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Problem 13 The last three changes in \( \mathrm{r},\mathrm{y} \), and \( \mathrm{{pi}} \) are given next. What are the predicted values of the next set of changes in these series? | 10 tail(TbGdpPi, n = 4) \( \begin{array}{lll} \mathrm{r} & \mathrm{y} & \mathrm{{pi}} \end{array} \) [233,] \( {0.07} \) 9.7 1.38 [234,] 0.04 9.7 0.31 [235,] 0.02 9.7 0.28 [236,] 0.07 9.7 -0.47\n\nNow fit a VAR(1) using the following commands.\n\n11 var1 = ar(del_dat, order.max=1) | No |
Problem 15 What do the elements of Phi_hat suggest about the relationships among the changes in \( r, y \), and \( {pi} \) ? | A VAR(1) process is stationary provided that the eigenvalues of \( \mathbf{\Phi } \) are less than one in magnitude. Compute the eigenvalues of \( \widehat{\mathbf{\Phi }} \):\n\n14 eigen.values \( = \) eigen(Phi_hat)\$values\n\n15 abs(eigen.values) | No |
Problem 16 Is the estimated process stationary? How does this result relate to the forecast calculations in Problem 14 above? | The dataset MacroVars.csv contains three US macroeconomic indicators from Quarter 1 of 1959 to Quarter 4 of 1997: Real Gross Domestic Product (a measure of economic activity), Consumer Price Index (a measure of inflation), and Federal Funds Rate (a proxy for monetary policy). Each series has been transformed to station... | No |
Problem 19 Plot DiffSqrtCpi and its ACF. Do you see any signs of long memory? If so, describe them. | Run the following code to estimate the amount of fractional differencing, fractionally difference DiffSqrtCpi appropriately, and check the ACF of the fractionally differenced series.\n\n4 library(\ | No |
Problem 21 Fit an ARIMA model to the fractionally differenced series using auto.arima(). Compare the models selected using AIC and BIC. | ## 13.8.6 Model-Based Bootstrapping of an ARIMA Process\n\nThis exercise uses the price of frozen orange juice. Run the following code to fit an ARIMA model.\n\n1 library(AER)\n\n2 library(forecast)\n\n3 data(\ | No |
Example 14.2. \( {AR}\left( 1\right) + {GARCH}\left( {1,1}\right) \) model fit to daily BMW stock log returns | This example uses the daily BMW stock log returns. The ugarchfit() function from R's rugarch package is used to fit an \( \mathrm{{AR}}\left( 1\right) + \mathrm{{GARCH}}\left( {1,1}\right) \) model to this series. Although ugarchfit () allows the white noise to have a nonGaussian distribution, we begin this example usi... | Yes |
In this example, an \( \mathrm{{AR}}\left( 1\right) + \mathrm{{APARCH}}\left( {1,1}\right) \) model with \( t \) -distributed errors is fit to the BMW log returns. The commands and abbreviated output from ugarchfit () is below. The estimate of \( \delta \) is 1.48 with a standard error of 0.14, so there is strong evide... | arma.aparch.t \( = \) ugarchspec(mean.model=list(arma0rder=c(1,0)), variance.model=list (model=\ | No |
In Example 9.9, we saw that a parsimonious model for the yearly log returns on the stock index diff (log(sp)) used diff (log(ip)) and diff (bnd) as predictors. Figure 14.9 contains ACF plots of the residuals [panel (a)] and squared residuals [panel (b)]. Externally studentized residuals were used, but the plots for the... | The auto.arima() function from R's forecast package selected an MA(1) model [i.e., ARIMA(0,0,1)] for the residuals. Next an MA(1)+ARCH(1) model was fit to the regression model's raw residuals. Sample ACF plots of the standardized residuals from the \( \mathrm{{MA}}\left( 1\right) + \mathrm{{ARCH}}\left( 1\right) \) mod... | Yes |
Problem 1 Plot both Tbill and Tbill.diff. Use both time series and ACF plots. Also, perform ADF and KPSS tests on both series. Which series do you think are stationary? Why? What types of heteroskedasticity can you see in the Tbill.diff series? | In the following code, the variable Tbill can be used if you believe that series is stationary. Otherwise, replace Tbill by Tbill.diff. This code will fit an ARMA+GARCH model to the series.\n\n8 library(rugarch)\n\n9 arma.garch.norm = ugarchspec(mean.model=list(armaOrder=c(1,0)), variance.model=list(garch0rder=c(1,1))\... | No |
Problem 7 Does the joint series exhibit conditional heteroskedasticity? Why? | Now fit and plot a EWMA model with the following R commands.\n\nEWMA.param \( = \) est.ewma(lambda.0=0.95, innov=TbPi.diff)\n\nEWMA.param\$lambda.hat\n\n\( \mathtt{{EWMA}.{Sigma}} = \mathtt{{sigma}.{ewma}}\left( {\mathtt{{lambda}} = \mathtt{{EWMA}.{param}} \oplus \mathtt{{lambda}.{hat}},\mathtt{{innov}} = \mathtt{{TbPi... | No |
1. Let \( Z \) have an \( N\left( {0,1}\right) \) distribution. Show that\n\n\[ E\left( \left| Z\right| \right) = {\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi }}\left| z\right| {e}^{-{z}^{2}/2}{dz} = 2{\int }_{0}^{\infty }\frac{1}{\sqrt{2\pi }}z{e}^{-{z}^{2}/2}{dz} = \sqrt{\frac{2}{\pi }}. \] | \[ E\left( \left| Z\right| \right) = {\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi }}\left| z\right| {e}^{-{z}^{2}/2}{dz} = 2{\int }_{0}^{\infty }\frac{1}{\sqrt{2\pi }}z{e}^{-{z}^{2}/2}{dz} = \sqrt{\frac{2}{\pi }}. \] | Yes |
Model (15.2)-(15.3) was simulated with \( {\phi }_{1} = {0.5},{\phi }_{2} = {0.55} \), and \( \lambda = 1 \) . A total of 5,000 observations were simulated, but, for visual clarity, only every 10th observation is plotted in Fig. 15.2. Neither \( {Y}_{1, t} \) nor \( {Y}_{2, t} \) is stationary, but \( {Y}_{1, t} - \lam... | 10 \( \mathrm{n} = {5000} \)\n\n11 set.seed(12345)\n\n12 a1 \( = {0.5} \)\n\na2 \( = {0.55} \)\n\nlambda \( = 1 \)\n\n\( \mathrm{y}1 = \operatorname{rep}\left( {0,\mathrm{n}}\right) \)\n\ny2 = y1\n\n\( \mathrm{e}1 = \operatorname{rnorm}\left( \mathrm{n}\right) \)\n\n\( \mathrm{e}2 = \operatorname{rnorm}\left( \mathrm{n... | No |
Problem 5 Does this difference series appear stationary? Why? | Run the following commands to conduct Johansen's cointegration test.\n\n4 library(urca)\n\n5 summary(ca.jo(CokePepsi)) | No |
Problem 7 Are these 10 stock price series cointegrated? If so, what is the rank of the cointegrating matrix, and what are the cointegrating vectors? | Rerun the Johansen’s cointegration test with \( \operatorname{lag}K = 8 \) . 10 summary(ca.jo(adjClose, K=8)) | No |
