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

To approximate \( e^{-0.98} \) using the given fifth Maclaurin polynomial for \( e^{2x} \) and its transformation, we need to follow these steps:

1. Construct the fifth Maclaurin polynomial for \( e^{2x} \).
2. Approximate \( e^{-0.98} \) by finding \( P(0.49) \) and then using the reciprocal \( \frac{1}{P(0.49)} \).
3. Calculate the absolute and relative errors.

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

The given polynomial form is:
\[ P(x) = (((x+3)x + )x + 2)x + 2)x + 1 \]

This appears to be incorrectly formatted, but we'll write out the correct fifth Maclaurin polynomial for \( e^{2x} \):
\[ P_5(x) = 1 + 2x + 2x^2 + \frac{8x^3}{6} + \frac{16x^4}{24} + \frac{32x^5}{120} \]
\[ P_5(x) = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \frac{2x^4}{3} + \frac{4x^5}{15} \]

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

Calculate \( P(0.49) \):
\[ P_5(0.49) = 1 + 2(0.49) + 2(0.49)^2 + \frac{4(0.49)^3}{3} + \frac{2(0.49)^4}{3} + \frac{4(0.49)^5}{15} \]

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

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

### Finding the Reciprocal \( \frac{1}{P(0.49)} \)

Using four-digit rounding:
\[ \frac{1}{2.6630} \approx 0.3755 \]

### Exact Value and Errors

The exact value of \( e^{-0.98} \) is:
\[ e^{-0.98} \approx 0.3753 \]

#### Absolute Error
\[ \text{Absolute Error} = | \text{Exact Value} - \text{Approximate Value} | = | 0.3753 - 0.3755 | = 0.0002 \]

#### Relative Error
\[ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Exact Value}} = \frac{0.0002}{0.3753} \approx 0.0005 \]

### Summary

- **Approximate Value using \( \frac{1}{P(0.49)} \):** \( 0.3755 \)
- **Exact Value of \( e^{-0.98} \):** \( 0.3753 \)
- **Absolute Error:** \( 0.0002 \)
- **Relative Error:** \( 0.0005 \)

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

Learn how to approximate e^-0.98 using Maclaurin polynomials for e^(2x). Step-by-step solution including calculating the fifth Maclaurin polynomial, finding P(0.49), and computing absolute and relative errors.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation