16.2: Potentials and Generators for General Markov Processes
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Our goal in this section is to continue the broad sketch of the general theory of Markov processes. As with the last section, some of the statements are not completely precise and rigorous, because we want to focus on the main ideas without being overly burdened by technicalities. If you are a new student of probability, or are primarily interested in applications, you may want to skip ahead to the study of discrete-time Markov chains.
Preliminaries
Basic Definitions
As usual, our starting point is a probability space \( (\Omega, \mathscr{F}, \P) \), so that \( \Omega \) is the set of outcomes, \( \mathscr{F} \) the \( \sigma \)-algebra of events, and \( \P \) the probability measure on the sample space \( (\Omega, \mathscr{F}) \). The set of times \( T \) is either \( \N \), discrete time with the discrete topology, or \( [0, \infty) \), continuous time with the usual Euclidean topology. The time set \( T \) is given the Borel \( \sigma \)-algebra \( \mathscr{T} \), which is just the power set if \( T = \N \), and then the time space \( (T, \mathscr{T}) \) is given the usual measure, counting measure in the discrete case and Lebesgue measure in the continuous case. The set of states \( S \) has an LCCB topology (locally compact, Hausdorff, with a countable base), and is also given the Borel \( \sigma \)-algebra \( \mathscr{S} \). Recall that to say that the state space is discrete means that \( S \) is countable with the discrete topology, so that \( \mathscr{S} \) is the power set of \( S \). The topological assumptions mean that the state space \( (S, \mathscr{S}) \) is nice enough for a rich mathematical theory and general enough to encompass the most important applications. There is often a natural Borel measure \( \lambda \) on \( (S, \mathscr{S}) \), counting measure \( \# \) if \( S \) is discrete, and for example, Lebesgue measure if \( S = \R^k \) for some \( k \in \N_+ \).
Recall also that there are several spaces of functions on \( S \) that are important. Let \( \mathscr{B} \) denote the set of bounded, measurable functions \( f: S \to \R \). Let \( \mathscr{C} \) denote the set of bounded, continuous functions \( f: S \to \R \), and let \( \mathscr{C}_0 \) denote the set of continuous functions \( f: S \to \R \) that vanish at \( \infty \) in the sense that for every \( \epsilon \gt 0 \), there exists a compact set \( K \subseteq S \) such \( \left|f(x)\right| \lt \epsilon \) for \( x \in K^c \). These are all vector spaces under the usual (pointwise) addition and scalar multiplication, and \( \mathscr{C}_0 \subseteq \mathscr{C} \subseteq \mathscr{B} \). The supremum norm, defined by \( \left\| f \right\| = \sup\{\left|f(x)\right|: x \in S\} \) for \( f \in \mathscr{B} \) is the norm that is used on these spaces.
Suppose now that \(\bs{X} = \{X_t: t \in T\}\) is a time-homogeneous Markov process with state space \( (S, \mathscr{S}) \) defined on the probability space \( (\Omega, \mathscr{F}, \P) \). As before, we also assume that we have a filtration \( \mathfrak{F} = \{\mathscr{F}_t: t \in T\} \), that is, an increasing family of sub \( \sigma \)-algebras of \( \mathscr{F} \), indexed by the time space, with the properties that \( X_t \) is measurable with repsect to \( \mathscr{F}_t \) for \( t \in T \). Intuitively, \( \mathscr{F}_t \) is the collection of events up to time \( t \in T \).
As usual, we let \( P_t \) denote the transition probability kernel for an increase in time of size \( t \in T \). Thus \[ P_t(x, A) = \P(X_t \in A \mid X_0 = x), \quad x \in S, \, A \in \mathscr{S} \] Recall that for \( t \in T \), the transition kernel \( P_t \) defines two operators, on the left with measures and on the right with functions. So, if \( \mu \) is a measure on \( (S, \mathscr{S}) \) then \( \mu P_t \) is the measure on \( (S, \mathscr{S}) \) given by \[ \mu P_t(A) = \int_S \mu(dx) P_t(x, A), \quad A \in \mathscr{S} \] If \( \mu \) is the distribution of \( X_0 \) then \( \mu P_t \) is the distribution of \( X_t \) for \( t \in T \). If \( f \in \mathscr{B} \) then \( P_t f \in \mathscr{B} \) is defined by \[ P_t f(x) = \int_S P_t(x, dy) f(y) = \E\left[f(X_t) \mid X_0 = x\right] \] Recall that the collection of transition operators \( \bs{P} = \{P_t: t \in T\} \) is a semigroup because \( P_s P_t = P_{s+t} \) for \( s, \, t \in T \). Just about everything in this section is defined in terms of the semigroup \( \bs{P} \), which is one of the main analytic tools in the study of Markov processes.
Feller Markov Processes
We make the same assumptions as in the Introduction. Here is a brief review:
We assume that the Markov process \( \bs{X} = \{X_t: t \in T\} \) satisfies the following properties (and hence is a Feller Markov process):
- For \( t \in T \) and \( y \in S \), the distribution of \( X_t \) given \( X_0 = x \) converges to the distribution of \( X_t \) given \( X_0 = y \) as \( x \to y \).
- Given \(X_0 = x \in S \), \( X_t \) converges in probability to \( x \) as \( t \downarrow 0 \).
Part (a) is an assumption on continuity in space , while part (b) is an assumption on continuity in time . If \( S \) is discrete then (a) automatically holds, and if \( T \) is discrete then (b) automatically holds. As we will see, the Feller assumptions are sufficient for a very nice mathematical theory, and yet are general enough to encompass the most important continuous-time Markov processes.
The process \( \bs{X} = \{X_t: t \in T\} \) has the following properties:
- There is a version of \( \bs{X} \) such that \( t \mapsto X_t \) is continuous from the right and has left limits.
- \( \bs{X} \) is a strong Markov process relative to the \( \mathfrak{F}^0_+ \), the right-continuous refinement of the natural filtration.
The Feller assumptions on the Markov process have equivalent formulations in terms of the transition semigroup.
The transition semigroup \( \bs{P} = \{P_t: t \in T\} \) has the following properties:
- If \( f \in \mathscr{C}_0 \) and \( t \in T \) then \( P_t f \in \mathscr{C}_0 \)
- If \( f \in \mathscr{C}_0 \) and \( x \in S \) then \( P_t f(x) \to f(x) \) as \( t \downarrow 0 \).
As before, part (a) is a condition on continuity in space , while part (b) is a condition on continuity in time . Once again, (a) is trivial if \( S \) is discrete, and (b) trivial if \( T \) is discrete. The first condition means that \( P_t \) is a linear operator on \( \mathscr{C}_0 \) (as well as being a linear operator on \( \mathscr{B} \)). The second condition leads to a stronger continuity result.
For \( f \in \mathscr{C}_0 \), the mapping \( t \mapsto P_t f \) is continuous on \( T \). That is, for \( t \in T \), \[ \|P_s f - P_t f\| = \sup\{\left|P_s f(x) - P_t f(x) \right|: x \in S\} \to 0 \text{ as } s \to t\]
Our interest in this section is primarily the continuous time case. However, we start with the discrete time case since the concepts are clearer and simpler, and we can avoid some of the technicalities that inevitably occur in continuous time.
Discrete Time
Suppose that \( T = \N \), so that time is discrete. Recall that the transition kernels are just powers of the one-step kernel. That is, we let \( P = P_1 \) and then \( P_n = P^n \) for \( n \in \N \).
Potential Operators
For \( \alpha \in (0, 1] \), the \( \alpha \)-potential kernel \( R_\alpha \) of \( \bs{X} \) is defined as follows: \[ R_\alpha(x, A) = \sum_{n=0}^\infty \alpha^n P^n(x, A), \quad x \in S, \, A \in \mathscr{S} \]
- The special case \( R = R_1 \) is simply the potential kernel of \( \bs{X} \).
- For \( x \in S \) and \( A \in \mathscr{S} \), \( R(x, A) \) is the expected number of visits of \( \bs{X} \) to \( A \), starting at \( x \).
Proof
The function \( x \mapsto R_\alpha(x, A) \) from \( S \) to \( [0, \infty) \) is measurable for \( A \in \mathscr{S} \) since \( x \mapsto P^n(x, A) \) is measurable for each \( n \in \N \). The mapping \( A \mapsto R_\alpha(x, A) \) is a positive measure on \( \mathscr{S} \) for \( x \in S \) since \( A \mapsto P^n(x, A) \) is a probability measure for each \( n \in \N \). Finally, the interpretation of \( R(x, A) \) for \( x \in S \) and \( A \in \mathscr{S} \) comes from interchanging sum and expected value, which is allowed since the terms are nonnegative: \[ R(x, A) = \sum_{n=0}^\infty P^n(x, A) = \sum_{n=0}^\infty \E[\bs{1}(X_n \in A) \mid X_0 = x] = \E\left( \sum_{n=0}^\infty \bs{1}(X_n \in A) \biggm| X_0 = x\right) = \E[\#\{n \in \N: X_n \in A\} \mid X_0 = x] \]
Note that it's quite possible that \( R(x, A) = \infty \) for some \( x \in S \) and \( A \in \mathscr{S} \). In fact, knowing when this is the case is of considerable importance in the study of Markov processes. As with all kernels, the potential kernel \( R_\alpha \) defines two operators, operating on the right on functions, and operating on the left on positive measures. For the right potential operator, if \( f: S \to \R \) is measurable then \[R_\alpha f(x) = \sum_{n=0}^\infty \alpha^n P^n f(x) = \sum_{n=0}^\infty \alpha^n \int_S P^n(x, dy) f(y) = \sum_{n=0}^\infty \alpha^n \E[f(X_n) \mid X_0 = x], \quad x \in S \] assuming as usual that the expected values and the infinite series make sense. This will be the case, in particular, if \( f \) is nonnegative or if \( p \in (0, 1) \) and \( f \in \mathscr{B} \).
If \( \alpha \in (0, 1) \), then \( R_\alpha(x, S) = \frac{1}{1 - \alpha} \) for all \( x \in S \).
Proof
Using geometric series, \[ R_\alpha(x, S) = \sum_{n=0}^\infty \alpha^n P^n(x, S) = \sum_{n=0}^\infty \alpha^n = \frac{1}{1 - \alpha} \]
It follows that for \( \alpha \in (0, 1) \), the right operator \( R_\alpha \) is a bounded, linear operator on \( \mathscr{B} \) with \(\left\|R_\alpha \right\| = \frac{1}{1 - \alpha}\). It also follows that \( (1 - \alpha) R_\alpha \) is a probability kernel. There is a nice interpretation of this kernel.
If \( \alpha \in (0, 1) \) then \( (1 - \alpha) R_\alpha(x, \cdot) \) is the conditional distribution of \( X_N \) given \( X_0 = x \in S \), where \( N \) is independent of \( \bs{X} \) and has the geometric distribution on \( \N \) with parameter \( 1 - \alpha \).