The expectation and variance of the return on a portfolio with two risky assets | Suppose that \( {\mu }_{1} = {0.14},{\mu }_{2} = {0.08},{\sigma }_{1} = {0.2},{\sigma }_{2} = {0.15} \), and \( {\rho }_{12} = 0 \) . Then\n\n\[ E\left( {R}_{P}\right) = {0.08} + {0.06w} \]\n\nand because \( {\rho }_{12} = 0 \) in this example,\n\n\[ {\sigma }_{{R}_{P}}^{2} = {\left( {0.2}\right) }^{2}{w}^{2} + {\left(... | Yes |
Example 16.3. The tangency portfolio with two risky assets\n\nSuppose as before that \( {\mu }_{1} = {0.14},{\mu }_{2} = {0.08},{\sigma }_{1} = {0.2},{\sigma }_{2} = {0.15} \), and \( {\rho }_{12} = 0 \) . Suppose as well that \( {\mu }_{f} = {0.06} \) . Then \( {V}_{1} = {0.14} - {0.06} = {0.08} \) and \( {V}_{2} = {0... | Plugging these values into formula (16.5), we get \( {w}_{T} = {0.693} \) and \( 1 - {w}_{t} = {0.307} \) . Therefore,\n\n\[ E\left( {R}_{T}\right) = \left( {0.693}\right) \left( {0.14}\right) + \left( {0.307}\right) \left( {0.08}\right) = {0.122}, \]\n\nand\n\n\[ {\sigma }_{T} = \sqrt{{\left( {0.693}\right) }^{2}{\lef... | Yes |
In this example, we will find the optimal investment with \( {\sigma }_{{R}_{p}} = {0.05} \) . | The maximum expected return with \( {\sigma }_{{R}_{p}} = {0.05} \) mixes the tangency portfolio and the risk-free asset such that \( {\sigma }_{{R}_{p}} = {0.05} \) . Since \( {\sigma }_{T} = {0.146} \), we have that \( {0.05} = {\sigma }_{{R}_{p}} = \omega {\sigma }_{T} = {0.146\omega } \), so that \( \omega = {0.05}... | Yes |
Finding the efficient frontier, tangency portfolio, and minimum variance portfolio with no short selling using quadratic programming | In this example, Example 16.6 is modified so that short sales are not allowed. Only three lines of code need to be changed. When short sales are prohibited, the target expected return on the portfolio must lie between the smallest and largest expected returns on the stocks. To prevent numerical errors, the target expec... | Yes |
Suppose that the risk-free rate of interest is \( {\mu }_{f} = {0.06} \), the expected return on the market portfolio is \( {\mu }_{M} = {0.15} \), and the risk of the market portfolio is \( {\sigma }_{M} = {0.22} \) . Then the slope of the CML is \( \left( {{0.15} - {0.06}}\right) /{0.22} = 9/{22} \) . The CML of this... | The CML is easy to derive. Consider an efficient portfolio that allocates a proportion \( w \) of its assets to the market portfolio and \( \left( {1 - w}\right) \) to the risk-free asset. Then\n\n\[ R = w{R}_{M} + \left( {1 - w}\right) {\mu }_{f} = {\mu }_{f} + w\left( {{R}_{M} - {\mu }_{f}}\right) . \]\n\n(17.2)\n\nT... | Yes |
Why are the Dow Jones stocks behaving differently compared to the equity funds? | The Dow Jones stocks are similar to each other since they are all large companies in the United States. Thus, we can expect that their returns will be highly correlated with each other and a few principal components will explain most of the variation. | No |
Estimating the covariance matrix of GE, IBM, and Mobil excess returns | The estimate of \( {\mathbf{\sum }}_{F} \) is the sample covariance matrix of the factors:\n\nMkt.RF SMB HML\n\nMkt.RF 21.1507 4.2326 -5.1045\n\nSMB 4.2326 8.1811 -1.0760\n\nHML -5.1045 -1.0760 7.1797\n\nThe estimate of \( \mathbf{\beta } \) is the matrix of regression coefficients (without the intercepts):\n\nMkt.RF S... | Yes |
Factor analysis of equity funds | The code for fitting a 4-factor model \( \\left( {p = 4}\\right) \) using factanal () is:\n\nequityFunds \( = \) read.csv(\ | No |
Problem 1 It is generally recommended that PCA be applied to time series that are stationary. Plot the first column of \( \mathtt{{yieldDat}} \) . (You can look at other columns as well. You will see that they are fairly similar.) Does the plot appear stationary? Why or why not? Include your plot with your work. | Another way to check for stationarity is to run the augmented Dickey-Fuller test. You can do that with the following code:\n\nlibrary(\ | No |
Problem 2 Based on the augmented Dickey-Fuller test, do you think the first column of yieldDat is stationary? Why or why not? | Run the following code to compute changes in the yield curves. Notice the use of \( \left\lbrack {-1,}\right\rbrack \) to delete the first row and similarly the use of \( \left\lbrack {-n,}\right\rbrack \) . \n\n\[ \n\text{n=dim(yieldDat)[1]} \n\] \n\n\[ \n\text{delta_yield = yieldDat[-1, ] - yieldDat[-n, ]} \n\] \n\nP... | No |
Problem 14 Does the likelihood ratio test suggest that two factors are enough? If not, what is the minimum number of factors that seems sufficient? | The following code will extract the loadings and uniquenesses.\n\nloadings \( = \operatorname{matrix}\left( \right. \) as.numeric \( \left( {\text{loadings}\left( \text{fact}\right) }\right) , \) ncol \( = 2 \) )\n\nunique \( = \) as.numeric(fact\$unique) | No |
Parametric VaR and ES for a position in an S&P 500 index fund | This example uses the same data set as in Example 19.2 so that parametric and nonparametric estimates can be compared. We will assume that the returns are i.i.d. with a \( t \) -distribution. Under this assumption, VaR is\n\n\[ \n{\widehat{\operatorname{VaR}}}^{\mathrm{t}}\left( \alpha \right) = - S \times \left\{ {\wi... | Yes |
Bootstrap confidence intervals for VaR and ES for a position in an S&P 500 index fund | In this example, we continue Examples 19.2 and 19.3 and find an approximate confidence interval for \( \operatorname{VaR}\left( \alpha \right) \) and \( \operatorname{ES}\left( \alpha \right) \) . We use \( \alpha = {0.05} \) as before and \( \gamma = {0.1}.B = 5,{000} \) resamples were taken.\n\nThe basic percentile c... | Yes |
Example 19.5. VaR and ES for a position in an S&P 500 index fund using a GARCH(1,1) model | An AR(1)+GARCH(1,1) model was fit to the log returns on the S&P 500. The \( \mathrm{{AR}}\left( 1\right) \) coefficient was small and not significantly different from 0, so a \( \mathrm{{GARCH}}\left( {1,1}\right) \) was used for estimation of \( \mathrm{{VaR}} \) and \( \mathrm{{ES}} \) . The \( \mathrm{{GARCH}}\left(... | No |
