16.13: Discrete-Time Birth-Death Chains
Basic Theory
Introduction
Suppose that \( S \) is an interval of integers (that is, a set of consecutive integers), either finite or infinite. A (discrete-time) birth-death chain on \( S \) is a discrete-time Markov chain \( \bs{X} = (X_0, X_1, X_2, \ldots) \) on \( S \) with transition probability matrix \( P \) of the form \[ P(x, x - 1) = q(x), \; P(x, x) = r(x), \; P(x, x + 1) = p(x); \quad x \in S \] where \( p \), \( q \), and \( r \) are nonnegative functions on \( S \) with \( p(x) + q(x) + r(x) = 1 \) for \( x \in S \).
If the interval \( S \) has a minimum value \( a \in \Z \) then of course we must have \( q(a) = 0 \). If \( r(a) = 1 \), the boundary point \( a \) is absorbing and if \( p(a) = 1 \), then \( a \) is reflecting . Similarly, if the interval \( S \) has a maximum value \( b \in \Z \) then of course we must have \( p(b) = 0 \). If \( r(b) = 1 \), the boundary point \( b \) is absorbing and if \( p(b) = 1 \), then \( b \) is reflecting . Several other special models that we have studied are birth-death chains; these are explored in below .
In this section, as you will see, we often have sums of products. Recall that a sum over an empty index set is 0, while a product over an empty index set is 1.
Recurrence and Transience
If \( S \) is finite, classification of the states of a birth-death chain as recurrent or transient is simple, and depends only on the state graph. In particular, if the chain is irreducible, then the chain is positive recurrent. So we will study the classification of birth-death chains when \( S = \N \). We assume that \( p(x) \gt 0 \) for all \( x \in \N \) and that \( q(x) \gt 0 \) for all \( x \in \N_+ \) (but of course we must have \( q(0) = 0 \)). Thus, the chain is irreducible.
Under these assumptions, the birth-death chain on \( \N \) is
- Aperiodic if \( r(x) \gt 0 \) for some \( x \in \N \).
- Periodic with period 2 if \( r(x) = 0 \) for all \( x \in \N \).
Proof
- If \( r(x) \gt 0 \) for some \( x \in \N \) then \( P(x, x) \gt 0 \) and hence the chain is aperiodic.
- If \( r(x) = 0 \) for every \( x \in \N \) then clearly the chain starting in \( x \) can be in state \( x \) again only at even times.
We will use the test for recurrence derived earlier with \( A = \N_+ \), the set of positive states. That is, we will compute the probability that the chain never hits 0, starting in a positive state.
The chain \( \bs{X} \) is recurrent if and only if \[ \sum_{x = 0}^\infty \frac{q(1) \cdots q(x)}{p(1) \cdots p(x)} = \infty \]
Proof
Let \( P_+ \) denote the restriction of \( P \) to \( \N_+ \times \N_+ \), and define \( u_+: \N_+ \to [0, 1] \) by \[ u_+(x) = \P(X_1 \gt 0, X_2 \gt 0, \ldots \mid X_0 = x), \quad x \in \N_+ \] So \( u_+(x) \) is the probability that chain never reaches 0, starting in \( x \in \N_+ \). From our general theory, we know that \( u_+ \) satisfies \( u_+ = P_+ u_+ \) and is the largest such function with values in \( [0, 1] \). Furthermore, we know that either \( u_+(x) = 0 \) for all \( x \in \N_+ \) or that \( \sup\{u_+(x): x \in [0, 1]\} = 1 \). In the first case the chain is recurrent, and in the second case the chain is transient.
The functional equation \( P_+ u = u \) for a function \( u: \N_+ \to [0, 1] \) is equivalent to the following system of equations: \begin{align} u(2) - u(1) & = \frac{q(1)}{p(1)} u(1) \\ u(x + 1) - u(x) & = \frac{q(x)}{p(x)}[u(x) - u(x - 1)], \quad x \in \{2, 3, \ldots\} \end{align} Solving this system of equations for the differences gives \[ u(x + 1) - u(x) = \frac{q(1) \cdots q(x)}{p(1) \cdots p(x)} u(1), \quad x \in \N_+ \] Solving this new systems gives \[ u(x) = u(1) \sum_{i=0}^{x-1} \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)}, \quad x \in \N_+ \] Note that \( u(x) \) is increasing in \( x \in \N_+ \) and so has a limit as \( x \to \infty \). Let \( A = \sum_{i=0}^\infty \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)} \).
- Suppose that \( A = \infty \). Letting \( x \to \infty \) in the displayed equation above for \( u(x) \) shows that \( u(1) = 0 \) and so \( u(x) = 0 \) for all \( x \). Hence the chain is recurrent.
- Suppose that \( A \lt \infty \). Define \( u(1) = 1/A \) and then more generally, \[ u(x) = \frac{1}{A} \sum_{i=0}^{x-1} \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)}, \quad x \in \N_+ \] The function \( u \) takes values in \( (0, 1) \) and satisfies the functional equation \( u = P_+ u \). Hence the chain is transient. Note that \( u(x) \to 1 \) as \( x \to \infty \) and so in fact, \( u = u_+ \), the function that we discussed above that gives the probability of staying in \( \N_+ \) for all time. We will return to this function below in our discussion of absorption .
Note that \( r \), the function that assigns to each state \( x \in \N \) the probability of an immediate return to \( x \), plays no direct role in whether the chain is transient or recurrent. Indeed all that matters are the ratios \( q(x) / p(x) \) for \( x \in \N_+ \).
Positive Recurrence and Invariant Distributions
Suppose again that we have a birth-death chain \( \bs{X} \) on \( \N \), with \( p(x) \gt 0 \) for all \( x \in \N \) and \( q(x) \gt 0 \) for all \( x \in \N_+\). Thus the chain is irreducible.
The function \( g: \N \to (0, \infty) \) defined by \[ g(x) = \frac{p(0) \cdots p(x - 1)}{q(1) \cdots q(x)}, \quad x \in \N \] is invariant for \( \bs{X} \), and is the only invariant function, up to multiplication by constants. Hence \( \bs{X} \) is positive recurrent if and only if \( B = \sum_{x = 0}^\infty g(x) \lt \infty \), in which case the (unique) invariant probability density function \( f \) is given by \( f(x) = \frac{1}{B} g(x) \) for \( x \in \N \).
Proof
Recall that by convention, a product over an empty index set is 1. So first, \begin{align*} (g P)(0) & = g(0) P(0, 0) + g(1) P(1, 0) = g(0) r(0) + g(1) q(1) \\ & = 1 r(0) + \frac{p(0)}{q(1)} q(1) = [1 - p(0)] + p(0) = 1 = g(0) \end{align*} Next, for \( y \in \N_+ \), \begin{align*} (g P)(y) & = g(y - 1) P(y - 1, y) + g(y) P(y, y) + g(y + 1) P(y + 1, y) \\ & = g(y - 1) p(y - 1) + g(y) r(y) + g(y + 1) q(y + 1) \\ & = g(y - 1) p(y - 1) + g(y) [1 - p(y) - q(y)] + g(y + 1) q(y + 1) \end{align*} But \begin{align*} g(y - 1) p(y - 1) & = g(y) q(y) = \frac{p(0) \cdots p(y - 1)}{q(1) \cdots q(y - 1)} \\ g(y + 1) q(y + 1) & = g(y) p(y) = \frac{p(0) \cdots p(y)}{q(1) \cdots q(y)} \end{align*} so \( (g P)(y) = g(y) \).
Conversely, suppose that \( h: \N \to \R \) is invariant for \( \bs{X} \). We will show by induction that \( h(x) = h(0) g(x) \) for all \( x \in \N \). The result is trivailly true for \( x = 0 \) since \( g(0) = 1 \). Next, \( (h P)(0) = h(0) \) gives \( h(0) P(0, 0) + h(1) P(1, 0) = h(0) \). But \( P(0, 0) = r(0) = [1 - p(0)] \) and \( P(1, 0) = q(1) \), so substituting and solving for \( h(1) \) gives \[ h(1) = h(0) \frac{p(0)}{q(1)} = h(0) g(1) \] so the result is true when \( x = 1 \). Assume now that \( y \in \N_+ \) and that the result is true for all \( x \in \N \) with \( x \le y \). Then \( (h P)(y) = h(y) \) gives \[ h(y - 1) P(y - 1, y) + h(y) P(y, y) + h(y + 1) P(y + 1, y) = h(y) \] But \( P(y - 1, y) = p(y - 1) \), \( P(y, y) = r(y) = 1 - p(y) - q(y) \), and \( P(y + 1, y) = q(y + 1) \). Also, by the induction hypothesis, \( h(y) = h(0) g(y) \) and \( h(y - 1) = h(0) g(y - 1) \) so substituting and using the definition of \( g \) gives \begin{align*} q(y + 1) h(y + 1) & = [p(y) + q(y)] h(0) \frac{p(0) \cdots p(y - 1)}{q(1) \cdots q(y)} - p(y - 1) h(0) \frac{p(0) \cdots p(y - 2)}{q(1) \cdots q(y - 1)} \\ & = h(0) \frac{p(0) \cdots p(y)}{q(1) \cdots q(y)} \end{align*} Finally, solving gives \[ h(y + 1) = h(0) \frac{p(0) \cdots p(y)}{q(1) \cdots q(y + 1)} = h(0) g(y + 1) \]
Here is a summary of the classification:
For the birth-death chain \( \bs X \), define \[A = \sum_{x = 0}^\infty \frac{q(1) \cdots q(x)}{p(1) \cdots p(x)}, \quad B = \sum_{x = 0}^\infty \frac{p(0) \cdots p(x - 1)}{q(1) \cdots q(x)}\]
- \( \bs X \) is transient if \( A \lt \infty \)
- \( \bs X \) is null recurrent if \( A = \infty \) and \( B = \infty \).
- \( \bs X \) is positive recurrent if \( B \lt \infty \).
Note again that \( r \), the function that assigns to each state \( x \in \N \) the probability of an immediate return to \( x \), plays no direct role in whether the chain is transient, null recurrent, or positive recurrent. Also, we know that an irreducible, recurrent chain has a positive invariant function that is unique up to multiplication by positive constants, but the birth-death chain gives an example where this is also true in the transient case.
Suppose now that \( n \in \N_+ \) and that \( \bs X = (X_0, X_1, X_2, \ldots) \) is a birth-death chain on the integer interval \( \N_n = \{0, 1, \ldots, n\} \). We assume that \( p(x) \gt 0 \) for \( x \in \{0, 1, \ldots, n - 1\} \) while \( q(x) \gt 0 \) for \( x \in \{1, 2, \ldots n\} \). Of course, we must have \( q(0) = p(n) = 0 \). With these assumptions, \( \bs X \) is irreducible, and since the state space is finite, positive recurrent. So all that remains is to find the invariant distribution. The result is essentially the same as when the state space is \( \N \).
The invariant probability density function \( f_n \) is given by \[ f_n(x) = \frac{1}{B_n} \frac{p(0) \cdots p(x - 1)}{q(1) \cdots q(x)} \text{ for } x \in \N_n \text{ where } B_n = \sum_{x=0}^n \frac{p(0) \cdots p(x - 1)}{q(1) \cdots q(x)} \]
Proof
Define \[ g_n(x) = \frac{p(0) \cdots p(x - 1)}{q(1) \cdots q(x)}, \quad x \in \N_n \] The proof thet \( g_n \) is invariant for \( \bs X \) is the same as before . The constant \( B_n \) is the normalizing constant.
Note that \( B_n \to B \) as \( n \to \infty \), and if \( B \lt \infty \), \( f_n(x) \to f(x) \) as \( n \to \infty \) for \( x \in \N \). We will see this type of behavior again. Results for the birth-death chain on \( \N_n \) often converge to the corresponding results for the birth-death chain on \( \N \) as \( n \to \infty \).
