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We first show that the generating function \( {h}_{n}\left( z\right) \) of the distribution of \( {Z}_{n} \) can be obtained from \( h\left( z\right) \) for any branching process. | We recall that the value of the generating function at the value \( z \) for any random variable \( X \) can be written as\n\n\[ h\left( z\right) = E\left( {z}^{X}\right) = {p}_{0} + {p}_{1}z + {p}_{2}{z}^{2} + \cdots . \]\n\nThat is, \( h\left( z\right) \) is the expected value of an experiment which has outcome \( {z... | Yes |
For the branching process of Example 10.8 we have\n\n\[ h\left( z\right) = 1/2 + \left( {1/4}\right) z + \left( {1/4}\right) {z}^{2}, \] | \[ {h}_{2}\left( z\right) = h\left( {h\left( z\right) }\right) = 1/2 + \left( {1/4}\right) \left\lbrack {1/2 + \left( {1/4}\right) z + \left( {1/4}\right) {z}^{2}}\right\rbrack \]\n\n\[ = + \left( {1/4}\right) {\left\lbrack 1/2 + \left( 1/4\right) z + \left( 1/4\right) {z}^{2}\right\rbrack }^{2} \]\n\n\[ = {11}/{16} + ... | Yes |
Assume that the probabilities \( {p}_{1},{p}_{2},\ldots \) form a geometric series: \( {p}_{k} = b{c}^{k - 1}, k = 1,2,\ldots \), with \( 0 < b \leq 1 - c \) and \( 0 < c < 1 \) . Then we have\n\n\[ \n{p}_{0} = 1 - {p}_{1} - {p}_{2} - \cdots \n\] | \[ \n= 1 - b - {bc} - b{c}^{2} - \cdots \n\]\n\n\[ \n= 1 - \frac{b}{1 - c}\text{.} \n\]\n\nThe generating function \( h\left( z\right) \) for this distribution is\n\n\[ \nh\left( z\right) = {p}_{0} + {p}_{1}z + {p}_{2}{z}^{2} + \cdots \n\]\n\n\[ \n= 1 - \frac{b}{1 - c} + {bz} + {bc}{z}^{2} + b{c}^{2}{z}^{3} + \cdots \n... | Yes |
Let us re-examine the Keyfitz data to see if a distribution of the type considered in Example 10.11 could reasonably be used as a model for this population. We would have to estimate from the data the parameters \( b \) and \( c \) for the formula \( {p}_{k} = b{c}^{k - 1} \) . | Solving Equation 10.6 and 10.7 for \( b \) and \( c \) gives\n\n\[ c = \frac{m - 1}{m - d} \]\n\nand\n\n\[ b = m{\left( \frac{1 - d}{m - d}\right) }^{2}. \]\n\nWe shall use the value 1.837 for \( m \) and .324 for \( d \) that we found in the Keyfitz example. Using these values, we obtain \( b = {.3666} \) and \( c = {... | Yes |
We now examine the random variable \( {Z}_{n} \) more closely for the case \( m < 1 \) (see Example 10.11). Fix a value \( t > 0 \) ; let \( \left\lbrack {t{m}^{n}}\right\rbrack \) be the integer part of \( t{m}^{n} \) . Then | \[ P\left( {{Z}_{n} = \left\lbrack {t{m}^{n}}\right\rbrack }\right) = {m}^{n}{\left( \frac{1 - d}{{m}^{n} - d}\right) }^{2}{\left( \frac{{m}^{n} - 1}{{m}^{n} - d}\right) }^{\left\lbrack {t{m}^{n}}\right\rbrack - 1} \] \[ = \frac{1}{{m}^{n}}{\left( \frac{1 - d}{1 - d/{m}^{n}}\right) }^{2}{\left( \frac{1 - 1/{m}^{n}}{1 -... | Yes |
Let us first assume that the buyer may sell the letter only to a single person. If you buy the letter you will want to compute your expected winnings. (We are ignoring here the fact that the passing on of chain letters through the mail is a federal offense with certain obvious resulting penalties.) Assume that each per... | Thus the chain in this situation is a highly unfavorable game. | No |
Example 10.15 Let \( X \) be a continuous random variable with range \( \\left\\lbrack {0,1}\\right\\rbrack \) and density function \( {f}_{X}\\left( x\\right) = 1 \) for \( 0 \\leq x \\leq 1 \) (uniform density). Then\n\n\[ \n{\\mu }_{n} = {\\int }_{0}^{1}{x}^{n}{dx} = \\frac{1}{n + 1}, \n\]\n\nand\n\n\[ \ng\\left( t\... | Here the series converges for all \( t \) . Alternatively, we have\n\n\[ \ng\\left( t\\right) = {\\int }_{-\\infty }^{+\\infty }{e}^{tx}{f}_{X}\\left( x\\right) {dx} \n\]\n\n\[ \n= {\\int }_{0}^{1}{e}^{tx}{dx} = \\frac{{e}^{t} - 1}{t}. \n\]\n\nThen (by L'Hôpital's rule)\n\n\[ \n{\\mu }_{0} = g\\left( 0\\right) = \\math... | Yes |
Let \( X \) have range \( \lbrack 0,\infty ) \) and density function \( {f}_{X}\left( x\right) = \lambda {e}^{-{\lambda x}} \) (exponential density with parameter \( \lambda \) ). In this case | \[ {\mu }_{n} = {\int }_{0}^{\infty }{x}^{n}\lambda {e}^{-{\lambda x}}{dx} = \lambda {\left( -1\right) }^{n}\frac{{d}^{n}}{d{\lambda }^{n}}{\int }_{0}^{\infty }{e}^{-{\lambda x}}{dx} = \lambda {\left( -1\right) }^{n}\frac{{d}^{n}}{d{\lambda }^{n}}\left\lbrack \frac{1}{\lambda }\right\rbrack = \frac{n!}{{\lambda }^{n}},... | Yes |
Let \( X \) have range \( \left( {-\infty , + \infty }\right) \) and density function \[ {f}_{X}\left( x\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2} \] (normal density). In this case we have \[ {\mu }_{n} = \frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{x}^{n}{e}^{-{x}^{2}/2}{dx} \] | \[ = \left\{ \begin{array}{ll} \frac{\left( {2m}\right) !}{{2}^{m}m!}, & \text{ if }n = {2m}, \\ 0, & \text{ if }n = {2m} + 1. \end{array}\right. \] (These moments are calculated by integrating once by parts to show that \( {\mu }_{n} = \) \( \left. {\left( {n - 1}\right) {\mu }_{n - 2}\text{, and observing that}{\mu }... | Yes |
Theorem 10.3 Suppose \( X \) is a continuous random variable with range contained in the interval \( \left\lbrack {-M, M}\right\rbrack \) . Then the series\n\n\[ g\left( t\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\mu }_{k}{t}^{k}}{k!} \]\n\nconverges for all \( t \) to an infinitely differentiable funct... | Proof. We have\n\[ {\mu }_{k} = {\int }_{-M}^{+M}{x}^{k}{f}_{X}\left( x\right) {dx} \]\n\nso\n\n\[ \left| {\mu }_{k}\right| \leq {\int }_{-M}^{+M}{\left| x\right| }^{k}{f}_{X}\left( x\right) {dx} \]\n\n\[ \leq {M}^{k}{\int }_{-M}^{+M}{f}_{X}\left( x\right) {dx} = {M}^{k}. \]\n\nHence, for all \( N \) we have\n\n\[ \mat... | Yes |
Theorem 10.4 If \( X \) is a bounded random variable, then the moment generating function \( {g}_{X}\left( t\right) \) of \( x \) determines the density function \( {f}_{X}\left( x\right) \) uniquely. | Sketch of the Proof. We know that\n\n\[ \n{g}_{X}\left( t\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\mu }_{k}{t}^{k}}{k!}\n\]\n\n\[ \n= {\int }_{-\infty }^{+\infty }{e}^{tx}f\left( x\right) {dx}\n\]\n\nIf we replace \( t \) by \( {i\tau } \), where \( \tau \) is real and \( i = \sqrt{-1} \), then the ser... | No |
