Math Problem Statement

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.)

Solution

We are given the Taylor polynomial T10(x)T_{10}(x) of degree 10 for the function f(x)=ln(1+x)f(x) = \ln(1 + x) around a=0a = 0, and we need to find the values of xx 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)f(x) = \ln(1 + x) around x=0x = 0 is given by:

f(x)=ln(1+x)=xx22+x33x44+f(x) = \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots

The 10th-degree Taylor polynomial T10(x)T_{10}(x) would be:

T10(x)=xx22+x33x44++x1010T_{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 R10(x)R_{10}(x), which represents the error when using T10(x)T_{10}(x) to approximate f(x)f(x), is approximately:

R10(x)x1111R_{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:

x11110.001\left| \frac{x^{11}}{11} \right| \leq 0.001

This simplifies to:

x11110.001\frac{x^{11}}{11} \leq 0.001

Multiplying both sides by 11:

x110.011x^{11} \leq 0.011

Step 4: Solving for xx

Now we solve for xx:

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

Using a calculator:

x(0.011)1/110.728x \leq (0.011)^{1/11} \approx 0.728

Final Answer:

The value of xx must be in the interval:

0<x0.7280 < x \leq 0.728

This means that for all positive values of xx up to approximately 0.728, the approximation using T10(x)T_{10}(x) will be within 0.001 of the actual value of ln(1+x)\ln(1 + x).


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(1x)\ln(1 - x) and how does it differ from ln(1+x)\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)\ln(1 + x)?
  4. How does the radius of convergence affect Taylor series approximations for ln(1+x)\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.

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Math Problem Analysis

Mathematical Concepts

Taylor Series
Approximation
Logarithmic Functions
Alternating Series

Formulas

f(x) = ln(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ...
T_{10}(x) = x - x^2/2 + x^3/3 - x^4/4 + ... + x^{10}/10
Error approximation: |x^{11}/11| ≤ 0.001

Theorems

Taylor series expansion
Alternating series error bound theorem

Suitable Grade Level

University-level Calculus or Advanced High School (Grades 11-12)