Absorption
Often when the state space \( S = \N \), the state of a birth-death chain represents a population of individuals of some sort (and so the terms birth and death have their usual meanings). In this case state 0 is absorbing and means that the population is extinct. Specifically, suppose that \( \bs X = (X_0, X_1, X_2, \ldots) \) is a birth-death chain on \( \N \) with \( r(0) = 1 \) and with \( p(x), \, q(x) \gt 0 \) for \( x \in \N_+ \). Thus, state 0 is absorbing and all positive states lead to each other and to 0. Let \( N = \min\{n \in \N: X_n = 0\} \) denote the time until absorption, where as usual, \( \min \emptyset = \infty \).
One of the following events will occur:
- Population extinction : \( N \lt \infty \) or equivalently, \( X_m = 0 \) for some \( m \in \N \) and hence \( X_n = 0 \) for all \( n \ge m\).
- Population explosion : \( N = \infty \) or equivalently \( X_n \to \infty \) as \( n \to \infty \).
Proof
Part (b) follows from the general theory, since 0 is absorbing, and all positive states lead to each other and to 0. Thus the positive states are transient and we know that with probability 1, a Markov chain will visit a transient state only finitely often. Thus \( N = \infty \) is equivalent to \( X_n \to \infty \) as \( n \to \infty \).
Naturally we would like to find the probability of these complementary events, and happily we have already done so in our study of recurrence above . Let \[ u(x) = \P(N = \infty) = \P(X_n \to \infty \text{ as } n \to \infty \mid X_0 = x), \quad x \in \N \] so the absorption probability is \[v(x) = 1 - u(x) = \P(N \lt \infty) = \P(X_n = 0 \text{ for some } n \in \N \mid X_0 = x), \quad x \in \N \]
For the birth-death chain \( \bs X \), \[ u(x) = \frac{1}{A} \sum_{i=0}^{x - 1} \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)} \text{ for } x \in \N_+ \text{ where } A = \sum_{i=0}^\infty \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)} \]
Proof
For \( x \in \N_+ \), note that \( u(x) = \P(X_n \in \N_+ \text{ for all } n \in \N \mid X_0 = x) \), the function that gives the probability of staying in the positive states for all time. The proof of the theorem on recurrence above has nothing to do with the transition probabilities in state 0, so the proof applies in this setting as well. In that proof we showed that \( u(x) \) as the form given above, where of course the value is 0 if \( A = \infty \). Trivially, \( u(0) = 0 \).
So if \( A = \infty \) then \( u(x) = 0 \) for all \( x \in S \). If \( A \lt \infty \) then \( u(x) \gt 0 \) for all \( x \in \N_+ \) and \( u(x) \to 1 \) as \( x \to \infty \). For the absorption probability, \( v(x) = 1 \) for all \( x \in \N \) if \( A = \infty \) and so absorption is certain. If \( A \lt \infty \) then \[v(x) = \frac{1}{A} \sum_{i=x}^\infty \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)}, \quad x \in \N \] Next we consider the mean time to absorption, so let \( m(x) = \E(N \mid X_0 = x) \) for \( x \in \N_+ \).
The mean absorption function is given by \[ m(x) = \sum_{j=1}^x \sum_{k=j-1}^\infty \frac{p(j) \cdots p(k)}{q(j) \cdots q(k+1)}, \quad x \in \N \]
Probabilisitic Proof
The number of steps required to go from state \( x \in \N_+ \) to \( x - 1 \) has the same distribution as the number of steps required to go from state 1 to 0, except with parameters \( p(y), \, q(y) \) for \( y \in \{x, x + 1, \ldots\} \) instead of parameters \( p(y), \, q(y) \) for \( y \in \{1, 2, \ldots\} \). So by the additivity of expected value, we just need to compute \( m(1) \) as a function of the parameters. Starting in state 1, the chain will be absorbed in state 0 after a random number of returns to state 1 without absorption. Whenever the chain is in state 1, absorption occurs at the next time with probability \( q(1) \) so it follows that the number of times that the chain is in state 1 before absorption has the geometric distribution on \( \N_+ \) with success parameter \( q(1) \). The mean of this distribution is \( 1 / q(1) \). On the other hand, starting in state 1, the number of steps until the chain is in state 1 again (without absorption) has the same distribution as the return time to state 0, starting in state 0 for the irreducible birth-death chain \( \bs{X}^\prime \) considered above but with birth and death functions \( p^\prime \) and \( q^\prime \) given by \( p^\prime(x) = p(x + 1) \) for \( x \in \N \) and \( q^\prime(x) = q(x + 1) \) for \( x \in \N_+ \). Thus, let \[ \mu = \sum_{k=0}^\infty \frac{p(1) \cdots p(k)}{q(2) \cdots q(k+1)} \] Then \( \mu \) is the mean return time to state 0 for the chain \( \bs{X}^\prime \). Specifically, note that if \( \mu = \infty \) then \( \bs{X}^\prime \) is either transient or null recurrent. If \( \mu \lt \infty \) then \( 1 / \mu \) is the invariant PDF at 0. So, it follows that \[ m(1) = \frac{1}{q(1)} \mu = \sum_{k=0}^\infty \frac{p(1) \cdots p(k)}{q(1) \cdots q(k + 1)} \] By our argument above, the mean time to go from state \( x \) to \( x - 1 \) is \[ \sum_{k=x-1}^\infty \frac{p(x) \cdots p(k)}{q(x) \cdots q(k + 1)} \]
Analytic Proof
Conditioning and using the Markov property, we have \[ m(x) = 1 + p(x) m(x + 1) + q(x) m(x - 1) + r(x) m(x), \quad x \in \N_+ \] with initial condition \( m(0) = 0 \). Equivalently, \[ m(x + 1) - m(x) = \frac{q(x)}{p(x)}[m(x) - m(x - 1)] - \frac{1}{p(x)}, \quad x \in \N_+ \] Solving gives \[ m(x + 1) - m(x) = \frac{q(1) \cdots q(x)}{p(1) \cdots p(x)} m(1) - \sum_{y=1}^x \frac{q(y+1) \cdots q(x)}{p(y) \cdots p(x)}, \quad x \in \N_+ \] Next, \( m(x) = \sum_{y=0}^{x-1} [m(y+1) - m(y)] \) for \( x \in \N \) which gives \[ m(x) = m(1) \sum_{y=0}^{x-1} \frac{q(1) \cdots q(y)}{p(1) \cdots p(y)} - \sum_{y=0}^{x-1} \sum_{z=1}^y \frac{q(z + 1) \cdots q(y)}{p(z) \cdots p(y)}, \quad x \in \N \] Finally, \( m(1) \) is given as in the first proof. The expression for \( m(x) \) is different, but equivalent, of course.
Next we will consider a birth-death chain on a finite integer interval with both endpoints absorbing. Our interest is in the probability of absorption in one endpoint rather than the other, and in the mean time to absorption. Thus suppose that \( n \in \N_+ \) and that \( \bs X = (X_0, X_1, X_2, \ldots) \) is a birth-death chain on \( \N_n = \{0, 1, \ldots, n\} \) with \( r(0) = r(n) = 1 \) and with \( p(x) \gt 0 \) and \( q(x) \gt 0 \) for \( x \in \{1, 2, \ldots, n - 1\} \). So the endpoints 0 and \( n \) are absorbing, and all other states lead to each other and to the endpoints. Let \( N = \min\{n \in \N: X_n \in \{0, n\}\} \), the time until absorption, and for \( x \in S \) let \( v_n(x) = \P(X_N = 0 \mid X_0 = x) \) and \( m_n(x) = \E(N \mid X_0 = x) \). The definitions make sense since \( N \) is finite with probability 1.
The absorption probability function for state 0 is given by \[ v_n(x) = \frac{1}{A_n} \sum_{i=x}^{n-1} \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)} \text{ for } x \in \N_n \text{ where } A_n = \sum_{i=0}^{n-1} \frac{q(1) \cdots q(i)}{p(1) \cdots p(i)} \]
Proof
Conditioning and using the Markov property, \( v_n \) satisfies the second-order linear difference equation \[ v_n(x) = p(x) v_n(x + 1) + q(x) v_n(x - 1) + r(x) v_n(x), \quad x \in \{1, 2, \ldots, n - 1\} \] with boundary conditions \( v_n(0) = 1 \), \( v_n(n) = 0 \). As we have seen before, the difference equation can be rewritten as \[v_n(x + 1) - v_n(x) = \frac{p(x)}{q(x)} [v_n(x) - v_n(x - 1)], \quad x \in \{1, 2, \ldots, n - 2\}\] Solving and applying the boundary conditions gives the result.
Note that \( A_n \to A \) as \( n \to \infty \) where \( A \) is the constant above for the absorption probability at 0 with the infinite state space \( \N \). If \( A \lt \infty \) then \( v_n(x) \to v(x) \) as \( n \to \infty \) for \( x \in \N \).
The mean absorption time is given by \[ m_n(x) = m_n(1) \sum_{y=0}^{x-1} \frac{q(1) \cdots q(y)}{p(1) \cdots p(y)} - \sum_{y=0}^{x-1} \sum_{z=1}^y \frac{q(z+1) \cdots q(y)}{p(z) \cdots p(y)}, \quad x \in \N_n \] where, with \( A_n \) as in the previous theorem, \[ m_n(1) = \frac{1}{A_n} \sum_{y=1}^{n-1} \sum_{z=1}^y \frac{q(z+1) \cdots q(y)}{p(z) \cdots p(y)} \]
Proof
The probabilistic proof above with state space \( \N \) and 0 absorbing does not work here, but the first part of the analytic proof does. So, \[ m_n(x) = m_n(1) \sum_{y=0}^{x-1} \frac{q(1) \cdots q(y)}{p(1) \cdots p(y)} - \sum_{y=0}^{x-1} \sum_{z=1}^y \frac{q(z + 1) \cdots q(y)}{p(z) \cdots p(y)}, \quad x \in \{1, 2, \ldots, n\} \] Substituting \( x = n \) and applying the boundary condition \( m_n(n) = 0 \), gives the result for \( m_n(1) \) in the theorem.
Time Reversal
Our next discussion is on the time reversal of a birth-death chain. Essentially, every recurrent birth-death chain is reversible.
Suppose that \( \bs X = (X_0, X_1, X_2, \ldots) \) is an irreducible, recurrent birth-death chain on an integer interval \( S \). Then \( \bs X \) is reversible.
Proof
We need to show that the Kolmogorov cycle condition is satisfied. That is, for every sequence of states \((x_0, x_1, x_2, \ldots, x_n) \) with \( x_0 = x_n \), \[ P(x_0, x_1) P(x_1, x_2) \cdots P(x_{n-1}, x_n) = P(x_n, x_{n-1}) P(x_{n-1}, x_{n-2}) \cdots P(x_1, x_0) \] We can restrict our attention to sequences where \( x_{i+1} \in \{x_i, x_i - 1, x_i + 1\} \) for each \( i \in \{1, 2, \ldots, n\} \). For such sequences, the cycle condition is trivially satisfied.
If \( S \) is finite and the chain \( \bs X \) is irreducible, then of course \( \bs X \) is recurrent (in fact positive recurrent), so by the previous result, \( \bs X \) is reversible. In the case \( S = \N \), we can use the invariant function above to show directly that the chain is reversible.
Suppose that \( \bs X = (X_0, X_1, X_2, \ldots) \) is a birth-death chain on \( \N \) with \( p(x) \gt 0 \) for \( x \in \N \) and \( q(x) \gt 0 \) for \( x \in \N_+ \). Then \( \bs X \) is reversible.