We consider the question of determining the probability that, given the chain is in state \( i \) today, it will be in state \( j \) two days from now. We denote this probability by \( {p}_{ij}^{\left( 2\right) } \) . | In Example 11.1, we see that if it is rainy today then the event that it is snowy two days from now is the disjoint union of the following three events: 1) it is rainy tomorrow and snowy two days from now, 2) it is nice tomorrow and snowy two days from now, and 3) it is snowy tomorrow and snowy two days from now. The p... | Yes |
Theorem 11.1 Let \( \mathbf{P} \) be the transition matrix of a Markov chain. The \( {ij} \) th entry \( {p}_{ij}^{\left( n\right) } \) of the matrix \( {\mathbf{P}}^{n} \) gives the probability that the Markov chain, starting in state \( {s}_{i} \), will be in state \( {s}_{j} \) after \( n \) steps. | Proof. The proof of this theorem is left as an exercise (Exercise 17). | No |
Consider again the weather in the Land of \( \mathrm{{Oz}} \). We know that the powers of the transition matrix give us interesting information about the process as it evolves. We shall be particularly interested in the state of the chain after a large number of steps. | The program MatrixPowers computes the powers of \( \mathbf{P} \). We have run the program MatrixPowers for the Land of Oz example to compute the successive powers of \( \mathbf{P} \) from 1 to 6. The results are shown in Table 11.1. We note that after six days our weather predictions are, to three-decimal-place accurac... | Yes |
Theorem 11.2 Let \( \mathbf{P} \) be the transition matrix of a Markov chain, and let \( \mathbf{u} \) be the probability vector which represents the starting distribution. Then the probability that the chain is in state \( {s}_{i} \) after \( n \) steps is the \( i \) th entry in the vector\n\n\[{\mathbf{u}}^{\left( n... | Proof. The proof of this theorem is left as an exercise (Exercise 18). | No |
In the Land of Oz example (Example 11.1) let the initial probability vector \( \mathbf{u} \) equal \( \left( {1/3,1/3,1/3}\right) \). Then we can calculate the distribution of the states after three days using Theorem 11.2 and our previous calculation of \( {\mathbf{P}}^{3} \). | \[ {\mathbf{u}}^{\left( 3\right) } = \mathbf{u}{\mathbf{P}}^{3} = \left( \begin{array}{lll} 1/3, & 1/3, & 1/3 \end{array}\right) \left( \begin{array}{lll} {.406} & {.203} & {.391} \\ {.406} & {.188} & {.406} \\ {.391} & {.203} & {.406} \end{array}\right) \] \[ = \left( \begin{array}{lll} {.401}, & {.198}, & {.401} \end... | Yes |
The President of the United States tells person A his or her intention to run or not to run in the next election. Then A relays the news to B, who in turn relays the message to \( \mathrm{C} \), and so forth, always to some new person. We assume that there is a probability \( a \) that a person will change the answer f... | \[ \mathbf{P} = \begin{array}{l} \text{ yes } \\ \text{ yes } \\ \text{ no } \end{array}\left( \begin{matrix} 1 - a & a \\ b & 1 - b \end{matrix}\right) . \] | No |
Each time a certain horse runs in a three-horse race, he has probability \( 1/2 \) of winning, \( 1/4 \) of coming in second, and \( 1/4 \) of coming in third, independent of the outcome of any previous race. We have an independent trials process, | but it can also be considered from the point of view of Markov chain theory. The transition matrix is \[ \mathbf{P} = \begin{array}{l} \mathrm{W} \\ \mathrm{W} \\ \mathrm{P} \\ \mathrm{S} \end{array}\left( \begin{matrix} \mathrm{P} & \mathrm{P} & \mathrm{S} \\ {.5} & {.25} & {.25} \\ {.5} & {.25} & {.25} \\ {.5} & {.25... | No |
Example 11.6 In the Dark Ages, Harvard, Dartmouth, and Yale admitted only male students. Assume that, at that time, 80 percent of the sons of Harvard men went to Harvard and the rest went to Yale, 40 percent of the sons of Yale men went to Yale, and the rest split evenly between Harvard and Dartmouth; and of the sons o... | \[ \mathbf{P} = \begin{array}{l} \mathrm{H} \\ \mathrm{H} \\ \mathrm{Y} \\ \mathrm{D} \end{array}\left( \begin{array}{lll} {.8} & {.2} & 0 \\ {.3} & {.4} & {.3} \\ {.3} & {.1} & {.7} \end{array}\right) . \] | No |
Example 11.8 (Ehrenfest Model) The following is a special case of a model, called the Ehrenfest model, \( {}^{3} \) that has been used to explain diffusion of gases. The general model will be discussed in detail in Section 11.5. We have two urns that, between them, contain four balls. At each step, one of the four ball... | \[ \mathbf{P} = \frac{1}{2}\left( \begin{matrix} 0 & 1 & 2 & 3 & 4 \\ 0 & 1 & 0 & 0 & 0 \\ 1/4 & 0 & 3/4 & 0 & 0 \\ 0 & 1/2 & 0 & 1/2 & 0 \\ 0 & 0 & 3/4 & 0 & 1/4 \\ 0 & 0 & 0 & 1 & 0 \end{matrix}\right) . \] | No |
Consider a process of continued matings. We start with an individual of known genetic character and mate it with a hybrid. We assume that there is at least one offspring. An offspring is chosen at random and is mated with a hybrid and this process repeated through a number of generations. The genetic type of the chosen... | The transition probabilities are\n\n\[ \mathbf{P} = \begin{matrix} \mathrm{{GG}} \\ \mathrm{{GG}} \\ \mathrm{{Gg}} \\ \mathrm{{gg}} \end{matrix}\left( \begin{matrix} \mathrm{G} & \mathrm{S} & \mathrm{{gg}} \\ {.5} & {.5} & 0 \\ {.25} & {.5} & {.25} \\ 0 & {.5} & {.5} \end{matrix}\right) . | No |
Example 11.11 We start with two animals of opposite sex, mate them, select two of their offspring of opposite sex, and mate those, and so forth. To simplify the example, we will assume that the trait under consideration is independent of sex. | Here a state is determined by a pair of animals. Hence, the states of our process will be: \( {s}_{1} = \left( {\mathrm{{GG}},\mathrm{{GG}}}\right) ,{s}_{2} = \left( {\mathrm{{GG}},\mathrm{{Gg}}}\right) ,{s}_{3} = \left( {\mathrm{{GG}},\mathrm{{gg}}}\right) ,{s}_{4} = \left( {\mathrm{{Gg}},\mathrm{{Gg}}}\right) ,{s}_{5... | Yes |
Example 11.12 (Stepping Stone Model) Our final example is another example that has been used in the study of genetics. It is called the stepping stone model. \( {}^{4} \) In this model we have an \( n \) -by- \( n \) array of squares, and each square is initially any one of \( k \) different colors. For each step, a sq... | This is an example of an absorbing Markov chain. This type of chain will be studied in Section 11.2. One of the theorems proved in that section, applied to the present example, implies that with probability 1 , the stones will eventually all be the same color. By watching the program run, you can see that territories a... | No |
