# Solve the given information Let f(x) = sin x. Determine an error bound for the approxiamtion f(0.34) = sin 0.34. and compare it to the actual error. Question
Polynomial arithmetic 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. 2020-12-04
We to find an error bound for the approximation and compare it to the actual error. The exact value of $$\displaystyle{\sin{{\left({0.34}\right)}}}={0.33348709214081}.$$ So, the actual error is $$\displaystyle{\left({0.33348709214081}-{0.33349}\right)}={2.91}\times{10}^{{-{6}}}$$ The bound for the error on [0.30, 0.35] is given by, $$\displaystyle{\left|{f{{\left({x}\right)}}}-{H}_{{5}}{\left({x}\right)}\right|}={\left|{\frac{{{{f}^{{6}}{\left(\xi\right)}}}}{{{6}!}}}{\left({x}={0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{\left({x}-{0.35}\right)}^{{2}}\right|}{N}{S}{K}={\left|{\frac{{-{\sin{{\left(\xi\right)}}}}}{{{720}}}}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{\left({x}-{0.35}\right)}^{{2}}\right|}$$ When $$\displaystyle\xi\in{\left[{0.30},{0.35}\right]}.$$ Evaluating this error term at x = 0.34 yields, |sin(0.34) - H_5 (0.34)| = |\frac{-sin(\xi)}{720}(0.04)^2 (0.02)^2 (-0.01)^2|
\leq |\frac{-sin(0.35)}{720}(0.04)^2 (0.02)^2 (-0.01)^2|
\leq 3.05 \times 10^{-14}ZSK This bound is not inconsistent with the actual error, because the approximation was computed using five-digit rounding arithmetic.

### Relevant Questions 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 $$\displaystyle{f{{\left({0.34}\right)}}}={\sin{{0.34}}}.$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}{\mid}{c}\right\rbrace}{x}&{f{{\left({x}\right)}}}&{f}'{\left({x}\right)}\backslash{h}{l}\in{e}{0.30}&{0.29552}&{0.95534}\backslash{0.32}&{0.31457}&{0.94924}\backslash{0.35}&{0.34290}&{0.93937}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ c. Let $$f(x) = \frac{e^{x}-e^{-x}}{x}$$.
The actual values values is $$f(0.1) = 2.003335000$$. Find the relative error for the values obtained in parts (b) and (c) An experiment on the probability is carried out, in which the sample space of the experiment is
$$S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.$$
Let event $$E={2, 3, 4, 5, 6, 7}, event$$
$$F={5, 6, 7, 8, 9}, event G={9, 10, 11, 12}, and event H={2, 3, 4}$$.
Assume that each outcome is equally likely. List the outcome s in For G.
Now find P( For G) by counting the number of outcomes in For G.
Determine P (For G ) using the General Addition Rule. 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). a) Use base b = 10, precision k = 4, idealized, chopping floating-point arithmetic to show that fl(g(1.015)) is inaccurate, where $$\displaystyle{g{{\left({x}\right)}}}={\frac{{{x}^{{\frac{{1}}{{4}}}}-{1}}}{{{x}-{1}}}}$$ b) Derive the second order (n = 2) quadratic Taylor polynomial approximation for $$\displaystyle{f{{\left({x}\right)}}}={x}^{{\wedge}}\frac{{1}}{{4}},$$ expanded about a = 1, and use it to get an accurate approximation to g(x) in part (a). c) Verify that your approximation in (b) is more accurate. The arithmetic mean (average) of two numbers c and d is given by $$\overline{x} = \frac{c+d}{2}$$
The value $$\overline{x}$$ is equidistant between c and d, so the sequence c,$$\overline{x}$$, d is an arithmetic sequence. inserting k equally spaced values between c and d, yields the arithmetic sequence $$c, \overline{X}_{1}, \overline{X}_{2}, \overline{X}_{3}, \overline{X}_{4}, ..., \overline{X}_{k}, d$$. Use this information for Exercise.
Insert four arithmetic means between 19 and 64. 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? Determine whether the following state-ments are true and give an explanation or counterexample.
a) All polynomials are rational functions, but not all rational functions are polynomials.
b) If f is a linear polynomial, then $$\displaystyle{f}\times{f}$$ is a quadratic polynomial.
c) If f and g are polynomials, then the degrees of $$\displaystyle{f}\times{g}$$ and $$\displaystyle{g}\times{f}$$ are equal.
d) To graph $$\displaystyle{g{{\left({x}\right)}}}={f{{\left({x}+{2}\right)}}}$$, shift the graph of f 2 units to the right. One type of Iodine disintegrates continuously at a constant rate of $$\displaystyle{8.6}\%$$ per day.
Suppose the original amount, $$\displaystyle{P}_{{0}}$$, is 10 grams, and let t be measured in days.
Because the Iodine is decaying continuously at a constant rate, we use the model $$\displaystyle{P}={P}_{{0}}{e}^{k}{t}$$ for the decay equation, where k is the rate of continuous decay. b. Let $$f(x) = \frac{e^{x}-e^{-x}}{x}$$.