Proof
With the function \( g \) defined
above
, it suffices to show the reversibility condition \( g(x)P(x, y) = g(y) P(y, x) \) for all \( x, \, y \in \N \). It then follows that \( g \) is invariant for \( \bs{X} \) and that \( \bs{X} \) is reversible with respect to \( g \). But since \( g \) is the only positive invariant function for \( \bs{X} \), up to multiplication by positive constants, we can omit the qualifying phrase
with respect to \( g \)
. For \( x \in \N \) and \( y = x + 1 \) we have \[g(x) P(x, y) = g(y) P(y, x) = \frac{p(0) \cdots p(x)}{q(1) \cdots q(x)}\] For \( x \in \N_+ \) and \( y = x - 1 \) we have \[ g(x) P(x, y) = g(y) P(y, x) = \frac{p(0) \cdots p(x - 1)}{q(1) \cdots q(x - 1)} \] In all other cases, the reversibility condition is trivially satisfied.
Thus, in the positive recurrent case, when the variables are given the invariant distribution, the transition matrix \( P \) describes the chain forward in time and backwards in time.
Examples and Special Cases
As always, be sure to try the problems yourself before looking at the solutions.
Constant Birth and Death Probabilities
Our first examples consider birth-death chains on \( \N \) with constant birth and death probabilities, except at the boundary points. Such chains are often referred to as random walks , although that term is used in a variety of different settings. The results are special cases of the general results above, but sometimes direct proofs are illuminating.
Suppose that \( \bs X = (X_0, X_1, X_2, \ldots) \) is the birth-death chain on \( \N \) with constant birth probability \( p \in (0, \infty) \) on \( \N \) and constant death probability \( q \in (0, \infty) \) on \( \N_+ \), with \( p + q \le 1 \). Then
- \( \bs X \) is transient if \( q \lt p \)
- \( \bs X \) is null recurrent if \( q = p \)
- \( \bs X \) is positive recurrent if \( q \gt p \), and the invariant distribution is the geometric distribution on \( \N \) with parameter \( p / q \) \[ f(x) = \left( 1 - \frac{p }{q} \right) \left( \frac{p}{q} \right)^x, \quad x \in \N \]
Next we consider the random walk on \( \N \) with 0 absorbing. As in the discussion of absorption above, \( v(x) \) denotes the absorption probability and \( m(x) \) the mean time to absorption, starting in state \( x \in \N \).
Suppose that \( \bs X = (X_0, X_1, \ldots) \) is the birth-death chain on \( \N \) with constant birth probability \( p \in (0, \infty)\) on \( \N_+ \) and constant death probability \( q \in (0, \infty) \) on \( \N_+ \), with \( p + q \le 1 \). Assume also that \( r(0) = 1 \), so that 0 is absorbing.
- If \( q \ge p \) then \( v(x) = 1 \) for all \( x \in \N \). If \( q \lt p \) then \( v(x) = (q/p)^x\) for \(x \in \N \).
- If \( q \le p \) then \( m(x) = \infty \) for all \( x \in \N_+ \). If \( q \gt p \) then \( m(x) = x / (q - p)\) for \(x \in \N\).
Proof
- This follows from the general result above for the absorption probability.
- This also follows from the general result above for the mean absorption time, but we will give a direct proof using the same ideas. If \( q \lt p \) then \( \P(N = \infty \mid X_0 = x) \gt 0 \) and hence \( m(x) = \infty \) for \( x \in \N_+ \). So suppose that \( q \ge p \) so that \( \P(N \lt \infty \mid X_0 = x) = 1 \) for \( x \in \N \). Because of the spatial homogeneity, the time required to reach state \( x - 1 \) starting in state \( x \in \N_+ \) has the same distribution as the time required to reach state 0 starting in state 1. By the additivity of expected value, it follows that \( m(x) = x \, m(1) \) for \( x \in \N \). So it remains for us to compute \( m(1) \). Starting in state 1, the chain will be absorbed into state 0 after a random number of intermediate returns to state 1 with absorption. In state 1, the probability of absorption at the next step is \( q \), so the number of times that the chain is in state 1 before absorption has the geometric distribution on \( \N_+ \) with success parameter \( q \). So the mean number of visits is \( 1 / q \). In state 1, the number of steps before a return to step 1 without absorption has the same distribution as the return time to state 0, starting in 0, for the recurrent chain considered in the previous exercise . The mean of this distribution is \( \infty \) if \( q = p \) and is \( 1 / f(0) \) if \( q \gt p \), were \( f \) is the invariant distribution. It follows that \[ m(1) = \frac{1}{q} \frac{1}{1 - p / q} = \frac{1}{q - p}\]
This chain is essentially the gambler's ruin chain . Consider a gambler who bets on a sequence of independent games, where \( p \) and \( q \) are the probabilities of winning and losing, respectively. The gambler receives one monetary unit when she wins a game and must pay one unit when she loses a game. So \( X_n \) is the gambler's fortune after playing \( n \) games.
Next we consider random walks on a finite interval.
Suppose that \( \bs X = (X_0, X_1, \ldots) \) is the birth-death chain on \( \N_n = \{0, 1, \ldots, n\} \) with constant birth probability \( p \in (0, \infty) \) on \( \{0, 1, \ldots, n - 1\} \) and constant death probability \( q \in (0, \infty) \) on \( \{1, 2, \ldots, n\} \), with \( p + q \le 1 \). Then \( \bs X \) is positive recurrent and the invariant probability density function \( f_n \) is given as follows:
- If \( p \ne q \) then \[ f_n(x) = \frac{(p/q)^x (1 - p/q)}{1 - (p/q)^{n+1}}, \quad x \in \N_n\]
- If \( p = q \) then \( f_n(x) = 1 / (n + 1) \) for \( x \in \N_n \).
Note that if \( p \lt q \) then the invariant distribution is a truncated geometric distribution, and \( f_n(x) \to f(x) \) for \( x \in \N \) where \( f \) is the invariant probability density function of the birth-death chain on \( \N \) considered above . If \( p = q \), the invariant distribution is uniform on \( \N_n \), certainly a reasonable result. Next we consider the chain with both endpoints absorbing. As before, \( v_n \) is the function that gives the probability of absorption in state 0, while \( m_n \) is the function that gives the mean time to absorption.
Suppose that \( \bs X = (X_0, X_1, \ldots) \) is the birth-death chain on \( \N_n = \{0, 1, \ldots, n\} \) with constant birth probability \( p \in (0, 1) \) and death probability \( q \in (0, \infty) \) on \( \{1, 2, \ldots, n - 1\} \), where \( p + q \le 1 \). Assume also that \( r(0) = r(n) = 1 \), so that \( 0 \) and \( n \) are absorbing.
- If \( p \ne q \) then \[ v_n(x) = \frac{(q/p)^x - (q/p)^n}{1 - (q/p)^n}, \quad x \in \N_n \]
- If \( p = q \) then \( v_n(x) = 1 - x / n \) for \( x \in \N_n \)
Note that if \( q \lt p \) then \( v_n(x) \to v(x) \) as \( n \to \infty \) for \( x \in \N \).
Suppose again that \( \bs X = (X_0, X_1, \ldots) \) is the birth-death chain on \( \N_n = \{0, 1, \ldots, n\} \) with constant birth probability \( p \in (0, 1) \) and death probability \( q \in (0, \infty) \) on \( \{1, 2, \ldots, n - 1\} \), where \( p + q \le 1 \). Assume also that \( r(0) = r(n) = 1 \), so that \( 0 \) and \( n \) are absorbing.
- If \( p \ne q \) then \[ m_n(x) = \frac{n}{p - q} \frac{1 - (q/p)^x}{1 - (q/p)^n} + \frac{x}{q - p}, \quad x \in \N_n \]
- If \( p = q \) then \[ m_n(x) = \frac{1}{2p}x(n - x), \quad x \in \N_n \]
Special Birth-Death Chains
Some of the random processes that we have studied previously are birth-death Markov chains.
Describe each of the following as a birth-death chain.
- The Ehrenfest chain.
- The modified Ehrenfest chain.
- The Bernoulli-Laplace chain
- The simple random walk on \( \Z \).
Answer
- The Ehrenfest chain with parameter \( m \in \N_+ \) is a birth death chain on \( S = \{0, 1, \ldots, m\} \) with \( q(x) = \frac{x}{m} \) and \( p(x) = \frac{m - x}{m} \) for \( x \in S \).
- The modified Ehrenfest chain with parameter \( m \in \N_+ \) is a birth death chain on \( S = \{0, 1, \ldots, m\} \) with \( q(x) = \frac{x}{2 m} \), \( r(x) = \frac{1}{2} \), and \( p(x) = \frac{m - x}{2 m} \) for \( x \in S \).
- The Bernoulli-Laplace chain with parameters \( j, \, k, \, r \in \N_+ \) with \(r \lt j + k \) is a birth-death chain on \( S = \{\max\{0, r - j\}, \ldots, \min\{k, r\}\} \) with \( q(x) = \frac{(j - r + x) x}{j k} \), \( r(x) = \frac{(r - x) x + (j - r + x)(k - x)}{j k} \), and \( p(x) = \frac{(r - x)(k - x)}{j k} \) for \( x \in S \).
- The simple random walk on \( \Z \) with parameter \( p \in (0, 1) \) is a birth-death chain on \( \Z \) with \( p(x) = p \) and \( q(x) = 1 - p \) for \( x \in \Z \).
Other Examples
Consider the birth-death process on \( \N \) with \( p(x) = \frac{1}{x + 1} \), \( q(x) = 1 - p(x) \), and \( r(x) = 0 \) for \( x \in S \).
- Find the invariant function \( g \).
- Classify the chain.
Answer
- Note that \( p(0) \cdots p(x - 1) = \frac{1}{x!} \) and \( q(1) \cdots q(x) = \frac{1}{x + 1} = p(x) \) for \( x \in \N \). Hence \( g(x) = \frac{x + 1}{x!} \).
- Note that \[ \sum_{x = 0}^\infty g(x) = \sum_{x = 1}^\infty \frac{1}{(x - 1)!} + \sum_{x = 0}^\infty \frac{1}{x!} = 2 e \] So the chain is positive recurrent, with invariant PDF \( f \) given by \[ f(x) = e^{-2} \frac{(x + 1)}{x!}, \quad x \in \N \] Also, the chain is periodic with period 2.


1.1: Assessing Slope of the Land
Here in the Pacific Northwest, we are fortunate to work in a landscape of varied landforms – from volcanic peaks to wide valleys; from steep, forested hillsides to gently rolling savannas; from rapidly cascading mountain streams to meandering river floodplains. Our varied topography is an integral part of our forest ecosystems, influencing our climate, soils, water, plant life and fish habitat (Figure 1.1). As natural resource technicians, we are often called upon to assess the topography, and one of the common elements we measure is the slope of the land. How steep is the hillside? Does the slope drain to a stream? Are there cliffs or bluffs present? Field data collected by technicians lead to informed decisions about land management activities such as providing shade for streams, building roads or trails, and prescribing timber management operations.
Defining Slope
Slope of the land is essentially the gradient or incline of the land. A steep slope refers to a sharp incline; a gentle slope refers to a slight incline. The steep, forested slopes in Figure 1.1 contrast with the gentler slope of the river’s path as it flows between them.
Driving down a highway you may see a road sign that reads “6% Grade” or “steep grade.” The grade of the road is essentially the slope of the road. The sign in Figure 1.2 indicates that the road descends at a 6% grade or a 6 % slope.
A 6% slope means that the road elevation changes 6 feet for every 100 feet of horizontal distance (Figure 1.3).
Mathematically, slope is defined as “the rise over the run” (or the rise divided by the run), where rise equals change in elevation and run equals horizontal distance:
or or in this case:
To express slope as a percent slope , we simply multiply the slope fraction by 100. So, .06 = 6%
%slope
6%
In our road example, the six foot change in elevation is the rise and the 100 foot horizontal distance of the road is the run . Driving uphill, we climb a +6% slope (Figure A below). Driving downhill, the “rise” is actually a “drop,” so we have a negative slope, or a downhill slope (Figure B below). When dealing with slope, a positive slope simply means uphill and a negative slope means downhill. A negative number does not mean “minus” as in algebraic expressions.
