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 P(0.49) and four-digit rounding arithmetic. Compute the absolute and relative error

To solve this problem, we need to construct the fifth Maclaurin polynomial for \( e^{2x} \) and \( e^{-2x} \), then use these polynomials to approximate \( e^{-0.98} \) using \( P(0.49) \), and finally compute the absolute and relative errors. 

### Maclaurin Polynomial for \( e^{2x} \)

The Maclaurin series for \( e^{2x} \) is given by:
\[ e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} + \frac{(2x)^5}{5!} + \cdots \]

The fifth Maclaurin polynomial is:
\[ P_5(x) = 1 + 2x + 2x^2 + \frac{8x^3}{3!} + \frac{16x^4}{4!} + \frac{32x^5}{5!} \]
\[ P_5(x) = 1 + 2x + 2x^2 + \frac{8x^3}{6} + \frac{16x^4}{24} + \frac{32x^5}{120} \]
\[ P_5(x) = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \frac{2x^4}{3} + \frac{4x^5}{15} \]

### Maclaurin Polynomial for \( e^{-2x} \)

The Maclaurin series for \( e^{-2x} \) is given by:
\[ e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = 1 - 2x + \frac{(2x)^2}{2!} - \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} - \frac{(2x)^5}{5!} + \cdots \]

The fifth Maclaurin polynomial is:
\[ P_5(x) = 1 - 2x + 2x^2 - \frac{8x^3}{3!} + \frac{16x^4}{4!} - \frac{32x^5}{5!} \]
\[ P_5(x) = 1 - 2x + 2x^2 - \frac{8x^3}{6} + \frac{16x^4}{24} - \frac{32x^5}{120} \]
\[ P_5(x) = 1 - 2x + 2x^2 - \frac{4x^3}{3} + \frac{2x^4}{3} - \frac{4x^5}{15} \]

### Approximation of \( e^{-0.98} \) using \( P(0.49) \)

To approximate \( e^{-0.98} \), we use:
\[ P_5(0.49) = 1 - 2(0.49) + 2(0.49)^2 - \frac{4(0.49)^3}{3} + \frac{2(0.49)^4}{3} - \frac{4(0.49)^5}{15} \]

Calculating each term separately:
\[ 1 \]
\[ -2(0.49) = -0.98 \]
\[ 2(0.49)^2 = 2(0.2401) = 0.4802 \]
\[ -\frac{4(0.49)^3}{3} = -\frac{4(0.117649)}{3} = -\frac{0.470596}{3} = -0.156865 \]
\[ \frac{2(0.49)^4}{3} = \frac{2(0.05764801)}{3} = \frac{0.11529602}{3} = 0.038432 \]
\[ -\frac{4(0.49)^5}{15} = -\frac{4(0.028236441)}{15} = -\frac{0.112945764}{15} = -0.007530 \]

Summing these terms:
\[ P_5(0.49) = 1 - 0.98 + 0.4802 - 0.156865 + 0.038432 - 0.007530 \]
\[ P_5(0.49) \approx 0.373 \]

### Exact Value and Errors

The exact value of \( e^{-0.98} \) can be found using a calculator:
\[ e^{-0.98} \approx 0.3753 \]

#### Absolute Error
\[ \text{Absolute Error} = | \text{Exact Value} - \text{Approximate Value} | = | 0.3753 - 0.373 | = 0.0023 \]

#### Relative Error
\[ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Exact Value}} = \frac{0.0023}{0.3753} \approx 0.0061 \]

### Summary

- **Approximate Value using \( P(0.49) \):** \( 0.373 \)
- **Exact Value of \( e^{-0.98} \):** \( 0.3753 \)
- **Absolute Error:** \( 0.0023 \)
- **Relative Error:** \( 0.0061 \)

Would you like more details or have any questions?

#### Related Questions
1. What is the general form of the Maclaurin series for \( e^x \)?
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 double-check your polynomial coefficients and calculations to ensure accuracy, especially when dealing with approximations.

Learn how to approximate e^-0.98 using Maclaurin polynomials for e^(2x) and e^(-2x), with detailed calculations and error analysis. This problem involves constructing fifth-degree polynomials, computing approximations, and analyzing absolute and relative errors.

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