Problem 20 - Essential Calculus

Question

Use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at 0 , for the function. Identify the interval of convergence. $$ f(x)=\ln \left(1-x^{2}\right)=\int \frac{1}{1+x} d x-\int \frac{1}{1-x} d x $$

Step-by-Step Solution

Answer

The power series representation of the function \(f(x)=\ln \left(1-x^{2}\right)\) is \(\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1} x^{n+1}\) and the interval of convergence is \(-1

Step 1: Power Series Expansion of the Integral Terms

Use power series representation to expand the integral terms in the given equation: \(\int \frac{1}{1+x} d x = \int \sum_{n=0}^{\infty}(-1)^{n} x^{n} d x\) and \(\int \frac{1}{1-x} d x = \int \sum_{n=0}^{\infty} x^{n} d x\)

Step 2: Integrate the Power Series

Integrate the expanded power series from step 1 term-by-term: \(\int \frac{1}{1+x} d x = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1} x^{n+1}\) and \(\int \frac{1}{1-x} d x = \sum_{n=0}^{\infty}\frac{1}{n+1} x^{n+1}\)

Step 3: Subtract the Series

Subtract the second power series from the first one, based on the representation of the function \(f(x)\): \(f(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1} x^{n+1}-\sum_{n=0}^{\infty}\frac{1}{n+1} x^{n+1} = \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1} x^{n+1}\)

Step 4: Determine the Interval of Convergence

The interval of convergence is the same as that of the original series, since the original series converges for \(-1

Key Concepts

Interval of Convergence

In the world of power series, determining the interval of convergence is crucial. Convergence essentially refers to where the infinite series settles to a specific value. For the given function, we start by recognizing the power series representation \(\frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^{n} x^{n}\). It converges for \(-1 < x < 1\). The same applies to \(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n}\). When integrating these series, their convergence intervals typically remain unchanged. Hence, for \(f(x) = \ln(1-x^2) = \int \frac{1}{1+x} dx - \int \frac{1}{1-x} dx\), the interval of convergence stays at \(-1 < x < 1\). This domain keeps the series valid and ensures that it sums to the true function value for any \(x\) in this range.To confirm the convergence, we also consider endpoints. However, no need to further explore them here as the terms potentially still behave differently at the boundaries.

Integration of Series

Integrating a series involves adding up infinitely many smaller integrals. This is often done term by term, especially with power series. In this scenario, we start with the function \(f(x)\) given by

• \(\int \frac{1}{1+x} dx\)
• \(\int \frac{1}{1-x} dx\)

Both have series expansions which are \(\sum_{n=0}^{\infty}(-1)^{n} x^{n}\) and \(\sum_{n=0}^{\infty}x^{n}\), respectively.The key here is integrating each term of the series independently:

• \(\int \sum_{n=0}^{\infty} (-1)^{n} x^{n} dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} x^{n+1}\)
• \(\int \sum_{n=0}^{\infty} x^{n} dx = \sum_{n=0}^{\infty} \frac{1}{n+1} x^{n+1}\)

By handling each term this way, we transform our function \(f(x)\) into a new power series where each element of the series is integrated and adjusted to reflect this operation.

Logarithmic Function

Logarithmic functions often look complex but are commonly derived from simpler functions using integration techniques.Here, we focus on \(f(x) = \ln(1-x^2)\). This can be decomposed using subtraction of integrals. We exploit the relation\[ f(x) = \int \frac{1}{1+x} dx - \int \frac{1}{1-x} dx \]This expression uses the idea that the difference of their integrals can simulate a logarithm. After integration,

• The first series becomes \(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} x^{n+1}\).
• The second transforms to \(\sum_{n=0}^{\infty} \frac{1}{n+1} x^{n+1}\).

Subtracting these two results, you receive the series representation for \(\ln(1-x^2)\). Power series thus provide a bridge from a simple fraction formula to a natural logarithmic form, revealing patterns and behaviors within these crucial mathematical constructs.

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Problem 21

Other exercises in this chapter

Problem 20

Use the binomial series to find the Maclaurin series for the function. $$ f(x)=\sqrt{1+x^{3}} $$

View solution

Problem 20

Approximate the sum of the series by using the first six terms. (See Example 4.) $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n}{2^{n}} $$

View solution

Problem 21

Verify that the infinite series converges. $$ \sum_{n=0}^{\infty} 2\left(\frac{3}{4}\right)^{n} $$

View solution

Problem 21

In Exercises 21-24, simplify the ratio of factorials. \(\frac{(n+1) !}{n !}\)

View solution

Key Concepts

Interval of Convergence

In the world of power series, determining the interval of convergence is crucial. Convergence essentially refers to where the infinite series settles to a specific value. For the given function, we start by recognizing the power series representation \(\frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^{n} x^{n}\). It converges for \(-1 < x < 1\). The same applies to \(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n}\). When integrating these series, their convergence intervals typically remain unchanged. Hence, for \(f(x) = \ln(1-x^2) = \int \frac{1}{1+x} dx - \int \frac{1}{1-x} dx\), the interval of convergence stays at \(-1 < x < 1\). This domain keeps the series valid and ensures that it sums to the true function value for any \(x\) in this range.To confirm the convergence, we also consider endpoints. However, no need to further explore them here as the terms potentially still behave differently at the boundaries.

Integration of Series

Integrating a series involves adding up infinitely many smaller integrals. This is often done term by term, especially with power series. In this scenario, we start with the function \(f(x)\) given by

• \(\int \frac{1}{1+x} dx\)
• \(\int \frac{1}{1-x} dx\)

Both have series expansions which are \(\sum_{n=0}^{\infty}(-1)^{n} x^{n}\) and \(\sum_{n=0}^{\infty}x^{n}\), respectively.The key here is integrating each term of the series independently:

• \(\int \sum_{n=0}^{\infty} (-1)^{n} x^{n} dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} x^{n+1}\)
• \(\int \sum_{n=0}^{\infty} x^{n} dx = \sum_{n=0}^{\infty} \frac{1}{n+1} x^{n+1}\)

By handling each term this way, we transform our function \(f(x)\) into a new power series where each element of the series is integrated and adjusted to reflect this operation.

Logarithmic Function

Logarithmic functions often look complex but are commonly derived from simpler functions using integration techniques.Here, we focus on \(f(x) = \ln(1-x^2)\). This can be decomposed using subtraction of integrals. We exploit the relation\[ f(x) = \int \frac{1}{1+x} dx - \int \frac{1}{1-x} dx \]This expression uses the idea that the difference of their integrals can simulate a logarithm. After integration,

• The first series becomes \(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} x^{n+1}\).
• The second transforms to \(\sum_{n=0}^{\infty} \frac{1}{n+1} x^{n+1}\).

Subtracting these two results, you receive the series representation for \(\ln(1-x^2)\). Power series thus provide a bridge from a simple fraction formula to a natural logarithmic form, revealing patterns and behaviors within these crucial mathematical constructs.