Note that the actual road distance is the hypotenuse of the illustrated slope triangle. Its length is called slope distance . Slope distance is always longer than the horizontal distance, or run. Applying the Pythagorean Theorem (a 2 + b 2 = c 2 ) to this triangle, we can calculate the slope distance, or hypotenuse (c).
a 2 + b 2 = c 2 where: 100 2 + 6 2 = c 2 10,036 = c 2 ft.
a = horizontal distance or run (in this example 100 ft.)
b= change in elevation or rise (in this example 6 ft.)
c = road distance or slope distance (in this example 100.2 ft.)
We calculated a slope distance of 100.2 ft. for a run of 100 ft. As you can see from this example, in a forest, a 6% slope would be considered a gentle slope.
Note that %slope is unitless and proportional. Therefore, it can be applied to any unit of measure (inches, yards, centimeters, etc.) and to any length. For example, a 25% slope is simply a 25:100 ratio. The 25% slope below shows that for every inch of horizontal distance, the slope rises .25 inches. For every 10 centimeters of horizontal distance, the slope rises 2.5 cm, and for every 5 inches of horizontal distance, it rises 1.25 inches.
Using Slope
When writing field notes about a site, we include information about the slope of the land. Sometimes a rough estimate of the average slope is sufficient; sometimes detailed measurements of slope are required. For example, a site description might read:
“A timber cruise was conducted on 20 acres of mixed conifer forest …………Approximately half of the acreage was flat, on slopes ranging from 3-7%. The other half of the acreage was steeper; with southwest slopes 40-60%.”
This tells us much more about the site than simply stating that there were 20 acres of mixed forest. We would expect different soil conditions and different vegetation to be present on the different slopes, and therefore, perhaps different management. Let’s say, for example, that these 20 acres are to be logged in the near future. If this is so, the forester will have to plan where to place any new access roads, where to locate landings for yarding the logs and what type of harvesting equipment to use. He knows that the slope of any new spur roads should not exceed 10%, and a cable system should be used to haul logs up to landings on slopes greater than 30% (Figure 1.4).
Prof iles may be run to get a detailed picture of the slope. When profiling a hillside, slope distances and %slope readings are taken where each major change in slope occurs (Figure 1.5). With this precise information, a logging system can be designed that will lift logs off the ground while yarding to reduce erosion.
Note that slopes can exceed 100%. When a slope equals 100%, it simply means that the rise is equal to the run . And although it certainly feels like you are climbing straight up on a 100% slope (pulling yourself up using roots and anything else you can grab), you are really walking up at a 45° angle, not a 90° angle.
%slope
If rise = run as illustrated above: 100%
100% slope = 45º angle slope
Other examples of where %slope is measured in natural resource settings include hiking trails and streams. Switchbacks on steep slopes reduce trail erosion and make hiking easier. Some trails are kept to 8% to comply with American with Disabilities Act (ADA) guidelines for wheelchair access.
Stream gradients vary, reflecting the terrain over which they flow at each stage of their journey to large rivers. Small tributaries often are the steepest, cascading down steep forested slopes at gradients of 60-100% slope or more. As streams merge downriver, the terrain often flattens out, and milder gradients of 3-10% slope may be measured (Figure 1.6).


28.3: Galaxy Mergers and Active Galactic Nuclei
By the end of this section, you will be able to:
- Explain how galaxies grow by merging with other galaxies and by consuming smaller galaxies (for lunch)
- Describe the effects that supermassive black holes in the centers of most galaxies have on the fate of their host galaxies
One of the conclusions astronomers have reached from studying distant galaxies is that collisions and mergers of whole galaxies play a crucial role in determining how galaxies acquired the shapes and sizes we see today. Only a few of the nearby galaxies are currently involved in collisions, but detailed studies of those tell us what to look for when we seek evidence of mergers in very distant and very faint galaxies. These in turn give us important clues about the different evolutionary paths galaxies have taken over cosmic time. Let’s examine in more detail what happens when two galaxies collide.
Mergers and Cannibalism
Figure 28.1 shows a dynamic view of two galaxies that are colliding. The stars themselves in this pair of galaxies will not be affected much by this cataclysmic event. (See the Astronomy Basics feature box Why Galaxies Collide but Stars Rarely Do.) Since there is a lot of space between the stars, a direct collision between two stars is very unlikely. However, the orbits of many of the stars will be changed as the two galaxies move through each other, and the change in orbits can totally alter the appearance of the interacting galaxies. A gallery of interesting colliding galaxies is shown in Figure 28.7. Great rings, huge tendrils of stars and gas, and other complex structures can form in such cosmic collisions. Indeed, these strange shapes are the signposts that astronomers use to identify colliding galaxies.
Throughout this book we have emphasized the large distances between objects in space. You might therefore have been surprised to hear about collisions between galaxies. Yet (except at the very cores of galaxies) we have not worried at all about stars inside a galaxy colliding with each other. Let’s see why there is a difference.
The reason is that stars are pitifully small compared to the distances between them. Let’s use our Sun as an example. The Sun is about 1.4 million kilometers wide, but is separated from the closest other star by about 4 light-years, or about 38 trillion kilometers. In other words, the Sun is 27 million of its own diameters from its nearest neighbor. If the Sun were a grapefruit in New York City, the nearest star would be another grapefruit in San Francisco. This is typical of stars that are not in the nuclear bulge of a galaxy or inside star clusters. Let’s contrast this with the separation of galaxies.
The visible disk of the Milky Way is about 100,000 light-years in diameter. We have three satellite galaxies that are just one or two Milky Way diameters away from us (and will probably someday collide with us). The closest major spiral is the Andromeda Galaxy (M31), about 2.4 million light-years away. If the Milky Way were a pancake at one end of a big breakfast table, M31 would be another pancake at the other end of the same table. Our nearest large galaxy neighbor is only 24 of our Galaxy’s diameters from us, and it will begin to crash into the Milky Way in about 4.5 billion years.
Galaxies in rich clusters are even closer together than those in our neighborhood (see The Distribution of Galaxies in Space). Thus, the chances of galaxies colliding are far greater than the chances of stars in the disk of a galaxy colliding. And we should note that the difference between the separation of galaxies and stars also means that when galaxies do collide, their stars almost always pass right by each other like smoke passing through a screen door.
The details of galaxy collisions are complex, and the process can take hundreds of millions of years. Thus, collisions are best simulated on a computer (Figure 28.8), where astronomers can calculate the slow interactions of stars, and clouds of gas and dust, via gravity. These calculations show that if the collision is slow, the colliding galaxies may coalesce to form a single galaxy.
When two galaxies of equal size are involved in a collision, we call such an interaction a merger (the term applied in the business world to two equal companies that join forces). But small galaxies can also be swallowed by larger ones—a process astronomers have called, with some relish, galactic cannibalism (Figure 28.9).
The very large elliptical galaxies we discussed in Galaxies probably form by cannibalizing a variety of smaller galaxies in their clusters. These “monster” galaxies frequently possess more than one nucleus and have probably acquired their unusually high luminosities by swallowing nearby galaxies. The multiple nuclei are the remnants of their victims (Figure 28.9). Many of the large, peculiar galaxies that we observe also owe their chaotic shapes to past interactions. Slow collisions and mergers can even transform two or more spiral galaxies into a single elliptical galaxy.
A change in shape is not all that happens when galaxies collide. If either galaxy contains interstellar matter, the collision can compress the gas and trigger an increase in the rate at which stars are being formed—by as much as a factor of 100. Astronomers call this abrupt increase in the number of stars being formed a starburst , and the galaxies in which the increase occurs are termed starburst galaxies (Figure 28.10). In some interacting galaxies, star formation is so intense that all the available gas is exhausted in only a few million years; the burst of star formation is clearly only a temporary phenomenon. While a starburst is going on, however, the galaxy where it is taking place becomes much brighter and much easier to detect at large distances.
When astronomers finally had the tools to examine a significant number of galaxies that emitted their light 11 to 12 billion years ago, they found that these very young galaxies often resemble nearby starburst galaxies that are involved in mergers: they also have multiple nuclei and peculiar shapes, they are usually clumpier than normal galaxies today, with multiple intense knots and lumps of bright starlight, and they have higher rates of star formation than isolated galaxies. They also contain lots of blue, young, type O and B stars, as do nearby merging galaxies.
Galaxy mergers in today’s universe are rare. Only about five percent of nearby galaxies are currently involved in interactions. Interactions were much more common billions of years ago (Figure 28.11) and helped build up the “more mature” galaxies we see in our time. Clearly, interactions of galaxies have played a crucial role in their evolution.
Active Galactic Nuclei and Galaxy Evolution
While galaxy mergers are huge, splashy events that completely reshape entire galaxies on scales of hundreds of thousands of light-years and can spark massive bursts of star formation, accreting black holes inside galaxies can also disturb and alter the evolution of their host galaxies. You learned in Active Galaxies, Quasars, and Supermassive Black Holes about a family of objects known as active galactic nuclei (AGN), all of them powered by supermassive black holes. If the black hole is surrounded by enough gas, some of the gas can fall into the black hole, getting swept up on the way into an accretion disk, a compact, swirling maelstrom perhaps only 100 AU across (the size of our solar system).
Within the disk the gas heats up until it shines brilliantly even in X-rays, often outshining the rest of the host galaxy with its billions of stars.  These powerful jets can extend far outside the starry edge of the galaxy.
AGN were much more common in the early universe, in part because frequent mergers provided a fresh gas supply for the black hole accretion disks. Examples of AGN in the nearby universe today include the one in galaxy M87 (see Figure 27.7), which sports a jet of material shooting out from its nucleus at speeds close to the speed of light, and the one in the bright galaxy NGC 5128, also known as Centaurus A (see Figure 28.12).
Many highly accelerated particles move with the jets in such galaxies. Along the way, the particles in the jets can plow into gas clouds in the interstellar medium, breaking them apart and scattering them. Since denser clouds of gas and dust are required for material to clump together to make stars, the disruption of the clouds can halt star formation in the host galaxy or cut it off before it even begins.
In this way, quasars and other kinds of AGN can play a crucial role in the evolution of their galaxies. For example, there is growing evidence that the merger of two gas-rich galaxies not only produces a huge burst of star formation, but also triggers AGN activity in the core of the new galaxy. That activity, in turn, could then slow down or shut off the burst of star formation—which could have significant implications for the apparent shape, brightness, chemical content, and stellar components of the entire galaxy. Astronomers refer to that process as AGN feedback , and it is apparently an important factor in the evolution of most galaxies.


22.10: Magnetic Force between Two Parallel Conductors
Learning Objectives
By the end of this section, you will be able to:
- Describe the effects of the magnetic force between two conductors.
- Calculate the force between two parallel conductors.
You might expect that there are significant forces between current-carrying wires, since ordinary currents produce significant magnetic fields and these fields exert significant forces on ordinary currents. But you might not expect that the force between wires is used to define the ampere. It might also surprise you to learn that this force has something to do with why large circuit breakers burn up when they attempt to interrupt large currents.
The force between two long straight and parallel conductors separated by a distance \(r\) can be found by applying what we have developed in preceding sections. Figure \(\PageIndex{1}\) shows the wires, their currents, the fields they create, and the subsequent forces they exert on one another. Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force \(F_{2}\)). The field due to \(I_{1}\) at a distance \(r\) is given to be
\[B_{1} = \frac{\mu_{0}I_{1}}{2\pi r}.\label{22.11.1}\]
This field is uniform along wire 2 and perpendicular to it, and so the force \(F_{2}\) it exerts on wire 2 is given by \(F = IlB sin\theta\) with \(sin \theta = 1\): \[F_{2} = I_{2}lB_{1}.\label{22.11.2}\] By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write \(F\) for the magnitude of \(F_{2}\). (Note that \(F_{1} = -F_{2}\).) Since the wires are very long, it is convenient to think in terms of \(F/l\), the force per unit length. Substituting the expression for \(B_{1}\) into the last equation and rearranging terms gives
\[\frac{F}{l} = \frac{\mu_{0}I_{1}I_{2}}{2\pi r}.\label{22.11.3}\]
\(F/l\) is the force per unit length between two parallel currents \(I_{1}\) and \(I_{2}\) separated by a distance \(r\). The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.