Example 19.6. VaR and ES for portfolios of the three stocks in the CRSPday data set | This example uses the data set CRSPday used earlier in Examples 7.1 and 7.4. There are four variables - returns on GE, IBM, Mobil, and the CRSP index and we found in Example 7.4 that their returns can be modeled as having a multivariate \( t \) -distribution with tail index 5.94 . In this example, we will only consider... | Yes |
Problem 2 For an investment of \$4,000, what are estimates of \( {\operatorname{VaR}}^{t}\left( {0.05}\right) \) and \( E{S}^{t}\left( {0.05}\right) \) ? | Now, fit a ARMA(0,0)+GARCH(1,1) model to the returns and calculate one step forecasts.\n\nlibrary(fGarch) # for qstd() function\n\nlibrary(rugarch)\n\ngarch.t \( = \) ugarchspec(mean.model=list(armaOrder=c \( \left( {0,0}\right) \) ),\n\nvariance.model=list(garchOrder=c(1,1)),\n\ndistribution.model \( = \) \ | No |
Updating the prior beliefs about the probability that a stock price will increase | We continue Example 20.1 but change the simple, but unrealistic, prior that said that \( \theta \) was either 0.4 or 0.6 to a more plausible prior where \( \theta \) could be any value in the interval \( \left\lbrack {0,1}\right\rbrack \), but with values near \( 1/2 \) more likely. Specifically, we use a \( \operatorn... | Yes |
Computing the posterior density of the probability that a stock price will increase-general case of a conjugate prior | Suppose now that the prior for \( \theta \) is \( \operatorname{Beta}\left( {\alpha ,\beta }\right) \) so that the prior density is\n\n\[ \pi \left( \theta \right) = {K}_{1}{\theta }^{\alpha - 1}{\left( 1 - \theta \right) }^{\beta - 1}, \]\n\nwhere \( {K}_{1} \) is a constant. Suppose that the stock price is observed o... | Yes |
Now suppose that \( {Y}_{1},\ldots ,{Y}_{n} \) are i.i.d. with a known mean \( \mu \) and an unknown variance \( {\sigma }^{2} \) and precision \( \tau = 1/{\sigma }^{2} \) . We will show that the conjugate priors for \( \tau \) are the gamma distributions and we will find the posterior distribution of \( \tau \) . Def... | Simple algebra shows that the likelihood is\n\n\[ f\left( {{Y}_{1},\ldots ,{Y}_{n} \mid \tau }\right) \propto \exp \left( {-\frac{1}{2}{n\tau }{s}^{2}}\right) {\tau }^{n/2}. \]\n\n(20.26)\n\nLet the prior distribution be the gamma distribution with shape parameter \( \alpha \) and scale parameter \( b \) which has dens... | Yes |
Posterior interval for a normal mean when the variance is known | This example continues Example 20.4. By (20.20) and (20.21), a \( \left( {1 - \alpha }\right) \) \( {100}\% \) posterior interval for \( \mu \) is\n\n\[ \frac{{\tau }_{\bar{Y}}\bar{Y} + {\tau }_{0}{\mu }_{0}}{{\tau }_{\bar{Y}} + {\tau }_{0}} \pm {z}_{\alpha /2}\sqrt{\frac{1}{\frac{n}{{\sigma }^{2}} + \frac{1}{{\sigma }... | Yes |
Using JAGS to fit a t-distribution to returns | In this example, a \( t \) -distribution will be fit to S&P 500 returns using JAGS called from R. The BUGS program below is in the file univt. bug. The program will run under any of OpenBUGS, WinBUGS, or JAGS. In this example, JAGS will be used.\n\nmodel\{\n\nfor(i in 1:N)\n\n\{\n\n\( r\left\lbrack i\right\rbrack \sim ... | Yes |
Example 20.11. \( {p}_{D} \) when estimating a normal mean with known precision | Suppose that \( \mathbf{Y} = \left( {{Y}_{1},\ldots ,{Y}_{n}}\right) \) are i.i.d. \( N\left( {\mu ,1}\right) \), so \( \mathbf{\theta } = \mu \) in this example. Then the log-likelihood is\n\n\[ \log \{ L\left( \mu \right) \} = - \frac{1}{2}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {Y}_{i} - \mu \right) }^{2} - \frac... | Yes |
In Example 7.4, the correlation matrix of the CRSPday returns data was estimated by maximum likelihood. In this example, the MLE will be compared to a Bayes estimate and the two estimates will be found to be very similar. | The BUGS program used in this example is\n\nmodel{\n\nfor \( \\left( {i\\text{in}1 : N}\\right) \)\n\n{\n\ny[i,1:m] ~ dmt(mu[], tau[,], df_likelihood)\n\n}\n\nmu[1:m] ~ dmt(muO[], Prec_mu[,], df_prior)\n\ntau[1:m,1:m] ~ dwish(Prec_tau[,], df_wishart)\n\nlambda[1:m,1:m] <- inverse(tau[,])\n\n} In the BUGS program, mu is... | Yes |
Fitting an ARMA(1,1) stochastic volatility model to the S&P 500 stock returns | A BUGS program to fit the stochastic volatility model is below.\n\nmodel\n\nfor \( \left( {i\text{in}1 : N}\right) \)\n\n\{\n\ny[i] ~ dt(mu, tau[i], nu)\n\}\n\n\( \operatorname{logh}\left\lbrack 1\right\rbrack \) daorm(0, \( {1.0}\mathrm{E} - 6 \) )\n\nfor \( \left( {i\text{in}2 : N}\right) \)\n\n\( \operatorname{logh}... | Yes |
Fitting a GARCH(1,1) model to the S&P 500 stock returns | The following \( R \) code fits a \( \operatorname{GARCH}\left( {1,1}\right) \) model to the data in Example 20.14.\n\nlibrary(rjags)\n\ndat \( = \) read.csv(\ | No |
Fitting a one factor model to stock returns | In this example, the factor will be the returns on the S&P 500 so this is a Bayesian version of the CAPM. The model that will be used is\n\n\[ \n{R}_{j, t} = {\beta }_{j}{R}_{M, t} + {\epsilon }_{j, t} \n\]\n\nwhere, as in Chap. \( {17},{R}_{j, t} \) is the return on the \( j \) stock at time \( t, j = 1,\ldots ,{10} \... | Yes |
Problem 3 Which posterior densities are most skewed? | The kurtosis of a \( t \) -distribution is \( 3\left( {\nu - 2}\right) /\left( {\nu - 4}\right) \) if \( \nu > 4 \) and is \( + \infty \) if \( \nu \leq 4 \) . Variables in \( \mathrm{R} \) can have infinite values: Inf is \( + \infty \) and \( - \operatorname{Inf} \) is \( - \infty \), so \( \mathrm{R} \) can handle i... | Yes |
Local polynomial estimation of forward rates. | The \( \mathrm{R} \) code in this section computes two estimates of the forward rate function, a parametric estimate based on the Nelson-Siegel model and a nonparametric estimates using local polynomial smoothing. The program is split into several chucks of code with discussion between the chucks.\n\nLine 1 reads in pr... | No |
Example 21.2. Estimating the drift and volatility for the evolution of the risk-free returns | In this example, we return to estimating the drift and squared volatility functions for the evolution of the risk-free returns. Three estimators will be used: local linear, local quadratic, and a penalized spline. The R code is:\n\nlibrary(Ecdat)\nlibrary(KernSmooth)\nlibrary(locfit)\nlibrary(mgcv)\n\ndata(Capm)\n\n\\(... | Yes |