Proof
Suppose that \( x \in S \) and \( A \in \mathscr{S} \). Conditioning on \( N \) gives \[ \P(X_N \in A \mid X_0 = x) = \sum_{n=0}^\infty \P(N = n) \P(X_N \in A \mid N = n, X_0 = x) \] But by the substitution rule and the assumption of independence, \[ \P(X_N \in A \mid N = n, X_0 = x) = \P(X_n \in A \mid N = n, X_0 = x) = \P(X_n \in A \mid X_0 = x) = P^n(x, A) \] Since \( N \) has the geometric distribution on \( N \) with parameter \( 1 - \alpha \) we have \( P(N = n) = (1 - \alpha) \alpha^n \) for \( n \in \N \). Substituting gives \[ \P(X_N \in A \mid X_0 = x) = \sum_{n=0}^\infty (1 - \alpha) \alpha^n P^n(x, A) = (1 - \alpha) R_\alpha(x, A)\]
So \( (1 - \alpha)R_\alpha \) is a transition probability kernel, just as \( P_n \) is a transition probability kernel, but corresponding to the random time \( N \) (with \( \alpha \in (0, 1) \) as a parameter), rather than the deterministic time \( n \in \N \). An interpretation of the potential kernel \( R_\alpha \) for \( \alpha \in (0, 1) \) can be also given in economic terms. Suppose that \( A \in \mathscr{S} \) and that we receive one monetary unit each time the process \( \bs{X} \) visits \( A \). Then as above, \( R(x, A) \) is the expected total amount of money we receive, starting at \( x \in S \). However, typically money that we will receive at times distant in the future has less value to us now than money that we will receive soon. Specifically suppose that a monetary unit received at time \( n \in \N \) has a present value of \( \alpha^n \), where \( \alpha \in (0, 1) \) is an inflation factor (sometimes also called a discount factor ). Then \( R_\alpha(x, A) \) gives the expected, total, discounted amount we will receive, starting at \( x \in S \). A bit more generally, if \( f \in \mathscr{B} \) is a reward function , so that \( f(x) \) is the reward (or cost, depending on the sign) that we receive when we visit state \( x \in S \), then for \( \alpha \in (0, 1) \), \( R_\alpha f(x) \) is the expected, total, discounted reward, starting at \( x \in S \).
For the left potential operator, if \( \mu \) is a positive measure on \( \mathscr{S} \) then \[\mu R_\alpha(A) = \sum_{n=0}^\infty \alpha^n \mu P^n(A) = \sum_{n=0}^\infty \alpha^n \int_S \mu(dx) P^n(x, A), \quad A \in \mathscr{S}\] In particular, if \( \mu \) is a probability measure and \( X_0 \) has distribution \( \mu \) then \( \mu P^n \) is the distribution of \( X_n \) for \( n \in \N \), so from the last result, \((1 - \alpha) \mu R_\alpha \) is the distribution of \( X_N \) where again, \( N \) is independent of \( \bs{X} \) and has the geometric distribution on \( \N \) with parameter \( 1 - \alpha \). The family of potential kernels gives the same information as the family of transition kernels.
The potential kernels \( \bs{R} = \{R_\alpha: \alpha \in (0, 1)\} \) completely determine the transition kernels \( \bs{P} = \{P_n: n \in \N\} \).
Proof
Note that for \( x \in S \) and \( A \in \mathscr{S} \), the function \( \alpha \mapsto R_\alpha(x, A) \) is a power series in \( \alpha \) with coefficients \( n \mapsto P^n(x, A) \). In the language of combinatorics, \( \alpha \mapsto R_\alpha(x, A) \) is the ordinary generating function of the sequence \( n \mapsto P^n(x, A) \). As noted above, this power series has radius of convergence at least 1, so we can extend the domain to \( \alpha \in (-1, 1) \). Thus, given the potential kernels, we can recover the transition kernels by taking derivatives and evaluating at 0: \[ P^n(x, A) = \frac{1}{n!}\left[\frac{d^n}{d\alpha^n} R_\alpha(x, A) \right]_{\alpha = 0} \]
Of course, it's really only necessary to determine \( P \), the one step transition kernel, since the other transition kernels are powers of \( P \). In any event, it follows that the kernels \( \bs{R} = \{R_\alpha: \alpha \in (0, 1)\} \), along with the initial distribution, completely determine the finite dimensional distributions of the Markov process \( \bs{X} \). The potential kernels commute with each other and with the transition kernels.
Suppose that \( \alpha, \, \beta \in (0, 1] \) and \( k \in \N \). Then (as kernels)
- \( P^k R_\alpha = R_\alpha P^k = \sum_{n=0}^\infty \alpha^n P^{n+k} \)
- \( R_\alpha R_\beta = R_\beta R_\alpha = \sum_{m=0}^\infty \sum_{n=0}^\infty \alpha^m \beta^n P^{m+n} \)
Proof
Suppose that \( f \in \mathscr{B} \) is nonnegative. The interchange of the sums with the kernel operation is allowed since the kernels are nonnegative. The other tool used is the semigroup property.
- Directly \[ R_\alpha P^k f = \sum_{n=0}^\infty \alpha^n P^n P^k f = \sum_{n=0}^\infty \alpha^n P^{n+k} f\] The other direction requires an interchange.
The same identities hold for the right operators on the entire space \( \mathscr{B} \), with the additional restrictions that \( \alpha \lt 1 \) and \( \beta \lt 1 \). The fundamental equation that relates the potential kernels is given next.
If \( \alpha, \, \beta \in (0, 1] \) with \( \alpha \le \beta \) then (as kernels), \[ \beta R_\beta = \alpha R_\alpha + (\beta - \alpha) R_\alpha R_\beta \]
Proof
If \( \alpha = \beta \) the equation is trivial, so assume \( \alpha \lt \beta \). Suppose that \( f \in \mathscr{B} \) is nonnegative.
The same identity holds holds for the right operators on the entire space \( \mathscr{B} \), with the additional restriction that \( \beta \lt 1 \).
If \( \alpha \in (0, 1] \), then (as kernels), \( I + \alpha R_\alpha P = I + \alpha P R_\alpha = R_\alpha \).
Proof
Suppose that \( f \in \mathscr{B} \) is nonnegative. From the result above, \[ (I + \alpha R_\alpha P) f = (I + \alpha P R_\alpha) f = f + \sum_{n=0}^\infty \alpha^{n+1} P^{n+1} f = \sum_{n = 0}^\infty \alpha^n P^n f = R_\alpha f \]
The same identity holds for the right operators on the entire space \( \mathscr{B} \), with the additional restriction that \( \alpha \lt 1 \). This leads to the following important result:
If \( \alpha \in (0, 1) \), then as operators on the space \( \mathscr{B} \),
- \( R_\alpha = (I - \alpha P)^{-1} \)
- \( P = \frac{1}{\alpha}\left(I - R_\alpha^{-1}\right) \)
Proof
The operators are bounded, so we can subtract. The identity \( I + \alpha R_\alpha P = R_\alpha \) leads to \( R_\alpha(I - \alpha P) = I \) and the identity \( I + \alpha P R_\alpha = R_\alpha \) leads to \( (I - \alpha P) R_\alpha = I \). Hence (a) holds. Part (b) follows from (a).
This result shows again that the potential operator \( R_\alpha \) determines the transition operator \( P \).
Examples and Applications
Our first example considers the binomial process as a Markov process.
Let \( \bs{I} = \{I_n: n \in \N_+\} \) be a sequence of Bernoulli trials with success parameter \( p \in (0, 1) \). Define the Markov process \( \bs{X} = \{X_n: n \in \N\} \) by \( X_n = X_0 + \sum_{k=1}^n I_k \) where \( X_0 \) takes values in \( \N \) and is independent of \( \bs{I} \).
- For \( n \in \N \), show that the transition probability matrix \( P^n \) of \( \bs{X} \) is given by \[ P^n(x, y) = \binom{n}{y - x} p^{y - x} (1 - p)^{n - y + x}, \quad x \in \N, \, y \in \{x, x + 1, \ldots, x + n\} \]
- For \( \alpha \in (0, 1] \), show that the potential matrix \( R_\alpha \) of \( \bs{X} \) is given by \[ R_\alpha(x, y) = \frac{1}{1 - \alpha + \alpha p} \left(\frac{\alpha p}{1 - \alpha + \alpha p}\right)^{y - x}, \quad x \in \N, \, y \in \{x, x + 1, \ldots\} \]
- For \( \alpha \in (0, 1) \) and \( x \in \N \), identify the probability distribution defined by \( (1 - \alpha) R_\alpha(x, \cdot) \).
- For \( x, \, y \in \N \) with \( x \le y \), interpret \( R(x, y) \), the expected time in \( y \) starting in \( x \), in the context of the process \( \bs{X} \).
Solutions
Recall that \( \bs{X} \) is a Markov process since it has stationary, independent increments.
- Note that for \( n, \, x \in \N \), \( P^n(x, \cdot) \) is the (discrete) PDF of \( x + \sum_{k=1}^n I_k \). The result follows since the sum of the indicator variables has the binomial distribution with parameters \( n \) and \( p \).
- Let \( \alpha \in (0, 1] \) and let \( x, \, y \in \N \) with \( x \le y \). Then \begin{align*} R_\alpha(x, y) & = \sum_{n=0}^\infty \alpha^n P^n(x, y) = \sum_{n = y - x}^\infty \alpha^n \binom{n}{y - x} p^{y-x} (1 - p)^{n - y + x} \\ & = (\alpha p)^{y - x} \sum_{n = y - x}^\infty \binom{n}{y - x}[\alpha (1 - p)]^{n - y + x} = \frac{(\alpha p)^{y - x}}{[1 - \alpha (1 - p)]^{n - x + 1}} \end{align*} Simplifying gives the result.
- For \( \alpha \in (0, 1) \), \[ (1 - \alpha) R_\alpha(x, y) = \frac{1 - \alpha}{1 - \alpha + \alpha p} \left(\frac{\alpha p}{1 - \alpha + \alpha p}\right)^{y - x} \] As a function of \( y \) for fixed \( x \), this is the PDF of \( x + Y_\alpha \) where \( Y_\alpha \) has the geometric distribution on \( N \) with parameter \( \frac{1 - \alpha}{1 - \alpha + \alpha p} \).
- Note that \( R(x, y) = 1 / p \) for \( x, \, y \in \N \) with \( x \le y \). Starting in state \( x \), the process eventually reaches \( y \) with probability 1. The process remains in state \( y \) for a geometrically distributed time, with parameter \( p \). The mean of this distribution is \( 1 / p \).
Continuous Time
With the discrete-time setting as motivation, we now turn the more important continuous-time case where \( T = [0, \infty) \).
Potential Kernels
For \( \alpha \in [0, \infty) \), the \( \alpha \)-potential kernel \( U_\alpha \) of \( \bs{X} \) is defined as follows: \[ U_\alpha(x, A) = \int_0^\infty e^{-\alpha t} P_t(x, A) \, dt, \quad x \in S, \, A \in \mathscr{S} \]
- The special case \( U = U_0 \) is simply the potential kerenl of \( \bs{X} \).
- For \( x \in S \) and \( A \in \mathscr{S} \), \( U(x, A) \) is the expected amount of time that \( \bs{X} \) spends in \( A \), starting at \( x \).
- The family of kernels \( \bs{U} = \{U_\alpha: \alpha \in (0, \infty)\} \) is known as the reolvent of \( \bs{X} \).
Proof
Since \( \bs{P} = \{P_t: t \in T\} \) is a Feller semigroup of transition operators, the mapping \((t, x) \mapsto P_t(x, A)\) from \( [0, \infty) \times S \) to \( [0, 1] \) is jointly measurable for \( A \in \mathscr{S} \). Thus, \( U_\alpha(x, A) \) makes sense for \( x \in S \) and \( A \in \mathscr{S} \) and \( x \mapsto U_\alpha(x, A) \) from \( S \) to \( [0, \infty) \) is measurable for \( A \in \mathscr{S} \). That \( A \mapsto U_\alpha(x, A) \) is a measure on \( \mathscr{S} \) follows from the usual interchange of sum and integral, via Fubini's theorem: Suppose that \( \{A_j: j \in J\} \) is a countable collection of disjoint sets in \( \mathscr{S} \), and let \( S = \bigcup_{j \in J} A_j \) \begin{align*} U_\alpha(x, A) & = \int_0^\infty e^{-\alpha t} P_t(x, A) \, dt = \int_0^\infty \left[\sum_{j \in J} e^{-\alpha t} P_t(x, A_j)\right] \, dt\\ & = \sum_{j \in J} \int_0^\infty e^{-\alpha t} P_t(x, A_j) \, dt = \sum_{j \in J} U_\alpha(x, A_j) \end{align*} Finally, the interpretation of \( U(x, A) \) for \( x \in S \) and \( A \in \mathscr{S} \) is another interchange of integrals: \[ U(x, A) = \int_0^\infty P_t(x, A) \, dt = \int_0^\infty \E[\bs{1}(X_t \in A) \mid X_0 = x] \, dt = \E\left( \int_0^\infty \bs{1}(X_t \in A) \, dt \biggm| X_0 = x\right) \] The inside integral is the Lebesgue measure of \( \{t \in [0, \infty): X_t \in A\} \).
