# Solve the given information Let f(x) = sin x. Use the following values and five-digit rounding arithmetic to construct the Hermite interpolating polynomial to approximate f(0.34) = sin 0.34. begin{array}{c|c} x & f(x) & f'(x) hline 0.30 & 0.29552 & 0.95534 0.32 & 0.31457 & 0.94924 0.35 & 0.34290 & 0.93937 end{array}

Question
Polynomial arithmetic
Solve the given information Let $$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}}.$$ Use the following values and five-digit rounding arithmetic to construct the Hermite interpolating polynomial to approximate $$\displaystyle{f{{\left({0.34}\right)}}}={\sin{{0.34}}}.$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}{\mid}{c}\right\rbrace}{x}&{f{{\left({x}\right)}}}&{f}'{\left({x}\right)}\backslash{h}{l}\in{e}{0.30}&{0.29552}&{0.95534}\backslash{0.32}&{0.31457}&{0.94924}\backslash{0.35}&{0.34290}&{0.93937}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$

2021-01-20
Consider the provided question, Given, $$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}}$$ We construct a divided- difference table, with each of the interpolation point included twice, that means, $$\displaystyle{x}_{{0}}={x}_{{1}}={0.30},{x}_{{2}}={x}_{{3}}={0.32}$$ and $$\displaystyle{x}_{{4}}={x}_{{5}}={0.35}.$$ each of the divided differences $$\displaystyle{f}{\left[{x}_{{i}},{x}_{{{i}+{1}}}\right]},$$ where $$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}}$$ and $$\displaystyle{x}_{{i}}={x}_{{{i}+{1}}}$$ is set equal to $$\displaystyle{f}'{\left({x}_{{i}}\right)}.$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\mathcal}\right\rbrace}{x}_{{i}}&{f}{\left[{x}_{{i}}\right]}&{f}'{\left[{x}_{{i}},{x}_{{{i}+{1}}}\right]}&&&&\backslash{h}{l}\in{e}{0.30}&{0.29552}\backslash&&{0.95534}\backslash&&&-{0.14200}\backslash&&&&-{1.0500}\backslash&&&&&{20.732}\backslash&&&&&&-{431.41}\backslash{0.30}&{0.29552}\backslash&&{0.95250}\backslash&&&-{0.16300}\backslash&&&&-{0.0134}\backslash&&&&&{0.83866}\backslash{0.32}&{0.31457}\backslash&&{0.94924}\backslash&&&-{0.16367}\backslash&&&&-{0.05533}\backslash{0.32}&{0.31457}\backslash&&{0.94433}\backslash&&&-{0.16533}\backslash{0.35}&{0.34290}\backslash&&{0.93937}\backslash{0.35}&{0.34290}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ It follows that Hermit interpolating polynomial is, $$\displaystyle{H}_{{5}}{\left({x}\right)}={0.29552}+{0.95532}{\left({x}-{0.30}\right)}-{0.142}{\left({x}-{0.30}\right)}^{{2}}{N}{S}{K}-{1.05}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}+{20.732}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{N}{S}{K}-{431.41}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{\left({x}-{0.35}\right)}$$ Now, evaluating this polynomial at x = 0.34 using five-digit rounding. $$\displaystyle{H}_{{5}}{\left({0.34}\right)}={0.29552}+{0.95534}{\left({0.34}-{0.30}\right)}-{0.142}{\left({0.34}-{0.30}\right)}^{{2}}{N}{S}{K}-{1.05}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}+{20.732}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}^{{2}}{N}{S}{K}-{431.41}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}^{{2}}{\left({0.34}-{0.35}\right)}{N}{S}{K}{H}_{{5}}{\left({0.34}\right)}={0.33349}$$

### Relevant Questions

An automobile tire manufacturer collected the data in the table relating tire pressure x​ (in pounds per square​ inch) and mileage​ (in thousands of​ miles). A mathematical model for the data is given by
$$\displaystyle​ f{{\left({x}\right)}}=-{0.554}{x}^{2}+{35.5}{x}-{514}.$$
$$\begin{array}{|c|c|} \hline x & Mileage \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}$$
​(A) Complete the table below.
$$\begin{array}{|c|c|} \hline x & Mileage & f(x) \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}$$
​(Round to one decimal place as​ needed.)
$$A. 20602060xf(x)$$
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,45), (30,51), (32,56), (34,50), and (36,46). A parabola opens downward and passes through the points (28,45.7), (30,52.4), (32,54.7), (34,52.6), and (36,46.0). All points are approximate.
$$B. 20602060xf(x)$$
Acoordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2.