Consider the observed time series, of length \( n = {25} \), reported in the last column of Tables A.2(a) and A.2(b) of the online supplementary material. Suppose we are interested in solving the single-lag testing problems for lags \( l = 2 \) and \( l = 3 \) . | If \( l = 3 \), the \( {n}_{3} = {22} \) pairs to be considered are highlighted in Table A.2(a). To classify these pairs in the \( k \times k \) contingency table reported in Table 1, the value of \( k \) first has to be determined. Based on (6) the value \( k = 2 \) is obtained, since, in this case, \( {k}_{s} = 2 \) ... | Yes |
Example 5 (Power of the test). By considering the data already presented in Example 2, we show how the multiple-lag ADF is able to detect the presence of serial dependence structures that are not captured by the Portmanteau ACF and by the Portmanteau ADF. | Figure 3 displays the rejection rates related to the test based on \( {\widehat{\delta }}_{L}, L \in \mathcal{P}\left( \mathcal{L}\right), L \neq \varnothing \), and \( \mathcal{L} = \) \( \{ 1,2,3,4,5\} \) . Similarly to Figure 1, which reports the results of the tests based on \( {\widehat{Q}}_{L} \) and \( {\widehat... | No |
Let \( {X}_{t} = \mathop{\sum }\limits_{{i = 1}}^{t}{\eta }_{i} \), where \( \left\{ {\eta }_{i}\right\} \) is i.i.d. with mean 0 and finite variance \( {\sigma }^{2} \), and \( t \in \mathbb{N} \) . Then its ACVF and ACF are | \[ {\gamma }_{t, s} = \operatorname{cov}\left( {{X}_{t},{X}_{s}}\right) = \left( {t \land s}\right) {\sigma }^{2};\;{\rho }_{t, s} = \frac{t \land s}{\sqrt{ts}},\;s, t \in \mathbb{N}, \] | Yes |
Show that \( {x}^{5} - 2{x}^{3} + 3{x}^{2} - 1 = 0 \) has a solution in the interval \( \left\lbrack {0,1}\right\rbrack \) . | Consider the function defined by \( f\left( x\right) = {x}^{5} - 2{x}^{3} + 3{x}^{2} - 1 \) . The function \( f \) is continuous on \( \left\lbrack {0,1}\right\rbrack \) . In addition,\n\n\[ f\left( 0\right) = - 1 < 0\;\text{ and }\;0 < 1 = f\left( 1\right) . \]\n\nThe Intermediate Value Theorem implies that a number \... | Yes |
Theorem 1.14 (Taylor's Theorem)\n\nSuppose \( f \in {C}^{n}\left\lbrack {a, b}\right\rbrack \), that \( {f}^{\left( n + 1\right) } \) exists on \( \left\lbrack {a, b}\right\rbrack \), and \( {x}_{0} \in \left\lbrack {a, b}\right\rbrack \) . For every \( x \in \left\lbrack {a, b}\right\rbrack \) , there exists a number ... | Here \( {P}_{n}\left( x\right) \) is called the \( n \) th Taylor polynomial for \( f \) about \( {x}_{0} \), and \( {R}_{n}\left( x\right) \) is called | Yes |
Example 3 Let \( f\left( x\right) = \cos x \) and \( {x}_{0} = 0 \) . Determine\n\n(a) the second Taylor polynomial for \( f \) about \( {x}_{0} \) ; and\n\n(b) the third Taylor polynomial for \( f \) about \( {x}_{0} \) . | Solution Since \( f \in {C}^{\infty }\left( \mathbb{R}\right) \), Taylor’s Theorem can be applied for any \( n \geq 0 \) . Also,\n\n\[ \n{f}^{\prime }\left( x\right) = - \sin x,{f}^{\prime \prime }\left( x\right) = - \cos x,{f}^{\prime \prime \prime }\left( x\right) = \sin x,\;\text{ and }\;{f}^{\left( 4\right) }\left(... | No |
Example 1 Determine the five-digit (a) chopping and (b) rounding values of the irrational number \( \pi \) . | Solution The number \( \pi \) has an infinite decimal expansion of the form \( \pi = {3.14159265}\ldots \) . Written in normalized decimal form, we have\n\n\[ \pi = {0.314159265}\ldots \times {10}^{1}. \]\n\n(a) The floating-point form of \( \pi \) using five-digit chopping is\n\n\[ {fl}\left( \pi \right) = {0.31415} \... | Yes |
Example 2 Determine the absolute and relative errors when approximating \( p \) by \( {p}^{ * } \) when\n\n(a) \( p = {0.3000} \times {10}^{1} \) and \( {p}^{ * } = {0.3100} \times {10}^{1} \) ; | (a) For \( p = {0.3000} \times {10}^{1} \) and \( {p}^{ * } = {0.3100} \times {10}^{1} \) the absolute error is 0.1, and the relative error is \( {0.333}\overline{3} \times {10}^{-1} \) . | Yes |
Example 3 Suppose that \( x = \frac{5}{7} \) and \( y = \frac{1}{3} \) . Use five-digit chopping for calculating \( x + y, x - y, x \times y \) , and \( x \div y \) . | Solution Note that\n\n\[ \nx = \frac{5}{7} = 0.\overline{714285}\;\text{ and }\;y = \frac{1}{3} = 0.\overline{3} \]\n\nimplies that the five-digit chopping values of \( x \) and \( y \) are\n\n\[ \n{fl}\left( x\right) = {0.71428} \times {10}^{0}\;\text{ and }\;{fl}\left( y\right) = {0.33333} \times {10}^{0}. \]\n\nThus... | Yes |
Determine the five-digit chopping values of \( x \ominus u,\left( {x \ominus u}\right) \oplus w,\left( {x \ominus u}\right) \otimes v \), and \( u \oplus v \) . | Solution These numbers were chosen to illustrate some problems that can arise with finite-digit arithmetic. Because \( x \) and \( u \) are nearly the same, their difference is small. The absolute error for \( x \ominus u \) is\n\n\[ \left| {\left( {x - u}\right) - \left( {x \ominus u}\right) }\right| = \left| {\left( ... | Yes |
Example 5 Let \( p = {0.54617} \) and \( q = {0.54601} \) . Use four-digit arithmetic to approximate \( p - q \) and determine the absolute and relative errors using (a) rounding and (b) chopping. | Solution The exact value of \( r = p - q \) is \( r = {0.00016} \) .\n\n(a) Suppose the subtraction is performed using four-digit rounding arithmetic. Rounding \( p \) and \( q \) to four digits gives \( {p}^{ * } = {0.5462} \) and \( {q}^{ * } = {0.5460} \), respectively, and \( {r}^{ * } = {p}^{ * } - {q}^{ * } = {0.... | Yes |
Example 6 Evaluate \( f\left( x\right) = {x}^{3} - {6.1}{x}^{2} + {3.2x} + {1.5} \) at \( x = {4.71} \) using three-digit arithmetic. | Solution Table 1.4 gives the intermediate results in the calculations.\n\n<table><thead><tr><th></th><th>\( x \)</th><th>\( {x}^{2} \)</th><th>\( {x}^{3} \)</th><th>\( {6.1}{x}^{2} \)</th><th>\( {3.2x} \)</th></tr></thead><tr><td>Exact</td><td>4.71</td><td>22.1841</td><td>104.487111</td><td>135.32301</td><td>15.072</td... | Yes |
Example 1 The \( N \) th Taylor polynomial for \( f\left( x\right) = \ln x \) expanded about \( {x}_{0} = 1 \) is\n\n\[ \n{P}_{N}\left( x\right) = \mathop{\sum }\limits_{{i = 1}}^{N}\frac{{\left( -1\right) }^{i + 1}}{i}{\left( x - 1\right) }^{i}, \]\n\nand the value of \( \ln {1.5} \) to eight decimal places is 0.40546... | Solution From calculus we know that if \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) is an alternating series with limit \( A \) whose terms decrease in magnitude, then \( A \) and the \( N \) th partial sum \( {A}_{N} = \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n} \) differ by less than the magnitude of the \( ... | Yes |