This force is responsible for the pinch effect in electric arcs and plasmas. The force exists whether the currents are in wires or not. In an electric arc, where currents are moving parallel to one another, there is an attraction that squeezes currents into a smaller tube. In large circuit breakers, like those used in neighborhood power distribution systems, the pinch effect can concentrate an arc between plates of a switch trying to break a large current, burn holes, and even ignite the equipment. Another example of the pinch effect is found in the solar plasma, where jets of ionized material, such as solar flares, are shaped by magnetic forces.
The operational definition of the ampere is based on the force between current-carrying wires. Note that for parallel wires separated by 1 meter with each carrying 1 ampere, the force per meter is
\[\frac{F}{l} = \frac{\left(4\pi \times 10^{-7} T \cdot m/A \right) \left( 1 A \right) ^{2}}{\left(2\pi\right)\left(1 m\right)} = 2 \times 10^{-7} N/m . \label{22.11.4}\]
Since \(\mu_{0}\) is exactly \(4\pi \times 10^{-7} T \cdot m/A\) by definition, and because \(1 T = 1 N/\left(A \cdot m\right)\), the force per meter is exaclty \(2 \times 10^{-7} N/m\). This is the basis of the operational definition of the ampere.
Definition: THE AMPERE
One ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space free of other magnetic fields, causes a force of exactly \(2 \times 10^{-7} N/m\) on each conductor.
Infinite-length straight wires are impractical and so, in practice, a current balance is constructed with coils of wire separated by a few centimeters. Force is measured to determine current. This also provides us with a method for measuring the coulomb. We measure the charge that flows for a current of one ampere in one second. That is, \(1 C = 1 A \cdot s\). For both the ampere and the coulomb, the method of measuring force between conductors is the most accurate in practice.
Summary
- The force between two parallel currents \(I_{1}\) and \(I_{2}\) separated by a distance \(r\), has a magnitude per unit length given by \[\frac{F}{l} = \frac{\mu_{0}I_{1}I_{2}}{2\pi r}.\]
- The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.


5.3: Compound Interest
Most banks, loans, credit cards, etc. charge you compound interest, not simple interest. Compound interest is interest paid both on the original principal and on all interest that has been added to the original principal. Interest on a mortgage or auto loan is compounded monthly. Interest on a savings account can be compounded quarterly (four times a year). Interest on a credit card can be compounded weekly or daily!
| Compounding type | Number of compounding periods per year |
|---|---|
| Annually | 1 |
| Semiannually | 2 |
| Quarterly | 4 |
| Monthly | 12 |
| Daily | 365 |
| Compound Interest: Interest paid on the principal AND the interest accrued. |
Suppose you invest $3000 into an account that pays you 7% interest per year for four years. Using compound interest, after the interest is calculated at the end of each year, then that amount is added to the total amount of the investment. Then the following year, the interest is calculated using the new total of the loan.
| Year | Interest Earned | Total of Loan |
|---|---|---|
| 1 | $3000*0.07 = $210 | $3000 + $210 = $3210 |
| 2 | $3210 *0.07 = $224.70 | $3210 + $224.70 = $3434.70 |
| 3 | $3434.70*0.07 = $240.43 | $3434.70 + $240.43 = $3675.13 |
| 4 | $3675.13 *0.07 = $257.26 | $3675.13 + $257.26 = $3932.39 |
| Total | $932.39 |
So, after four years, you have earned $932.39 in interest for a total of $3932.39.
\[F=P\left(1+\frac{r}{n}\right)^{nt} \nonumber \]
where
- F = Future value
- P = Present value
- r = Annual percentage rate (APR) changed into a decimal
- t = Number of years
- n = Number of compounding periods per year
Example \(\PageIndex{2}\): Comparing Simple Interest versus Compound Interest
Let’s compare a savings plan that pays 6% simple interest versus another plan that pays 6% annual interest compounded quarterly. If we deposit $8,000 into each savings account, how much money will we have in each account after three years?
6% Simple Interest : P = $8,000, r = 0.06, t = 3
Thus, we have $9440.00 in the simple interest account after three years.
6% Interest Compounded Quarterly: P = $8,000, r = 0.06, t = 3, n=4
So, we have $9564.95 in the compounded quarterly account after three years.
With simple interest we earn $1440.00 on our investment, while with compound interest we earn $1564.95.
Example \(\PageIndex{3}\): Compound Interest—Compounded Monthly
In comparison with Example \(\PageIndex{2}\) consider another account with 6% interest compounded monthly. If we invest $8000 in this account, how much will there be in the account after three years?
P = $8,000, r = 0.06, t = 3, n = 12
Thus, we will have $9573.44 in the compounded monthly account after three years.
Interest compounded monthly earns you $9573.44 - $9564.95 = $8.49 more than interest compounded quarterly.
Example \(\PageIndex{4}\): Compound Interest—Savings Bond
Sophia’s grandparents bought her a savings bond for $200 when she was born. The interest rate was 3.28% compounded semiannually, and the bond would mature in 30 years. How much will Sophia’s bond be worth when she turns 30?
P = $200, r = 0.0328, t = 30, n = 2
Sophia’s savings bond will be worth $530.77 after 30 years.
| Continuous Compounding: Interest is compounded infinitely many times per year. |
|
Continuous Compounding Interest Formula : where, F = Future value P = Present value r = Annual percentage rate (APR) changed into a decimal t = Number of years |
Example \(\PageIndex{5}\): Continuous Compounding Interest
Isabel invested her inheritance of $100,000 into an account earning 5.7% interest compounded continuously for 20 years. What will her balance be after 20 years?
P = $100,000, r = 0.057, t = 20
Isabel’s balance will be $312,676.84 after 20 years.
| Annual Percentage Yield (APY) : the actual percentage by which a balance increases in one year. |
Example \(\PageIndex{6}\): Annual Percentage Yield (APY)
Find the Annual Percentage Yield for an investment account with
a. 7.7% interest compounded monthly
b. 7.7% interest compounded daily
c. 7.7% interest compounded continuously.
To find APY, it is easiest to examine an investment of $1 for one year.
- P = $1, r = 0.077, t = 1, n = 12
The percentage the $1 was increased was 7.9776%. The APY is 7.9776%.
- P = $1, r = 0.077, t = 1, n = 365
The percentage the $1 was increased was 8.0033%. The APY is 8.0033%.
- P = $1, r = 0.077, t = 1
The percentage the $1 was increased was 8.0042%. The APY is 8.0042%.


11.8: Discussion Questions
What role do customers play in innovation? Is innovation possible without customer support?
2 .Do you believe that teamwork is important to entrepreneurship? Or do you prefer solopreneurship? Is one approach “better” than the other? Why or why not?
3 .What are the advantages of using a canvas to plan your business?
4 .What types of businesses fit best with each respective canvas type?
5 .Why would impact measures be important in social entrepreneurship ventures?
6 .How do the arts relate to the business model creation process?
7 .What are the advantages of using a lean model canvas compared to the traditional business model canvas?
8 .What components are in the breakeven analysis? How would you calculate your breakeven point?
9 .Why is feasibility analysis important?
10 .What are the different components of a feasibility analysis?
11 .Explain the purpose and importance of business plan.
12 .Describe the components of a business plan.
13 .Should you write a business plan even if you do not need outside financing? Why or why not?


22.4: Radiation from an Electrically-Short Dipole
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The simplest distribution of radiating current that is encountered in common practice is the electrically-short dipole (ESD) . This current distribution is shown in Figure \(\PageIndex{1}\).
Figure \(\PageIndex{1}\): Current distribution of the electrically-short dipole (ESD). ( CC BY-SA 4.0 ; C. Wang)The two characteristics that define the ESD are (1) the current is aligned along a straight line, and (2) the length \(L\) of the line is much less than one-half of a wavelength; i.e., \(L\ll\lambda/2\). The latter characteristic is what we mean by “electrically-short.” 1
The current distribution of an ESD is approximately triangular in magnitude, and approximately constant in phase. How do we know this? First, note that distributions of current cannot change in a complex or rapid way over such distances which are much less than a wavelength. If this is not immediately apparent, recall the behavior of transmission lines: The current standing wave on a transmission line exhibits a period of \(\lambda/2\), regardless the source or termination. For the ESD, \(L\ll\lambda/2\) and so we expect an even simpler variation. Also, we know that the current at the ends of the dipole must be zero, simply because the dipole ends there. These considerations imply that the current distribution of the ESD is well-approximated as triangular in magnitude. 2 Expressed mathematically:
\[\widetilde{I}(z) \approx I_0 \left(1-\frac{2}{L}\left|z\right|\right) \nonumber \]
where \(I_0\) (SI base units of A) is a complex-valued constant indicating the maximum current magnitude and phase.
There are two approaches that we might consider in order to find the electric field radiated by an ESD. The first approach is to calculate the magnetic vector potential \(\widetilde{\bf A}\) by integration over the current distribution, calculate \(\widetilde{\bf H}=(1/\mu)\nabla\times\widetilde{\bf A}\), and finally calculate \(\widetilde{\bf E}\) from \(\widetilde{\bf H}\) using Ampere’s law. We shall employ a simpler approach, shown in Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\): Current distribution of the electricallyshort dipole (ESD) approximated as a large number of Hertzian dipoles. ( CC BY-SA 4.0 ; C. Wang)Imagine the ESD as a collection of many shorter segments of current that radiate independently. The total field is then the sum of these short segments. Because these segments are very short relative to the length of the dipole as well as being short relative to a wavelength, we may approximate the current over each segment as approximately constant. In other words, we may interpret each of these segments as being, to a good approximation, a Hertzian dipole.
The advantage of this approach is that we already have a solution for each of the segments. In Section 9.4, it is shown that a \(\hat{\bf z}\)-directed Hertzian dipole at the origin radiates the electric field
\[\widetilde{\bf E}({\bf r}) \approx \hat{\bf \theta} j\eta \frac{\widetilde{I}\cdot\beta \Delta l}{4\pi}~\left(\sin\theta\right) \frac{e^{-j\beta r}}{r} \nonumber \]
where \(\widetilde{I}\) and \(\Delta l\) may be interpreted as the current and length of the dipole, respectively. In this expression, \(\eta\) is the wave impedance of medium in which the dipole radiates (e.g., \(\approx 377~\Omega\) for free space), and we have presumed lossless media such that the attenuation constant \(\alpha\approx 0\) and the phase propagation constant \(\beta=2\pi/\lambda\). This expression also assumes field points far from the dipole; specifically, distances \(r\) that are much greater than \(\lambda\). Repurposing this expression for the present problem, the segment at the origin radiates the electric field:
\[\widetilde{\bf E}({\bf r};z'=0) \approx \hat{\bf \theta} j\eta \frac{I_0\cdot\beta \Delta l}{4\pi}~\left(\sin\theta\right) \frac{e^{-j\beta r}}{r} \nonumber \]
where the notation \(z'=0\) indicates the Hertzian dipole is located at the origin. Letting the length \(\Delta l\) of this segment shrink to differential length \(dz'\), we may describe the contribution of this segment to the field radiated by the ESD as follows:
\[d\widetilde{\bf E}({\bf r};z'=0) \approx \hat{\bf \theta} j\eta \frac{I_0\cdot\beta dz'}{4\pi}~\left(\sin\theta\right) \frac{e^{-j\beta r}}{r} \nonumber \]
Using this approach, the electric field radiated by any segment can be written:
\[d\widetilde{\bf E}({\bf r};z') \approx \hat{\bf \theta}' j\eta\beta \frac{\widetilde{I}(z')}{4\pi}\left(\sin\theta'\right) \frac{e^{-j\beta \left|{\bf r}-\hat{\bf z}z'\right|}}{\left|{\bf r}-\hat{\bf z}z'\right|}~dz' \nonumber \]
Note that \(\theta\) is replaced by \(\theta'\) since the ray \({\bf r}-\hat{\bf z}z'\) forms a different angle (i.e., \(\theta'\)) with respect to \(\hat{\bf z}\). Similarly, \(\hat{\bf \theta}\) is replaced by \(\hat{\bf \theta}'\), since it also varies with \(z'\). The electric field radiated by the ESD is obtained by integration over these contributions:
\[\widetilde{\bf E}({\bf r}) \approx \int_{-L/2}^{+L/2} d\widetilde{\bf E}(\hat{\bf r};z') \nonumber \]
yielding:
\[\widetilde{\bf E}({\bf r}) \approx j \frac{\eta\beta}{4\pi} \int_{-L/2}^{+L/2} \hat{\bf \theta}' \widetilde{I}(z')~\left(\sin\theta'\right) \frac{e^{-j\beta \left|{\bf r}-\hat{\bf z}z'\right|}}{\left|{\bf r}-\hat{\bf z}z'\right|} dz' \nonumber \]
Given some of the assumptions we have already made, this expression can be further simplified. For example, note that \(\theta'\approx\theta\) since \(L\ll r\). For the same reason, \(\hat{\bf \theta}'\approx\hat{\bf \theta}\). Since these variables are approximately constant over the length of the dipole, we may move them outside the integral, yielding:
\[\widetilde{\bf E}({\bf r}) \approx \hat{\bf \theta} j \frac{\eta\beta}{4\pi}~\left(\sin\theta\right) \int_{-L/2}^{+L/2} \widetilde{I}(z') \frac{e^{-j\beta \left|{\bf r}-\hat{\bf z}z'\right|}}{\left|{\bf r}-\hat{\bf z}z'\right|} dz' \label{m0198_eE1} \]
It is also possible to simplify the expression \(\left|{\bf r}-\hat{\bf z}z'\right|\). Consider Figure \(\PageIndex{3}\).