As with discrete time, it's quite possible that \( U(x, A) = \infty \) for some \( x \in S \) and \( A \in \mathscr{S} \), and knowing when this is the case is of considerable interest. As with all kernels, the potential kernel \( U_\alpha \) defines two operators, operating on the right on functions, and operating on the left on positive measures. If \( f: S \to \R \) is measurable then, giving the right potential operator in its many forms, \begin{align*} U_\alpha f(x) & = \int_S U_\alpha(x, dy) f(y) = \int_0^\infty e^{-\alpha t} P_t f(x) \, dt \\ & = \int_0^\infty e^{-\alpha t} \int_S P_t(x, dy) f(y) = \int_0^\infty e^{-\alpha t} \E[f(X_t) \mid X_0 = x] \, dt, \quad x \in S \end{align*} assuming that the various integrals make sense. This will be the case in particular if \( f \) is nonnegative, or if \( f \in \mathscr{B} \) and \( \alpha \gt 0 \).
If \( \alpha \gt 0 \), then \( U_\alpha(x, S) = \frac{1}{\alpha} \) for all \( x \in S \).
Proof
For \( x \in S \), \[ U_\alpha(x, S) = \int_0^\infty e^{-\alpha t} P_t(x, S) \, dt = \int_0^\infty e^{-\alpha t} dt = \frac{1}{\alpha} \]
It follows that for \( \alpha \in (0, \infty) \), the right potential operator \( U_\alpha \) is a bounded, linear operator on \( \mathscr{B} \) with \( \|U_\alpha\| = \frac{1}{\alpha} \). It also follows that \( \alpha U_\alpha \) is a probability kernel. This kernel has a nice interpretation.
If \( \alpha \gt 0 \) then \( \alpha U_\alpha (x, \cdot) \) is the conditional distribution of \( X_\tau \) where \( \tau \) is independent of \( \bs{X} \) and has the exponential distribution on \( [0, \infty) \) with parameter \( \alpha \).
Proof
Suppose that \( x \in S \) and \( A \in \mathscr{S} \). The random time \( \tau \) has PDF \( f(t) = \alpha e^{-\alpha t} \) for \( t \in [0, \infty) \). Hence, conditioning on \( \tau \) gives \[ \P(X_\tau \in A \mid X_0 = x) = \int_0^\infty \alpha e^{-\alpha t} \P(X_\tau \in A \mid \tau = t, X_0 = x) \, dt \] But by the substitution rule and the assumption of independence, \[ \P(X_\tau \in A \mid \tau = t, X_0 = x) = \P(X_t \in A \mid \tau = t, X_0 = x) = \P(X_t \in A \mid X_0 = x) = P_t(x, A) \] Substituting gives \[ \P(X_\tau \in A \mid X_0 = x) = \int_0^\infty \alpha e^{-\alpha t} P_t(x, A) \, dt = \alpha U_\alpha(x, A)\]
So \( \alpha U_\alpha \) is a transition probability kernel, just as \( P_t \) is a transition probability kernel, but corresponding to the random time \( \tau \) (with \( \alpha \in (0, \infty) \) as a parameter), rather than the deterministic time \( t \in [0, \infty) \). As in the discrete case, the potential kernel can also be interpreted in economic terms. Suppose that \( A \in \mathscr{S} \) and that we receive money at a rate of one unit per unit time whenever the process \( \bs{X} \) is in \( A \). Then \( U(x, A) \) is the expected total amount of money that we receive, starting in state \( x \in S \). But again, money that we receive later is of less value to us now than money that we will receive sooner. Specifically, suppose that one monetary unit at time \( t \in [0, \infty) \) has a present value of \( e^{-\alpha t} \) where \( \alpha \in (0, \infty) \) is the inflation factor or discount factor . The \( U_\alpha(x, A) \) is the total, expected, discounted amount that we receive, starting in \( x \in S \). A bit more generally, suppose that \( f \in \mathscr{B} \) and that \( f(x) \) is the reward (or cost, depending on the sign) per unit time that we receive when the process is in state \( x \in S \). Then \( U_\alpha f(x) \) is the expected, total, discounted reward, starting in state \( x \in S \).
For the left potential operator, if \( \mu \) is a positive measure on \( \mathscr{S} \) then \begin{align*}\mu U_\alpha(A) & = \int_S \mu(dx) U_\alpha(x, A) = \int_0^\infty e^{-\alpha t} \mu P_t (A) \, dt\\ & = \int_0^\infty e^{-\alpha t} \left[\int_S \mu(dx) P_t(x, A)\right] dt = \int_0^\infty e^{-\alpha t} \left[\int_S \mu(dx) \P(X_t \in A) \right] dt, \quad A \in \mathscr{S} \end{align*} In particular, suppose that \( \alpha \gt 0 \) and that \( \mu \) is a probability measure and \( X_0 \) has distribution\( \mu \). Then \( \mu P_t \) is the distribution of \( X_t \) for \( t \in [0, \infty) \), and hence from the last result, \( \alpha \mu U_\alpha \) is the distribution of \( X_\tau \), where again, \( \tau \) is independent of \( \bs{X} \) and has the exponential distribution on \( [0, \infty) \) with parameter \( \alpha \). The family of potential kernels gives the same information as the family of transition kernels.
The resolvent \(\bs{U} = \{U_\alpha: \alpha \in (0, \infty)\} \) completely determines the family of transition kernels \( \bs{P} = \{P_t: t \in (0, \infty)\} \).
Proof
Note that for \( x \in S \) and \( A \in \mathscr{S} \), the function \( \alpha \mapsto U_\alpha(x, A) \) on \( (0, \infty) \) is the Laplace transform of the function \( t \mapsto P_t(x, A) \) on \( [0, \infty) \). The Laplace transform of a function determines the function completely.
It follows that the resolvent \( \{U_\alpha: \alpha \in [0, \infty)\} \), along with the initial distribution, completely determine the finite dimensional distributions of the Markov process \( \bs{X} \). This is much more important here in the continuous-time case than in the discrete-time case, since the transition kernels \( P_t \) cannot be generated from a single transition kernel. The potential kernels commute with each other and with the transition kernels.
Suppose that \( \alpha, \, \beta, \, t \in [0, \infty) \). Then (as kernels),
- \( P_t U_\alpha = U_\alpha P_t = \int_0^\infty e^{-\alpha s} P_{s+t} ds\)
- \( U_\alpha U_\beta = \int_0^\infty \int_0^\infty e^{-\alpha s} e^{-\beta t} P_{s+t} ds \, dt \)
Proof
Suppose that \( f \in \mathscr{B} \) is nonnegative. The interchanges of operators and integrals below are interchanges of integrals, and are justified since the integrands are nonnegative. The other tool used is the semigroup property of \( \bs{P} = \{P_t: t \in [0, \infty)\} \).
- Directly, \[ U_\alpha P_t f = \int_0^\infty e^{-\alpha s} P_s P_t f \, ds = \int_0^\infty e^{-\alpha s} P_{s+t} f \, ds \] The other direction involves an interchange.
The same identities hold for the right operators on the entire space \( \mathscr{B} \) under the additional restriction that \( \alpha \gt 0 \) and \( \beta \gt 0 \). The fundamental equation that relates the potential kernels, known as the resolvent equation , is given in the next theorem:
If \( \alpha, \, \beta \in [0, \infty) \) with \( \alpha \le \beta \) then (as kernels) \( U_\alpha = U_\beta + (\beta - \alpha) U_\alpha U_\beta \).
Proof
If \( \alpha = \beta \) the equation is trivial, so assume \( \alpha \lt \beta \). Suppose that \( f \in \mathscr{B} \) is nonnegative.  The inverse transformation is \( s = v, \, t = u - v \) with Jacobian \( -1 \). Hence we have \begin{align*} U_\alpha U_\beta f & = \int_0^\infty \int_0^u e^{-\alpha v} e^{-\beta(u - v)} P_u f \, dv \, du = \int_0^\infty \left(\int_0^u e^{(\beta - \alpha) v} dv\right) e^{-\beta u} P_u f \, du \\ & = \frac{1}{\beta - \alpha} \int_0^\infty \left[e^{(\beta - \alpha) u} - 1\right] e^{-\beta u} P_u f du\\ & = \frac{1}{\beta - \alpha}\left(\int_0^\infty e^{-\alpha u} P_u f \, du - \int_0^\infty e^{-\beta u} P_u f \, du\right) = \frac{1}{\beta - \alpha}\left(U_\alpha f - U_\beta f\right) \end{align*} Simplifying gives the result. Note that \( U_\beta f \) is finite since \( \beta \gt 0 \).
The same identity holds for the right potential operators on the entire space \( \mathscr{B} \), under the additional restriction that \( \alpha \gt 0 \). For \( \alpha \in (0, \infty) \), \( U_\alpha \) is also an operator on the space \( \mathscr{C}_0 \).
If \( \alpha \in (0, \infty) \) and \( f \in \mathscr{C}_0 \) then \( U_\alpha f \in \mathscr{C}_0 \).
Proof
Suppose that \( f \in \mathscr{C}_0 \) and that \( (x_1, x_2, \ldots) \) is a sequence in \( S \). Then \( P_t f \in \mathscr{C}_0 \) for \( t \in [0, \infty) \). Hence if \( x_n \to x \in S \) as \( n \to \infty \) then \( e^{-\alpha t} P_t f(x_n) \to e^{-\alpha t} P_t f(x) \) as \( n \to \infty \) for each \( t \in [0, \infty) \). By the dominated convergence theorem, \[ U_\alpha f(x_n) = \int_0^\infty e^{-\alpha t} P_t f(x_n) \, dt \to \int_0^\infty e^{-\alpha t} P_t f(x) \, dt = U_\alpha f(x) \text{ as } n \to \infty \] Hence \( U_\alpha f \) is continuous. Next suppose that \( x_n \to \infty \) as \( n \to \infty \). This means that for every compact \( C \subseteq S\), there exist \( m \in \N_+ \) such that \( x_n \notin C \) for \( n \gt m \). Them \( e^{-\alpha t} P_t f(x_n) \to 0 \) as \( n \to \infty \) for each \( t \in [0, \infty) \). Again by the dominated convergence theorem, \[ U_\alpha f(x_n) = \int_0^\infty e^{-\alpha t} P_t f(x_n) \, dt \to 0 \text{ as } n \to \infty \] So \( U_\alpha f \in \mathscr{C}_0 \).
If \( f \in \mathscr{C}_0 \) then \( \alpha U_\alpha f \to f \) as \( \alpha \to \infty \).
Proof
Convergence is with respect to the supremum norm on \( \mathscr{C}_0 \), of course. Suppose that \( f \in \mathscr{C}_0 \).
Infinitesimal Generator
In continuous time, it's not at all clear how we could construct a Markov process with desired properties, say to model a real system of some sort. Stated mathematically, the existential problem is how to construct the family of transition kernels \( \{P_t: t \in [0, \infty)\} \) so that the semigroup property \(P_s P_t = P_{s + t}\) is satisfied for all \( s, \, t \in [0, \infty) \). The answer, as for similar problems in the deterministic world, comes essentially from calculus, from a type of derivative.