Example 2 Suppose that, for \( n \geq 1 \) ,\n\n\[ \n{\alpha }_{n} = \frac{n + 1}{{n}^{2}}\;\text{ and }\;{\widehat{\alpha }}_{n} = \frac{n + 3}{{n}^{3}}. \n\] \n\nBoth \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\alpha }_{n} = 0 \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\widehat{\alpha }}_{n} = ... | Solution Define the sequences \( {\beta }_{n} = 1/n \) and \( {\widehat{\beta }}_{n} = 1/{n}^{2} \) . Then \n\n\[ \n\left| {{\alpha }_{n} - 0}\right| = \frac{n + 1}{{n}^{2}} \leq \frac{n + n}{{n}^{2}} = 2 \cdot \frac{1}{n} = 2{\beta }_{n} \n\] \n\nwhich require bounds both above and below the sequence or function under... | Yes |
Example 3 Use the third Taylor polynomial about \( h = 0 \) to show that \( \cos h + \frac{1}{2}{h}^{2} = 1 + O\left( {h}^{4}\right) \) . | Solution In Example 3(b) of Section 1.1 we found that this polynomial is\n\n\[ \cos h = 1 - \frac{1}{2}{h}^{2} + \frac{1}{24}{h}^{4}\cos \widetilde{\xi }\left( h\right) ,\]\n\nfor some number \( \widetilde{\xi }\left( h\right) \) between zero and \( h \) . This implies that\n\n\[ \cos h + \frac{1}{2}{h}^{2} = 1 + \frac... | Yes |
Example 1 Show that \( f\left( x\right) = {x}^{3} + 4{x}^{2} - {10} = 0 \) has a root in \( \left\lbrack {1,2}\right\rbrack \), and use the Bisection method to determine an approximation to the root that is accurate to at least within \( {10}^{-4} \) . | Solution Because \( f\left( 1\right) = - 5 \) and \( f\left( 2\right) = {14} \) the Intermediate Value Theorem 1.11 ensures that this continuous function has a root in \( \left\lbrack {1,2}\right\rbrack \). For the first iteration of the Bisection method we use the fact that at the midpoint of \( \left\lbrack {1,2}\rig... | Yes |
Theorem 2.1 Suppose that \( f \in C\left\lbrack {a, b}\right\rbrack \) and \( f\left( a\right) \cdot f\left( b\right) < 0 \) . The Bisection method generates a sequence \( {\left\{ {p}_{n}\right\} }_{n = 1}^{\infty } \) approximating a zero \( p \) of \( f \) with\n\n\[ \left| {{p}_{n} - p}\right| \leq \frac{b - a}{{2}... | Proof For each \( n \geq 1 \), we have\n\n\[ {b}_{n} - {a}_{n} = \frac{1}{{2}^{n - 1}}\left( {b - a}\right) \;\text{ and }\;p \in \left( {{a}_{n},{b}_{n}}\right) . \]\n\nSince \( {p}_{n} = \frac{1}{2}\left( {{a}_{n} + {b}_{n}}\right) \) for all \( n \geq 1 \), it follows that\n\n\[ \left| {{p}_{n} - p}\right| \leq \fra... | Yes |
Determine the number of iterations necessary to solve \( f\left( x\right) = {x}^{3} + 4{x}^{2} - {10} = 0 \) with accuracy \( {10}^{-3} \) using \( {a}_{1} = 1 \) and \( {b}_{1} = 2 \) . | Solution We we will use logarithms to find an integer \( N \) that satisfies\n\n\[ \left| {{p}_{N} - p}\right| \leq {2}^{-N}\left( {b - a}\right) = {2}^{-N} < {10}^{-3}. \]\n\nLogarithms to any base would suffice, but we will use base-10 logarithms because the tolerance is given as a power of 10 . Since \( {2}^{-N} < {... | Yes |
Determine any fixed points of the function \( g\left( x\right) = {x}^{2} - 2 \) . | Solution A fixed point \( p \) for \( g \) has the property that\n\n\[ p = g\left( p\right) = {p}^{2} - 2\;\text{ which implies that }\;0 = {p}^{2} - p - 2 = \left( {p + 1}\right) \left( {p - 2}\right) . \]\n\nA fixed point for \( g \) occurs precisely when the graph of \( y = g\left( x\right) \) intersects the graph o... | Yes |
Theorem 2.3 (i) If \( g \in C\left\lbrack {a, b}\right\rbrack \) and \( g\left( x\right) \in \left\lbrack {a, b}\right\rbrack \) for all \( x \in \left\lbrack {a, b}\right\rbrack \), then \( g \) has at least one fixed point in \( \left\lbrack {a, b}\right\rbrack \) . | (i) If \( g\left( a\right) = a \) or \( g\left( b\right) = b \), then \( g \) has a fixed point at an endpoint. If not, then \( g\left( a\right) > a \) and \( g\left( b\right) < b \) . The function \( h\left( x\right) = g\left( x\right) - x \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), with\n\n\[ h\left( ... | Yes |
Example 2 Show that \( g\left( x\right) = \left( {{x}^{2} - 1}\right) /3 \) has a unique fixed point on the interval \( \left\lbrack {-1,1}\right\rbrack \) . | Solution The maximum and minimum values of \( g\left( x\right) \) for \( x \) in \( \left\lbrack {-1,1}\right\rbrack \) must occur either when \( x \) is an endpoint of the interval or when the derivative is 0 . Since \( {g}^{\prime }\left( x\right) = {2x}/3 \), the function \( g \) is continuous and \( {g}^{\prime }\l... | Yes |
Example 3 Show that Theorem 2.3 does not ensure a unique fixed point of \( g\left( x\right) = {3}^{-x} \) on the interval \( \left\lbrack {0,1}\right\rbrack \), even though a unique fixed point on this interval does exist. | Solution \( {g}^{\prime }\left( x\right) = - {3}^{-x}\ln 3 < 0 \) on \( \left\lbrack {0,1}\right\rbrack \), the function \( g \) is strictly decreasing on \( \left\lbrack {0,1}\right\rbrack \) . So\n\n\[ g\left( 1\right) = \frac{1}{3} \leq g\left( x\right) \leq 1 = g\left( 0\right) ,\;\text{ for }\;0 \leq x \leq 1.\]\n... | Yes |
Corollary 2.5 If \( g \) satisfies the hypotheses of Theorem 2.4, then bounds for the error involved in using \( {p}_{n} \) to approximate \( p \) are given by\n\n\[ \left| {{p}_{n} - p}\right| \leq {k}^{n}\max \left\{ {{p}_{0} - a, b - {p}_{0}}\right\} \]\n\n(2.5)\n\nand\n\n\[ \left| {{p}_{n} - p}\right| \leq \frac{{k... | Proof Because \( p \in \left\lbrack {a, b}\right\rbrack \), the first bound follows from Inequality (2.4):\n\n\[ \left| {{p}_{n} - p}\right| \leq {k}^{n}\left| {{p}_{0} - p}\right| \leq {k}^{n}\max \left\{ {{p}_{0} - a, b - {p}_{0}}\right\} .\n\nFor \( n \geq 1 \), the procedure used in the proof of Theorem 2.4 implies... | Yes |