Figure \(\PageIndex{3}\): Parallel ray approximation for an ESD. ( CC BY-SA 4.0 ; C. Wang)Since we have already assumed that \(r\gg L\) (i.e., the distance to field points is much greater than the length of the dipole), the vector \({\bf r}\) is approximately parallel to the vector \({\bf r}-\hat{\bf z}z'\). Subsequently, it must be true that
\[\left|{\bf r}-\hat{\bf z}z'\right| \approx r-\hat{\bf r}\cdot\hat{\bf z}z' \label{m0198_ePRA} \]
Note that the magnitude of \(r-\hat{\bf r}\cdot\hat{\bf z}z'\) must be approximately equal to \(r\), since \(r\gg L\). So, insofar as \(\left|{\bf r}-\hat{\bf z}z'\right|\) determines the magnitude of \(\widetilde{\bf E}({\bf r})\), we may use the approximation:
\[\left|{\bf r}-\hat{\bf z}z'\right| \approx r ~~~\mbox{(magnitude)} \nonumber \]
Insofar as \(\left|{\bf r}-\hat{\bf z}z'\right|\) determines phase , we have to be a bit more careful. The part of the integrand of Equation \ref{m0198_eE1} that exhibits varying phase is \(e^{-j\beta \left|{\bf r}-\hat{\bf z}z'\right|}\). Using Equation \ref{m0198_ePRA}, we find
\[e^{-j\beta\left|{\bf r}-\hat{\bf z}z'\right|} \approx e^{-j\beta r} e^{+j\beta\hat{\bf r}\cdot\hat{\bf z}z'} \nonumber \]
The worst case in terms of phase variation within the integral is for field points along the \(z\) axis. For these points, \(\hat{\bf r}\cdot\hat{\bf z}=\pm 1\) and subsequently \(\left|{\bf r}-\hat{\bf z}z'\right|\) varies from \(z-L/2\) to \(z+L/2\) where \(z\) is the location of the field point. However, since \(L\ll\lambda\) (i.e., because the dipole is electrically short), this difference in lengths is much less than \(\lambda/2\). Therefore, the phase \(\beta \hat{\bf r}\cdot\hat{\bf z}z'\) varies by much less than \(\pi\) radians, and subsequently \(e^{-j\beta\hat{\bf r}\cdot\hat{\bf z}z'}\approx 1\). We conclude that under these conditions,
\[e^{-j\beta \left|{\bf r}-\hat{\bf z}z'\right|} \approx e^{-j\beta r} ~~~\mbox{(phase)} \nonumber \]
Applying these simplifications for magnitude and phase to Equation \ref{m0198_eE1}, we obtain:
\[\widetilde{\bf E}({\bf r}) \approx \hat{\bf \theta} j \frac{\eta\beta}{4\pi} \left(\sin\theta\right) \frac{e^{-j\beta r}}{r}~\int_{-L/2}^{+L/2} \widetilde{I}(z') dz' \nonumber \]
The integral in this equation is very easy to evaluate; in fact, from inspection (Figure \(\PageIndex{1}\)), we determine it is equal to \(I_0L/2\). Finally, we obtain:
\[\boxed{ \widetilde{\bf E}({\bf r}) \approx \hat{\bf \theta} j \eta \frac{I_0\cdot\beta L}{8\pi} ~\left(\sin\theta\right) ~\frac{e^{-j\beta r}}{r} } \label{m0198_eESDE} \]
Summarizing:
The electric field intensity radiated by an ESD located at the origin and aligned along the \(z\) axis is given by Equation \ref{m0198_eESDE}. This expression is valid for \(r\gg\lambda\).
It is worth noting that the variation in magnitude, phase, and polarization of the ESD with field point location is identical to that of a single Hertzian dipole having current moment \(\hat{\bf z} I_0 L/2\) (Section 9.4). However, the magnitude of the field radiated by the ESD is exactly one-half that of the Hertzian dipole. Why one-half? Simply because the integral over the triangular current distribution assumed for the ESD is one-half the integral over the uniform current distribution that defines the Hertzian dipole. This similarly sometimes causes confusion between Hertzian dipoles and ESDs. Remember that ESDs are physically realizable, whereas Hertzian dipoles are not.
It is common to eliminate the factor of \(\beta\) in the magnitude using the relationship \(\beta=2\pi/\lambda\), yielding:
\[\widetilde{\bf E}({\bf r}) \approx \hat{\bf \theta} j \frac{\eta I_0}{4} \frac{L}{\lambda} ~\left(\sin\theta\right) ~\frac{e^{-j\beta r}}{r} \nonumber \]
At field points \(r\gg\lambda\), the wave appears to be locally planar. Therefore, we are justified using the plane wave relationship \(\widetilde{\bf H} = \frac{1}{\eta} \hat{\bf r} \times \widetilde{\bf E}\) to calculate \(\widetilde{\bf H}\). The result is:
\[\widetilde{\bf H}({\bf r}) \approx \hat{\bf \phi} j \frac{I_0}{4} \frac{L}{\lambda} ~\left(\sin\theta\right) ~\frac{e^{-j\beta r}}{r} \label{m0198_eESDH} \]
Finally, let us consider the spatial characteristics of the radiated field. Figures \(\PageIndex{4}\) and \(\PageIndex{5}\) show the result in a plane of constant \(\phi\). Figures \(\PageIndex{6}\) and \(\PageIndex{7}\) show the result in the \(z=0\) plane. Note that the orientations of the electric and magnetic field vectors indicate a Poynting vector \(\widetilde{\bf E}\times\widetilde{\bf H}\) that is always directed radially outward from the location of the dipole. This confirms that power flow is always directed radially outward from the dipole. Due to the symmetry of the problem, Figures \(\PageIndex{4}\) – \(\PageIndex{7}\) provide a complete characterization of the relative magnitudes and orientations of the radiated fields.
Figure \(\PageIndex{4}\): Magnitude of the radiated field in any plane of constant \(\phi\). ( CC BY-SA 4.0 ; S. Lally) Figure \(\PageIndex{5}\): Orientation of the electric and magnetic fields in any plane of constant \(\phi\). ( CC BY-SA 4.0 ; S. Lally) Figure \(\PageIndex{6}\): Magnitude of the radiated field in any plane of constant \(z\). ( CC BY-SA 4.0 ; S. Lally) Figure \(\PageIndex{7}\): Orientation of the electric and magnetic fields in the \(z=0\) plane. ( CC BY-SA 4.0 ; S. Lally)- A potential source of confusion is that the Hertzian dipole is also a “dipole” which is “electrically-short.” The distinction is that the current comprising a Hertzian dipole is constant over its length. This condition is rarely and only approximately seen in practice, whereas the triangular magnitude distribution is a relatively good approximation to a broad class of commonly-encountered electrically-short wire antennas. Thus, the term “electrically-short dipole,” as used in this book, refers to the triangular distribution unless noted otherwise.↩
- A more rigorous analysis leading to the same conclusion is possible, but is beyond the scope of this book.↩


3.7: Summary
3.1 Factors Influencing the Health of Persons Assigned Female at Birth
The health of a person AFAB is influenced by a great number of factors. Access to health care plays a large part in the health of the individual, family, and community. Health-care education regarding STIs, contraception, mental health, and general overall health influence choices regarding health and safety.
SDOH also contribute greatly to the health of the individual and community.
Poverty, limited education, unemployment, and inadequate access to health care impact nutrition, housing conditions, environmental risks, and overall well-being. Nurses play a crucial role in identifying and addressing these social determinants. Nurses provide resources and education to help improve health outcomes and reduce health disparities.
3.2 Leading Causes of Death and Health Screenings
Understanding the leading causes of morbidity and mortality among persons AFAB is paramount for public health efforts. These insights allow policymakers, nurses, physicians, and communities to develop targeted strategies and interventions to address health disparities. By addressing these health concerns, implementing effective health-care strategies, and promoting proactive health screening, health-care providers can work toward reducing mortality rates and improving the overall well-being of persons on a global scale.
Nursing plays a critical role in preventive screening, as it involves identifying and addressing health risks and potential health problems before they become more serious. Preventive screening aims to detect diseases or conditions early, providing an opportunity for timely intervention and treatment, which can lead to better health outcomes and reduced health-care costs.
3.3 Health Promotion
Health care of persons AFAB encompasses a range of components to promote and maintain health and well-being throughout the lifespan. Nurses are crucial in providing reproductive health services, including family planning and contraceptive counseling, preconception care, prenatal and postnatal care, and menopause management. Nurses provide education, assist with screenings and tests, and offer support and guidance during various reproductive stages.
Nurses are at the forefront of health promotion and education. They are trained to provide information on healthy self-care practices, including nutrition, exercise, and stress management advice. In addition, nurses also educate persons about the importance of regular checkups, vaccinations, and healthy behaviors to prevent chronic diseases. Nurses engage in health counseling to empower persons to take an active role in their health care. They guide healthy self-care choices and practices and disease prevention. Nurses collaborate with other health-care professionals to ensure holistic and coordinated care. Nurses play a key role in raising awareness about health issues, promoting evidence-based practices, and advocating for policies that support individuals’ health-care needs.
3.4 Well-Person Care
Nurses play a crucial role in reproductive wellness history and physical exams, as they are often the first health-care professionals to interact with patients during these assessments. Reproductive wellness history and physical exams are essential components of health and focus on evaluating and addressing reproductive health concerns. Nurses' involvement in reproductive wellness history and physicals is essential for providing comprehensive care, empowering patients with knowledge, and supporting their reproductive health needs throughout their lifespan.
Nursing is integral to trauma-informed care . Nurses have been trained to be compassionate and have a sensitive approach to providing health care for persons who may have experienced trauma in the past. They acknowledge the potential impact of trauma on a person's physical and emotional health and aim to create a safe and supportive environment for the patient during the health-care encounter.
3.5 Preconceptual Care
Preconception care is critical to having a healthy pregnancy. Risk factors or conditions that can increase pregnancy complications and impact the fetus's health are identified during the preconception visit. Common risk factors include advanced maternal age, chronic medical conditions, obesity, smoking, alcohol and substance use, inadequate nutrition, and infections. Important considerations include lack of immunizations, genetic factors, environmental exposures, and inadequate prenatal care. Addressing these risks through preconception counseling, self-care practices, and proper medical management can optimize maternal and fetal health during pregnancy. Nurses are an integral part in providing education to persons considering conception.


