1|=3$$ $$|10|>|{-2}|+|3|=5$$

So system is diagonally dominant. Gauss-Seidel will converge.

c) With $x_0=y_0=z_0=0$:

After four iterations:

$$x \approx 1.15493,\qquad y \approx 0.83090,\qquad z \approx 1.38172$$

---

IUS Preps - Your Academic Success Partner