Example 11.13 A man walks along a four-block stretch of Park Avenue (see Figure 11.3). If he is at corner \( 1,2 \), or 3, then he walks to the left or right with equal probability. He continues until he reaches corner 4 , which is a bar, or corner 0 , which is his home. If he reaches either home or the bar, he stays t... | We form a Markov chain with states \( 0,1,2,3 \), and 4 . States 0 and 4 are absorbing states. The transition matrix is then\n\n\[ \mathbf{P} = \frac{0}{2}\left( \begin{matrix} 0 & 1 & 2 & 3 & 4 \\ 1 & 0 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 & 0 \\ 0 & 1/2 & 0 & 1/2 & 0 \\ 0 & 0 & 1/2 & 0 & 1/2 \\ 0 & 0 & 0 & 0 & 1 \end{mat... | Yes |
Theorem 11.3 In an absorbing Markov chain, the probability that the process will be absorbed is 1 (i.e., \( {\mathbf{Q}}^{n} \rightarrow \mathbf{0} \) as \( n \rightarrow \infty \) ). | Proof. From each nonabsorbing state \( {s}_{j} \) it is possible to reach an absorbing state. Let \( {m}_{j} \) be the minimum number of steps required to reach an absorbing state, starting from \( {s}_{j} \). Let \( {p}_{j} \) be the probability that, starting from \( {s}_{j} \), the process will not reach an absorbin... | Yes |
Theorem 11.4 For an absorbing Markov chain the matrix \( \mathbf{I} - \mathbf{Q} \) has an inverse \( \mathbf{N} \) and \( \mathbf{N} = \mathbf{I} + \mathbf{Q} + {\mathbf{Q}}^{2} + \cdots \) . The \( {ij} \) -entry \( {n}_{ij} \) of the matrix \( \mathbf{N} \) is the expected number of times the chain is in state \( {s... | Proof. Let \( \left( {\mathbf{I} - \mathbf{Q}}\right) \mathbf{x} = 0 \) ; that is \( \mathbf{x} = \mathbf{Q}\mathbf{x} \) . Then, iterating this we see that \( \mathbf{x} = {\mathbf{Q}}^{n}\mathbf{x} \) . Since \( {\mathbf{Q}}^{n} \rightarrow \mathbf{0} \), we have \( {\mathbf{Q}}^{n}\mathbf{x} \rightarrow \mathbf{0} \... | Yes |
Theorem 11.5 Let \( {t}_{i} \) be the expected number of steps before the chain is absorbed, given that the chain starts in state \( {s}_{i} \), and let \( \mathbf{t} \) be the column vector whose \( i \) th entry is \( {t}_{i} \) . Then\n\n\[ \mathbf{t} = \mathbf{N}\mathbf{c} \]\n\nwhere \( \mathbf{c} \) is a column v... | Proof. If we add all the entries in the \( i \) th row of \( \mathbf{N} \), we will have the expected number of times in any of the transient states for a given starting state \( {s}_{i} \), that is, the expected time required before being absorbed. Thus, \( {t}_{i} \) is the sum of the entries in the \( i \) th row of... | Yes |
Theorem 11.6 Let \( {b}_{ij} \) be the probability that an absorbing chain will be absorbed in the absorbing state \( {s}_{j} \) if it starts in the transient state \( {s}_{i} \) . Let \( \mathbf{B} \) be the matrix with entries \( {b}_{ij} \) . Then \( \mathbf{B} \) is an \( t \) -by- \( r \) matrix, and\n\n\[ \mathbf... | Proof. We have\n\n\[ {\mathbf{B}}_{ij} = \mathop{\sum }\limits_{n}\mathop{\sum }\limits_{k}{q}_{ik}^{\left( n\right) }{r}_{kj} \]\n\n\[ = \mathop{\sum }\limits_{k}\mathop{\sum }\limits_{n}{q}_{ik}^{\left( n\right) }{r}_{kj} \]\n\n\[ = \mathop{\sum }\limits_{k}{n}_{ik}{r}_{kj} \]\n\n\[ = {\left( \mathbf{{NR}}\right) }_{... | Yes |
In the Drunkard's Walk example, we found that\n\n\[ \mathbf{N} = \frac{1}{2}\left( \begin{matrix} 1 & 2 & 3 \\ 3/2 & 1 & 1/2 \\ 1 & 2 & 1 \\ 1/2 & 1 & 3/2 \end{matrix}\right) \] | Hence,\n\n\[ \mathbf{t} = \mathbf{N}\mathbf{c} = \left( \begin{matrix} 3/2 & 1 & 1/2 \\ 1 & 2 & 1 \\ 1/2 & 1 & 3/2 \end{matrix}\right) \left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) \]\n\n\[ = \left( \begin{array}{l} 3 \\ 4 \\ 3 \end{array}\right) . \]\n\nThus, starting in states \( 1,2 \), and 3, the expected ... | Yes |
Theorem 11.7 Let \( \mathbf{P} \) be the transition matrix for a regular chain. Then, as \( n \rightarrow \) \( \infty \), the powers \( {\mathbf{P}}^{n} \) approach a limiting matrix \( \mathbf{W} \) with all rows the same vector \( \mathbf{w} \) . The vector \( \mathbf{w} \) is a strictly positive probability vector ... | In the next section we give two proofs of this fundamental theorem. We give here the basic idea of the first proof.\n\nWe want to show that the powers \( {\mathbf{P}}^{n} \) of a regular transition matrix tend to a matrix with all rows the same. This is the same as showing that \( {\mathbf{P}}^{n} \) converges to a mat... | No |
Recall that for the Land of Oz example of Section 11.1, the sixth power of the transition matrix \( \mathbf{P} \) is, to three decimal places, \[ {\mathbf{P}}^{6} = \begin{matrix} \mathrm{R} & \mathrm{N} & \mathrm{S} \\ \mathrm{R} & \left( \begin{array}{lll} {.4} & {.2} & {.4} \\ {.4} & {.2} & {.4} \\ {.4} & {.2} & {.4... | Theorem 11.7 predicts that, for large \( n \), the rows of \( \mathbf{P} \) approach a common vector. It is interesting that this occurs so soon in our example. | No |
Theorem 11.8 Let \( \mathbf{P} \) be a regular transition matrix, let\n\n\[ \mathbf{W} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{P}}^{n} \]\n\nlet \( \mathbf{w} \) be the common row of \( \mathbf{W} \), and let \( \mathbf{c} \) be the column vector all of whose components are 1 . Then\n\n(a) \( \mathbf{... | Proof. To prove part (a), we note that from Theorem 11.7,\n\n\[ {\mathbf{P}}^{n} \rightarrow \mathbf{W}\text{.} \]\n\nThus,\n\n\[ {\mathbf{P}}^{n + 1} = {\mathbf{P}}^{n} \cdot \mathbf{P} \rightarrow \mathbf{W}\mathbf{P} \]\n\nBut \( {\mathbf{P}}^{n + 1} \rightarrow \mathbf{W} \), and so \( \mathbf{W} = \mathbf{{WP}} \)... | Yes |
By Theorem 11.7 we can find the limiting vector \( \mathbf{w} \) for the Land of \( \mathrm{{Oz}} \) from the fact that\n\n\[ \n{w}_{1} + {w}_{2} + {w}_{3} = 1 \n\]\n\nand\n\n\[ \n\left( \begin{array}{lll} {w}_{1} & {w}_{2} & {w}_{3} \end{array}\right) \left( \begin{matrix} 1/2 & 1/4 & 1/4 \\ 1/2 & 0 & 1/2 \\ 1/4 & 1/4... | These relations lead to the following four equations in three unknowns:\n\n\[ \n{w}_{1} + {w}_{2} + {w}_{3} = 1 \n\]\n\n\[ \n\left( {1/2}\right) {w}_{1} + \left( {1/2}\right) {w}_{2} + \left( {1/4}\right) {w}_{3} = {w}_{1}, \n\]\n\n\[ \n\left( {1/4}\right) {w}_{1} + \left( {1/4}\right) {w}_{3} = {w}_{2}, \n\]\n\n\[ \n\... | Yes |