The infinitesimal generator of the Markov process \( \bs{X} \) is the operator \( G: \mathscr{D} \to \mathscr{C}_0 \) defined by \[ G f = \lim_{t \downarrow 0} \frac{P_t f - f}{t} \] on the domain \( \mathscr{D} \subseteq \mathscr{C}_0 \) for which the limit exists.
As usual, the limit is with respect to the supremum norm on \( \mathscr{C}_0 \), so \( f \in \mathscr{D} \) and \( G f = g \) means that \( f, \, g \in \mathscr{C}_0 \) and \[ \left\|\frac{P_t f - f}{t} - g \right\| = \sup\left\{\left| \frac{P_t f(x) - f(x)}{t} - g(x) \right|: x \in S \right\} \to 0 \text{ as } t \downarrow 0 \] So in particular, \[ G f(x) = \lim_{t \downarrow 0} \frac{P_t f(x) - f(x)}{t} = \lim_{t \downarrow 0} \frac{\E[f(X_t) \mid X_0 = x] - f(x)}{t}, \quad x \in S \]
The domain \( \mathscr{D} \) is a subspace of \( \mathscr{C}_0 \) and the generator \( G \) is a linear operator on \( \mathscr{D} \)
- If \( f \in \mathscr{D} \) and \( c \in \R \) then \( c f \in \mathscr{D} \) and \( G(c f) = c G f \).
- If \( f, \, g \in \mathscr{D} \) then \( f + g \in \mathscr{D} \) and \( G(f + g) = G f + G g \).
Proof
These are simple results that depend on the linearity of \( P_t \) for \( t \in [0, \infty) \) and basic results on convergence.
- If \( f \in \mathscr{D} \) then \[ \frac{P_t (c f) - (c f)}{t} = c \frac{P_t f - f}{t} \to c G f \text{ as } t \downarrow 0\]
- If \( f, \, g \in \mathscr{D} \) then \[ \frac{P_t(f + g) - (f + g)}{t} = \frac{P_t f - f}{t} + \frac{P_t g - g}{t} \to G f + G g \text{ as } t \downarrow 0\]
Note \( G \) is the (right) derivative at 0 of the function \( t \mapsto P_t f \). Because of the semigroup property, this differentiability property at \( 0 \) implies differentiability at arbitrary \( t \in [0, \infty) \). Moreover, the infinitesimal operator and the transition operators commute:
If \( f \in \mathscr{D} \) and \( t \in [0, \infty) \), then \( P_t f \in \mathscr{D} \) and the following derivative rules hold with respect to the supremum norm.
 All limits and statements about derivatives and continuity are with respect to the supremum norm.
- By assumption, \[ \frac{1}{h}(P_h f - f) \to G f \text{ as } h \downarrow 0 \] Since \( P_t \) is a bounded, linear operator on the space \( \mathscr{C}_0 \), it preserves limits, so \[ \frac{1}{h}(P_t P_h f - P_t f) = \frac{1}{h}(P_{t+h} f - P_t f) \to P_t G f \text{ as } h \downarrow 0 \] This proves the result for the derivative from the right. But since \( t \mapsto P_t f \) is continuous, the the result is also true for the two-sided derivative.

The last result gives a possible solution to the dilema that motivated this discussion in the first place. If we want to construct a Markov process with desired properties, to model a a real system for example, we can start by constructing an appropriate generator \( G \) and then solve the initial value problem \[P^\prime_t = G P_t, \quad P_0 = I \] to obtain the transition operators \( \bs{P} = \{P_t: t \in [0, \infty)\} \). The next theorem gives the relationship between the potential operators and the infinitesimal operator, which in some ways is better. This relationship is analogous to the relationship between the potential operators and the one-step operator given above in discrete time
Suppose \( \alpha \in (0, \infty) \).

Proof
- By definition, if \( f \in \mathscr{D} \) then \( G f \in \mathscr{C}_0 \).  The last term is \( \alpha U_\alpha f \).
- Suppose that \( f \in \mathscr{C}_0 \).  The second term converges to \(\alpha U_\alpha f\) as \( t \downarrow 0 \).
For \( \alpha \gt 0 \), the operators \( U_\alpha \) and \( G \) have an inverse relationship.
Suppose again that \( \alpha \in (0, \infty) \).
- \( U_\alpha = (\alpha I - G)^{-1}: \mathscr{C}_0 \to \mathscr{D}\)
- \( G = \alpha I - U_\alpha^{-1} : \mathscr{D} \to \mathscr{C}_0\)
Proof
Recall that \( U_\alpha: \mathscr{C}_0 \to \mathscr{D} \) and \( G: \mathscr{D} \to \mathscr{C}_0 \)
- By part(a) the previous result we have \( \alpha U_\alpha - U_\alpha G = I \) so \( U_\alpha(\alpha I - G) = I \). By part (b) we have \(\alpha U_\alpha - G U_\alpha = I\) so \( (\alpha I - G) U_\alpha = I \).
- This follows from (a).
So, from the generator \( G \) we can determine the potential operators \( \bs{U} = \{U_\alpha: \alpha \in (0, \infty)\} \), which in turn determine the transition operators \( \bs{P} = \{P_t: t \in (0, \infty)\} \). In continuous time, transition operators \( \bs{P} = \{P_t: t \in [0, \infty)\} \) can be obtained from the single, infinitesimal operator \( G \) in a way that is reminiscent of the fact that in discrete time, the transition operators \( \bs{P} = \{P^n: n \in \N\} \) can be obtained from the single, one-step operator \( P \).
Examples and Applications
Our first example is essentially deterministic.
Consider the Markov process \( \bs{X} = \{X_t: t \in [0, \infty)\} \) on \( \R \) satisfying the ordinary differential equation \[ \frac{d}{dt} X_t = g(X_t), \quad t \in [0, \infty) \] where \( g: \R \to \R \) is Lipschitz continuous. The infinitesimal operator \( G \) is given by \( G f(x) = f^\prime(x) g(x)\) for \( x \in \R \) on the domain \( \mathscr{D} \) of functions \( f: \R \to \R \) where \( f \in \mathscr{C}_0\) and \(f^\prime \in \mathscr{C}_0 \).
Proof
Recall that the only source of randomness in this process is the initial sate \( X_0 \). By the continuity assumptions on \( g \), there exists a unique solution \( X_t(x) \) to the differential equation with initial value \( X_0 = x \), defined for all \( t \in [0, \infty) \). The transition operator \( P_t \) for \( t \in [0, \infty) \) is defined on \( \mathscr{B} \) by \( P_t f(x) = f[X_t(x)] \) for \( x \in \R \). By the ordinary chain rule, if \( f \) is differentiable, \[ \frac{P_t f(x) - f(x)}{t} = \frac{f[X_t(x)] - f(x)}{t} \to f^\prime(x) g(x) \text{ as } t \downarrow 0 \]
Our next example considers the Poisson process as a Markov process. Compare this with the binomial process above.
Let \( \bs{N} = \{N_t: t \in [0, \infty)\} \) denote the Poisson process on \( \N \) with rate \( \beta \in (0, \infty) \). Define the Markov process \( \bs{X} = \{X_t: t \in [0, \infty)\} \) by \( X_t = X_0 + N_t \) where \( X_0 \) takes values in \( \N \) and is independent of \( \bs{N} \).
- For \( t \in [0, \infty) \), show that the probability transition matrix \( P_t \) of \( \bs{X} \) is given by \[ P_t(x, y) = e^{-\beta t} \frac{(\beta t)^{y - x}}{(y - x)!}, \quad x, \, y \in \N, \, y \ge x \]
- For \( \alpha \in [0, \infty) \), show that the potential matrix \( U_\alpha \) of \( \bs{X} \) is given by \[ U_\alpha(x, y) = \frac{1}{\alpha + \beta} \left(\frac{\beta}{\alpha + \beta}\right)^{y - x}, \quad x, \, y \in \N, \, y \ge x \]
- For \( \alpha \gt 0 \) and \( x \in \N \), identify the probability distribution defined by \( \alpha U_\alpha(x, \cdot) \).
- Show that the infinitesimal matrix \( G \) of \( \bs{X} \) is given by \( G(x, x) = -\beta \), \( G(x, x + 1) = \beta \) for \( x \in \N \).
Solutions
- Note that for \( t \in [0, \infty) \) and \( x \in \N \), \( P_t(x, \cdot) \) is the (discrete) PDF of \( x + N_t \) since \( N_t \) has the Poisson distribution with parameter \( \beta t \).
- Let \( \alpha \in [0, \infty) \) and let \( x, \, y \in \N \) with \( x \le y \). Then \begin{align*} U_\alpha(x, y) & = \int_0^\infty e^{-\alpha t} P_t(x, y) \, dt = \int_0^\infty e^{-\alpha t} e^{-\beta t} \frac{(\beta t)^{y - x}}{(y - x)!} dt \\ & = \frac{\beta^{y - x}}{(y - x)!} \int_0^\infty e^{-(\alpha + \beta) t} t^{y - x} \, dt \end{align*} The change of variables \( s = (\alpha + \beta)t \) gives \[ U_\alpha(x, y) = \frac{\beta^{y-x}}{(y - x)! (\alpha + \beta)^{y - x + 1}} \int_0^\infty e^{-s} s^{y-x} \, ds \] But the last integral is \( \Gamma(y - x + 1) = (y - x)! \). Simplifying gives the result.
- For \( \alpha \gt 0 \), \[ \alpha U_\alpha(x, y) = \frac{\alpha}{\alpha + \beta} \left(\frac{\beta}{\alpha + \beta}\right)^{y - x}, \quad x, \, y \in \N, \, y \ge x \] As a function of \( y \) for fixed \( x \), this is the PDF of \( x + Y_\alpha \) where \( Y_\alpha \) has the geometric distribution with parameter \( \frac{\alpha}{\alpha + \beta} \).
- Note that for \( x, \, y \in \N \), \( G(x, y) = \frac{d}{dt} P_t(x, y) \bigm|_{t=0} \). By simple calculus, this is \( -\beta\) if \( y = x \), \( \beta \) if \( y = x + 1 \), and 0 otherwise.


11.2: Assess and Analyze the Impact of Nutrition on the Cardiovascular System
By the end of this section, you should be able to:
- 11.1.1 Recognize cues that indicate the impact of nutrition on the cardiovascular system.
- 11.1.2 Analyze cues to determine the impact of nutrition on the cardiovascular system.
Normal Function of the Cardiovascular System
The cardiovascular system includes the heart, blood, blood vessels, lymph, lymphatic vessels, and glands that move blood and lymph through the body. Assessment of this system involves the mechanical and electrical functions of the heart and blood vessels and their effectiveness at circulating blood throughout the body. Assessment should also include an examination of a client’s nutritional status.
The cardiovascular system’s primary role is to circulate blood throughout the body. The components of this system include the heart, known as the cardiac muscle, and blood vessels. The blood vessels include the arteries , which carry oxygenated blood from the heart to other parts of the body; veins , which carry deoxygenated blood from the body to the heart; and capillaries , tiny blood vessels with thin walls where arteries and veins exchange blood supply.
The heart has two components—the electrical conduction system and the mechanical working component—that work together to effectively pump blood and maintain proper perfusion , or movement of blood, oxygen, and nutrients to tissues in the body. The electrical system sends an impulse to the excitable cardiac tissue cells and triggers them to mechanically contract in a specific pattern to push blood throughout the body.
Nutrition can heavily impact the heart’s ability to complete the task of pumping blood. Like other muscles in the body, the heart requires electrolytes to produce action-potential impulses in muscles. The electrical charge that sends an impulse to the heart to trigger contraction is highly dependent on the involuntary process of potassium and sodium ion exchange (Balchem, 2021). Likewise, muscles, including the heart, are dependent on calcium and magnesium ion exchange for physical contraction or movement (Balchem, 2021). Low levels of these electrolytes can result in dysrhythmias and ineffective cardiac output and perfusion.