Data points are plotted at (43,30), (45,36), (47,41), (49,35), and (51,31). A parabola opens downward and passes through the points (43,30.7), (45,37.4), (47,39.7), (49,37.6), and (51,31). All points are approximate.
$$C. 20602060xf(x)$$
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (43,45), (45,51), (47,56), (49,50), and (51,46). A parabola opens downward and passes through the points (43,45.7), (45,52.4), (47,54.7), (49,52.6), and (51,46.0). All points are approximate.
$$D.20602060xf(x)$$
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,30), (30,36), (32,41), (34,35), and (36,31). A parabola opens downward and passes through the points (28,30.7), (30,37.4), (32,39.7), (34,37.6), and (36,31). All points are approximate.
​(C) Use the modeling function​ f(x) to estimate the mileage for a tire pressure of 29
$$\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$ and for 35
$$\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$
The mileage for the tire pressure $$\displaystyle{29}\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$ is
The mileage for the tire pressure $$\displaystyle{35}\frac{{{l}{b}{s}}}{{{s}{q}}}$$ in. is
(Round to two decimal places as​ needed.)
(D) Write a brief description of the relationship between tire pressure and mileage.
A. As tire pressure​ increases, mileage decreases to a minimum at a certain tire​ pressure, then begins to increase.
B. As tire pressure​ increases, mileage decreases.
C. As tire pressure​ increases, mileage increases to a maximum at a certain tire​ pressure, then begins to decrease.
D. As tire pressure​ increases, mileage increases.
First, construct the sixth degree Taylor polynomial $$\displaystyle{P}_{{6}}{\left({x}\right)}$$ for function $$\displaystyle{f{{\left({x}\right)}}}={\sin{{\left({x}^{{2}}\right)}}}$$ about $$\displaystyle{x}_{{0}}={0}$$ The use $$\displaystyle{\int_{{{0}}}^{{{1}}}}{P}_{{6}}{\left({x}\right)}{\left.{d}{x}\right.}$$ to approximate the integral $$\displaystyle\ {\int_{{{0}}}^{{{1}}}}{\sin{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}.$$ Use 4-digit rounding arithmetic in all calculations. What is the approximate value?
For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{1}&{2}&{3}&{4}&{5}&{6}&{7}&{8}&{9}&{10}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{409.4}&{260.7}&{170.4}&{110.6}&{74}&{44.7}&{32.4}&{19.5}&{12.7}&{8.1}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Given the following function: $$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}-{1.99}$$ a)Use three-digit rounding frithmetic, the assumption that $$\displaystyle{e}^{{{1.53}}}={4.62}$$, and the fact that $$\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}$$ to evaluate $$\displaystyle{f{{\left({1.53}\right)}}}$$ b)Redo the same calculation by first rewriting the equation using the polynomial factoring technique c)Calculate the percentage relative errors in both part a) and b) to the true result $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
Given the following function:
$$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}$$
a) Use three-digit rounding frithmetic, the assumption that $$\displaystyle{e}^{{{1.53}}}={4.62}$$, and the fact that $$\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}$$ to evaluate $$\displaystyle{f{{\left({1.53}\right)}}}$$
b) Redo the same calculation by first rewriting the equation using the polynomial factoring technique
c) Calculate the percentage relative errors in both part a) and b) to the true result $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
Solve the given information Let $$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}}.$$ Determine an error bound for the approxiamtion $$\displaystyle{f{{\left({0.34}\right)}}}={\sin{{0.34}}}.$$ and compare it to the actual error.
b. Let $$f(x) = \frac{e^{x}-e^{-x}}{x}$$.
