Chapter 9: Problem 2
In each part, obtain the Maclaurin series for the function by making an appropriate substitution in the Maclaurin series for ln(1 + x). Include the general term in your answer, and state the radius of convergence of the series. \(\begin{array}{ll}{\text { (a) } \ln (1-x)} & {\text { (b) } \ln \left(1+x^{2}\right)} \\ {\text { (c) } \ln (1+2 x)} & {\text { (d) } \ln (2+x)}\end{array}\)
Short Answer
Step by step solution
Introduction to Maclaurin Series for ln(1+x)
Obtain Maclaurin Series for ln(1-x)
Obtain Maclaurin Series for ln(1+x²)
Obtain Maclaurin Series for ln(1+2x)
Obtain Maclaurin Series for ln(2+x)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radius of Convergence
For example, with the standard Maclaurin series for \( \ln(1+x) \), the radius of convergence is 1 because our series converges for \( |x| < 1 \).
In the given exercise, we are asked to find the radius of convergence after substituting different expressions into \( \ln(1+x) \). Substitutions can impact the radius of convergence:
- When substituting \( x \rightarrow -x \), the radius remains 1 as the substitution doesn’t affect magnitude.
- Substituting \( x \rightarrow x^2 \) also keeps the radius 1 because \( |x^2|<1 \) implies \( |x|<1 \).
- However, substituting \( x \rightarrow 2x \) changes the radius to \( \frac{1}{2} \) because the series can only handle values where \( |2x| < 1 \), meaning \( |x| < \frac{1}{2} \).
- For \( \ln(2+x) \), utilizing \( \ln\left(1+\frac{x}{2}\right) \) gives a radius of convergence of 2, since \( |x/2| < 1 \) implies \( |x| < 2 \).
Logarithmic Functions
The function \( \ln(x) \) is defined for \( x > 0 \) and exhibits several key properties:
- \( \ln(1) = 0 \)
- \( \ln(xy) = \ln(x) + \ln(y) \)
- \( \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \)
- \( \ln(x^n) = n\ln(x) \)
Such manipulations allow us to approximate logarithmic functions that can be tough to calculate directly, using a series expansion within a determined radius of convergence.
Power Series
For the function \( \ln(1+x) \), the Maclaurin series can be written as \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n} \] Here, each term is increasingly smaller, resulting in convergence when \( x \) is within the radius of convergence.
By using substitutions into this series, such as \( x^2 \) or \( 2x \), we can generate new series for related logarithmic functions. Each substitution creates a new series with possibly different coefficients and an adjusted radius of convergence, while still maintaining the structure of a power series.
- The flexibility of the power series allows it to adapt and approximate various functions, providing insights into their behavior.
- With transformations like these, we can calculate or approximate values for more complex logarithmic functions without direct computation.