Math Problem Statement

The fifth Maclaurin polynomial for e^(2x) and e^-(2x) are P(x) = = (((x+3) x + )x+2)x + 2)x + 1 Approximate e-0.98 using 1/( P(0.49)) and four-digit rounding arithmetic. Compute the absolute and relative error

Solution

To solve the problem, we need to correctly interpret the fifth Maclaurin polynomial for e2xe^{2x} given in a compact form and then approximate e0.98e^{-0.98} using 1P(0.49)\frac{1}{P(0.49)}. Finally, we will calculate the absolute and relative errors.

Fifth Maclaurin Polynomial for e2xe^{2x}

Given the polynomial: P(x)=(((x+3)x+)x+2)x+2)x+1P(x) = (((x+3)x + )x + 2)x + 2)x + 1

It seems the given expression is incorrect or typographically unclear. Let's start with the correct Maclaurin series expansion for e2xe^{2x}:

e2x=n=0(2x)nn!=1+2x+(2x)22!+(2x)33!+(2x)44!+(2x)55!+e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} + \frac{(2x)^5}{5!} + \cdots

The fifth Maclaurin polynomial is: P5(x)=1+2x+2x2+8x36+16x424+32x5120P_5(x) = 1 + 2x + 2x^2 + \frac{8x^3}{6} + \frac{16x^4}{24} + \frac{32x^5}{120} P5(x)=1+2x+2x2+4x33+2x43+4x515P_5(x) = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \frac{2x^4}{3} + \frac{4x^5}{15}

Calculation of P(0.49)P(0.49)

P5(0.49)=1+2(0.49)+2(0.49)2+4(0.49)33+2(0.49)43+4(0.49)515P_5(0.49) = 1 + 2(0.49) + 2(0.49)^2 + \frac{4(0.49)^3}{3} + \frac{2(0.49)^4}{3} + \frac{4(0.49)^5}{15}

Calculating each term: 11 +2(0.49)=0.98+ 2(0.49) = 0.98 +2(0.49)2=2(0.2401)=0.4802+ 2(0.49)^2 = 2(0.2401) = 0.4802 +4(0.49)33=4(0.117649)3=0.4705963=0.156865+ \frac{4(0.49)^3}{3} = \frac{4(0.117649)}{3} = \frac{0.470596}{3} = 0.156865 +2(0.49)43=2(0.05764801)3=0.115296023=0.038432+ \frac{2(0.49)^4}{3} = \frac{2(0.05764801)}{3} = \frac{0.11529602}{3} = 0.038432 +4(0.49)515=4(0.028236441)15=0.11294576415=0.007530+ \frac{4(0.49)^5}{15} = \frac{4(0.028236441)}{15} = \frac{0.112945764}{15} = 0.007530

Summing these terms: P5(0.49)=1+0.98+0.4802+0.156865+0.038432+0.007530P_5(0.49) = 1 + 0.98 + 0.4802 + 0.156865 + 0.038432 + 0.007530 P5(0.49)2.6630P_5(0.49) \approx 2.6630

Approximation Using 1P(0.49)\frac{1}{P(0.49)}

Using four-digit rounding arithmetic: 1P(0.49)=12.66300.3755\frac{1}{P(0.49)} = \frac{1}{2.6630} \approx 0.3755

Exact Value and Errors

The exact value of e0.98e^{-0.98} is: e0.980.3753e^{-0.98} \approx 0.3753

Absolute Error

Absolute Error=Exact ValueApproximate Value=0.37530.3755=0.0002\text{Absolute Error} = | \text{Exact Value} - \text{Approximate Value} | = | 0.3753 - 0.3755 | = 0.0002

Relative Error

Relative Error=Absolute ErrorExact Value=0.00020.37530.0005\text{Relative Error} = \frac{\text{Absolute Error}}{\text{Exact Value}} = \frac{0.0002}{0.3753} \approx 0.0005

Summary

  • Approximate Value using 1P(0.49)\frac{1}{P(0.49)}: 0.37550.3755
  • Exact Value of e0.98e^{-0.98}: 0.37530.3753
  • Absolute Error: 0.00020.0002
  • Relative Error: 0.00050.0005

Would you like more details or have any questions?

Related Questions

  1. What is the general form of the Maclaurin series for exe^x?
  2. How does the Taylor series differ from the Maclaurin series?
  3. What are some applications of Maclaurin series in real-world problems?
  4. How can we improve the accuracy of our polynomial approximations?
  5. What are some limitations of using Maclaurin series for approximations?
  6. How do we determine the radius of convergence for a power series?
  7. What is the significance of the fifth-degree polynomial in approximations?
  8. How can we extend these concepts to multivariable functions?

Tip

Always use the exact value and keep intermediate results in higher precision to minimize rounding errors.

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Math Problem Analysis

Mathematical Concepts

Maclaurin series
Exponential functions
Polynomial approximation

Formulas

Maclaurin series expansion for e^(2x)
Approximation using polynomial

Theorems

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Suitable Grade Level

Advanced High School / College