Example 11.20 (Example 11.19 continued) We set \( {w}_{1} = 1 \), and then solve the first and second linear equations from \( \mathbf{w}\mathbf{P} = \mathbf{w} \) . We have\n\n\[ \left( {1/2}\right) + \left( {1/2}\right) {w}_{2} + \left( {1/4}\right) {w}_{3} = 1, \]\n\n\[ \left( {1/4}\right) + \left( {1/4}\right) {w}_... | If we solve these, we obtain\n\n\[ \left( \begin{array}{lll} {w}_{1} & {w}_{2} & {w}_{3} \end{array}\right) = \left( \begin{array}{lll} 1 & 1/2 & 1 \end{array}\right) . \]\n\nNow we divide this vector by the sum of the components, to obtain the final answer:\n\n\[ \mathbf{w} = \left( \begin{array}{lll} {.4} & {.2} & {.... | Yes |
Theorem 11.9 Let \( \mathbf{P} \) be the transition matrix for a regular chain and \( \mathbf{v} \) an arbitrary probability vector. Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{{vP}}}^{n} = \mathbf{w} \]\n\nwhere \( \mathbf{w} \) is the unique fixed probability vector for \( \mathbf{P} \) . | Proof. By Theorem 11.7,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{P}}^{n} = \mathbf{W} \]\n\nHence,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathbf{{vP}}}^{n} = \mathbf{{vW}} \]\n\nBut the entries in \( \mathbf{v} \) sum to 1, and each row of \( \mathbf{W} \) equals \( \mathbf{w} \) . ... | Yes |
Theorem 11.10 For an ergodic Markov chain, there is a unique probability vector \( \mathbf{w} \) such that \( \mathbf{{wP}} = \mathbf{w} \) and \( \mathbf{w} \) is strictly positive. Any row vector such that \( \mathbf{{vP}} = \mathbf{v} \) is a multiple of \( \mathbf{w} \) . Any column vector \( \mathbf{x} \) such tha... | Proof. This theorem states that Theorem 11.8 is true for ergodic chains. The result follows easily from the fact that, if \( \mathbf{P} \) is an ergodic transition matrix, then \( \overline{\mathbf{P}} = \left( {1/2}\right) \mathbf{I} + \left( {1/2}\right) \mathbf{P} \) is a regular transition matrix with the same fixe... | No |
In the Land of \( \mathrm{{Oz}} \), there are 525 days in a year. We have simulated the weather for one year in the Land of \( \mathrm{{Oz}} \), using the program SimulateChain. The results are shown in Table 11.2. | We note that the simulation gives a proportion of times in each of the states not too different from the long run predictions of \( {.4},{.2} \), and .4 assured by Theorem 11.7 . To get better results we have to simulate our chain for a longer time. We do this for 10,000 days without printing out each day's weather. Th... | No |
A white rat is put into the maze of Figure 11.4. There are nine compartments with connections between the compartments as indicated. The rat moves through the compartments at random. That is, if there are \( k \) ways to leave a compartment, it chooses each of these with equal probability. We can represent the travels ... | That this chain is not regular can be seen as follows: From an odd-numbered state the process can go only to an even-numbered state, and from an even-numbered state it can go only to an odd number. Hence, starting in state \( i \) the process will be alternately in even-numbered and odd-numbered states. Therefore, odd ... | Yes |
Example 11.23 (Example 11.8 continued) We recall the Ehrenfest urn model of Example 11.8. The transition matrix for this chain is as follows:\n\n\[ \n\\mathbf{P} = \\begin{array}{l} 0 \\\\ 1 \\\\ 2 \\\\ 3 \\\\ 4 \\\\ 5 \\end{array}\\left( \\begin{matrix} 0 & 1 & 2 & 3 & 4 \\\\ {.000} & {1.000} & {.000} & {.000} & {.000... | If we run the program FixedVector for this chain, we obtain the vector\n\n\[ \n\\mathbf{w} = \\left( \\begin{matrix} 0 & 1 & 2 & 3 & 4 \\\\ {.0625} & {.2500} & {.3750} & {.2500} & {.0625} \\end{matrix}\\right)\n\]\n\nBy Theorem 11.12, we can interpret these values for \( {w}_{i} \) as the proportion of times the proces... | Yes |
Lemma 11.1 Let \( \mathbf{P} \) be an \( r \) -by- \( r \) transition matrix with no zero entries. Let \( d \) be the smallest entry of the matrix. Let \( \mathbf{y} \) be a column vector with \( r \) components, the largest of which is \( {M}_{0} \) and the smallest \( {m}_{0} \) . Let \( {M}_{1} \) and \( {m}_{1} \) ... | Proof. In the discussion following Theorem11.7, it was noted that each entry in the vector \( \mathbf{{Py}} \) is a weighted average of the entries in \( \mathbf{y} \) . The largest weighted average that could be obtained in the present case would occur if all but one of the entries of \( \mathbf{y} \) have value \( {M... | Yes |
Example 11.24 Let us return to the maze example (Example 11.22). We shall make this ergodic chain into an absorbing chain by making state 5 an absorbing state. For example, we might assume that food is placed in the center of the maze and once the rat finds the food, he stays to enjoy it (see Figure 11.5). | The new transition matrix in canonical form is\n\n\[ \mathbf{P} = \begin{matrix} 1 & 2 & 3 & 4 & 6 & 7 & 8 & 9 & 5 & & \\ 1 & 2 & 0 & 1/2 & 0 & 0 & 1/2 & 0 & 0 & 0 & 0 \\ 2 & 1/3 & 0 & 1/3 & 0 & 0 & 0 & 0 & 0 & 1/3 & \\ 3 & 0 & 1/2 & 0 & 1/2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 1/3 & 0 & 0 & 1/3 & 0 & 1/3 & 1/3 & \\ ... | Yes |
Theorem 11.15 For an ergodic Markov chain, the mean recurrence time for state \( {s}_{i} \) is \( {r}_{i} = 1/{w}_{i} \), where \( {w}_{i} \) is the \( i \) th component of the fixed probability vector for the transition matrix. | Proof. Multiplying both sides of Equation 11.6 by \( \mathbf{w} \) and using the fact that\n\n\[ \mathbf{w}\left( {\mathbf{I} - \mathbf{P}}\right) = \mathbf{0} \]\n\ngives\n\n\[ \mathbf{w}\mathbf{C} - \mathbf{w}\mathbf{D} = \mathbf{0}. \]\n\nHere \( \mathbf{{wC}} \) is a row vector with all entries 1 and \( \mathbf{{wD... | Yes |
Corollary 11.1 For an ergodic Markov chain, the components of the fixed probability vector \( \mathbf{w} \) are strictly positive. | Proof. We know that the values of \( {r}_{i} \) are finite and so \( {w}_{i} = 1/{r}_{i} \) cannot be 0. | Yes |
Proposition 11.1 Let \( \\mathbf{P} \) be the transition matrix of an ergodic chain, and let \( \\mathbf{W} \) be the matrix all of whose rows are the fixed probability row vector for \( \\mathbf{P} \). Then the matrix \[ \\mathbf{I} - \\mathbf{P} + \\mathbf{W} \] has an inverse. | Proof. Let \( \\mathbf{x} \) be a column vector such that \[ \\left( {\\mathbf{I} - \\mathbf{P} + \\mathbf{W}}\\right) \\mathbf{x} = \\mathbf{0}. \] To prove the proposition, it is sufficient to show that \( \\mathbf{x} \) must be the zero vector. Multiplying this equation by \( \\mathbf{w} \) and using the fact that \... | Yes |