For the mechanical working component of the heart to function properly, the cardiac muscle must be supplied with oxygen and nutrients from the coronary arteries. The cardiac muscle must also be in good physical working condition and able to receive impulses from the heart’s electrical conduction system. The mechanical working component is not necessary for the electrical conduction system to function; but, without the mechanical working component, no blood is pumped through the body—and if the electrical conduction system is functioning independently, there will only be pulseless electrical activity (PEA) , or electrical conduction throughout the heart without physical movement of the heart muscle.
Another consideration is the vascular portion of the cardiovascular system. While the heart requires electrolytes to create electrical impulses to trigger contractions strong enough to circulate blood throughout the body, it may still be unable to adequately perfuse the body if the vessels are blocked or become hardened, as in arteriosclerosis . The primary cause of this disease is a nutritional intake high in unhealthy fats and cholesterol. When low-density lipid (LDL) cholesterol levels are chronically high, fatty deposits known as plaques form in artery walls and reduce blood flow through that vessel (Icahn School of Medicine at Mount Sinai, 2023). When the heart is forced to push against the increased pressure from the less-elastic and smaller diameter of the diseased vessels, it causes the heart muscle to continually be overworked. The heart muscle can become damaged, and the eventual result can be heart failure.
The other consideration in normal function of the heart is the pressure it exerts against the vessel walls when contracting and relaxing, known as blood pressure . When blood pressure is chronically high, the client has an increased risk of heart failure. Blood pressure is largely regulated in the body by the renin-angiotensin-aldosterone system (RAAS) that utilizes fluid and electrolytes, especially sodium (Martyniak and Tomasik, 2022). This system works to change the sodium concentration in the blood to raise or lower blood pressure as needed by the body (Martyniak and Tomasik, 2022). When the concentration of sodium is chronically high due to excessive intake and is not able to be excreted by the kidneys, the RAAS system is unable to adequately control blood pressure.
There are two main circulatory systems of interconnected blood vessels that flow through the body: the systemic circulatory system and the pulmonary circulatory system. The systemic system provides blood to the body’s tissues, organs, and cells, and the pulmonary system transports blood between the heart and lungs allowing for oxygen and carbon dioxide exchange to occur (Sherrell, 2021).
Through the circulation of blood, the cardiovascular system performs several other functions. Blood is considered a “fluid connective tissue” (Visible Body, 2023), and it supports, protects, and gives structure to the blood vessels. Blood provides numerous other functions, including:
- Clotting for tissue repair
- Protection from infection
- Transportation of body wastes to the kidneys for excretion
- Exchange of oxygen and carbon dioxide between body tissues
- Regulating body temperature
- Transporting nutrients and hormones throughout the body for use and functional regulation (American Society of Hematology, 2023)
Assessment of Nutrition and the Function of the Cardiovascular System
A thorough assessment of the nutritional status of a client’s cardiovascular system involves examining each component. Many diseases are associated with poor nutrition or malnutrition. The most effective method of determining whether nutrition is a contributing factor is to evaluate each component individually. Note that the electrical conduction system and the mechanical working component should be assessed separately.
Assessing Electrical Conduction System of the Heart
The normal electrical pattern of the heart (Figure 11.2) starts with the initiation of an impulse at the sinoatrial node (SA node) , the pacemaker of the heart. The impulse then travels through the atria , the top two heart chambers, causing atrial contraction, down to the atrioventricular node (AV node) , an electrical gatekeeper that controls a delay between the atrial and ventricular impulse, allowing blood to fill the ventricle. From the AV node, the impulse travels to the bundle of His , a bundle of conducting fibers that branch to each ventricle, then down both bundle branches to the Purkinje fibers , or terminal-conducting fibers in the ventricles. This action causes the ventricles , or bottom chambers of the heart, to contract (Johns Hopkins Medicine, 2023b).
An Electrocardiogram (ECG/EKG) or telemetry can be used to evaluate the electrical conduction of the heart. An ECG is a recording of a heart’s electrical activity through electrodes placed on the chest to capture signals. Telemetry is the continual measurement of a heart’s ECG through a portable device that automatically transmits the reading to a monitor.
A nurse should know several important measurement facts of a telemetry wave (Figure 11.3):
- Each small square on the telemetry strip represents 0.04 seconds.
- Each portion of the wave indicates a different part of the cardiac polarization (relaxation) and depolarization (contraction) cycle.
- Each wave is expected to have a certain appearance or shape and specific deflection: positive being upward and negative being downward.
For the heart to be in optimal health, the electrical conductivity should be a normal sinus rhythm (NSR) . An NSR is a rhythm that originates from the sinus node and is characteristic of a healthy heart’s cardiac rhythm. With an NSR, the wave tracing should be within normal measurements, and the deflections and appearances or shapes of all wave components should have a pulse that is calculated between 60 and 100 beats per minute. In addition, the waves should be consistent and resemble each other. Table 11.1 outlines the telemetry wave components, normal measurement, existence in the cardiac cycle, and the expected shape and deflection.
| Wave/Segment | Normal Measurement of Segment | Part of Cardiac Cycle | Wave/Segment Appearance/Shape and Deflection |
|---|---|---|---|
| P | Atrial depolarization | Positive deflection/smooth hill shape | |
| Q | Start of ventricular depolarization | Negative deflection/sharp downstroke if shown (Some normal telemetry readings do not have a true Q wave.) | |
| R | Ventricular depolarization | Positive deflection/sharp mountain peak upward | |
| S | Ventricular depolarization | Negative deflection/sharp downward end of the mountain | |
| T | Ventricular repolarization | Positive deflection/ smooth hill shape | |
| U | Atrial repolarization: not an expected finding on every client | Positive deflection/smooth hill shape (Most normal telemetry readings do not have a U wave.) | |
| QRS complex | 0.06–0.12 seconds | Ventricular depolarization | Cluster of QRS waves together |
| PR interval | 0.12–0.20 seconds | Beginning of atrial depolarization to beginning of ventricular depolarization | P wave and segment between P wave and Q wave |
| QT interval | 0.40–0.44 seconds | Beginning of ventricular depolarization through ventricular repolarization | QRS complex, ST segment, and to end of T wave |
| ST segment | Isoelectric line, not measured separately | Time in-between ventricular depolarization and repolarization | End of ST segment to beginning of T wave—must be on isoelectric line, no deflection |
There are certain telemetry changes in which nutritional issues and electrolyte abnormalities are often the primary cause. Common examples include:
- Long QT syndrome —when the QT interval is greater than 0.47 seconds in males and 0.48 seconds in females. This can lead to life-threatening dysrhythmias and can be caused by low levels of potassium, magnesium, or calcium.
- Torsades de pointes —a life-threatening dysrhythmia known as a multifocal ventricular tachycardia, caused by low levels of magnesium (Cohagan & Brandis, 2022).
- A peaked or mountain T wave illustrates hyperkalemia (high levels of potassium in the blood) (Bord, 2022).
Assessing the Mechanical Working Component of the Heart and the Blood Vessels
A healthy heart represents the flow of an adequate blood supply, proper stimulation through the electrical system, and effectiveness at physically pumping blood through the body. Normal blood flow through the heart starts when deoxygenated blood is returned to the heart from the body via the superior vena cava. It travels through the right atrium, through the tricuspid valve, and into the right ventricle. The blood is then pumped through the pulmonary valve and into the pulmonary artery to be re-oxygenated in the lungs (Figure 11.4). Once the blood has gone through the exchanges needed and is oxygenated, it returns to the heart through the pulmonary veins into the left atrium. It will then travel through the mitral valve into the left ventricle, through the aortic valve into the aorta, and out to the body for use (Pediatric Heart Specialists, 2023).
The oxygenated blood is returned to the body through two types of arteries . Elastic arteries are located closer to the heart and are more flexible than muscular arteries . Muscular arteries contain smoother muscle and are farther away from the heart. Both types of arteries must have strong walls, because arteries handle higher pressures from the blood being pumped out of the heart (Cleveland Clinic, 2023b). The main function of arteries is to distribute the oxygenated blood to all body organs and tissues, including the heart itself, through the coronary arteries. Refer to Figure 11.5 for a visual of the artery wall.
As blood travels through the body via the arteries, the arteries get smaller in diameter until they terminate into arterioles , the smallest arteries. These arterioles link with capillaries and then connect to venules , the smallest veins (Gupta and Shea, 2022). Capillaries are thin-walled vessels that connect arterioles and venules. As shown in Figure 11.6, they are the vehicle for the final exchange of oxygen and nutrients into tissues and wastes into the blood (Gupta & Shea, 2022).
After passing through the capillaries and venules, blood continues into the veins on the way back to the heart. Veins have thinner walls than arteries, because pressures are lower in this portion of the cardiovascular system. Veins also dilate in response to fluid. Some veins, like the large veins in the legs, have valves in them to prevent backflow of blood (Gupta & Shea, 2022). Once blood has traveled through the veins, it ends up in the heart to start the cycle all over again.
Assessment
The nurse should complete a client’s assessment by recording manual blood pressure and apical pulse (a pulse point on the chest at the apex of the heart). In addition, the nurse should look for other signs and symptoms of adequate perfusion , including:
- Quality and rate of peripheral pulses (carotid auscultation and palpation, brachial, radial, ulnar, femoral, popliteal, and pedal)
- Skin color check for pallor or cyanosis
- Skin temperature for degree of warmth and presence of moisture (Malik & Goyal, 2022)
- Skin turgor (elasticity) to check for edema and venous stasis sores or discoloration
- Capillary refill, which should be less than 2 seconds
Abnormal findings and their relationship to specific nutritional deficiencies are discussed in the next section.
Read the following clinical scenario and then answer the questions that follow. This case study will evolve throughout the chapter.
Mr. Thompson reports to the provider and nurse that he has had difficulty achieving and maintaining an erection and has noticed worrisome leg swelling. He also states that he was told by a nurse at a health fair a month ago that he should have his blood pressure checked, because it was 139/93 mm Hg. Today his blood pressure is 126/88 mm Hg. The swelling in his legs appears to be dependent edema that is only present when his legs are downward for periods of time and appears to be due to insufficiency of blood return in the lower extremities.
-
The client has hypertension, peripheral arterial disease, and erectile dysfunction.
-
The client has hypertension, peripheral vascular disease, and erectile dysfunction.
-
The client has elevated blood pressure, peripheral vascular disease, and erectile dysfunction.
-
Electrocardiogram
-
Echocardiogram
-
Troponin
-
Chest X-ray
The provider orders an ECG that is done at bedside and shows no abnormalities. The provider also orders a fasting lipid panel (serum cholesterol levels) to help determine the client’s risk for heart disease. Because Mr. Thompson is not fasting, the provider instructs him to return in the morning for a blood draw and to return in one week for follow-up. The provider also explains to Mr. Thompson that erectile dysfunction can be directly related to the decreased blood flow caused by hypertension. The client’s hypertension will have to be managed before the erectile dysfunction can be fully evaluated for cause.
(source: Antipolis, S. (2020).  European Society of Cardiology . Retrieved from www.escardio.org/The-ESC/Pre...-your-sex-life)
Analysis of Nutrition and the Cardiovascular System
A client’s nutritional status and needs can be identified with a physical assessment and lab results. Potassium, cholesterol, and sodium levels can be monitored to best track nutritional needs for cardiovascular health.