19.0: Introduction
Becky, a public health nurse, identifies increasing rates of sudden infant death syndrome (SIDS) across the country within various populations. Upon review of local community assessment data, Becky notes that community SIDS rates match the national average, which both have increased over the past 5 years. Concerned, Becky begins planning for a SIDS prevention education program. She identifies interested partners and forms an interdisciplinary team to assist in planning. The team outlines available resources and decides on an education plan and interventions to implement within the community. Part of this plan is to partner with community advocacy groups to evaluate infant sleep conditions and provide community education using multiple methods. The program will initially run for 3 months, followed by an evaluation of the plan and its execution to determine its effectiveness.
Planning community health education effectively is a multifaceted process encompassing various health disciplines. Becky seeks the support of various community partners to help ensure the SIDS prevention education program’s success. Public health nurses are crucial in this process as they have expertise in client care and advocating for clients. This chapter will discuss principles of planning in health education practice, developing a health education plan, and steps involved in planning health education activities.


1.9: Arguments with Missing Premises
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Quite often, an argument will not explicitly state a premise that we can see is needed in order for the argument to be valid. In such a case, we can supply the premise(s) needed in order so make the argument valid. Making missing premises explicit is a central part of reconstructing arguments in standard form. We have already dealt in part with this in the section on paraphrasing, but now that we have introduced the concept of validity, we have a useful tool for knowing when to supply missing premises in our reconstruction of an argument. In some cases, the missing premise will be fairly obvious, as in the following:
Gary is a convicted sex-offender, so Gary is not allowed to work with children. The premise and conclusion of this argument are straightforward:
1. Gary is a convicted sex-offender
2. Therefore, Gary is not allowed to work with children (from 1)
However, as stated, the argument is invalid. (Before reading on, see if you can provide a counterexample for this argument. That is, come up with an imaginary scenario in which the premise is true and yet the conclusion is false.) Here is just one counterexample (there could be many): Gary is a convicted sex-offender but the country in which he lives does not restrict convicted sex-offenders from working with children. I don’t know whether there are any such countries, although I suspect there are (and it doesn’t matter for the purpose of validity whether there are or aren’t). In any case, it seems clear that this argument is relying upon a premise that isn’t explicitly stated. We can and should state that premise explicitly in our reconstruction of the standard form argument. But what is the argument’s missing premise? The obvious one is that no sex-offenders are allowed to work with children, but we could also use a morecarefully statement like this one:
Where Gary lives, no convicted sex-offenders are allowed to work with children.
It should be obvious why this is a more “careful” statement. It is more careful because it is not so universal in scope, which means that it is easier for the statement to be made true. By relativizing the statement that sex-offenders are not allowed to work with children to the place where Gary lives, we leave open the possibility that other places in the world don’t have this same restriction. So even if there are other places in the world where convicted sex-offenders are allowed to work with children, our statements could still be true since in this place (the place where Gary lives) they aren’t. (For more on strong and weak statements, see section 1.10). So here is the argument in standard form:
1. Gary is a convicted sex-offender.
2. Where Gary lives, no convicted sex-offenders are allowed to work with children.
3. Therefore, Gary is not allowed to work with children. (from 1-2)
This argument is now valid: there is no way for the conclusion to be false, assuming the truth of the premises. This was a fairly simple example where the missing premise needed to make the argument valid was relatively easy to see. As we can see from this example, a missing premise is a premise that the argument needs in order to be as strong as possible. Typically, this means supplying the statement(s) that are needed to make the argument valid. But in addition to making the argument valid, we want to make the argument plausible. This is called “the principle of charity.” The principle of charity states that when reconstructing an argument, you should try to make that argument (whether inductive or deductive) as strong as possible. When it comes to supplying missing premises, this means supplying the most plausible premises needed in order to make the argument either valid (for deductive arguments) or inductively strong (for inductive arguments).
Although in the last example figuring out the missing premise was relatively easy
to do, it is not always so easy. Here is an argument whose missing premises are
not as easy to determine:
Since children who are raised by gay couples often have psychological and emotional problems, the state should discourage gay couples from raising children.
The conclusion of this argument, that the state should not allow gay marriage, is apparently supported by a single premise, which should be recognizable from the occurrence of the premise indicator, “since.” Thus, our initial reconstruction of the standard form argument looks like this:
1. Children who are raised by gay couples often have psychological and emotional problems.
2. Therefore, the state should discourage gay couples from raising children.
However, as it stands, this argument is invalid because it depends on certain missing premises. The conclusion of this argument is a normative statement —a statement about whether something ought to be true, relative to some standard of evaluation. Normative statements can be contrasted with descriptive statements , which are simply factual claims about what is true. For example, “Russia does not allow gay couples to raise children” is a descriptive statement. That is, it is simply a claim about what is in fact the case in Russia today. In contrast, “Russia should not allow gay couples to raise children” is a normative statement since it is not a claim about what is true, but what ought to be true, relative to some standard of evaluation (for example, a moral or legal standard). An important idea within philosophy, which is often traced back to the Scottish philosopher David Hume (1711-1776), is that statements about what ought to be the case (i.e., normative statements) can never be derived from statements about what is the case (i.e., descriptive statements). This is known within philosophy as the is-ought gap . The problem with the above argument is that it attempts to infer a normative statement from a purely descriptive statement, violating the is-ought gap. We can see the problem by constructing a counterexample. Suppose that in society x it is true that children raised by gay couples have psychological problems. However, suppose that in that society people do not accept that the state should do what it can to decrease harm to children. In this case, the conclusion, that the state should discourage gay couples from raising children, does not follow. Thus, we can see that the argument depends on a missing or assumed premise that is not explicitly stated. That missing premise must be a normative statement, in order that we can infer the conclusion, which is also a normative statement. There is an important general lesson here: Many times an argument with a normative conclusion will depend on a normative premise which is not explicitly stated. The missing normative premise of this particular argument seems to be something like this:
The state should always do what it can to decrease harm to children.
Notice that this is a normative statement, which is indicated by the use of the word “should.” There are many other words that can be used to capture normative statements such as: good, bad, and ought. Thus, we can reconstruct the argument, filling in the missing normative premise like this:
1. Children who are raised by gay couples often have psychological and emotional problems.
2. The state should always do what it can to decrease harm to children.
3. Therefore, the state should discourage gay couples from raising children. (from 1-2)
However, although the argument is now in better shape, it is still invalid because it is still possible for the premises to be true and yet the conclusion false. In order to show this, we just have to imagine a scenario in which both the premises are true and yet the conclusion is false. Here is one counterexample to the argument (there are many). Suppose that while it is true that children of gay couples often have psychological and emotional problems, the rate of psychological problems in children raised by gay couples is actually lower than in children raised by heterosexual couples. In this case, even if it were true that the state should always do what it can to decrease harm to children, it does not follow that the state should discourage gay couples from raising children. In fact, in the scenario I’ve described, just the opposite would seem to follow: the state should discourage heterosexual couples from raising children.
But even if we suppose that the rate of psychological problems in children of gay couples is higher than in children of heterosexual couples, the conclusion still doesn’t seem to follow. For example, it could be that the reason that children of gay couples have higher rates of psychological problems is that in a society that is not yet accepting of gay couples, children of gay couples will face more teasing, bullying and general lack of acceptance than children of heterosexual couples. If this were true, then the harm to these children isn’t so much due to the fact that their parents are gay as it is to the fact that their community does not accept them. In that case, the state should not necessarily discourage gay couples from raising children. Here is an analogy: At one point in our country’s history (if not still today) it is plausible that the children of black
Americans suffered more psychologically and emotionally than the children of white Americans. But for the government to discourage black Americans from raising children would have been unjust, since it is likely that if there was a higher incidence of psychological and emotional problems in black Americans, then it was due to unjust and unequal conditions, not to the black parents, per se. So, to return to our example, the state should only discourage gay couples from raising children if they know that the higher incidence of psychological problems in children of gay couples isn’t the result of any kind of injustice, but is due to the simple fact that the parents are gay.
Thus, one way of making the argument (at least closer to) valid would be to add the following two missing premises:
A. The rate of psychological problems in children of gay couples is higher than in children of heterosexual couples.
B. The higher incidence of psychological problems in children of gay couples is not due to any kind of injustice in society, but to the fact that the parents are gay.
So the reconstructed standard form argument would look like this:
1. Children who are raised by gay couples often have psychological and emotional problems.
2. The rate of psychological problems in children of gay couples is higher than in children of heterosexual couples.
3. The higher incidence of psychological problems in children of gay couples is not due to any kind of injustice in society, but to the fact that the parents are gay.
4. The state should always do what it can to decrease harm to children.
5. Therefore, the state should discourage gay couples from raising children. (from 1-4)
In this argument, premises 2-4 are the missing or assumed premises. Their addition makes the argument much stronger, but making them explicit enables us to clearly see what assumptions the argument relies on in order for the argument to be valid. This is useful since we can now clearly see which premises of the argument we may challenge as false. Arguably, premise 4 is false, since the state shouldn’t always do what it can to decrease harm to children. Rather, it should only do so as long as such an action didn’t violate other rights that the state has to protect or create larger harms elsewhere.
The important lesson from this example is that supplying the missing premises of an argument is not always a simple matter. In the example above, I have used the principle of charity to supply missing premises. Mastering this skill is truly an art (rather than a science) since there is never just one correct way of doing it (cf. section 1.5) and because it requires a lot of skilled practice.
Exercise
Supply the missing premise or premises needed in order to make the following arguments valid. Try to make the premises as plausible as possible while making the argument valid (which is to apply the principle of charity).
1. Ed rides horses. Therefore, Ed is a cowboy.
2. Tom was driving over the speed limit. Therefore, Tom was doing something wrong.
3. If it is raining then the ground is wet. Therefore, the ground must be wet.
4. All elves drink Guinness, which is why Olaf drinks Guinness.
5. Mark didn’t invite me to homecoming. Instead, he invited his friend Alexia. So he must like Alexia more than me.
6. The watch must be broken because every time I have looked at it, the hands have been in the same place.
7. Olaf drank too much Guinness and fell out of his second story apartment window. Therefore, drinking too much Guinness caused Olaf to injure himself.
8. Mark jumped into the air. Therefore, Mark landed back on the ground.
9. In 2009 in the United States, the net worth of the median white household was $113,149 a year, whereas the net worth of the median black household was $5,677.
10. The temperature of the water is 212 degrees Fahrenheit. Therefore, the water is boiling.
11. Capital punishment sometimes takes innocent lives, such as the lives of individuals who were later found to be not guilty. Therefore, we should not allow capital punishment.
12. Allowing immigrants to migrate to the U.S. will take working-class jobs away from working-class folks. Therefore, we should not allow immigrants to migrate to the U.S.
13. Prostitution is a fair economic exchange between two consenting adults. Therefore, prostitution should be allowed.
14. Colleges are more interested in making money off of their football athletes than in educating them. Therefore, college football ought to be banned.
15. Edward received an F in college Algebra. Therefore, Edward should have studied more.


18.17: Sound
Check Your Understanding
17.1. Sound and light both travel at definite speeds, and the speed of sound is slower than the speed of light. The first shell is probably very close by, so the speed difference is not noticeable. The second shell is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears.
17.2. 10 dB: rustle of leaves; 50 dB: average office; 100 dB: noisy factory
17.3. Amplitude is directly proportional to the experience of loudness. As amplitude increases, loudness increases.
17.4. In the example, the two speakers were producing sound at a single frequency. Music has various frequencies and wavelengths.
17.5. Regular headphones only block sound waves with a physical barrier. Noise-canceling headphones use destructive interference to reduce the loudness of outside sounds.