Use three-digit rounding arithmetic to evaluate f(0.1).
The following table shows the approximate average household income in the United States in 1990, 1995, and 2003. ($$\displaystyle{t}={0}$$ represents 1990.)
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{t(Year)}&{0}&{5}&{13}\backslash{h}{l}\in{e}\text{H(Household Income in}\ \{1},{000}{)}&{30}&{35}&{43}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Which of the following kinds of models would best fit the given data?
Explain your choice of model. ( a, b, c, and m are constants.)
a) Linear: $$\displaystyle{H}{\left({t}\right)}={m}{b}\ +\ {b}$$
b) Quadratic: $$\displaystyle{H}{\left({t}\right)}={a}{t}^{{{2}}}\ +\ {b}{t}\ +\ {c}$$
c) Exponential: $$\displaystyle{H}{\left({t}\right)}={A}{b}^{{{t}}}$$
a) Evaluate the polynomial $$\displaystyle{y}={x}^{{3}}-{7}{x}^{{2}}+{8}{x}-{0.35}$$
at $$\displaystyle{x}={1.37}$$ . Use 3-digit arithmetic with chopping. Evaluate the percent relative error. b) Repeat (a) but express y as $$\displaystyle{y}={\left({\left({x}-{7}\right)}{x}+{8}\right)}{x}-{0.35}$$ Evaluate the error and compare with part (a).
The accompanying two-way table was constructed using data in the article “Television Viewing and Physical Fitness in Adults” (Research Quarterly for Exercise and Sport, 1990: 315–320). The author hoped to determine whether time spent watching television is associated with cardiovascular fitness. Subjects were asked about their television-viewing habits and were classified as physically fit if they scored in the excellent or very good category on a step test. We include MINITAB output from a chi-squared analysis. The four TV groups corresponded to different amounts of time per day spent watching TV (0, 1–2, 3–4, or 5 or more hours). The 168 individuals represented in the first column were those judged physically fit. Expected counts appear below observed counts, and MINITAB displays the contribution to $$\displaystyle{x}^{{{2}}}$$ from each cell.
State and test the appropriate hypotheses using $$\displaystyle\alpha={0.05}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{a}\mp,\ {1}&{a}\mp,\ {2}&{a}\mp,\ {T}{o}{t}{a}{l}\backslash{h}{l}\in{e}{1}&{a}\mp,\ {35}&{a}\mp,\ {147}&{a}\mp,\ {182}\backslash{h}{l}\in{e}&{a}\mp,\ {25.48}&{a}\mp,\ {156.52}&{a}\mp,\backslash{h}{l}\in{e}{2}&{a}\mp,\ {101}&{a}\mp,\ {629}&{a}\mp,\ {730}\backslash{h}{l}\in{e}&{a}\mp,\ {102.20}&{a}\mp,\ {627.80}&{a}\mp,\backslash{h}{l}\in{e}{3}&{a}\mp,\ {28}&{a}\mp,\ {222}&{a}\mp,\ {250}\backslash{h}{l}\in{e}&{a}\mp,\ {35.00}&{a}\mp,\ {215.00}&{a}\mp,\backslash{h}{l}\in{e}{4}&{a}\mp,\ {4}&{a}\mp,\ {34}&{a}\mp,\ {38}\backslash{h}{l}\in{e}&{a}\mp,\ {5.32}&{a}\mp,\ {32.68}&{a}\mp,\backslash{h}{l}\in{e}{T}{o}{t}{a}{l}&{a}\mp,\ {168}&{a}\mp,\ {1032}&{a}\mp,\ {1200}\backslash{h}{l}\in{e}$$
$$\displaystyle{C}{h}{i}{s}{q}={a}\mp,\ {3.557}\ +\ {0.579}\ +\ {a}\mp,\ {0.014}\ +\ {0.002}\ +\ {a}\mp,\ {1.400}\ +\ {0.228}\ +\ {a}\mp,\ {0.328}\ +\ {0.053}={6.161}$$
$$\displaystyle{d}{f}={3}$$