Example 1 Consider the function \( f\left( x\right) = \cos x - x = 0 \) . Approximate a root of \( f \) using (a) a fixed-point method, and (b) Newton's Method | Solution (a) A solution to this root-finding problem is also a solution to the fixed-point problem \( x = \cos x \), and the graph in Figure 2.9 implies that a single fixed-point \( p \) lies in \( \left\lbrack {0,\pi /2}\right\rbrack \) .\n\nFigure 2.9\n\n be such that \( g\left( x\right) \in \left\lbrack {a, b}\right\rbrack \), for all \( x \in \left\lbrack {a, b}\right\rbrack \) . Suppose, in addition, that \( {g}^{\prime } \) is continuous on \( \left( {a, b}\right) \) and a positive constant \( k < 1 \) ex... | Proof We know from the Fixed-Point Theorem 2.4 in Section 2.2 that the sequence converges to \( p \) . Since \( {g}^{\prime } \) exists on \( \left( {a, b}\right) \), we can apply the Mean Value Theorem to \( g \) to show that for any \( n \) ,\n\n\[ {p}_{n + 1} - p = g\left( {p}_{n}\right) - g\left( p\right) = {g}^{\p... | Yes |
Theorem 2.9 Let \( p \) be a solution of the equation \( x = g\left( x\right) \) . Suppose that \( {g}^{\prime }\left( p\right) = 0 \) and \( {g}^{\prime \prime } \) is continuous with \( \left| {{g}^{\prime \prime }\left( x\right) }\right| < M \) on an open interval \( I \) containing \( p \) . Then there exists a \( ... | Proof Choose \( k \) in \( \left( {0,1}\right) \) and \( \delta > 0 \) such that on the interval \( \left\lbrack {p - \delta, p + \delta }\right\rbrack \), contained in \( I \), we have \( \left| {{g}^{\prime }\left( x\right) }\right| \leq k \) and \( {g}^{\prime \second } \) continuous. Since \( \left| {{g}^{\prime }\... | Yes |
Theorem 2.11 The function \( f \in {C}^{1}\left\lbrack {a, b}\right\rbrack \) has a simple zero at \( p \) in \( \left( {a, b}\right) \) if and only if \( f\left( p\right) = 0 \), but \( {f}^{\prime }\left( p\right) \neq 0 \) . | Proof If \( f \) has a simple zero at \( p \), then \( f\left( p\right) = 0 \) and \( f\left( x\right) = \left( {x - p}\right) q\left( x\right) \), where \( \mathop{\lim }\limits_{{x \rightarrow p}}q\left( x\right) \neq 0 \) . Since \( f \in {C}^{1}\left\lbrack {a, b}\right\rbrack \) ,\n\n\[ \n{f}^{\prime }\left( p\rig... | Yes |
Example 1 Let \( f\left( x\right) = {e}^{x} - x - 1 \) . (a) Show that \( f \) has a zero of multiplicity 2 at \( x = 0 \) . (b) Show that Newton’s method with \( {p}_{0} = 1 \) converges to this zero but not quadratically. | Solution (a) We have\n\n\[ f\left( x\right) = {e}^{x} - x - 1,\;{f}^{\prime }\left( x\right) = {e}^{x} - 1\;\text{ and }\;{f}^{\prime \prime }\left( x\right) = {e}^{x}, \]\n\nso\n\n\[ f\left( 0\right) = {e}^{0} - 0 - 1 = 0,\;{f}^{\prime }\left( 0\right) = {e}^{0} - 1 = 0\;\text{ and }\;{f}^{\prime \prime }\left( 0\righ... | Yes |
In Example 1 it was shown that \( f\left( x\right) = {e}^{x} - x - 1 \) has a zero of multiplicity 2 at \( x = 0 \) and that Newton’s method with \( {p}_{0} = 1 \) converges to this zero but not quadratically. Show that the modification of Newton's method as given in Eq. (2.13) improves the rate of convergence. | Solution Modified Newton's method gives\n\n\[ \n{p}_{1} = {p}_{0} - \frac{f\left( {p}_{0}\right) {f}^{\prime }\left( {p}_{0}\right) }{{f}^{\prime }{\left( {p}_{0}\right) }^{2} - f\left( {p}_{0}\right) {f}^{\prime \prime }\left( {p}_{0}\right) } = 1 - \frac{\left( {e - 2}\right) \left( {e - 1}\right) }{{\left( e - 1\rig... | Yes |
The sequence \( {\left\{ {p}_{n}\right\} }_{n = 1}^{\infty } \), where \( {p}_{n} = \cos \left( {1/n}\right) \), converges linearly to \( p = 1 \) . Determine the first five terms of the sequence given by Aitken’s \( {\Delta }^{2} \) method. | Solution In order to determine a term \( {\widehat{p}}_{n} \) of the Aitken’s \( {\Delta }^{2} \) method sequence we need to have the terms \( {p}_{n},{p}_{n + 1} \), and \( {p}_{n + 2} \) of the original sequence. So to determine \( {\widehat{p}}_{5} \) we need the first 7 terms of \( \left\{ {p}_{n}\right\} \) . Thes... | No |
Theorem 2.16 (Fundamental Theorem of Algebra)\n\nIf \( P\left( x\right) \) is a polynomial of degree \( n \geq 1 \) with real or complex coefficients, then \( P\left( x\right) = 0 \) has at least one (possibly complex) root. | Although the Fundamental Theorem of Algebra is basic to any study of elementary functions, the usual proof requires techniques from the study of complex function theory. The reader is referred to [SaS], p. 155, for the culmination of a systematic development of the topics needed to prove the Theorem. | No |
Example 1 Determine all the zeros of the polynomial \( P\\left( x\\right) = {x}^{3} - 5{x}^{2} + {17x} - {13} \) . | Solution It is easily verified that \( P\\left( 1\\right) = 1 - 5 + {17} - {13} = 0 \) . so \( x = 1 \) is a zero of \( P \) and\n\n\( \\left( {x - 1}\\right) \) is a factor of the polynomial. Dividing \( P\\left( x\\right) \) by \( x - 1 \) gives\n\n\[ P\\left( x\\right) = \\left( {x - 1}\\right) \\left( {{x}^{2} - {4... | No |
If \( P\left( x\right) \) is a polynomial of degree \( n \geq 1 \) with real or complex coefficients, then there exist which gives the complex zeros unique constants \( {x}_{1},{x}_{2},\ldots ,{x}_{k} \), possibly complex, and unique positive integers \( {m}_{1},{m}_{2},\ldots \) , \( {m}_{k} \), such that \( \mathop{\... | By Corollary 2.17 the collection of zeros of a polynomial is unique and, if each zero \( {x}_{i} \) is counted as many times as its multiplicity \( {m}_{i} \), a polynomial of degree \( n \) has exactly \( n \) zeros. | Yes |
Theorem 2.19\n\n## (Horner's Method)\n\nLet\n\n\\[ P\\left( x\\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \\cdots + {a}_{1}x + {a}_{0}. \\]\n\nDefine \\( {b}_{n} = {a}_{n} \\) and\n\n\\[ {b}_{k} = {a}_{k} + {b}_{k + 1}{x}_{0},\\;\\text{ for }k = n - 1, n - 2,\\ldots ,1,0. \\]\n\nThen \\( {b}_{0} = P\\left( {x}_... | Proof By the definition of \\( Q\\left( x\\right) \\) ,\n\n\\[ \\left( {x - {x}_{0}}\\right) Q\\left( x\\right) + {b}_{0} = \\left( {x - {x}_{0}}\\right) \\left( {{b}_{n}{x}^{n - 1} + \\cdots + {b}_{2}x + {b}_{1}}\\right) + {b}_{0} \\]\n\n\\[ = \\left( {{b}_{n}{x}^{n} + {b}_{n - 1}{x}^{n - 1} + \\cdots + {b}_{2}{x}^{2}... | Yes |
Use Horner’s method to evaluate \( P\left( x\right) = 2{x}^{4} - 3{x}^{2} + {3x} - 4 \) at \( {x}_{0} = - 2 \) . | Solution When we use hand calculation in Horner's method, we first construct a table, which suggests the synthetic division name that is often applied to the technique. For this problem, the table appears as follows:\n\n<table><thead><tr><th></th><th>Coefficient of \( {x}^{4} \)</th><th>Coefficient of \( {x}^{3} \)</th... | No |