Nutritional Analysis and the Impact on the Cardiovascular System
Nutritional issues can lead to changes in the electrical conduction of the heart. As a result, the client’s telemetry rhythm can become abnormal. Rhythm changes include atrial fibrillation , atrial flutter , T-wave abnormalities , ST segment changes related to AMI, U-wave presence or prominence, torsades de pointes , prolonged QT interval, and supraventricular tachycardia. Rhythm changes can relate to one or more deficiencies. See Table 11.2 for a list of telemetry changes, the nutritional relation, and client symptoms.
| Telemetry Change | Nutritional Relation(s) | Client Symptom(s) |
|---|---|---|
| Atrial fibrillation/atrial flutter |
|
|
| Torsade de pointes |
|
|
| Supraventricular tachycardia |
|
|
| T-wave changes |
|
|
| U-wave presence/prominence |
|
|
| ST segment changes/AMI |
|
|
| QT interval |
|
|
| Many other rhythm disturbances |
|
|
Numerous nutritional factors create or impact cardiovascular disease s such as coronary heart disease , cardiomyopathy (a disease of the heart muscle), and hypertension . Table 11.3 outlines a handful of cardiovascular diseases, the impact of nutrition, and the effect on the body.
| Disease | Nutrition Impact | Effects on the Body |
|---|---|---|
| Coronary artery disease (CAD) |
Dietary intake that is high in cholesterol,
sodium, trans fats, added sugars, and low potassium and high fluid intake (for CHF and cardiomyopathy and especially with increased sodium specifically) |
Can lead to hypertension, cardiomyopathy, and AMI |
| Peripheral vascular disease (PVD) | Slows return of blood to the heart, leading to pooling of deoxygenated blood in the lower extremities | |
| Peripheral arterial disease (PAD) | Slows flow of blood to the body causing ischemia | |
| Chronic heart failure (CHF) | Causes weakening of the heart muscle | |
| Cardiomyopathy | Causes enlargement and weakening of the heart muscle | |
| Hypertension | Increases workload on the heart leading to cardiomyopathy and CHF |


8.13: What is a Function?
Learning Outcomes
- Define functions in Microsoft Excel
A function is a preexisting formula in Excel that does calculations based on specific values. In previous modules, you learned a few Excel functions: COUNTIF, IF, and SUM. Excel offers hundreds of functions from broadly used to highly specialized. For now, we will focus on financial, logical and look-up functions and formulas.
A few general tips to remember about functions and formulas:
- Functions begin with an equal sign. Excel will begin suggesting functions as soon as you begin typing the equal sign.
-
Functions take one or more arguments, with cell references or numbers, enclosed in parentheses.
- When a function needs more than one argument, enter a comma between the arguments without a space and close the formula inside of parentheses.
- There are multiple ways to enter functions in Excel; type in the formula bar under the ribbon (1), clicking on the fx button by the formula bar (2), select a cell and type = (3), select the Formulas tab Insert Function button (4) or select a specific function based on a category in the Formulas tab (5).
These next functions and formulas build out your skills in Excel and assist you with business, science, statistics and many other types of data analysis. Let’s get started.
Contributors and Attributions
CC licensed content, Original
- What is a Function?. Authored by : Sherri Pendleton. Provided by : Lumen Learning. License : CC BY: Attribution


11.2: Use the Rectangular Coordinate System (Part 2)
Verify Solutions to an Equation in Two Variables
All the equations we solved so far have been equations with one variable. In almost every case, when we solved the equation we got exactly one solution. The process of solving an equation ended with a statement such as x = 4. Then we checked the solution by substituting back into the equation.
Here’s an example of a linear equation in one variable, and its one solution.
\[\begin{split} 3x + 5 &= 17 \\ 3x &= 12 \\ x &= 4 \end{split}\]
But equations can have more than one variable. Equations with two variables can be written in the general form Ax + By = C. An equation of this form is called a linear equation in two variables.
An equation of the form Ax + By = C, where A and B are not both zero, is called a linear equation in two variables.
Notice that the word “line” is in linear.
Here is an example of a linear equation in two variables, x and y:
\[\begin{split} \textcolor{red}{A} x + \textcolor{blue}{B} y &= \textcolor{green}{C} \\ x + \textcolor{blue}{4} y &= \textcolor{green}{8} \\ \textcolor{red}{A = 1},\; \textcolor{blue}{B = 4},\; &\textcolor{green}{C = 8} \end{split}\]
Is y = −5x + 1 a linear equation? It does not appear to be in the form Ax + By = C. But we could rewrite it in this form.
| $$y = -5x + 1$$ | |
| Add 5x to both sides. | $$y + 5x = -5x + 1 + 5x$$ |
| Simplify. | $$y + 5x = 1$$ |
| Use the Commutative Property to put it in Ax + By = C. | $$\begin{split} \textcolor{red}{A} x + \textcolor{blue}{B} y &= C \\ 5 +\quad y &= 1 \end{split}$$ |
By rewriting y = −5x + 1 as 5x + y = 1, we can see that it is a linear equation in two variables because it can be written in the form Ax + By = C.
Linear equations in two variables have infinitely many solutions. For every number that is substituted for x, there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair (x, y). When we substitute these values of x and y into the equation, the result is a true statement because the value on the left side is equal to the value on the right side.
An ordered pair (x, y) is a solution to the linear equation Ax + By = C, if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.
Determine which ordered pairs are solutions of the equation x + 4y = 8: (a) (0, 2) (b) (2, −4) (c) (−4, 3)
Solution
Substitute the x- and y-values from each ordered pair into the equation and determine if the result is a true statement.
| (a) (0, 2) | (b) (2, −4) | (c) (−4, 3) |
| $$\begin{split} x = \textcolor{blue}{0},\; y &= \textcolor{red}{2} \\ x + 4y &= 8 \\ \textcolor{blue}{0} + 4 \cdot \textcolor{red}{2} &\stackrel{?}{=} 8 \\ 0 + 8 &\stackrel{?}{=} 8 \\ 8 &= 8\; \checkmark \end{split}$$ | $$\begin{split} x = \textcolor{blue}{2},\; y &= \textcolor{red}{-4} \\ x + 4y &= 8 \\ \textcolor{blue}{2} + 4(\textcolor{red}{-4}) &\stackrel{?}{=} 8 \\ 2 + (-16) &\stackrel{?}{=} 8 \\ -14 &\neq 8 \end{split}$$ | $$\begin{split} x = \textcolor{blue}{-4},\; y &= \textcolor{red}{3} \\ x + 4y &= 8 \\ \textcolor{blue}{-4} + 4 \cdot \textcolor{red}{3} &\stackrel{?}{=} 8 \\ -4 + 12 &\stackrel{?}{=} 8 \\ 8 &= 8\; \checkmark \end{split}$$ |
| (0, 2) is a solution. | (2, −4) is not a solution. | (−4, 3) is a solution. |
Determine which ordered pairs are solutions to the given equation: 2x + 3y = 6
(a) (3, 0) (b) (2, 0) (c) (6, −2)
- Answer
-
(a), (c)
Determine which ordered pairs are solutions to the given equation: 4x − y = 8
(a)(0, 8) (b) (2, 0) (c) (1, −4)
- Answer
-
(b), (c)
Determine which ordered pairs are solutions of the equation. y = 5x − 1: (a) (0, −1) (b) (1, 4) (c) (−2, −7)
Solution
Substitute the x- and y-values from each ordered pair into the equation and determine if it results in a true statement.
| (a) (0, -1) | (b) (1, 4) | (c) (−2, -7) |
| $$\begin{split} x &= \textcolor{blue}{0},\; y = \textcolor{red}{-1} \\ y &= 5x - 1 \\ \textcolor{red}{-1} &\stackrel{?}{=} 5(\textcolor{blue}{0}) - 1 \\ -1 &\stackrel{?}{=} 0 - 1 \\ -1 &= -1\; \checkmark \end{split}$$ | $$\begin{split} x &= \textcolor{blue}{1},\; y = \textcolor{red}{4} \\ y &= 5x - 1 \\ \textcolor{red}{4} &\stackrel{?}{=} 5(\textcolor{blue}{1}) - 1 \\ 4 &\stackrel{?}{=} 5 - 1 \\ 4 &= 4\; \checkmark \end{split}$$ | $$\begin{split} x &= \textcolor{blue}{-2},\; y = \textcolor{red}{-7} \\ y &= 5x - 1 \\ \textcolor{red}{-7} &\stackrel{?}{=} 5(\textcolor{blue}{-2}) - 1 \\ -7 &\stackrel{?}{=} -10 - 1 \\ -7 &\neq -11 \end{split}$$ |
| (0, -1) is a solution. | (1, 4) is a solution. | (−2, -7) is not a solution. |
Determine which ordered pairs are solutions of the given equation: y = 4x − 3
(a) (0, 3) (b) (1, 1) (c) (1, 1)
- Answer
-
(b)
Determine which ordered pairs are solutions of the given equation: y = −2x + 6
(a)(0, 6) (b) (1, 4) (c) (−2, −2)
- Answer
-
(a), (b)
Complete a Table of Solutions to a Linear Equation
In the previous examples, we substituted the x- and y-values of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do we find the ordered pairs if they are not given? One way is to choose a value for x and then solve the equation for y. Or, choose a value for y and then solve for x.
We’ll start by looking at the solutions to the equation y = 5x − 1 we found in Example \(\PageIndex{1}\). We can summarize this information in a table of solutions.
| y = 5x − 1 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | -1 | (0, -1) |
| 1 | 4 | (1, 4) |
To find a third solution, we’ll let x = 2 and solve for y.
| $$y = 5x − 1$$ | |
| Substitute x = \(\textcolor{blue}{2}\). | $$y = 5(\textcolor{blue}{2}) - 1$$ |
| Multiply. | $$y = 10 − 1$$ |
| Simplify. | $$y = 9$$ |
The ordered pair is a solution to y = 5x - 1. We will add it to the table.
| y = 5x − 1 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | -1 | (0, -1) |
| 1 | 4 | (1, 4) |
| 2 | 9 | (2, 9) |
We can find more solutions to the equation by substituting any value of x or any value of y and solving the resulting equation to get another ordered pair that is a solution. There are an infinite number of solutions for this equation.
Complete the table to find three solutions to the equation y = 4x − 2:
| y = 4x - 2 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | ||
| -1 | ||
| 2 |
Solution
Substitute x = 0, x = −1, and x = 2 into y = 4x − 2.
| x = \(\textcolor{blue}{0}\) | x = \(\textcolor{blue}{-1}\) | x = \(\textcolor{blue}{2}\) |
| y = 4x − 2 | y = 4x − 2 | y = 4x − 2 |
| y = 4 • \(\textcolor{blue}{0}\) − 2 | y = 4(\(\textcolor{blue}{-1}\)) − 2 | y = 4 • \(\textcolor{blue}{2}\) − 2 |
| y = 0 − 2 | y = −4 − 2 | y = 8 − 2 |
| y = −2 | y = −6 | y = 6 |
| (0, −2) | (−1, −6) | (2, 6) |
The results are summarized in the table.
| y = 4x - 2 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | -2 | (0, -2) |
| -1 | -6 | (-1, -6) |
| 2 | 6 | (2, 6) |
Complete the table to find three solutions to the equation: y = 3x − 1.
| y = 3x − 1 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | ||
| -1 | ||
| 2 |
- Answer
-
y = 3x − 1 x y (x, y) 0 -1 (0, -1) -1 -4 (-1, -4) 2 5 (2, 5)
Complete the table to find three solutions to the equation: y = 6x + 1.
| y = 6x + 1 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | ||
| 1 | ||
| -2 |
- Answer
-
y = 6x + 1 x y (x, y) 0 1 (0, 1) -1 7 (1, 7) 2 -11 (-2, -11)
Complete the table to find three solutions to the equation 5x − 4y = 20:
| 5x − 4y = 20 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | ||
| 0 | ||
| 5 |
Solution
| $$\begin{split} x &= \textcolor{blue}{0} \\ 5x - 4y &= 20 \\ 5 \cdot \textcolor{blue}{0} - 4y &= 20 \\ \textcolor{blue}{0} - 4y &= 20 \\ -4y &= 20 \\ y &= -5 \\ (0,\; &-5) \end{split}$$ | $$\begin{split} y &= \textcolor{red}{0} \\ 5x - 4y &= 20 \\ 5x - 4 \cdot \textcolor{red}{0} &= 20 \\ 5x - 0 &= 20 \\ 5x &= 20 \\ x &= 4 \\ (&4,\; 0) \end{split}$$ | $$\begin{split} y &= \textcolor{red}{5} \\ 5x - 4y &= 20 \\ 5x - 4 \cdot \textcolor{red}{5} &= 20 \\ 5x - 20 &= 20 \\ 5x &= 40 \\ x &= 8 \\ (&8,\; 5) \end{split}$$ |
The results are summarized in the table.
| 5x − 4y = 20 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | -5 | (0, -5) |
| 4 | 0 | (-4, 0) |
| 8 | 5 | (8, 5) |
Complete the table to find three solutions to the equation: 2x − 5y = 20.
| 2x - 5y = 20 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | ||
| 0 | ||
| -5 |
- Answer
-
2x - 5y = 20 x y (x, y) 0 -4 (0, -4) 10 0 (10, 0) -5 -6 (-5, -6)
Complete the table to find three solutions to the equation: 3x − 4y = 12.
| 3x - 4y = 12 | ||
| x | y | (x, y) |
| 0 | ||
| 0 | ||
| -4 |
- Answer
-
3x - 4y = 12 x y (x, y) 0 -3 (0, -3) 4 0 (4, 0) -4 -6 (-4, -6)
Find Solutions to Linear Equations in Two Variables
To find a solution to a linear equation, we can choose any number we want to substitute into the equation for either x or y. We could choose 1, 100, 1,000, or any other value we want. But it’s a good idea to choose a number that’s easy to work with. We’ll usually choose 0 as one of our values.