Example 11.26 Let \( \mathbf{P} \) be the transition matrix for the weather in the Land of \( \mathrm{Oz} \). Then | \[ \mathbf{I} - \mathbf{P} + \mathbf{W} = \left( \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) - \left( \begin{matrix} 1/2 & 1/4 & 1/4 \\ 1/2 & 0 & 1/2 \\ 1/4 & 1/4 & 1/2 \end{matrix}\right) + \left( \begin{matrix} 2/5 & 1/5 & 2/5 \\ 2/5 & 1/5 & 2/5 \\ 2/5 & 1/5 & 2/5 \end{matrix}\right) \] ... | Yes |
Lemma 11.2 Let \( \mathbf{Z} = {\left( \mathbf{I} - \mathbf{P} + \mathbf{W}\right) }^{-1} \), and let \( \mathbf{c} \) be a column vector of all 1 ’s. Then\n\n\[ \mathbf{Z}\mathbf{c} = \mathbf{c} \]\n\n\[ \mathbf{{wZ}} = \mathbf{w} \]\n\nand\n\n\[ \mathbf{Z}\left( {\mathbf{I} - \mathbf{P}}\right) = \mathbf{I} - \mathbf... | Proof. Since \( \mathbf{{Pc}} = \mathbf{c} \) and \( \mathbf{{Wc}} = \mathbf{c} \) ,\n\n\[ \mathbf{c} = \left( {\mathbf{I} - \mathbf{P} + \mathbf{W}}\right) \mathbf{c} \]\n\nIf we multiply both sides of this equation on the left by \( \mathbf{Z} \), we obtain\n\n\[ \mathbf{Z}\mathbf{c} = \mathbf{c} \]\n\nSimilarly, sin... | Yes |
Theorem 11.16 The mean first passage matrix \( \mathbf{M} \) for an ergodic chain is determined from the fundamental matrix \( \mathbf{Z} \) and the fixed row probability vector \( \mathbf{w} \) by\n\n\[ \n{m}_{ij} = \frac{{z}_{jj} - {z}_{ij}}{{w}_{j}}.\n\] | Proof. We showed in Equation 11.6 that\n\n\[ \n\left( {\mathbf{I} - \mathbf{P}}\right) \mathbf{M} = \mathbf{C} - \mathbf{D}\n\]\n\nThus,\n\n\[ \n\mathbf{Z}\left( {\mathbf{I} - \mathbf{P}}\right) \mathbf{M} = \mathbf{Z}\mathbf{C} - \mathbf{Z}\mathbf{D}\n\]\n\nand from Lemma 11.2,\n\n\[ \n\mathbf{Z}\left( {\mathbf{I} - \... | Yes |
In the Land of Oz example, we find that \[ \mathbf{Z} = {\left( \mathbf{I} - \mathbf{P} + \mathbf{W}\right) }^{-1} = \left( \begin{matrix} {86}/{75} & 1/{25} & - {14}/{75} \\ 2/{25} & {21}/{25} & 2/{25} \\ - {14}/{75} & 1/{25} & {86}/{75} \end{matrix}\right) . \] | We have also seen that \( \mathbf{w} = \left( {2/5,1/5,2/5}\right) \) . So, for example, \[ {m}_{12} = \frac{{z}_{22} - {z}_{12}}{{w}_{2}} \] \[ = \frac{{21}/{25} - 1/{25}}{1/5} \] \[ = 4\text{,} \] by Theorem 11.16. Carrying out the calculations for the other entries of \( \mathbf{M} \), we obtain \[ \mathbf{M} = \lef... | Yes |
Let us consider the Ehrenfest model (see Example 11.8) for gas diffusion for the general case of \( {2n} \) balls. Every second, one of the \( {2n} \) balls is chosen at random and moved from the urn it was in to the other urn. If there are \( i \) balls in the first urn, then with probability \( i/{2n} \) we take one ... | This defines the transition matrix of an ergodic, non-regular Markov chain (see Exercise 15). Here the physicist is interested in long-term predictions about the state occupied. In Example 11.23, we gave an intuitive reason for expecting that the fixed vector \( \mathbf{w} \) is the binomial distribution with parameter... | Yes |
Theorem 12.2 For \( n \geq 1 \), the probabilities \( \left\{ {u}_{2k}\right\} \) and \( \left\{ {f}_{2k}\right\} \) are related by the equation\n\n\[ \n{u}_{2n} = {f}_{0}{u}_{2n} + {f}_{2}{u}_{{2n} - 2} + \cdots + {f}_{2n}{u}_{0}.\n\] | Proof. There are \( {u}_{2n}{2}^{2n} \) paths of length \( {2n} \) which have endpoints \( \left( {0,0}\right) \) and \( \left( {{2n},0}\right) \) . The collection of such paths can be partitioned into \( n \) sets, depending upon the time of the first return to the origin. A path in this collection which has a first r... | Yes |
Theorem 12.3 For \( m \geq 1 \), the probability of a first return to the origin at time \( {2m} \) is given by\n\n\[ \n{f}_{2m} = \frac{{u}_{2m}}{{2m} - 1} = \frac{\left( \begin{matrix} {2m} \\ m \end{matrix}\right) }{\left( {{2m} - 1}\right) {2}^{2m}}.\n\] | Proof. We begin by defining the generating functions\n\n\[ \nU\left( x\right) = \mathop{\sum }\limits_{{m = 0}}^{\infty }{u}_{2m}{x}^{m}\n\]\n\nand\n\n\[ \nF\left( x\right) = \mathop{\sum }\limits_{{m = 0}}^{\infty }{f}_{2m}{x}^{m}\n\]\n\nTheorem 12.2 says that\n\n\[ \nU\left( x\right) = 1 + U\left( x\right) F\left( x\... | Yes |
Example 12.1 (Eventual Return in \( {\mathbf{R}}^{1} \) ) One has to approach the idea of eventual return with some care, since the sample space seems to be the set of all walks of infinite length, and this set is non-denumerable. To avoid difficulties, we will define \( {w}_{n} \) to be the probability that a first re... | In terms of the \( {f}_{n} \) probabilities, we see that\n\n\[ \n{w}_{2n} = \mathop{\sum }\limits_{{i = 1}}^{n}{f}_{2i} \n\]\n\nThus,\n\n\[ \n{w}_{ * } = \mathop{\sum }\limits_{{i = 1}}^{\infty }{f}_{2i} \n\]\n\nIn the proof of Theorem 12.3, the generating function\n\n\[ \nF\left( x\right) = \mathop{\sum }\limits_{{m =... | Yes |
Example 12.2 (Eventual Return in \( {\mathbf{R}}^{m} \) ) We now turn our attention to the case that the random walk takes place in more than one dimension. We define \( {f}_{2n}^{\left( m\right) } \) to be the probability that the first return to the origin in \( {\mathbf{R}}^{m} \) occurs at time \( {2n} \) . The qua... | \[ {u}_{2n}^{\left( m\right) } = {f}_{0}^{\left( m\right) }{u}_{2n}^{\left( m\right) } + {f}_{2}^{\left( m\right) }{u}_{{2n} - 2}^{\left( m\right) } + \cdots + {f}_{2n}^{\left( m\right) }{u}_{0}^{\left( m\right) }. \] (12.2) | No |
Consider the function that maps non-negative real numbers to their positive square root. This function is denoted by\n\n\[ f\left( x\right) = \sqrt{x} \] | Note, since this is a function, and its range consists of the non-negative real numbers, we have that\n\n\[ \sqrt{{x}^{2}} = \left| x\right| \] | No |
What does the inverse of this function tell you? What is the inverse of this function? | Solution While \( v\left( t\right) \) tells you how many gallons of water are in the pool after a period of time, the inverse of \( v\left( t\right) \) tells you how much time must be spent to obtain a given volume. To compute the inverse function, first set \( v = v\left( t\right) \) and write\n\n\[ v = {700t} + {200}... | Yes |