Find an approximation to a zero of\n\n\\[ P\\left( x\\right) = 2{x}^{4} - 3{x}^{2} + {3x} - 4, \\]\n\nusing Newton’s method with \\( {x}_{0} = - 2 \\) and synthetic division to evaluate \\( P\\left( {x}_{n}\\right) \\) and \\( {P}^{\\prime }\\left( {x}_{n}\\right) \\) for each iterate \\( {x}_{n} \\). | Solution With \\( {x}_{0} = - 2 \\) as an initial approximation, we obtained \\( P\\left( {-2}\\right) \\) in Example 1 by \n\nUsing Theorem 2.19 and Eq. (2.16),\n\n\\[ Q\\left( x\\right) = 2{x}^{3} - 4{x}^{2} + {5x}... | No |
Determine the linear Lagrange interpolating polynomial that passes through the points \( \left( {2,4}\right) \) and \( \left( {5,1}\right) \) . | In this case we have\n\n\[ \n{L}_{0}\left( x\right) = \frac{x - 5}{2 - 5} = - \frac{1}{3}\left( {x - 5}\right) \;\text{ and }\;{L}_{1}\left( x\right) = \frac{x - 2}{5 - 2} = \frac{1}{3}\left( {x - 2}\right) , \n\] \n\nso \n\n\[ \nP\left( x\right) = - \frac{1}{3}\left( {x - 5}\right) \cdot 4 + \frac{1}{3}\left( {x - 2}\... | Yes |
The interpolation formula named for Joseph Louis Lagrange (1736-1813) was likely known by Isaac Newton around 1675, but it appears to first have been published in 1779 by Edward Waring (1736-1798). Lagrange wrote extensively on the subject of interpolation and his work had significant influence on later mathematicians.... | \[ f\left( {x}_{k}\right) = P\left( {x}_{k}\right) ,\;\text{ for each }k = 0,1,\ldots, n. \] \[ P\left( x\right) = f\left( {x}_{0}\right) {L}_{n,0}\left( x\right) + \cdots + f\left( {x}_{n}\right) {L}_{n, n}\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{n}f\left( {x}_{k}\right) {L}_{n, k}\left( x\right) , \] (3.1) ... | Yes |
Example 2 (a) Use the numbers (called nodes) \( {x}_{0} = 2,{x}_{1} = {2.75} \), and \( {x}_{2} = 4 \) to find the second Lagrange interpolating polynomial for \( f\left( x\right) = 1/x \) . | Solution (a) We first determine the coefficient polynomials \( {L}_{0}\left( x\right) ,{L}_{1}\left( x\right) \), and \( {L}_{2}\left( x\right) \) . In nested form they are\n\n\[ \n{L}_{0}\left( x\right) = \frac{\left( {x - {2.75}}\right) \left( {x - 4}\right) }{\left( {2 - {2.5}}\right) \left( {2 - 4}\right) } = \frac... | Yes |
Example 3 In Example 2 we found the second Lagrange polynomial for \( f\left( x\right) = 1/x \) on \( \left\lbrack {2,4}\right\rbrack \) using the nodes \( {x}_{0} = 2,{x}_{1} = {2.75} \), and \( {x}_{2} = 4 \) . Determine the error form for this polynomial, and the maximum error when the polynomial is used to approxim... | Solution Because \( f\left( x\right) = {x}^{-1} \), we have\n\n\[ \n{f}^{\prime }\left( x\right) = - {x}^{-2},\;{f}^{\prime \prime }\left( x\right) = 2{x}^{-3},\;\text{ and }\;{f}^{\prime \prime \prime }\left( x\right) = - 6{x}^{-4}.\n\]\n\nAs a consequence, the second Lagrange polynomial has the error form\n\n\[ \n\fr... | Yes |
Example 1 Suppose that \( {x}_{0} = 1,{x}_{1} = 2,{x}_{2} = 3,{x}_{3} = 4,{x}_{4} = 6 \), and \( f\left( x\right) = {e}^{x} \) . Determine the interpolating polynomial denoted \( {P}_{1,2,4}\left( x\right) \), and use this polynomial to approximate \( f\left( 5\right) \) . | Solution This is the Lagrange polynomial that agrees with \( f\left( x\right) \) at \( {x}_{1} = 2,{x}_{2} = 3 \), and \( {x}_{4} = 6 \) . Hence\n\n\[ \n{P}_{1,2,4}\left( x\right) = \frac{\left( {x - 3}\right) \left( {x - 6}\right) }{\left( {2 - 3}\right) \left( {2 - 6}\right) }{e}^{2} + \frac{\left( {x - 2}\right) \le... | Yes |
Theorem 3.5 Let \( f \) be defined at \( {x}_{0},{x}_{1},\ldots ,{x}_{k} \), and let \( {x}_{j} \) and \( {x}_{i} \) be two distinct numbers in this set. Then\n\n\[ P\left( x\right) = \frac{\left( {x - {x}_{j}}\right) {P}_{0,1,\ldots, j - 1, j + 1,\ldots, k}\left( x\right) - \left( {x - {x}_{i}}\right) {P}_{0,1,\ldots,... | Proof For ease of notation, let \( Q \equiv {P}_{0,1,\ldots, i - 1, i + 1,\ldots, k} \) and \( \widehat{Q} \equiv {P}_{0,1,\ldots, j - 1, j + 1,\ldots, k} \) . Since \( Q\left( x\right) \) and \( \widehat{Q}\left( x\right) \) are polynomials of degree \( k - 1 \) or less, \( P\left( x\right) \) is of degree at most \( ... | Yes |
Apply Neville's method to the data by constructing a recursive table of the form shown in Table 3.4. | Solution Let \( {x}_{0} = {1.0},{x}_{1} = {1.3},{x}_{2} = {1.6},{x}_{3} = {1.9} \), and \( {x}_{4} = {2.2} \), then \( {Q}_{0,0} = f\left( {1.0}\right) \) , \( {Q}_{1,0} = f\left( {1.3}\right) ,{Q}_{2,0} = f\left( {1.6}\right) ,{Q}_{3,0} = f\left( {1.9}\right) \), and \( {Q}_{4,0} = f\left( {2.2}\right) \) . These are ... | Yes |
Use Neville’s method and four-digit rounding arithmetic to approximate \( f\left( {2.1}\right) = \ln {2.1} \) by completing the Neville table. | Solution Because \( x - {x}_{0} = {0.1}, x - {x}_{1} = - {0.1}, x - {x}_{2} = - {0.2} \), and we are given \( {Q}_{0,0} = {0.6931},{Q}_{1,0} = {0.7885} \), and \( {Q}_{2,0} = {0.8329} \), we have\n\n\[ \n{Q}_{1,1} = \frac{1}{0.2}\left\lbrack {\left( {0.1}\right) {0.7885} - \left( {-{0.1}}\right) {0.6931}}\right\rbrack ... | Yes |
Complete the divided difference table for the data used in Example 1 of Section 3.2, and construct the interpolating polynomial that uses all this data. | Solution The first divided difference involving \( {x}_{0} \) and \( {x}_{1} \) is\n\n\[ f\left\lbrack {{x}_{0},{x}_{1}}\right\rbrack = \frac{f\left\lbrack {x}_{1}\right\rbrack - f\left\lbrack {x}_{0}\right\rbrack }{{x}_{1} - {x}_{0}} = \frac{{0.6200860} - {0.7651977}}{{1.3} - {1.0}} = - {0.4837057}. \]\n\nThe remainin... | Yes |
Theorem 3.6 Suppose that \( f \in {C}^{n}\left\lbrack {a, b}\right\rbrack \) and \( {x}_{0},{x}_{1},\ldots ,{x}_{n} \) are distinct numbers in \( \left\lbrack {a, b}\right\rbrack \) . Then a number \( \xi \) exists in \( \left( {a, b}\right) \) with\n\n\[ f\left\lbrack {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\rbrack = \... | Proof Let\n\n\[ g\left( x\right) = f\left( x\right) - {P}_{n}\left( x\right) \]\n\nSince \( f\left( {x}_{i}\right) = {P}_{n}\left( {x}_{i}\right) \) for each \( i = 0,1,\ldots, n \), the function \( g \) has \( n + 1 \) distinct zeros in \( \left\lbrack {a, b}\right\rbrack \) . Generalized Rolle’s Theorem 1.10 implies ... | Yes |