Find a solution to the equation 3x + 2y = 6.
Solution
| Step 1 : Choose any value for one of the variables in the equation. |
We can substitute any value we want for x or any value for y. Let's pick x = 0. What is the value of y if x = 0? |
|
| Step 2 : Substitute that value into the equation. Solve for the other variable. |
Substitute 0 for x. Simplify. Divide both sides by 2. |
$$\begin{split} 3x + 2y &= 6 \\ 3 \cdot \textcolor{blue}{0} + 2y &= 6 \\ 0 + 2y &= 6 \\ 2y &= 6 \\ y &= 3 \end{split}$$ |
| Step 3 : Write the solution as an ordered pair. | So, when x = 0, y = 3. | This solution is represented by the ordered pair (0, 3). |
| Step 4 : Check. |
Substitute x = \(\textcolor{blue}{0}\), y = \(\textcolor{red}{3}\) into the equation 3x + 2y = 6. Is the result a true equation? Yes! |
$$\begin{split} 3x + 2y &= 6\\ 3 \cdot \textcolor{blue}{0} + 2 \cdot \textcolor{red}{3} &\stackrel{?}{=} 6 \\ 0 + 6 &\stackrel{?}{=} 6 \\ 6 &= 6\; \checkmark \end{split}$$ |
Find a solution to the equation: 4x + 3y = 12.
- Answer
-
Answers will vary
Find a solution to the equation: 2x + 4y = 8.
- Answer
-
Answers will vary
We said that linear equations in two variables have infinitely many solutions, and we’ve just found one of them. Let’s find some other solutions to the equation 3x + 2y = 6.
Find three more solutions to the equation 3x + 2y = 6.
Solution
To find solutions to 3x + 2y = 6, choose a value for x or y. Remember, we can choose any value we want for x or y. Here we chose 1 for x, and 0 and −3 for y.
| Substitute it into the equation. | $$\begin{split} y &= \textcolor{red}{0} \\ 3x + 2y &= 6 \\ 3x + 2(\textcolor{red}{0}) &= 6 \end{split}$$ | $$\begin{split} x &= \textcolor{blue}{1} \\ 3x + 2y &= 6 \\ 3(\textcolor{blue}{1}) + 2y &= 6 \end{split}$$ | $$\begin{split} y &= \textcolor{red}{-3} \\ 3x + 2y &= 6 \\ 3x + 2(\textcolor{red}{-3}) &= 6 \end{split}$$ |
| Simplify. Solve. | $$\begin{split} 3x + 0 &= 6 \\ 3x &= 6 \end{split}$$ | $$\begin{split} 3 + 2y &= 6 \\ 2y &= 3 \end{split}$$ | $$\begin{split} 3x - 6 &= 6 \\ 3x &= 12 \end{split}$$ |
| x = 2 | y = \(\dfrac{3}{2}\) | x = 4 | |
| Write the ordered pair. | (2, 0) | (1, \(\dfrac{3}{2}\)) | (4, −3) |
Check your answers.
| (2, 0) | (1, \(\dfrac{3}{2}\)) | (4, −3) |
| $$\begin{split} 3x + 2y &= 6 \\ 3 \cdot \textcolor{blue}{2} + 2 \cdot \textcolor{red}{0} &\stackrel{?}{=} 6 \\ 6 + 0 &\stackrel{?}{=} 6 \\ 6 &= 6\; \checkmark \end{split}$$ | $$\begin{split} 3x + 2y &= 6 \\ 3 \cdot \textcolor{blue}{1} + 2 \cdot \textcolor{red}{\dfrac{3}{2}} &\stackrel{?}{=} 6 \\ 3 + 3 &\stackrel{?}{=} 6 \\ 6 &= 6\; \checkmark \end{split}$$ | $$\begin{split} 3x + 2y &= 6 \\ 3 \cdot \textcolor{blue}{4} + 2 \cdot (\textcolor{red}{-3}) &\stackrel{?}{=} 6 \\ 12 + (-6) &\stackrel{?}{=} 6 \\ 6 &= 6\; \checkmark \end{split}$$ |
So (2, 0), (1, \(\dfrac{3}{2}\)) and (4, −3) are all solutions to the equation 3x + 2y = 6. In the previous example, we found that (0, 3) is a solution, too. We can list these solutions in a table.
| 3x + 2y = 6 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | 3 | (0, 3) |
| 2 | 0 | (2, 0) |
| 1 | \(\dfrac{3}{2}\) | (1, \(\dfrac{3}{2}\)) |
| 4 | -3 | (4, -3) |
Find three solutions to the equation: 2x + 3y = 6.
- Answer
-
Answers will vary
Find three solutions to the equation: 4x + 2y = 8.
- Answer
-
Answers will vary
Let’s find some solutions to another equation now.
Find three solutions to the equation x − 4y = 8.
Solution
| Choose a value for x or y. | x = \(\textcolor{blue}{0}\) | y = \(\textcolor{red}{0}\) | y = \(\textcolor{red}{3}\) |
| Substitute it into the equation. | $$\textcolor{blue}{0} - 4y = 8$$ | $$x - 4 \cdot \textcolor{red}{0} = 8$$ | $$x - 4 \cdot \textcolor{red}{3} = 8$$ |
| Solve. | $$\begin{split} -4y &= 8 \\ y &= -2 \end{split}$$ | $$\begin{split} x - 0 &= 8 \\ x &= 8 \end{split}$$ | $$\begin{split} x - 12 &= 8 \\ x &= 20 \end{split}$$ |
| Write the ordered pair. | (0, −2) | (8, 0) | (20, 3) |
So (0, −2), (8, 0), and (20, 3) are three solutions to the equation x − 4y = 8.
| x − 4y = 8 | ||
|---|---|---|
| x | y | (x, y) |
| 0 | -2 | (0, -2) |
| 8 | 0 | (8, 0) |
| 20 | 3 | (20, 3) |
Remember, there are an infinite number of solutions to each linear equation. Any point you find is a solution if it makes the equation true.
Find three solutions to the equation: 4x + y = 8.
- Answer
-
Answers will vary
Find three solutions to the equation: x + 5y = 10.
- Answer
-
Answers will vary
Plotting Points
Identifying Quadrants
Verifying Solution to Linear Equation
Practice Makes Perfect
Plot Points on a Rectangular Coordinate System
In the following exercises, plot each point on a coordinate grid.
- (3, 2)
- (4, 1)
- (1, 5)
- (3, 4)
- (4, 1), (1, 4)
- (3, 2), (2, 3)
- (3, 4), (4, 3)
In the following exercises, plot each point on a coordinate grid and identify the quadrant in which the point is located.
- (a) (−4, 2) (b) (−1, −2) (c) (3, −5) (d) \(\left(2, \dfrac{5}{2}\right)\)
- (a) (−2, −3) (b) (3, −3) (c) (−4, 1) (d) \(\left(1, \dfrac{3}{2}\right)\)
- (a) (−1, 1) (b) (−2, −1) (c) (1, −4) (d) \(\left(3, \dfrac{7}{2}\right)\)
- (a) (3, -2) (b) (−3, 2) (c) (-3, −2) (d) (3, 2)
- (a) (4, -1) (b) (−4, 1) (c) (-4, -1) (d) (4, 1)
- (a) (−2, 0) (b) (−3, 0) (c) (0, 4) (d) (0, 2)
Identify Points on a Graph
In the following exercises, name the ordered pair of each point shown.
Verify Solutions to an Equation in Two Variables
In the following exercises, determine which ordered pairs are solutions to the given equation.
-
2x + y = 6
- (1, 4)
- (3, 0)
- (2, 3)
-
x + 3y = 9
- (0, 3)
- (6, 1)
- (-3, -3)
-
4x − 2y = 8
- (3, 2)
- (1, 4)
- (0, -4)
-
3x − 2y = 12
- (4, 0)
- (2, -3)
- (1, 6)
-
y = 4x + 3
- (4, 3)
- (-1, -1)
- (\(\dfrac{1}{2}\), 5)
-
y = 2x − 5
- (0, -5)
- (2, 1)
- (\(\dfrac{1}{2}\), -4)
-
y = \(\dfrac{1}{2}\)x − 1
- (2, 0)
- (-6, -4)
- (-4, -1)
-
y = \(\dfrac{1}{3}\) x + 1
- (-3, 0)
- (9, 4)
- (-6, -1)
Find Solutions to Linear Equations in Two Variables
In the following exercises, complete the table to find solutions to each linear equation.
y = 2x − 4
| x | y | (x, y) |
|---|---|---|
| -1 | ||
| 0 | ||
| 2 |
- y = 3x − 1
| x | y | (x, y) |
|---|---|---|
| -1 | ||
| 0 | ||
| 2 |
- y = − x + 5
| x | y | (x, y) |
|---|---|---|
| -2 | ||
| 0 | ||
| 3 |
- y = \(\dfrac{1}{3}\)x + 1
| x | y | (x, y) |
|---|---|---|
| 0 | ||
| 3 | ||
| 6 |
- y = \(− \dfrac{3}{2}\)x − 2
| x | y | (x, y) |
|---|---|---|
| -2 | ||
| 0 | ||
| 2 |
- x + 2y = 8
| x | y | (x, y) |
|---|---|---|
| 0 | ||
| 4 | ||
| 0 |
Everyday Math
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Weight of a baby
Mackenzie recorded her baby’s weight every two months. The baby’s age, in months, and weight, in pounds, are listed in the table, and shown as an ordered pair in the third column.
- Plot the points on a coordinate grid.
- Why is only Quadrant I needed?
| Age | Weight | (x, y) |
|---|---|---|
| 0 | 7 | (0, 7) |
| 2 | 11 | (2, 11) |
| 4 | 15 | (4, 15) |
| 6 | 16 | (6, 16) |
| 8 | 19 | (8, 19) |
| 10 | 20 | (10, 20) |
| 12 | 21 | (12, 21) |
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Weight of a child Latresha recorded her son’s height and weight every year. His height, in inches, and weight, in pounds, are listed in the table, and shown as an ordered pair in the third column.
- Plot the points on a coordinate grid.
