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
asked 2020-11-01
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 \(\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}\)

Answers (1)

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%
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