Numerical Method Practice Questions

Simpson's 1/3rd rule is a numerical method for approximating the value of a definite integral. The distance covered is the integral of the velocity function over the time interval [0, 20]. The rule approximates this integral using the formula:
$\int_{a}^{b} f(x) dx \approx \frac{h}{3} [y_0 + 4(y_1+y_3+...) + 2(y_2+y_4+...) + y_n]$
Here, the time interval $h=2$ minutes. Applying this formula to the provided table of velocity values results in an approximate distance of 309.33 kilometers.

The secant method is an iterative root-finding algorithm that approximates the derivative used in Newton's method. It uses a line (the secant line) through two points on the function's curve to estimate the root. The formula for the next estimate, $x_t$, based on the previous two estimates, $x_a$ and $x_b$, is derived from the x-intercept of this secant line. The equation is $x_t = x_b - f(x_b) \frac{x_b - x_a}{f(x_b) - f(x_a)}$. This expression matches the one provided in option C.

To determine the truthfulness of the statements, we apply the Newton-Raphson method.
The function is $f(x) = 0.75x^3 - 2x^2 - 2x + 4$.
Its derivative is $f'(x) = 2.25x^2 - 4x - 2$.
The Newton-Raphson formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

Starting with $x_0 = 2$:
$f(x_0) = f(2) = 0.75(2)^3 - 2(2)^2 - 2(2) + 4 = 6 - 8 - 4 + 4 = -2$.
$f'(x_0) = f'(2) = 2.25(2)^2 - 4(2) - 2 = 9 - 8 - 2 = -1$.
So, $x_1 = 2 - \frac{-2}{-1} = 2 - 2 = 0$.

Next, for $x_1 = 0$:
$f(x_1) = f(0) = 0.75(0)^3 - 2(0)^2 - 2(0) + 4 = 4$.
$f'(x_1) = f'(0) = 2.25(0)^2 - 4(0) - 2 = -2$.
So, $x_2 = 0 - \frac{4}{-2} = 0 - (-2) = 2$.

Next, for $x_2 = 2$:
$f(x_2) = f(2) = -2$.
$f'(x_2) = f'(2) = -1$.
So, $x_3 = 2 - \frac{-2}{-1} = 2 - 2 = 0$.

From the calculations, $x_3 = 0$. So, statement (I) is TRUE.
The sequence obtained is $2, 0, 2, 0, ...$, which means the method oscillates and does not converge to a solution. Therefore, statement (II) is FALSE.

The function $f(x) = x^2$ is a convex function, meaning its graph curves upwards.
For a convex function, the trapezoidal rule, which approximates the area under the curve using trapezoids, will always overestimate or be equal to the exact definite integral. Thus, statement (I) is TRUE.
Simpson's rule approximates the area using parabolic segments. Since $f(x) = x^2$ is a polynomial of degree 2, Simpson's rule provides an exact value for its integral. Thus, statement (II) is TRUE.

To compute the definite integral $\int_{0}^{3}f(x)dx$ using the trapezoidal rule, we use the formula:
$\int_{a}^{b}f(x)dx \approx \frac{h}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)]$
From the given data, the interval is $h = 0.3 - 0 = 0.3$.
The function values are $f(x_0)=0, f(x_1)=0.09, f(x_2)=0.36, f(x_3)=0.81, f(x_4)=1.44, f(x_5)=2.25, f(x_6)=3.24, f(x_7)=4.41, f(x_8)=5.76, f(x_9)=7.29, f(x_{10})=9.00$.
Substitute these values into the trapezoidal rule formula:
$\int_{0}^{3}f(x)dx \approx \frac{0.3}{2} [0 + 2(0.09 + 0.36 + 0.81 + 1.44 + 2.25 + 3.24 + 4.41 + 5.76 + 7.29) + 9.00]$
Summing the intermediate $f(x)$ values: $0.09 + 0.36 + \dots + 7.29 = 25.65$.
So, the calculation becomes:
$\int_{0}^{3}f(x)dx \approx 0.15 [0 + 2(25.65) + 9.00] = 0.15 [0 + 51.30 + 9.00] = 0.15 [60.30]$
$\int_{0}^{3}f(x)dx \approx 9.045$

To find the number of iterations for the bisection method to converge to a solution, we evaluate the function at the endpoints and the midpoint of the interval.

First, calculate $f(x)$ at the initial interval endpoints:
$f(1) = 1^4 - 1^3 - 1^2 - 4 = 1 - 1 - 1 - 4 = -5$
$f(9) = 9^4 - 9^3 - 9^2 - 4 = 6561 - 729 - 81 - 4 = 5747$
Since $f(1) < 0$ and $f(9) > 0$, a root exists in $[1, 9]$.

Iteration 1:
Midpoint $x_0 = (1+9)/2 = 5$.
$f(5) = 5^4 - 5^3 - 5^2 - 4 = 625 - 125 - 25 - 4 = 471$.
Since $f(5) > 0$, the root is in $[1, 5]$.

Iteration 2:
Midpoint $x_1 = (1+5)/2 = 3$.
$f(3) = 3^4 - 3^3 - 3^2 - 4 = 81 - 27 - 9 - 4 = 41$.
Since $f(3) > 0$, the root is in $[1, 3]$.

Iteration 3:
Midpoint $x_2 = (1+3)/2 = 2$.
$f(2) = 2^4 - 2^3 - 2^2 - 4 = 16 - 8 - 4 - 4 = 0$.
Since $f(2) = 0$, we have found the exact root. The method converges after 3 iterations.

The Newton-Raphson method is an iterative technique for finding the roots of a real-valued function, $f(x)$. The formula for the next approximation, $x_{n+1}$, based on the current approximation, $x_n$, is:

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

In this problem, we need to find a root for the equation $x^2 - 13 = 0$. So, our function is $f(x) = x^2 - 13$.

The first derivative of this function is $f'(x) = 2x$.

We are given the initial value $x_0 = 3.5$. We need to find the approximation after one iteration, which is $x_1$.

First, let's calculate the values of the function and its derivative at $x_0 = 3.5$:
$f(3.5) = (3.5)^2 - 13 = 12.25 - 13 = -0.75$
$f'(3.5) = 2 \times 3.5 = 7$

Now, substitute these values into the Newton-Raphson formula:
$x_1 = 3.5 - \frac{-0.75}{7}$
$x_1 = 3.5 + \frac{0.75}{7}$
$x_1 \approx 3.5 + 0.10714$
$x_1 \approx 3.60714$

Thus, the approximation after one iteration is approximately 3.607.

Let the series converge to a limit $L$. Then, as $n \to \infty$, $x_n \to L$ and $x_{n+1} \to L$.
Substitute $L$ into the given recurrence relation:
$L = \frac{L}{2} + \frac{9}{8L}$
Subtract $\frac{L}{2}$ from both sides:
$\frac{L}{2} = \frac{9}{8L}$
Multiply both sides by $2L$:
$L^2 = \frac{9 \times 2}{8}$
$L^2 = \frac{18}{8}$
$L^2 = \frac{9}{4}$
Taking the square root of both sides:
$L = \pm\sqrt{\frac{9}{4}}$
$L = \pm\frac{3}{2}$
$L = \pm 1.5$
Since $x_0 = 0.5 > 0$ and all terms in the sequence will be positive, the series converges to the positive limit.
Therefore, $L = 1.5$.