Textbook Question
Approximating ln 2 Consider the following three ways to approximate
ln 2.
e. Using four terms of the series, which of the three series derived in parts (a)–(d) gives the best approximation to ln 2? Can you explain why?
Ch. 11 - Power Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.R.65b
Approximating ln 2 Consider the following three ways to approximate
ln 2.
b. Use the Taylor series for ln (1 - x) centered at 0 and the identity ln 2 = -ln 1/2. Write the resulting infinite series.
Verified step by step guidance1
Recall the Taylor series expansion for \( \ln(1 - x) \) centered at 0, which is given by the infinite series:
\[ \ln(1 - x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} \quad \text{for} \quad |x| < 1. \]
Use the identity \( \ln 2 = -\ln \frac{1}{2} \). This means we can write \( \ln 2 \) as the negative of \( \ln(1 - x) \) by choosing \( x \) such that \( 1 - x = \frac{1}{2} \).
Solve for \( x \) in the equation \( 1 - x = \frac{1}{2} \), which gives \( x = \frac{1}{2} \).
Substitute \( x = \frac{1}{2} \) into the Taylor series for \( \ln(1 - x) \) to express \( \ln \frac{1}{2} \) as:
\[ \ln \frac{1}{2} = \ln(1 - \frac{1}{2}) = -\sum_{n=1}^{\infty} \frac{(\frac{1}{2})^n}{n}. \]
Finally, apply the negative sign from the identity \( \ln 2 = -\ln \frac{1}{2} \) to write \( \ln 2 \) as the infinite series:
\[ \ln 2 = -\ln \frac{1}{2} = \sum_{n=1}^{\infty} \frac{(\frac{1}{2})^n}{n}. \]
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Taylor Series Expansion
A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. For ln(1 - x) centered at 0, the series expresses ln(1 - x) as a power series in x, which converges for |x| < 1. This allows approximation of ln(1 - x) by summing a finite number of terms.
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Logarithm Properties and Identities
Logarithm identities, such as ln(a/b) = ln a - ln b and ln(1/x) = -ln x, help transform expressions into more convenient forms. Using ln 2 = -ln(1/2) leverages these properties to rewrite ln 2 in terms of ln(1 - x), enabling the use of the Taylor series for ln(1 - x) to approximate ln 2.
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Convergence of Infinite Series
Understanding the convergence criteria of the Taylor series is essential to ensure the approximation is valid. The series for ln(1 - x) converges when |x| < 1, so choosing x = 1/2 fits within this radius. Recognizing convergence guarantees that the infinite series accurately represents ln(1 - x) and thus ln 2.
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Related Practice
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