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We are given the Taylor polynomial \( T_{10}(x) \) of degree 10 for the function \( f(x) = \ln(1 + x) \) around \( a = 0 \), and we need to find the values of \( x \) such that the approximation error is within 0.001.

To solve this, let's first review the key details of the problem and break it down:

### Step 1: Taylor Series Expansion
The Taylor series expansion of \( f(x) = \ln(1 + x) \) around \( x = 0 \) is given by:

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

The 10th-degree Taylor polynomial \( T_{10}(x) \) would be:

\[
T_{10}(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots + \frac{x^{10}}{10}
\]

### Step 2: Error Bound for Alternating Series
The problem suggests using the **alternating series approximation** to determine the error. For alternating series, the error in truncating the series at the nth term is at most the absolute value of the first omitted term. Therefore, the remainder \( R_{10}(x) \), which represents the error when using \( T_{10}(x) \) to approximate \( f(x) \), is approximately:

\[
R_{10}(x) \approx \left| \frac{x^{11}}{11} \right|
\]

### Step 3: Setting Up the Error Condition
We are given that the error must be less than or equal to 0.001:

\[
\left| \frac{x^{11}}{11} \right| \leq 0.001
\]

This simplifies to:

\[
\frac{x^{11}}{11} \leq 0.001
\]

Multiplying both sides by 11:

\[
x^{11} \leq 0.011
\]

### Step 4: Solving for \( x \)
Now we solve for \( x \):

\[
x \leq \left( 0.011 \right)^{1/11}
\]

Using a calculator:

\[
x \leq (0.011)^{1/11} \approx 0.728
\]

### Final Answer:
The value of \( x \) must be in the interval:

\[
0 < x \leq 0.728
\]

This means that for all positive values of \( x \) up to approximately 0.728, the approximation using \( T_{10}(x) \) will be within 0.001 of the actual value of \( \ln(1 + x) \).

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Would you like a more detailed explanation of any of the steps or further clarification?

Here are 5 related questions to explore:
1. What is the Taylor series for \( \ln(1 - x) \) and how does it differ from \( \ln(1 + x) \)?
2. Why does the alternating series approximation provide such a good error bound for series with alternating signs?
3. What is the error bound if we only use a Taylor polynomial of degree 5 for \( \ln(1 + x) \)?
4. How does the radius of convergence affect Taylor series approximations for \( \ln(1 + x) \)?
5. What are some other common functions that can be approximated using Taylor series?

**Tip:** When using Taylor series, higher-degree polynomials provide better approximations, but calculating the error is crucial to ensure accuracy within a desired range.

Let T_{10}(x) be the Taylor polynomial of degree 10 of the function f(x) = ln(1 + x) at a = 0. Suppose you approximate f(x) by T_{10}(x), find all positive values of x for which this approximation is within 0.001 of the right answer. (Hint: use the alternating series approximation.)

Explore how to approximate the function f(x) = ln(1 + x) using the 10th-degree Taylor polynomial T_{10}(x), and determine the range of x for which the error is within 0.001. Learn key concepts of alternating series and error bounds.

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