Example 2 Consider the table of data given in the previous examples. Use Stirling's formula to approximate \( f\left( {1.5}\right) \) with \( {x}_{0} = {1.6} \) . | Solution To apply Stirling's formula we use the underlined entries in the difference Table 3.14.\n\nTable 3.14\n\n<table><thead><tr><th>\( x \)</th><th>\( f\left( x\right) \)</th><th>First divided differences</th><th>Second divided differences</th><th>Third divided differences</th><th>Fourth divided differences</th></t... | No |
The osculating polynomial approximating \( f \) is the polynomial \( P\left( x\right) \) of least degree such that\n\n\[ \frac{{d}^{k}P\left( {x}_{i}\right) }{d{x}^{k}} = \frac{{d}^{k}f\left( {x}_{i}\right) }{d{x}^{k}},\;\text{ for each }i = 0,1,\ldots, n\;\text{ and }\;k = 0,1,\ldots ,{m}_{i}. \] | Proof First recall that\n\n\[ {L}_{n, j}\left( {x}_{i}\right) = \left\{ \begin{array}{ll} 0, & \text{ if }i \neq j \\ 1, & \text{ if }i = j \end{array}\right. \]\n\nHence when \( i \neq j \) ,\n\n\[ {H}_{n, j}\left( {x}_{i}\right) = 0 \]\n\n\[ \text{and}\;{\widehat{H}}_{n, j}\left( {x}_{i}\right) = 0\text{,} \]\n\nwher... | Yes |
Example 1 Use the Hermite polynomial that agrees with the data listed in Table 3.15 to find an approximation of \( f\left( {1.5}\right) \) . | Solution We first compute the Lagrange polynomials and their derivatives. This gives\n\n\[ \n{L}_{2,0}\left( x\right) = \frac{\left( {x - {x}_{1}}\right) \left( {x - {x}_{2}}\right) }{\left( {{x}_{0} - {x}_{1}}\right) \left( {{x}_{0} - {x}_{2}}\right) } = \frac{50}{9}{x}^{2} - \frac{175}{9}x + \frac{152}{9},\;{L}_{2,0}... | Yes |
Use the data given in Example 1 and the divided difference method to determine the Hermite polynomial approximation at \( x = {1.5} \) . | The underlined entries in the first three columns of Table 3.17 are the data given in Example 1. The remaining entries in this table are generated by the standard divided-difference formula (3.9).\n\nFor example, for the second entry in the third column we use the second 1.3 entry in the second column and the first 1.6... | Yes |
Construct a natural cubic spline that passes through the points \( \\left( {1,2}\\right) ,\\left( {2,3}\\right) \), and \( \\left( {3,5}\\right) \) . | This spline consists of two cubics. The first for the interval \( \\left\\lbrack {1,2}\\right\\rbrack \), denoted\n\n\[ \n{S}_{0}\\left( x\\right) = {a}_{0} + {b}_{0}\\left( {x - 1}\\right) + {c}_{0}{\\left( x - 1\\right) }^{2} + {d}_{0}{\\left( x - 1\\right) }^{3}, \n\] \nand the other for \( \\left\\lbrack {2,3}\\rig... | Yes |
Theorem 3.11 If \( f \) is defined at \( a = {x}_{0} < {x}_{1} < \cdots < {x}_{n} = b \), then \( f \) has a unique natural spline interpolant \( S \) on the nodes \( {x}_{0},{x}_{1},\ldots ,{x}_{n} \) ; that is, a spline interpolant that satisfies the natural boundary conditions \( {S}^{\prime \prime }\left( a\right) ... | Proof The boundary conditions in this case imply that \( {c}_{n} = {S}^{\prime \prime }\left( {x}_{n}\right) /2 = 0 \) and that\n\n\[ 0 = {S}^{\prime \prime }\left( {x}_{0}\right) = 2{c}_{0} + 6{d}_{0}\left( {{x}_{0} - {x}_{0}}\right) ,\]\n\nso \( {c}_{0} = 0 \) . The two equations \( {c}_{0} = 0 \) and \( {c}_{n} = 0 ... | Yes |
Theorem 3.12 If \( f \) is defined at \( a = {x}_{0} < {x}_{1} < \cdots < {x}_{n} = b \) and differentiable at \( a \) and \( b \), then \( f \) has a unique clamped spline interpolant \( S \) on the nodes \( {x}_{0},{x}_{1},\ldots ,{x}_{n} \) ; that is, a spline interpolant that satisfies the clamped boundary conditio... | Proof Since \( {f}^{\prime }\left( a\right) = {S}^{\prime }\left( a\right) = {S}^{\prime }\left( {x}_{0}\right) = {b}_{0} \), Eq. (3.20) with \( j = 0 \) implies\n\n\[ \n{f}^{\prime }\left( a\right) = \frac{1}{{h}_{0}}\left( {{a}_{1} - {a}_{0}}\right) - \frac{{h}_{0}}{3}\left( {2{c}_{0} + {c}_{1}}\right) .\n\]\n\nConse... | Yes |
Determine the clamped spline \( s\left( x\right) \) that uses the data points \( \left( {0,1}\right) ,\left( {1, e}\right) ,\left( {2,{e}^{2}}\right) \), and \( \left( {3,{e}^{3}}\right) \) and the additional information that, since \( {f}^{\prime }\left( x\right) = {e}^{x} \), so \( {f}^{\prime }\left( 0\right) = 1 \)... | Solution As in Example 2, we have \( n = 3,{h}_{0} = {h}_{1} = {h}_{2} = 1,{a}_{0} = 0,{a}_{1} = e,{a}_{2} = {e}^{2} \) , and \( {a}_{3} = {e}^{3} \). This together with the information that \( {f}^{\prime }\left( 0\right) = 1 \) and \( {f}^{\prime }\left( 3\right) = {e}^{3} \) gives the the matrix \( A \) and the vect... | Yes |
Construct a pair of Lagrange polynomials to approximate the curve shown in Figure 3.15, using the data points shown on the curve. | Solution There is flexibility in choosing the parameter, and we will choose the points \( {\left\{ {t}_{i}\right\} }_{i = 0}^{4} \) equally spaced in \( \left\lbrack {0,1}\right\rbrack \), which gives the data in Table 3.20.\n\nTable 3.20\n\n<table><thead><tr><th>\( i \)</th><th>0</th><th>1</th><th>2</th><th>3</th><th>... | Yes |
Determine the graph of the parametric curve generated Eq. (3.23) and (3.24) when the end points are \( \\left( {{x}_{0},{y}_{0}}\\right) = \\left( {0,0}\\right) \) and \( \\left( {{x}_{1},{y}_{1}}\\right) = \\left( {1,0}\\right) \), and respective guide points, as shown in Figure 3.18 | Solution The endpoint information implies that \( {x}_{0} = 0,{x}_{1} = 1,{y}_{0} = 0 \), and \( {y}_{1} = 0 \), and the guide points at \( \\left( {1,1}\\right) \) and \( \\left( {0,1}\\right) \) imply that \( {\\alpha }_{0} = 1,{\\alpha }_{1} = 1,{\\beta }_{0} = 1 \), and \( {\\beta }_{1} = - 1 \) . Note that the slo... | Yes |
Use the forward-difference formula to approximate the derivative of \( f\left( x\right) = \ln x \) at \( {x}_{0} = {1.8} \) using \( h = {0.1}, h = {0.05} \), and \( h = {0.01} \), and determine bounds for the approximation errors. | The forward-difference formula\n\n\[ \frac{f\left( {{1.8} + h}\right) - f\left( {1.8}\right) }{h} \]\n\nwith \( h = {0.1} \) gives\n\n\[ \frac{\ln {1.9} - \ln {1.8}}{0.1} = \frac{{0.64185389} - {0.58778667}}{0.1} = {0.5406722}. \]\n\nBecause \( {f}^{\prime \prime }\left( x\right) = - 1/{x}^{2} \) and \( {1.8} < \xi < {... | Yes |
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