Logarithms Questions and Answers

Answer: C

An interval containing one endpoint but not the other is called half open - equivalently half closed, since it is open at one end and closed at the other.

\[ [a,b) \quad\text{and}\quad (a,b] \]

The two names describe the same thing from opposite viewpoints.

Answer: A

In \( F(x, y, y', \ldots, y^{(n)}) = 0 \), the variable \( x \) is independent and \( y \) is the unknown function of \( x \).

\[ y = y(x), \qquad y' = \frac{dy}{dx} \]

The primes are derivatives with respect to \( x \), which is what fixes \( x \) as the independent variable.

Answer: C

Growth proportional to the amount present gives exponential growth, and doubling every week means the population multiplies by 2 each week.

\[ A(4) = A_{0} \times 2^{4} = 16\,A_{0} \]

The doublings compound rather than add: four weeks give \( 2 \times 2 \times 2 \times 2 = 16 \) times, not \( 2 \times 4 = 8 \) times.

Answer: C

An interval that includes one endpoint but not the other is half closed (equally, half open).

\[ [a,b) = \{\, x \mid a \leq x < b \,\} \]

A closed interval includes both ends and an open one neither.

Answer: C

The parenthesis at \( b \) excludes that endpoint, so the interval is open; it extends to \( +\infty \) with no upper endpoint, so it is unbounded.

\[ (b, +\infty) = \{\, x \mid x > b \,\} \]

Hence open and unbounded. Infinity is never an included endpoint, so that side always carries a parenthesis.

Answer: C

The whole point of the transform is that the subsidiary equation is algebraic: derivatives in \( t \) have become powers of \( s \).

\[ (s^{2} + as + b)Y(s) = G(s) \quad\Longrightarrow\quad Y(s) = \frac{G(s)}{s^{2}+as+b} \]

Solving it needs only algebraic manipulation - no integration or differentiation at all. The calculus returns only in the inverse transform.

Answer: A

A variable is a quantity that can change during an experiment - the thing being varied or measured.

A function is a rule relating variables, and an equation a statement of equality between expressions; neither is the changing quantity itself.

Answer: B

Write the hyperbolic sine in exponential form and use the shift theorem.

\[ \sinh t \cos t = \frac{1}{2}\left(e^{t} - e^{-t}\right)\cos t \]

With \( \mathcal{L}\{e^{at}\cos t\} = \dfrac{s-a}{(s-a)^{2}+1} \):

\[ \mathcal{L} = \frac{1}{2}\left[\frac{s-1}{s^{2}-2s+2} - \frac{s+1}{s^{2}+2s+2}\right] \]

Over the common denominator \( (s^{2}-2s+2)(s^{2}+2s+2) = s^{4}+4 \):

\[ \text{numerator} = (s-1)(s^{2}+2s+2) - (s+1)(s^{2}-2s+2) = 2s^{2} - 4 \]

\[ \mathcal{L}\{\sinh t \cos t\} = \frac{s^{2}-2}{(s^{2}+2s+2)(s^{2}-2s+2)} \]

As a check, \( \sinh t \cos t \approx t \) for small \( t \), and the expression behaves like \( \tfrac{1}{s^{2}} \) for large \( s \) - which is \( \mathcal{L}\{t\} \).

Answer: A

Faà di Bruno's formula generalises the chain rule to higher derivatives, giving \( \tfrac{d^{n}}{dx^{n}}f(g(x)) \) as a sum over the partitions of \( n \).

The first case is the ordinary chain rule; the second is

\[ (f \circ g)'' = f''(g)\,(g')^{2} + f'(g)\,g'' \]

Leibniz's formula is the analogous generalisation for the product rule, not the chain rule.