Suppose you are standing on a bridge that is 60 meters above sea-level. You toss a ball up into the air with an initial velocity of 30 meters per second. If \( t \) is the time (in seconds) after we toss the ball, then the height at time \( t \) is approximately \( h\left( t\right) = - 5{t}^{2} + {30t} + {60} \) . What... | While \( h\left( t\right) \) tells you how the height the ball is above sea-level at an instant of time, the inverse of \( h\left( t\right) \) tells you what time it is when the ball is at a given height. There is only one problem: There is no function that is the inverse of \( h\left( t\right) \) . Consider Figure 7, ... | Yes |
Does \( f\left( x\right) \) have an inverse? If so what is it? If not, attempt to restrict the domain of \( f\left( x\right) \) and find an inverse on the restricted domain. | In this case \( f\left( x\right) \) is one-to-one and \( {f}^{-1}\left( x\right) = \sqrt[3]{x} \) . See Figure 9. | Yes |
Does \( f\left( x\right) \) have an inverse? If so what is it? If not, attempt to restrict the domain of \( f\left( x\right) \) and find an inverse on the restricted domain. | Solution In this case \( f\left( x\right) \) is not one-to-one. However, it is one-to-one on the interval \( \lbrack 0,\infty ) \) . Hence we can find an inverse of \( f\left( x\right) = {x}^{2} \) on this interval, and it is our familiar function \( \sqrt{x} \) . See Figure 10. | Yes |
Example 1.1.1 Let \( f\left( x\right) = \lfloor x\rfloor \) . Explain why the limit\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 2}}f\left( x\right) \]\n\ndoes not exist. | Solution The function \( \lfloor x\rfloor \) is the function that returns the greatest integer less than or equal to \( x \) . Since \( f\left( x\right) \) is defined for all real numbers, one might be tempted to think that the limit above is simply \( f\left( 2\right) = 2 \) . However, this is not the case. If \( x < ... | Yes |
Let \( f\left( x\right) = \sin \left( \frac{1}{x}\right) \) . Explain why the limit\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}f\left( x\right) \]\ndoes not exist. | Solution In this case \( f\left( x\right) \) oscillates \ | No |
Example 1.1.3 Let \( f\left( x\right) = \lfloor x\rfloor \) . Discuss\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 2 - }}f\left( x\right) ,\;\mathop{\lim }\limits_{{x \rightarrow 2 + }}f\left( x\right) ,\;\text{ and }\;\mathop{\lim }\limits_{{x \rightarrow 2}}f\left( x\right) .\n\] | Solution From the plot of \( f\left( x\right) \), see Figure 1.3, we see that\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 2 - }}f\left( x\right) = 1,\;\text{ and }\;\mathop{\lim }\limits_{{x \rightarrow 2 + }}f\left( x\right) = 2.\n\]\n\nSince these limits are different, \( \mathop{\lim }\limits_{{x \rightarrow 2}}f\le... | Yes |
Show that \( \mathop{\lim }\limits_{{x \rightarrow 2}}{x}^{2} = 4 \) . | Solution We want to show that for any given \( \varepsilon > 0 \), we can find a \( \delta > 0 \) such that\n\n\[ \left| {{x}^{2} - 4}\right| < \varepsilon \]\n\nwhenever \( 0 < \left| {x - 2}\right| < \delta \) . Start by factoring the left-hand side of the inequality above\n\n\[ \left| {x + 2}\right| \left| {x - 2}\r... | Yes |
Theorem 1.2.2 (Limit Product Law) Suppose \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) and \( \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = M \) . Then \[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) g\left( x\right) = {LM} \] | Proof Given any \( \varepsilon \) we need to find a \( \delta \) such that \[ 0 < \left| {x - a}\right| < \delta \] implies \[ \left| {f\left( x\right) g\left( x\right) - {LM}}\right| < \varepsilon . \] Here we use an algebraic trick, add \( 0 = - f\left( x\right) M + f\left( x\right) M \) : We will use this same trick... | No |
Theorem 1.2.3 (Limit Composition Law) Suppose that \( \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = M \) and This is sometimes written as \( \mathop{\lim }\limits_{{x \rightarrow M}}f\left( x\right) = f\left( M\right) \) . Then \[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( {g\left( x\right) }\right)... | \[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( {g\left( x\right) }\right) = \mathop{\lim }\limits_{{g\left( x\right) \rightarrow M}}f\left( {g\left( x\right) }\right) . \] | No |
Theorem 1.2.4 (Limit Root Law) Suppose that \( n \) is a positive integer. Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\sqrt[n]{x} = \sqrt[n]{a} \]\n\nprovided that \( a \) is positive if \( n \) is even. | This theorem is not too difficult to prove from the definition of limit. | No |
Theorem 1.3.1 (Limit Laws) Suppose that \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L,\mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = M, k \) is some constant, and \( n \) is a positive integer. | Constant Law \( \mathop{\lim }\limits_{{x \rightarrow a}}{kf}\left( x\right) = k\mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = {kL} \) . Sum Law \( \mathop{\lim }\limits_{{x \rightarrow a}}\left( {f\left( x\right) + g\left( x\right) }\right) = \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) + \ma... | Yes |
Example 1.3.2 Compute \( \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{{x}^{2} - {3x} + 5}{x - 2} \) . | Solution Using limit laws,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{{x}^{2} - {3x} + 5}{x - 2} = \frac{\mathop{\lim }\limits_{{x \rightarrow 1}}{x}^{2} - {3x} + 5}{\mathop{\lim }\limits_{{x \rightarrow 1}}\left( {x - 2}\right) } \]\n\n\[ = \frac{\mathop{\lim }\limits_{{x \rightarrow 1}}{x}^{2} - \mathop{\li... | Yes |
Example 1.3.3 Compute \( \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{{x}^{2} + {2x} - 3}{x - 1} \) . | Solution We can’t simply plug in \( x = 1 \) because that makes the denominator Limits allow us to examine functions where they are zero. However, when taking limits we assume \( x \neq 1 \) : not defined.\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 1}}\frac{{x}^{2} + {2x} - 3}{x - 1} = \mathop{\lim }\limits_{{x \right... | Yes |
Example 1.3.4 Compute \( \mathop{\lim }\limits_{{x \rightarrow - 1}}\frac{\sqrt{x + 5} - 2}{x + 1} \) . | Solution Using limit laws,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow - 1}}\frac{\sqrt{x + 5} - 2}{x + 1} = \mathop{\lim }\limits_{{x \rightarrow - 1}}\frac{\sqrt{x + 5} - 2}{x + 1}\frac{\sqrt{x + 5} + 2}{\sqrt{x + 5} + 2} \]\n\n\[ = \mathop{\lim }\limits_{{x \rightarrow - 1}}\frac{x + 5 - 4}{\left( {x + 1}\right) \le... | Yes |
