Computational Methods: 25 MCQs with answers

$f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$ implies there exists $c \in (a,b)$ such that $f'(c)=0$.

Rolle's theorem states that if a function is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists at least one point $c$ in $(a,b)$ where the derivative is zero.

Secant method

The secant method is specifically mentioned in the problem for finding the root of $x^3 - 2x - 5 = 0$.

Choosing two initial guesses

Newton-Raphson requires only one initial guess, not two. The secant method requires two initial guesses.

The first derivative of $f$ at $x$.

The forward difference $\Delta f(x) = f(x+h) - f(x)$ approximates the first derivative when divided by $h$.

True

Rolle's theorem is indeed a special case of the Mean Value Theorem where $f(a)=f(b)$.

To approximate solutions to mathematical problems

Numerical methods are designed to approximate solutions to mathematical problems that may not have exact solutions.

Estimating values between known data points

Interpolation is the process of estimating unknown values that fall within the range of known data points.

$\Delta f(x) = f(x) - f(x-h)$

The backward difference is defined as $\Delta f(x) = f(x) - f(x-h)$, representing the change in function value over a backward step.

It approximates the area using trapezoids

The trapezoidal rule approximates the area under a curve by dividing it into trapezoids rather than rectangles.

To construct a polynomial that passes through given points

The Lagrange interpolation formula is used to construct a polynomial that exactly passes through a given set of points.

Forward uses future points, backward uses past points

The forward difference uses future points ($f(x+h)$), while the backward difference uses past points ($f(x-h)$).

Simulating physical systems

Numerical methods are widely used to simulate physical systems where exact solutions are difficult or impossible to obtain.

Numerical error

Numerical error refers to the difference between the true value and the approximated value obtained through numerical methods.

The values of the function at specific points

In interpolation, 'nodes' refer to the specific points at which the function values are known and used to estimate other values.

There exists $c \in (0, \pi)$ such that $f'(c) = 0$.

Since $f(0) = \frac{\sin 0}{e^0} = 0$ and $f(\pi) = \frac{\sin \pi}{e^{\pi}} = 0$, and $f$ is continuous and differentiable on the interval, Rolle's theorem guarantees a point $c$ where $f'(c)=0$.

—

The third forward difference of a cubic polynomial is constant and equals $6 \times$ leading coefficient $\times h^3$. Here leading coefficient is $-6$, so $\Delta^3 = -36$.

$2.884$

Solving $x^3 - 24 = 0$ by Newton-Raphson gives $x \approx 2.884$ after a few iterations.

$0.7833$

Applying Simpson's 1/3 rule with $h=0.5$ gives $\frac{0.5}{3}[f(0)+4f(0.5)+f(1)] = \frac{0.5}{3}[1+4(0.8)+0.5] = 0.7833$.

$21$

Using interpolation on the cumulative frequency, the number of students between 40 and 45 is approximately 21.

It is more accurate for smooth functions.

Simpson's 1/3 rule uses quadratic interpolation and is generally more accurate than the trapezoidal rule for smooth functions.

False

Newton-Raphson may fail to converge if the initial guess is far from the root or if the derivative is zero or near zero.

$27$

The minimum occurs at $x=3$, giving $f(3) = 9 + 18 = 27$.

$-10.5$

The minimum occurs at $x \approx 2.5$, giving $f(2.5) = 0.25(39.0625) - 6.25 - 12.5 + 1 = -10.5$.

$\frac{5}{2}$

The partial fraction decomposition gives $\frac{5/2}{x-1} + \frac{-15}{x-2} + \frac{31/2}{x-3}$, so the coefficient of $\frac{1}{x-1}$ is $\frac{5}{2}$.

$0.693$

Romberg's method applied to $\int_0^1 \frac{dx}{1+x}$ gives $\ln 2 \approx 0.693$.