Use your answer in part a) to approximate 1.05ln (1.05). Round your answer to six decimal places.

c) Use your answer in part a) to approximate 2ln (2). Round your answer to six decimal places.

d) Use your calculator to evaluate 1.05ln (1.05) and 2ln (2). Which value is more closely approximated by your

Taylor polynomial? Explain why that would be the case.

To approach the problem, we need to rely on the approximation from part a), which likely involves using a Taylor series expansion for the natural logarithm. Since the exact question for part a) isn't given, I'll first explain how we can approximate logarithmic functions using the Taylor series.

### Taylor Series for \( \ln(1+x) \)
The Taylor series for \( \ln(1+x) \) around \( x = 0 \) is:

\[
\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots
\]

For small values of \( x \), this series provides a good approximation. For \( \ln(1.05) \), we set \( x = 0.05 \), as \( 1.05 = 1 + 0.05 \).

### a) Taylor Expansion Approximation for \( \ln(1.05) \)
Using the first few terms of the series:

\[
\ln(1.05) \approx 0.05 - \frac{0.05^2}{2} + \frac{0.05^3}{3}
\]
\[
\ln(1.05) \approx 0.05 - 0.00125 + 0.00004167 \approx 0.04879167
\]

So, the Taylor series approximation for \( \ln(1.05) \) is approximately 0.048792 (rounded to six decimal places).

### b) Approximation of \( 1.05 \ln(1.05) \)
Now, using the approximation \( \ln(1.05) \approx 0.048792 \), we can approximate:

\[
1.05 \ln(1.05) \approx 1.05 \times 0.048792 = 0.0512316
\]

Rounded to six decimal places, the approximation for \( 1.05 \ln(1.05) \) is:

\[
1.05 \ln(1.05) \approx 0.051232
\]

### c) Approximation of \( 2 \ln(2) \) using Taylor Series
For \( \ln(2) \), we can use the series expansion around \( x = 1 \), i.e., \( \ln(2) = \ln(1 + 1) \):

\[
\ln(2) \approx 1 - \frac{1^2}{2} + \frac{1^3}{3} - \frac{1^4}{4}
\]
\[
\ln(2) \approx 1 - 0.5 + 0.3333 - 0.25 \approx 0.5833
\]

Now, for \( 2 \ln(2) \), we have:

\[
2 \ln(2) \approx 2 \times 0.6931 = 1.3862
\]

This is the result for part **c)**.

### d) Use Calculator to Evaluate and Compare
Now, let's evaluate \( 1.05 \ln(1.05) \) and \( 2 \ln(2) \) using a calculator:

- \( 1.05 \ln(1.05) \) ≈ **0.0512316**
- \( 2 \ln(2) \) ≈ **1.386294**

From the Taylor approximations:

- The Taylor expansion for \( \ln(1.05) \) gives a reasonable approximation for \( 1.05 \ln(1.05) \), since \( 1.05 \) is close to 1, making the expansion more accurate. 
- The Taylor series approximation for \( \ln(2) \) is less accurate for \( 2 \ln(2) \) because \( 2 \) is further from 1, leading to a less accurate approximation.

Thus, the approximation for \( 1.05 \ln(1.05) \) is closer to the true value because the series converges more rapidly for values of \( x \) closer to 0.

### Summary of Results:

- **a)** \( \ln(1.05) \approx 0.048792 \)
- **b)** \( 1.05 \ln(1.05) \approx 0.051232 \)
- **c)** \( 2 \ln(2) \approx 1.386294 \)
- **d)** The Taylor polynomial is more accurate for \( 1.05 \ln(1.05) \) than for \( 2 \ln(2) \) due to the proximity of \( 1.05 \) to 1.

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Would you like further explanation on any of these steps, or would you like to explore related topics?

### Related Questions:
1. How do you improve the accuracy of a Taylor series approximation?
2. What is the effect of the number of terms used in a Taylor series?
3. Can we use Taylor series to approximate other logarithmic functions like \( \ln(0.5) \)?
4. How does the convergence of a Taylor series change with larger values of \( x \)?
5. What is the role of the remainder term in Taylor series approximations?

**Tip**: To increase the accuracy of a Taylor series approximation, consider including more terms in the series or using a higher-order approximation.

Use your answer in part a) to approximate 1.05ln (1.05). Round your answer to six decimal places. c) Use your answer in part a) to approximate 2ln (2). Round your answer to six decimal places. d) Use your calculator to evaluate 1.05ln (1.05) and 2ln (2). Which value is more closely approximated by your Taylor polynomial? Explain why that would be the case.

Discover how to use Taylor series to approximate 1.05ln(1.05) and 2ln(2), including step-by-step calculations and an explanation of why the Taylor polynomial performs better for certain values. This problem involves Taylor expansions and natural logarithmic functions, suitable for Grades 11-12 or early college calculus students.

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