Example 1.3.6 Compute\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow 0}}\\frac{\\sin \\left( x\\right) }{x}\n\\] | Solution To compute this limit, use the Squeeze Theorem, Theorem 1.3.5. First note that we only need to examine \\( x \\in \\left( {\\frac{-\\pi }{2},\\frac{\\pi }{2}}\\right) \\) and for the present time, we’ll assume that \\( x \\) is positive-consider the diagrams below:\n\n![f0c2bb2c-af41-4ae2-8326-b33faa659faa_33_... | Yes |
Example 2.1.1 Find the vertical asymptotes of\n\n\\[ \nf\\left( x\\right) = \\frac{{x}^{2} - {9x} + {14}}{{x}^{2} - {5x} + 6} \n\\]\n | Solution Start by factoring both the numerator and the denominator:\n\n\\[ \n\\frac{{x}^{2} - {9x} + {14}}{{x}^{2} - {5x} + 6} = \\frac{\\left( {x - 2}\\right) \\left( {x - 7}\\right) }{\\left( {x - 2}\\right) \\left( {x - 3}\\right) } \n\\]\n\nUsing limits, we must investigate when \\( x \\rightarrow 2 \\) and \\( x \... | Yes |
Example 2.2.1 Compute\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}\\frac{{6x} - 9}{x - 1} \n\\] | Solution Write\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}\\frac{{6x} - 9}{x - 1} = \\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}\\frac{{6x} - 9}{x - 1}\\frac{1/x}{1/x} \n\\]\n\n\\[ \n= \\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}\\frac{\\frac{6x}{x} - \\frac{9}{x}}{\\frac{x}{x} - \\fra... | Yes |
Example 2.2.2 Compute\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow - \\infty }}\\frac{x + 1}{\\sqrt{{x}^{2}}} \n\\] | Solution In this case we multiply the numerator and denominator by \\( - 1/x \\) , which is a positive number as since \\( x \\rightarrow - \\infty, x \\) is a negative number.\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow - \\infty }}\\frac{x + 1}{\\sqrt{{x}^{2}}} = \\mathop{\\lim }\\limits_{{x \\rightarrow - \\i... | Yes |
Example 2.2.3 Compute\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}\\frac{\\sin \\left( {7x}\\right) }{x} + 4 \n\\] | Solution We can bound our function\n\n\\[ \n- 1/x + 4 \\leq \\frac{\\sin \\left( {7x}\\right) }{x} + 4 \\leq 1/x + 4. \n\\]\n\nSince\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }} - 1/x + 4 = 4 = \\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}1/x + 4 \n\\]\n\nwe conclude by the Squeeze Theorem, Theo... | Yes |
Example 2.2.4 Give the horizontal asymptotes of\n\n\[ f\left( x\right) = \frac{{6x} - 9}{x - 1} \] | Solution From our previous work, we see that \( \mathop{\lim }\limits_{{x \rightarrow \infty }}f\left( x\right) = 6 \), and upon further inspection, we see that \( \mathop{\lim }\limits_{{x \rightarrow - \infty }}f\left( x\right) = 6 \) . Hence the horizontal asymptote of \( f\left( x\right) \) is the line \( y = 6 \) ... | Yes |
Example 2.2.5 Give a horizontal asymptote of\n\n\[ f\left( x\right) = \frac{\sin \left( {7x}\right) }{x} + 4 \] | Solution Again from previous work, we see that \( \mathop{\lim }\limits_{{x \rightarrow \infty }}f\left( x\right) = 4 \) . Hence \( y = 4 \) is a horizontal asymptote of \( f\left( x\right) \) . | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\ln \left( x\right) \] | The function \( \ln \left( x\right) \) grows very slowly, and seems like it may have a horizontal asymptote, see Figure 2.6. However, if we consider the definition of the natural log\n\n\[ \ln \left( x\right) = y\; \Leftrightarrow \;{e}^{y} = x \]\n\nSince we need to raise \( e \) to higher and higher values to obtain ... | Yes |
Find the discontinuities (the values for \( x \) where a function is not continuous) for the function given in Figure 2.7. | Solution From Figure 2.7 we see that \( \mathop{\lim }\limits_{{x \rightarrow 4}}f\left( x\right) \) does not exist as\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 4 - }}f\left( x\right) = 1\;\text{ and }\;\mathop{\lim }\limits_{{x \rightarrow 4 + }}f\left( x\right) \approx {3.5} \]\n\nHence \( \mathop{\lim }\limits_{{x... | Yes |
Example 2.3.2 Consider the function\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} \sqrt[5]{x}\sin \left( \frac{1}{x}\right) & \text{ if }x \neq 0 \\ 0 & \text{ if }x = 0 \end{array}\right. \]\n\nsee Figure 2.8. Is this function continuous? | Solution Considering \( f\left( x\right) \), the only issue is when \( x = 0 \) . We must show that \( \mathop{\lim }\limits_{{x \rightarrow 0}}f\left( x\right) = 0 \) . Note\n\n\[ - \left| \sqrt[5]{x}\right| \leq f\left( x\right) \leq \left| \sqrt[5]{x}\right| \]\n\nSince\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}... | Yes |
Example 2.3.4 Explain why the function \( f\left( x\right) = {x}^{3} + 3{x}^{2} + x - 2 \) has a root between 0 and 1 . | Solution By Theorem 1.3.1, \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = f\left( a\right) \), for all real values of \( a \), and hence \( f \) is continuous. Since \( f\left( 0\right) = - 2 \) and \( f\left( 1\right) = 3 \), and 0 is between -2 and 3, by the Intermediate Value Theorem, Theorem 2.3.3, ... | Yes |
Example 2.3.5 Approximate a root of \( f\left( x\right) = {x}^{3} + 3{x}^{2} + x - 2 \) to one decimal place. | Solution If we compute \( f\left( {0.1}\right), f\left( {0.2}\right) \), and so on, we find that \( f\left( {0.6}\right) < 0 \) and \( f\left( {0.7}\right) > 0 \), so by the Intermediate Value Theorem, \( f \) has a root between 0.6 and 0.7 . Repeating the process with \( f\left( {0.61}\right), f\left( {0.62}\right) \)... | Yes |
Example 3.1.1 Compute\n\[ \frac{d}{dx}\left( {{x}^{3} + 1}\right) \] | Solution Using the definition of the derivative,\n\n\[ \frac{d}{dx}f\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{\left( x + h\right) }^{3} + 1 - \left( {{x}^{3} + 1}\right) }{h} \]\n\n\[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{x}^{3} + 3{x}^{2}h + {3x}{h}^{2} + {h}^{3} + 1 - {x}^{3} - 1}{... | Yes |
Compute \[ \frac{d}{dt}\frac{1}{t} \] | Solution Using the definition of the derivative, \[ \frac{d}{dt}\frac{1}{t} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\frac{1}{t + h} - \frac{1}{t}}{h} \] \[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\frac{t}{t\left( {t + h}\right) } - \frac{t + h}{t\left( {t + h}\right) }}{h} \] \[ = \mathop{\lim }\limit... | Yes |
Theorem 3.1.3 (Differentiability Implies Continuity) If \( f\left( x\right) \) is a differentiable function at \( x = a \), then \( f\left( x\right) \) is continuous at \( x = a \) . | Proof We want to show that \( f\left( x\right) \) is continuous at \( x = a \), hence we must show that\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = f\left( a\right) \]\n\nConsider\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\left( {f\left( x\right) - f\left( a\right) }\right) = \mathop{\lim }\l... | Yes |
