# Given the following function: f(x)=1.01e^{4x}-4.62e^{3x}-3.11e^{2x}+12.2e^{x} - 1.99 a)Use three-digit rounding frithmetic, the assumption that e^{1.53}=4.62, and the fact that e^{nx}=(e^{x})^{n} to evaluate f(1.53) 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 f(1.53)=-7.60787

Question
Polynomial arithmetic
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 $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$

2020-11-02
a) $$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{3}}{x}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{x}}-{1.99}{N}{S}{K}{f{{\left({x}\right)}}}={1.01}{\left({e}^{{x}}\right)}^{{4}}-{4.62}{\left({e}^{{x}}\right)}^{{3}}-{3.11}{\left({e}^{{{x}}}\right)}^{{2}}-{12.2}{e}^{{x}}-{1.99}{\left({a}{s}{e}^{{{n}{x}}}={\left({e}^{{x}}\right)}^{{n}}\right)}{N}{S}{K}{f{{\left({1.53}\right)}}}={1.01}{\left({e}^{{{1.53}}}\right)}^{{4}}-{4.62}{\left({e}^{{{1.53}}}\right)}^{{3}}-{3.11}{\left({e}^{{{1.53}}}\right)}^{{2}}+{12.2}{e}^{{{1.53}}}-{1.99}{N}{S}{K}={1.01}{\left({4.62}\right)}^{{4}}-{4.62}{\left({4.62}\right)}^{{3}}-{3.11}{\left({4.62}\right)}^{{2}}+{12.2}{\left({4.62}\right)}-{1.99}{\left({a}{s}{e}^{{{1.53}={4.62}}}\right)}{N}{S}{K}={1.01}{\left({455.583}\right)}-{4.62}{\left({98.611}\right)}-{3.11}{\left({21.344}\right)}+{56.364}-{1.99}{N}{S}{K}={460.139}-{455.583}-{66.380}+{54.374}{N}{S}{K}=-{7.45}$$ Therefore, the value of f(1.53) abtained by this method is -7.45. b) The given function can be factorized sa follows. $$\displaystyle{f{{\left({x}\right)}}}={1.01}{\left({e}^{{x}}+{1.715}\right)}{\left({e}^{{x}}-{0.173}\right)}{\left({e}^{{x}}-{1.415}\right)}{\left({e}^{{x}}-{4.702}\right)}$$ On substituting $$\displaystyle{x}={1.53}$$ and using $$\displaystyle{e}^{{{1.53}}}={4.62},$$ we get $$\displaystyle{f{{\left({1.53}\right)}}}={1.01}{\left({4.62}+{1.715}\right)}{\left({4.62}-{0.173}\right)}{\left({4.62}-{1.415}\right)}{\left({4.62}-{4.702}\right)}{N}{S}{K}={1.01}{\left({6.335}\right)}{\left({4.447}\right)}{\left({3.205}\right)}{\left(-{0.082}\right)}{N}{S}{K}=-{7.478}$$ Therefore, the value of $$\displaystyle{f{{\left({1.53}\right)}}}$$ obtained by this method is -7.478. c) Percentage error $$\displaystyle\delta$$ is given by $$\displaystyle\delta={\left|{\frac{{\nu_{{A}}-\nu_{{E}}}}{{\nu_{{E}}}}}\right|}{100}\%$$ Here $$\displaystyle\nu_{{A}}$$ is the actual value and $$\displaystyle\nu_{{E}}$$ is the expected value which is -7.60787 in this case. For the value obtained in part (a), the percentage error is $$\displaystyle\delta_{{a}}={\left|{\frac{{-{7.45}+{7.60787}}}{{-{7.60787}}}}\right|}{100}\%{N}{S}{K}={2.057}\%$$ For the value obtained in part (b), the percentage error is $$\displaystyle\delta_{{d}}={\left|{\frac{{-{7.478}+{7.60787}}}{{-{7.60787}}}}\right|}{100}\%{N}{S}{K}={1.707}\%$$ Therefore, the percentage error in part (a) is 2.075% and that is part(b) is 1.707%

### Relevant Questions

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 $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same $$\displaystyle{P}{\left({x}\right)}={3}{x}^{{4}}-{5}{x}^{{3}}+{x}^{{2}}-{3}{x}+{5}{N}{S}{K}{Q}{\left({x}\right)}={\left({\left({\left({3}{x}-{5}\right)}{x}+{1}\right)}{x}-{3}\right)}{x}+{5}$$ Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial $$\displaystyle{R}{\left({x}\right)}={x}^{{5}}—{2}{x}^{{4}}+{3}{x}^{{3}}—{2}{x}^{{3}}+{3}{x}+{4}$$ in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head. Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division?
DISCOVER: Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same $$P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x +5$$
$$Q(x) = (((3x - 5)x + 1)x 3)x + 5$$
Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial
R(x) =x^{5} - 2x^{4} + 3x^{3} - 2x^{2} + 3x + 4\) in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head.
Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division?
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).
1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance.
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?
b. Let $$f(x) = \frac{e^{x}-e^{-x}}{x}$$.
Consider the quantity$$a^{2}\ -\ b^{2}$$ where a and b are real numbers.
(a) Under what conditions should one expect an unusually large relative error in the computed value of $$a^{2}\ -\ b^{2}$$ when this expression is evaluated in finite precision arithmetic?
(b)cWs 4-digit (decimal) rounding arithmetic to evaluate both $$a^{2}\ -\ b^{2}\ and\ (a\ +\ b)(a\ -\ b)\ with\ a\ = 995.1\ and\ b = 995.0.$$ Calculate th relative error in each result.
(c) The expression $$(a\ +\ b)(a\ -\ b)\ is\ algebraically\ equivalent\ to\ a^{2}\ -\ b^{2},$$ but it is a more accurate way to calculate this quantity if both a and b have exact floating point representations. Why?
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.