- Why is only Quadrant I needed?
| Height | Weight | (x, y) |
| 28 | 22 | (28, 22) |
| 31 | 27 | (31, 27) |
| 33 | 33 | (33, 33) |
| 37 | 35 | (37, 35) |
| 40 | 41 | (40, 41) |
| 42 | 45 | (42, 45) |
Writing Exercises
- Have you ever used a map with a rectangular coordinate system? Describe the map and how you used it.
- How do you determine if an ordered pair is a solution to a given equation?
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


6.10: Current Issues- Stereotypes in Policing
-
- Last updated
- Save as PDF
- Alison S. Burke, David Carter, Brian Fedorek, Tiffany Morey, Lore Rutz-Burri, & Shanell Sanchez
- Southern Oregon University via OpenOregon
Human beings are infamous for stereotyping. A first impression when meeting a new person takes only seven seconds. According to Pitts (2013), smiling, shaking hands, introductions, speaking clearly, maintaining eye contact, looking smart, and not sitting down, are sure-fire ways to ensure a more positive stereotype; however regardless, the seven seconds is irrefutable.
Stereotyping in policing is almost a foregone conclusion. Citizens expect their police to protect them by being not only reactive but proactive. One of the most popular policing methods is to view a situation and proactively make a quick decision on whether or not a crime is about to occur, and if it is, stop it from happening. One of the ways police proactively operate is through stereotyping. “Police officers spend a great deal of time working their beats…one thing is common to all police officer working personalities: in an effort to know who or what is ‘wrong’ on their beat, police officers must know who is ‘right’ or who belongs.” [1]
When officers cross over the line is when they leave out the step of asking the who, what, where, when, why, and how after the stereotyping occurs, to confirm their thoughts. It is at this point that the officer is engaging in a type of implicit bias policing and this opens many doors to corruption. It is another slippery slope that officers must always be aware of while performing their many duties.
“Stereotyping or Terry Stop” Example
No matter how controversial policing is today, one common thread is that citizens want police
to be proactive, not reactive to violent crime. Whether it is stopping an active shooter in a school, a burglary, or even a robbery, proactivity in policing is necessary to halt horrendous crimes from occurring. In 1968 the United States Supreme Court decided
Terry v. Ohio
, which further explained reasonable searches under the 4th amendment and played a vital role in the below story.
I was one of six officers with my department, chosen to work in our first ever problem-solving unit (PSU). Our substation was placed in a neighborhood where 21 murders had occurred in just a few months. We were tasked with being proactive and working with the community to stop the bloodshed. The two-mile radius neighborhood were composed predominantly of minority residents, and their distrust of our presence was apparent immediately. The residents had stereotyped us as rotten apple police officers and at first, did not want our assistance, they did not trust us. This was further agitated by the fact that we were there to be proactive and stop any future murders from occurring. Plus, we had the Terry v. Ohio decision to assist with our proactive actions. The citizens did not understand what that meant. They also did not understand that as one of the officers pledged to protect the area, I was feeling like an unwanted officer in their neighborhood. They did not know that I got into policing to change the world, if even just a little. They did not know how frustrated it made me feel when no matter how hard I tried, no one would not trust me.  I cried that night when I got home from work, wondering how I could help this neighborhood if I couldn’t even get one mom to trust me? Whether or not it was true, stereotyping had occurred in this neighborhood. Then on top of it, I had to utilize the Terry v. Ohio decision to be proactive, to keep the murders from occurring. It felt like a Catch-22 with no solution. The answer did not come to me instantly. All I could think to do was my job and stop the murders (as I was directed to do). I remembered that four-year-old in my everyday actions. I knew my efforts could make him safer, but would he ever understand? I utilized Terry v. Ohio , by learning the neighborhood, recognizing the residents, and learning who belonged, and who did not. I was either on foot or on a bicycle in this neighborhood due to the small size. This allowed for a lot of interaction with the residents.  Ohio gave me the right to investigate further. At first, this angered the neighborhood. They felt we were harassing them, stereotyping them. I could understand how they felt that way, and instead of trying to make this neighborhood safe overnight, I decided to begin to change how the neighborhood perceived us, slowly. The way I did this was through education. Through my daily interactions, I talked to the citizens in the neighborhood about what I was doing and why. Instead of speaking in ‘general’ I spoke about only one incident at a time.
One night, at 2:00 a.m., I was walking with my partner through the neighborhood. I had to head back to the sub-station and as I rounded one building, I saw two citizens looking through a window of an apartment. I stopped and just watched. Everything ran through my mind. Had they lost their keys? Was this their apartment? Or, were they looking to break-in and burglarize the apartment, possibly even commit a home invasion and hurt those inside? All of this happened in seconds, not minutes. Because of Terry v. Ohio , I legally investigated. The two citizens did not live in the apartment and they were trying to burglarize it. One of the suspects had a gun. Because of Terry v. Ohio , I was able to be pro-active and stop this from occurring. A single mom and three children under six years of age lived in that apartment and were home. The mom did not trust banks and kept her savings in between her mattresses. I do not know what I stopped that night. I do not know if the suspects would have used the gun, or if they would have found the mom’s large cash savings or what else they might have taken; however, I did stop a burglary from occurring and that felt good. The next day while speaking to some of the citizens in the neighborhood, I explained this. I used this one example to explain why my unit was there. How we stopped this crime and how we all wanted to make a difference. This one story did not change how the neighborhood saw us; however, after many more stories such as this, I began to see a change.
Michelle, one of the citizens in the neighborhood got my cellular telephone number (yes we had cell phones back in the day!). She began calling me when she heard talk about a possible crime occurring. We hosted many events in the neighborhood as well. From ice cream socials, back-to-school fairs, and we even worked hard to find donations and get every child in the neighborhood a bicycle (or scooter). After three years, our substation closed. We had gone a year without a murder and the crime rate dropped 98%. To this day Michelle still calls me and we chat about what is currently occurring in our lives. Michelle is my friend and I dare say I think she thinks of me as her friend too. There was a lot of stereotyping that went on in that neighborhood during those years. I found my way through it all and I think the neighborhood did as well. In the end, we worked together through good ole community policing and made the area safe again.
-
Perez, D.W. (2011).
The paradoxes of police work
. Florence, KY: Cengage Publ.
↵


4.2: Mechanical Cooling – Principles
Click play on the following audio player to listen along as you read this section.
Mechanical cooling, or refrigeration , refers to any method that uses energy to actively cool an area. Examples include refrigerators and freezers, air conditioner units and heat pumps. Regardless of the scale of the cooling system, whether air conditioning a building or keeping the inside of your refrigerator cold, all cooling systems are governed by the first two laws of thermodynamics :
- Energy may be neither created nor destroyed, but may be changed from one form to another
Accordingly, thermal energy or heat may not be created to warm an area nor destroyed to cool an area. Instead we use heaters to convert either stored chemical energy (gas) or electrical energy into heat, or we use cooling systems to remove heat from a given area.
- Thermal energy always flows spontaneously in the form of heat from regions of higher temperature to regions of lower temperature, increasing the entropy of the system.
An object or area that is hotter than the ambient temperature surrounding it will naturally release heat until equilibrium is established and there is no longer a temperature gradient. A bowl of hot soup will cool to room temperature over time. Conversely, objects that are colder than the ambient temperature will absorb thermal energy until they reach ambient temperature. This fundamental relationship underlies the principle behind many styles of mechanical cooling systems , including DX, or direct expansion cooling.


1.9: Continuity
The angle measure is also assumed to be continuous. Namely, the following property of angle measure will become a part of the axioms:
The function
\[\measuredangle: (A, O, B) \mapsto \measuredangle AOB\]
is continuous at any triple of poitns \((A, O, B)\) such that \(O \ne A\) and \(O \ne B\) and \(\measuredangle AOB \ne \pi\).
To explain this property, we need to extend the notion of continuity to the functions between metric spaces. The definition is a straightforward generalization of the standard definition for the real-to-real functions.
Further, let \(\mathcal{X}\) and \(\mathcal{Y}\) be two metric spaces, and \(d_{\mathcal{X}}, d_{\mathcal{Y}}\) be their metrics.
A map \(f: \mathcal{X} \to \mathcal{Y}\) is called continuous at point \(A \in \mathcal{X}\) if for any \(\varepsilon > 0\) there is \(\delta > 0\), such that
\(d_{\mathcal{X}} (A, A') < \delta \Rightarrow d_{\mathcal{Y}}(f(A), f(A')) < \varepsilon.\)
(Informally it means that sufficiently small changes of \(A\) result in arbitrarily small changes of \(f(A)\).)
A map \(f: \mathcal{X} \to \mathcal{Y}\) is called continuous if it is continuous at every point \(A \in \mathcal{X}\).
One may define a continuous map of several variables the same way. Assume \(f(A, B, C)\) is a function which returns a point in the space \(\mathcal{Y}\) for a triple of points \((A, B, C)\) in the space \(\mathcal{X}\). The map \(f\) might be defined only for some triples in \(\mathcal{X}\).
Assume \(f(A, B, C)\) is defined. Then, we say that \(f\) is continuous at the triple \((A, B, C)\) if for any \(\varepsilon > 0\) there is \(\delta > 0\) such that
\(d_{\mathcal{Y}} (f(A, B, C), f(A',B',C')) < \varepsilon.\)
if \(d_{\mathcal{X}} (A, A') < \delta, d_{\mathcal{X}} (B, B') < \delta\), and \(d_{\mathcal{X}} (C, C') < \delta\).
Exercise \(\PageIndex{1}\)
Let \(\mathcal{X}\) be a metric space.
(a) Let \(A \in \mathcal{X}\) be a fixed point. Show that the function
\[f(B):= d_{\mathcal{X}} (A, B)\]
is continuous at any point \(B\).
(b) Show that \(d_{\mathcal{X}} (A, B)\) is continuous at any pair \(A, B \in \mathcal{X}\).
- Hint
-
(a). By the triangle inequality, \(|f(A') - f(A)| \le d(A', A)\). Therefore, we can take \(\delta = \varepsilon\).
(b). By the triangle inequality,
\[\begin{array} {rcl} {|f(A',B') - f(A, B)|} & \le & {|f(A',B') - F(A, B')| + |F(A, B') - F(A, B)|} \\ {} & \le & {d(A',A) + d(B',B).} \end{array}\]
Therefore, we can take \(\delta = \dfrac{\varepsilon}{2}\).
Exercise \(\PageIndex{2}\)
Let \(\mathcal{X}, \mathcal{Y}\), and \(\mathcal{Z}\) be metric spaces. Assume that the functions \(f: \mathcal{X} \to \mathcal{Y}\) and \(g: \mathcal{Y} \to \mathcal{Z}\) are continuous at any point, and \(h = g \circ f\) is their composition; that is, \(h(A) = g(f(A))\) for any \(A \in \mathcal{X}\). Show that \(h: \mathcal{X} \to \mathcal{Z}\) is continuous at any point.
- Hint
-
Fix \(A \in \mathcal{X}\) and \(B \in \mathcal{Y}\) such that \(f(A) = B\).
Fix \(\varepsilon > 0\). Since \(g\) is continuous at \(B\), there is a positive value \(\delta_1\) such that
\(d_{\mathcal{Z}} (g(B'), g(B)) < \varepsilon\) if \(d_{\mathcal{Y}} (B', B) < \delta_1\).
Since \(f\) is continuous at \(A\), there is \(\delta_2 > 0\) such that
\(d_{\mathcal{Y}} (f(A'), f(A)) < \delta_1\) if \(d_{\mathcal{X}} (A', A) < \delta_2\).
Since \(f(A) = B\), we get that
\(d_{\mathcal{Z}} (h(A'), h(A)) < \varepsilon\) if \(d_{\mathcal{X}} (A',A) < \delta_2\).
Hence the result.
Exercise \(\PageIndex{3}\)
Show that any distance-preserving map is continuous at any point.