17.6. When the tube resonates at its natural frequency, the wave’s node is located at the closed end of the tube, and the antinode is located at the open end. The length of the tube is equal to one-fourth of the wavelength of this wave. Thus, if we know the wavelength of the wave, we can determine the length of the tube.
17.7. Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.
17.8. An easy way to understand this event is to use a graph, as shown below. It appears that beats are produced, but with a more complex pattern of interference.
17.9. If I am driving and I hear Doppler shift in an ambulance siren, I would be able to tell when it was getting closer and also if it has passed by. This would help me to know whether I needed to pull over and let the ambulance through.
Conceptual Questions
1. Sound is a disturbance of matter (a pressure wave) that is transmitted from its source outward. Hearing is the human perception of sound.
3. Consider a sound wave moving through air. The pressure of the air is the equilibrium condition, it is the change in pressure that produces the sound wave.
5. The frequency does not change as the sound wave moves from one medium to another. Since the speed changes and the frequency does not, the wavelength must change. This is similar to the driving force of a harmonic oscillator or a wave on the string.
7. The transducer sends out a sound wave, which reflects off the object in question and measures the time it takes for the sound wave to return. Since the speed of sound is constant, the distance to the object can found by multiplying the velocity of sound by half the time interval measured.
9. The ear plugs reduce the intensity of the sound both in water and on land, but Navy researchers have found that sound under water is heard through vibrations mastoid, which is the bone behind the ear.
11. The fundamental wavelength of a tube open at each end is 2L, where the wavelength of a tube open at one end and closed at one end is 4L. The tube open at one end has the lower fundamental frequency, assuming the speed of sound is the same in both tubes.
13. The wavelength in each is twice the length of the tube. The frequency depends on the wavelength and the speed of the sound waves. The frequency in room B is higher because the speed of sound is higher where the temperature is higher.
15. When resonating at the fundamental frequency, the wavelength for pipe C is 4L, and for pipes A and B is 2L. The frequency is equal to f = \(\frac{v}{\lambda}\). Pipe C has the lowest frequency and pipes A and B have equal frequencies, higher than the one in pipe C.
17. Since the boundary conditions are both symmetric, the frequencies are f n = \(\frac{nv}{2L}\). Since the speed is the same in each, the frequencies are the same. If the wave speed were doubled in the string, the frequencies in the string would be twice the frequencies in the tube.
19. The frequency of the unknown fork is 255 Hz. No, if only the 250 Hz fork is used, listening to the beat frequency could only limit the possible frequencies to 245 Hz or 255 Hz.
21. The beat frequency is 0.7 Hz.
23. Observer 1 will observe the highest frequency. Observer 2 will observe the lowest frequency. Observer 3 will hear a higher frequency than the source frequency, but lower than the frequency observed by observer 1, as the source approaches and a lower frequency than the source frequency, but higher than the frequency observed by observer 1, as the source moves away from observer 3.
25. Doppler radar can not only detect the distance to a storm, but also the speed and direction at which the storm is traveling.
27. The speed of sound decreases as the temperature decreases. The Mach number is equal to M = \(\frac{v_{s}}{v}\), so the plane should slow down.
Problems
29. s max = 4.00 nm, \(\lambda\) = 1.72 m, f = 200 Hz, v = 343.17 m/s
31. a. \(\lambda\) = 68.60 \(\mu\)m
b. \(\lambda\) = 360.00 \(\mu\)m
33. a. k = 183.09 m −1
b. \(\Delta\)P = −1.11 Pa
35. s 1 = 7.00 nm, s 2 = 3.00 nm, kx 1 + \(\phi\) = 0 rad; kx 2 + \(\phi\) = 1.128 rad; k(x 2 − x 1 ) = 1.128 rad, k = 5.64 m −1 ; \(\lambda\) = 1.11 m, f = 306.31 Hz
37. k = 5.28 x 10 3 m; s(x, t) = 4.50 nm cos(5.28 x 10 3 m −1 x − 2\(\pi\)(5.00 MHz)t)
39. \(\lambda\) = 3.43 mm
41. \(\lambda\) = 6.00 m; s max = 2.00 mm; v = 600 m/s; T = 0.01 s
43. (a) f = 100 Hz, (b) \(\lambda\) = 3.43 m
45. f = 3400 Hz
47. a. v = 5.96 x 10 3 m/s
b. steel (from value in Table 17.1 )
49. v = 363 m s
51. \(\Delta\)x = 924 m
53. V = 0.05 m 3 ; m = 392.5 kg; \(\rho\) = 7850 kg/m 3 ; v = 5047.54 m/s
55. T C = 35 °C, v = 351.58 m/s; \(\Delta\)x 1 = 35.16 m, \(\Delta\)x 2 = 52.74 m \(\Delta\)x = 63.39 m
57. a. t 5.00 °C = 0.0180 s, t 35.0 °C = 0.0171 s
b. % uncertainty = 5.00%
c. This uncertainty could definitely cause difficulties for the bat, if it didn’t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey.
59. 1.26 x 10 −3 W/m 2
61. 85 dB
63. a. 93 dB
b. 83 dB
65. 1.58 x 10 −13 W/m 2
67. A decrease of a factor of 10 in intensity corresponds to a reduction of 10 dB in sound level: 120 dB − 10 dB = 110 dB.
69. We know that 60 dB corresponds to a factor of 106 increase in intensity. Therefore, I \(\propto\) X 2 \(\Rightarrow \frac{I_{2}}{I_{1}} = \left(\dfrac{X_{2}}{X_{1}}\right)^{2}\), so that X 2 = 10 −6 atm. 120 dB corresponds to a factor of 10 12 increase \(\Rightarrow\) 10 −9 atm(10 12 ) 1/2 = 10 −3 atm.
71. 28.2 dB
73. 1 x 10 6 km
75. 73 dB − 70 dB = 3 dB; Such a change in sound level is easily noticed.
77. 2.5; The 100-Hz tone must be 2.5 times more intense than the 4000-Hz sound to be audible by this person.
79. 0.974 m
81. 11.0 kHz; The ear is not particularly sensitive to this frequency, so we don’t hear overtones due to the ear canal.
83.  a. 22.0 °C
b. 1.01 m
89. First overtone = 180 Hz; second overtone = 270 Hz; third overtone = 360 Hz
91. 1.56 m
93. The pipe has symmetrical boundary conditions; \[\begin{split} \lambda_{n} & = \frac{2}{n} L, \quad f_{n} = \frac{nv}{2L}, \quad n = 1, 2, 3 \\ \lambda_{1} & = 6.00\; m, \quad \lambda_{2} = 3.00\; m, \quad \lambda_{3} = 2.00\; m \\ f_{1} & = 57.17\; Hz, \quad f_{2} = 114.33\; Hz, \quad f_{3} = 171.50\; Hz \end{split}\]
95. \(\lambda_{6}\) = 0.5 m; v = 1000 m/s; F T = 6500 N
97. f = 6.40 kHz
99. 1.03 or 3%
101. \[\begin{split} f_{B} & = |f_{1} − f_{2}| \\ |128.3\; Hz − 128.1\; Hz| & = 0.2\; Hz; \\ |128.3\; Hz − 127.8\; Hz| & = 0.5\; Hz; |128.1\; Hz − 127.8\; Hz| & = 0.3\; Hz \end{split}\]
103. v A = 135.87 m/s, v B = 141.42 m/s, \(\lambda_{A}\) = \(\lambda_{B}\) = 0.40 m, \(\Delta\)f = 15.00 Hz
105. v = 155.54 m/s, f string = 971.17 Hz, n = 16.23; f string = 1076.83 Hz, n = 18.00
The frequency is 1076.83 Hz and the wavelength is 0.14 m.
107. f 2 = f 1 ± fB = 260.00 Hz ± 1.50 Hz, so that f 2 = 261.50 Hz or f 2 = 258.50 Hz
109. \[\begin{split} f_{ace} & = \frac{f_{1} + f_{2}}{2}; f_{B} = f_{1} − f_{2}\; (assume\; f_{1} \geq f_{2}); \\ f_{ace} & = \frac{(f_{B} + f_{2}) + f_{2}}{2} \Rightarrow f_{2} = 4099.750\; Hz, f_{1} = 4100.250\; Hz \end{split}\]
111. a. 878 Hz
b. 735 Hz
113. 3.79 x 10 3 Hz
115. a. 12.9 m/s
b. 193 Hz
117. The first eagle hears 4.23 x 10 3 Hz. The second eagle hears 3.56 x 10 3 Hz.
119. v s = 31.29 m/s; f o = 1.12 kHz
121. An audible shift occurs when \(\frac{f_{obs}}{f_{s}}\) ≥ 1.003; \[\begin{split} f_{obs} & = f_{s} \frac{v}{v − v_{s}} \Rightarrow \frac{f_{obs}}{f_{s}} = \frac{v}{v − v_{s}} \Rightarrow \\ v_{s} & = 0.990\; m/s \end{split}\]
123. \(\theta\) = 30.02°; v s = 680.00 m/s; tan \(\theta\) = \(\frac{y}{v_{s} t}\), t = 21.65 s
125. sin \(\theta\) = 1 M, \(\theta\) = 56.47°; y = 9.31 km
127. s 1 = 6.34 nm; s 2 = 2.30 nm; kx 1 + \(\phi\) = 0 rad; kx 2 + \(\phi\) = 1.20 rad; k(x 2 − x 1 ) = 1.20 rad; k = 3.00 m −1 ; \(\omega\) = 1019.62 s −1 ; s 1 = s max cos(kx 1 − \(\phi\)); \(\phi\) = 5.66 rad; s(x, t) = (6.30 nm)cos(3.00 m −1 x − 1019.62 s −1 t + 5.66)
Additional Problems
129. v s = 346.40 m/s; \[\begin{split} \lambda_{n} & = \frac{2}{n} L \quad f_{n} = \frac{v_{s}}{\lambda_{n}} \\ \lambda_{1} & = 1.60\; m \quad f_{1} = 216.50\; Hz \\ \lambda_{2} & = 0.80\; m \quad f_{2} = 433.00\; Hz \end{split}\]
131. a.  I = 31.62 \(\mu\)W/m 2
b. I = 0.16 \(\mu\)W/m 2
c. s max = 104.39 \(\mu\)m
d. s max = 7.43 \(\mu\)m
137. \(\frac{f_{A}}{f_{D}} = \frac{v + v_{s}}{v − v_{s}}\), (v − v s )\(\frac{f_{A}}{f_{D}}\) = v + v s , v = 347.39 m/s; T C = 27.70 °C
Challenge Problems
139. \(\sqrt{x^{2} + d^{2}} − x = \lambda\), x 2 + d 2 = \(\lambda\) + x) 2 ; x 2 + d 2 = \(\lambda^{2}\) + 2x\(\lambda\) + x 2 , d 2 = \(\lambda^{2}\) + 2x\(\lambda\); x = \(\frac{d^{2} − \left(\dfrac{v}{f}\right)^{2}}{2 \frac{v}{f}}\)
141. a. For maxima: \(\Delta\)r = d sin \(\theta\); d sin \(\theta\) = n\(\lambda\) n = 0, ± 1, ± 2 ..., \(\theta\) = sin −1 \(\left(\dfrac{n \lambda}{d}\right)\) n = 0, ± 1, ± 2 ...
b. For minima: \(\Delta\)r = d sin \(\theta\); d sin \(\theta\) = \(\left(n + \dfrac{1}{2}\right) \lambda\) n = 0, ± 1, ± 2 ... \(\theta\) = sin −1 \(\left(\left(n + \dfrac{1}{2}\right) \dfrac{\lambda}{d}\right)\) n = 0, ± 1, ± 2 ...
143. a.  a. f = 268.62 Hz
b. \(\Delta\) f ≈ \(\frac{1}{2} \frac{\Delta F_{T}}{F_{T}}\)f = 1.34 Hz
147. a.  This work is licensed by OpenStax University Physics under a
Creative Commons Attribution License (by 4.0)
.