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

### 1. Evaluating the given polynomial \(P(x)\)

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

We'll first simplify the polynomial to a more understandable form. Let's rewrite and expand it step by step:

\[ P(x) = (((x + 3)x + 2)x + 2)x + 1 \]

Starting from the innermost expression:
\[ (x + 3)x = x^2 + 3x \]
\[ (x^2 + 3x + 2)x = x^3 + 3x^2 + 2x \]
\[ (x^3 + 3x^2 + 2x + 2)x = x^4 + 3x^3 + 2x^2 + 2x \]
\[ (x^4 + 3x^3 + 2x^2 + 2x + 1) = x^4 + 3x^3 + 2x^2 + 2x + 1 \]

Therefore, the polynomial is:
\[ P(x) = x^4 + 3x^3 + 2x^2 + 2x + 1 \]

### 2. Evaluating \(P(x)\) at \(x = 0.49\)

We will now approximate \(P(0.49)\) using four-digit rounding arithmetic:

1. **Calculate \(x^2\):**
   \[ x^2 = (0.49)^2 = 0.2401 \]

2. **Calculate \(x^3\):**
   \[ x^3 = (0.49)^3 = 0.117649 \approx 0.1176 \]

3. **Calculate \(x^4\):**
   \[ x^4 = (0.49)^4 = 0.05764801 \approx 0.05765 \]

4. **Substitute and evaluate the polynomial:**
   \[ P(0.49) = 0.05765 + 3(0.1176) + 2(0.2401) + 2(0.49) + 1 \]
   \[ P(0.49) = 0.05765 + 0.3528 + 0.4802 + 0.98 + 1 \]

Summing these terms while rounding to four decimal places at each step:
   \[ P(0.49) \approx 0.0577 + 0.3528 + 0.4802 + 0.98 + 1 \]
   \[ P(0.49) \approx 0.4105 + 0.4802 + 0.98 + 1 \]
   \[ P(0.49) \approx 0.8907 + 0.98 + 1 \]
   \[ P(0.49) \approx 1.8707 + 1 \]
   \[ P(0.49) \approx 2.8707 \]

### 3. Approximate \(e^{-0.98}\)

The given problem asks to approximate \(e^{-0.98}\) using the polynomial evaluated at \(x = 0.49\):
\[ e^{-0.98} \approx P(0.49) = 2.8707 \]

### 4. Compute the Absolute and Relative Error

Let's find the actual value of \(e^{-0.98}\) using a calculator:
\[ e^{-0.98} \approx 0.3753 \]

#### Absolute Error
\[ \text{Absolute Error} = | \text{Approximate Value} - \text{Actual Value} | \]
\[ \text{Absolute Error} = |2.8707 - 0.3753| = 2.4954 \]

#### Relative Error
\[ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Actual Value}} \]
\[ \text{Relative Error} = \frac{2.4954}{0.3753} \approx 6.65 \]

### Summary
- Approximation of \(e^{-0.98}\) using the given polynomial: \(2.8707\)
- Absolute error: \(2.4954\)
- Relative error: \(6.65\) (or 665%)

Would you like more details or have any questions? Here are some related questions:

1. How is the Maclaurin series derived from the Taylor series?
2. What are the conditions for a Maclaurin series to converge to its function?
3. Can you derive the Maclaurin series for \(\sin(x)\) and \(\cos(x)\)?
4. How does the error term in a Maclaurin series approximation impact its accuracy?
5. What are some real-world applications of Maclaurin series approximations?
6. How do you generalize the Maclaurin series to higher dimensions?
7. Can you explain the significance of the radius of convergence for series?
8. How would you compute higher-order Maclaurin polynomials for more complex functions?

**Tip:** When using polynomial approximations, make sure to evaluate the polynomial and its terms accurately to avoid large errors, especially for exponential functions where small changes can have significant impacts.

Learn how to approximate e^-0.98 using the fifth Maclaurin polynomial for e^(2x). This detailed solution involves polynomial expansion, four-digit rounding arithmetic, and error analysis. Suitable for advanced high school students and those interested in mathematical approximation techniques.

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