Example 3.1.4 Compute\n\\[ \n\\frac{d}{dx}\\left| x\\right| \n\\] | Solution Using the definition of the derivative,\n\n\\[ \n\\frac{d}{dx}\\left| x\\right| = \\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\frac{\\left| {x + h}\\right| - \\left| x\\right| }{h}. \n\\]\n\nIf \\( x \\) is positive we may assume that \\( x \\) is larger than \\( h \\), as we are taking the limit as \\( h \\... | Yes |
Theorem 3.2.1 (The Constant Rule) Given a constant \( c \) ,\n\n\[ \frac{d}{dx}c = 0 \] | Proof From the limit definition of the derivative, write\n\n\[ \frac{d}{dx}c = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{c - c}{h} \]\n\n\[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{0}{h} \]\n\n\[ = \mathop{\lim }\limits_{{h \rightarrow 0}}0 = 0\text{.} \] | Yes |
Theorem 3.2.2 (The Power Rule) For any real number \( n \) ,\n\n\[ \frac{d}{dx}{x}^{n} = n{x}^{n - 1} \] | Proof At this point we will only prove this theorem for n being a positive integer. Later in Section 6.3, we will give the complete proof. From the limit definition of the derivative, write\n\n\[ \frac{d}{dx}{x}^{n} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{\left( x + h\right) }^{n} - {x}^{n}}{h}. \]\n\nStart b... | No |
Example 3.2.3 Compute\n\n\[ \n\frac{d}{dx}{x}^{13} \n\] | Solution Applying the power rule, we write\n\n\[ \n\frac{d}{dx}{x}^{13} = {13}{x}^{12} \n\] | Yes |
Example 3.2.4 Compute\n\n\[ \n\frac{d}{dx}\frac{1}{{x}^{4}} \n\] | Solution Applying the power rule, we write\n\n\[ \n\frac{d}{dx}\frac{1}{{x}^{4}} = \frac{d}{dx}{x}^{-4} = - 4{x}^{-5}. \n\] | Yes |
Compute\n\n\[ \frac{d}{dx}\sqrt[5]{x}. \] | Solution Applying the power rule, we write\n\n\[ \frac{d}{dx}\sqrt[5]{x} = \frac{d}{dx}{x}^{1/5} = \frac{{x}^{-4/5}}{5} \] | Yes |
Theorem 3.2.6 (The Sum Rule) If \( f\left( x\right) \) and \( g\left( x\right) \) are differentiable and \( c \) is a constant, then\n\n(a) \( \frac{d}{dx}\left( {f\left( x\right) + g\left( x\right) }\right) = {f}^{\prime }\left( x\right) + {g}^{\prime }\left( x\right) \) | Proof We will only prove part (a) above, the rest are similar. Write\n\n\[ \frac{d}{dx}\left( {f\left( x\right) + g\left( x\right) }\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {x + h}\right) + g\left( {x + h}\right) - \left( {f\left( x\right) + g\left( x\right) }\right) }{h} \]\n\n\[ = \mathop{\lim... | Yes |
Compute\n\n\[ \frac{d}{dx}\left( {{x}^{5} + \frac{1}{x}}\right) \] | Solution Write\n\n\[ \frac{d}{dx}\left( {{x}^{5} + \frac{1}{x}}\right) = \frac{d}{dx}{x}^{5} + \frac{d}{dx}{x}^{-1} \]\n\n\[ = 5{x}^{4} - {x}^{-2}\text{. } \] | Yes |
Compute\n\n\[ \frac{d}{dx}\left( {\frac{3}{\sqrt[3]{x}} - 2\sqrt{x} + \frac{1}{{x}^{7}}}\right) \] | Solution Write\n\n\[ \frac{d}{dx}\left( {\frac{3}{\sqrt[3]{x}} - 2\sqrt{x} + \frac{1}{{x}^{7}}}\right) = 3\frac{d}{dx}{x}^{-1/3} - 2\frac{d}{dx}{x}^{1/2} + \frac{d}{dx}{x}^{-7} \]\n\n\[ = - {x}^{-4/3} - {x}^{-1/2} - 7{x}^{-8} \] | Yes |
Theorem 3.2.9 (The Derivative of \( {e}^{x} \) )\n\n\[ \frac{d}{dx}{e}^{x} = {e}^{x} \] | Proof From the limit definition of the derivative, write\n\n\[ \frac{d}{dx}{e}^{x} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{e}^{x + h} - {e}^{x}}{h} \]\n\n\[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{e}^{x}{e}^{h} - {e}^{x}}{h} \]\n\n\[ = \mathop{\lim }\limits_{{h \rightarrow 0}}{e}^{x}\frac{{e}^{h} - ... | Yes |
Example 3.2.10 Compute:\n\n\[ \n\frac{d}{dx}\left( {8\sqrt{x} + 7{e}^{x}}\right) \n\] | Solution Write:\n\n\[ \n\frac{d}{dx}\left( {8\sqrt{x} + 7{e}^{x}}\right) = 8\frac{d}{dx}{x}^{1/2} + 7\frac{d}{dx}{e}^{x} \n\]\n\n\[ \n= 4{x}^{-1/2} + 7{e}^{x}\text{.} \n\] | Yes |
Example 4.1.2 Find all local maximum and minimum points for the function \( f\left( x\right) = {x}^{3} - x. \) | Solution Write\n\n\[ \frac{d}{dx}f\left( x\right) = 3{x}^{2} - 1 \]\n\nThis is defined everywhere and is zero at \( x = \pm \sqrt{3}/3 \) . Looking first at \( x = \sqrt{3}/3 \) , we see that\n\n\[ f\left( {\sqrt{3}/3}\right) = - 2\sqrt{3}/9. \]\n\nNow we test two points on either side of \( x = \sqrt{3}/3 \), making s... | Yes |
Consider the function\n\n\[ f\left( x\right) = \frac{{x}^{4}}{4} + \frac{{x}^{3}}{3} - {x}^{2} \]\n\nFind the intervals on which \( f\left( x\right) \) is increasing and decreasing and identify the local extrema of \( f\left( x\right) \). | Solution Start by computing\n\n\[ \frac{d}{dx}f\left( x\right) = {x}^{3} + {x}^{2} - {2x} \]\n\nNow we need to find when this function is positive and when it is negative. To do this, solve\n\n\[ {f}^{\prime }\left( x\right) = {x}^{3} + {x}^{2} - {2x} = 0. \]\n\nFactor \( {f}^{\prime }\left( x\right) \)\n\n\[ {f}^{\pri... | Yes |
Example 4.3.2 Describe the concavity of \( f\left( x\right) = {x}^{3} - x \) . | Solution To start, compute the first and second derivative of \( f\left( x\right) \) with respect\nto \( x \) ,\n\n\[ \n{f}^{\prime }\left( x\right) = 3{x}^{2} - 1\;\text{ and }\;{f}^{\prime \prime }\left( x\right) = {6x}.\n\]\n\nSince \( {f}^{\prime \prime }\left( 0\right) = 0 \), there is potentially an inflection po... | Yes |
Theorem 4.4.1 (Second Derivative Test) Suppose that \( {f}^{\prime \prime }\left( x\right) \) is continuous on an open interval and that \( {f}^{\prime }\left( a\right) = 0 \) for some value of \( a \) in that interval. | - If \( {f}^{\prime \prime }\left( a\right) < 0 \), then \( f\left( x\right) \) has a local maximum at \( a \) . \n- If \( {f}^{\prime \prime }\left( a\right) > 0 \), then \( f\left( x\right) \) has a local minimum at \( a \) . \n- If \( {f}^{\prime \prime }\left( a\right) = 0 \), then the test is inconclusive. In this... | Yes |
Use the second derivative test, Theorem 4.4.1, to locate the local extrema of \( f\left( x\right) = \frac{{x}^{4}}{4} + \frac{{x}^{3}}{3} - {x}^{2} \). | Solution Start by computing\n\n\[ {f}^{\prime }\left( x\right) = {x}^{3} + {x}^{2} - {2x}\;\text{ and }\;{f}^{\prime \prime }\left( x\right) = 3{x}^{2} + {2x} - 2. \]\n\nUsing the same technique as used in the solution of Example 4.2.2, we find that\n\n\[ {f}^{\prime }\left( {-2}\right) = 0,\;{f}^{\prime }\left( 0\righ... | Yes |
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