Problem 68 - Calculus Early Transcendentals

Question

Replace \(x\) by \(x-1\) in the series \(\ln (1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^{k}}{k}\) to obtain a power series for \(\ln x\) centered at \(x=1 .\) What is the interval of convergence for the new power series?

Step-by-Step Solution

Answer

In summary, we have found the power series representation for the function \(\ln x\), centered at \(x=1\), to be: $$\ln x=\sum_{k=1}^{\infty}\frac{(-1)^{k+1} (x-1)^{k}}{k}$$ The interval of convergence is \((0,2)\).

Step 1: Replace \(x\) by \(x-1\) in the given series

We replace \(x\) by \(x-1\) in the given series, \(\ln (1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1} x^{k}}{k}\). This substitution leads to the following expression: $$\ln (1+(x-1))=\sum_{k=1}^{\infty}\frac{(-1)^{k+1} (x-1)^{k}}{k}$$

Step 2: Simplify the new power series

Here, we notice that the simplified expression inside the natural logarithm function is merely \(x\). With this simplification, we have: $$\ln x=\sum_{k=1}^{\infty}\frac{(-1)^{k+1} (x-1)^{k}}{k}$$ Thus, we have found the power series representation for the function \(\ln x\) centered at \(x=1\).

Step 3: Determine the interval of convergence

To determine the interval of convergence, we'll use the ratio test, which states that the series converges if: $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| < 1$$ In our case, we have \(a_k=\frac{(-1)^{k+1} (x-1)^{k}}{k}\). Therefore: $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty} \left|\frac{(-1)^{k+2}(x-1)^{k+1}}{k+1}\cdot \frac{k}{(-1)^{k+1}(x-1)^k}\right|$$ The negative terms will cancel out, and we're left with: $$\lim_{k\to\infty} \frac{k}{k+1} \cdot |x-1|$$ Now, the limit of \(\frac{k}{k+1}\) as \(k\) goes to infinity is \(1\). Consequently, we have: $$1\cdot|x-1|<1 \Rightarrow -1

Key Concepts

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm (ln), are fundamental in mathematics. The natural logarithm of a number \( x \), written as \( \ln x \), is the power to which the base \( e \) must be raised to produce \( x \). This is expressed as:
\[ e^{\ln x} = x. \]
Logarithmic functions have properties that make them particularly useful in various mathematical fields:

• Product Rule: \( \ln(ab) = \ln a + \ln b \)
• Quotient Rule: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• Power Rule: \( \ln(a^b) = b \cdot \ln a \)

In this exercise, we derive a power series for \( \ln x \) by manipulating a known series for \( \ln(1+x) \). By substituting \( x \) with \( x-1 \), we can represent the natural logarithm function in terms of a series centered at \( x=1 \). This transformation allows for a more flexible way to compute logarithmic functions around different centers.

Interval of Convergence

The interval of convergence of a power series is the range of values for which the series converges to a finite sum. Understanding this concept is crucial when working with series, as it determines where the series behaves properly.
In our given series, \( \ln x = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} (x-1)^{k}}{k} \), we applied certain tests to establish the interval on which it converges.
By using the Ratio Test, we found that the series converges when \( x \) is between 0 and 2, or \( 0 < x < 2 \). This interval is crucial since it tells us the values of \( x \) for which our derived series
accurately represents the natural logarithm function. Outside this interval, the series may diverge, leading to incorrect results.

Ratio Test

The Ratio Test is a powerful tool used to examine the convergence of infinite series. It helps determine whether a series will converge or diverge by examining the limit of successive terms.
Consider a series with terms \( a_k \). The test states:

• If \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \), the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

In our exercise, \( a_k = \frac{(-1)^{k+1} (x-1)^k}{k} \). Applying the Ratio Test leads to:
\[ \lim_{k\to\infty} \left|\frac{(-1)^{k+2}(x-1)^{k+1}}{k+1} \cdot \frac{k}{(-1)^{k+1}(x-1)^k}\right| = \lim_{k\to\infty} \frac{k}{k+1} \cdot |x-1|. \]
Simplifying gives us \( 1 \cdot |x-1| < 1 \), which translates to the condition \( 0 < x < 2 \). This elegant test quickly helps identify the interval we need to focus on when working with various series.

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Key Concepts

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm (ln), are fundamental in mathematics. The natural logarithm of a number \( x \), written as \( \ln x \), is the power to which the base \( e \) must be raised to produce \( x \). This is expressed as:
\[ e^{\ln x} = x. \]
Logarithmic functions have properties that make them particularly useful in various mathematical fields:

• Product Rule: \( \ln(ab) = \ln a + \ln b \)
• Quotient Rule: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• Power Rule: \( \ln(a^b) = b \cdot \ln a \)

In this exercise, we derive a power series for \( \ln x \) by manipulating a known series for \( \ln(1+x) \). By substituting \( x \) with \( x-1 \), we can represent the natural logarithm function in terms of a series centered at \( x=1 \). This transformation allows for a more flexible way to compute logarithmic functions around different centers.

Interval of Convergence

The interval of convergence of a power series is the range of values for which the series converges to a finite sum. Understanding this concept is crucial when working with series, as it determines where the series behaves properly.
In our given series, \( \ln x = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} (x-1)^{k}}{k} \), we applied certain tests to establish the interval on which it converges.
By using the Ratio Test, we found that the series converges when \( x \) is between 0 and 2, or \( 0 < x < 2 \). This interval is crucial since it tells us the values of \( x \) for which our derived series
accurately represents the natural logarithm function. Outside this interval, the series may diverge, leading to incorrect results.

Ratio Test

The Ratio Test is a powerful tool used to examine the convergence of infinite series. It helps determine whether a series will converge or diverge by examining the limit of successive terms.
Consider a series with terms \( a_k \). The test states:

• If \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \), the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

In our exercise, \( a_k = \frac{(-1)^{k+1} (x-1)^k}{k} \). Applying the Ratio Test leads to:
\[ \lim_{k\to\infty} \left|\frac{(-1)^{k+2}(x-1)^{k+1}}{k+1} \cdot \frac{k}{(-1)^{k+1}(x-1)^k}\right| = \lim_{k\to\infty} \frac{k}{k+1} \cdot |x-1|. \]
Simplifying gives us \( 1 \cdot |x-1| < 1 \), which translates to the condition \( 0 < x < 2 \). This elegant test quickly helps identify the interval we need to focus on when working with various series.