Problem 41 - Calculus Early Transcendentals

Question

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for \(f\) (perhaps more than once). Give the interval of convergence for the resulting series. $$g(x)=\frac{1}{(1-x)^{2}} \text { using } f(x)=\frac{1}{1-x}$$

Step-by-Step Solution

Answer

Question: Find the power series representation for the function g(x) where g(x) is the second derivative of f(x) =1/(1-x), and find the interval of convergence for the resulting series. Answer: The power series representation for g(x) is given by g(x) = Σ(n=1 to ∞) n(n-1)x^(n-2) and the interval of convergence is -1

Step 1: Find the power series representation of f(x)

First, we write the power series representation for \(f(x)=\frac{1}{1-x}\) using the geometric series formula for \(|x|<1\): $$f(x) = \sum_{n=0}^{\infty} x^n$$

Step 2: Differentiate the power series to get f'(x)

Now, we will differentiate the power series representation of f(x) with respect to x to obtain the power series of its first derivative f'(x): $$f'(x) = \frac{d}{dx}\left(\sum_{n=0}^{\infty} x^n\right) = \sum_{n=0}^{\infty} \frac{d}{dx}(x^n) = \sum_{n=1}^{\infty} nx^{n-1}$$ We drop the n=0 term since its derivative is 0.

Step 3: Differentiate f'(x) to get g(x)

Since g(x) is the second derivative of f(x), we must differentiate f'(x) with respect to x: $$g(x) = \frac{d}{dx}\left(\sum_{n=1}^{\infty} nx^{n-1}\right) = \sum_{n=1}^{\infty} \frac{d}{dx}(nx^{n-1}) = \sum_{n=1}^{\infty} n(n-1)x^{n-2}$$

Step 4: Find the interval of convergence

The original series converges if \(|x|<1\), because it is a geometric series. The interval of convergence of the derived series remains the same because differentiating or integrating a convergent power series retains the region of convergence. Thus, the interval of convergence for g(x) is \(-1<\)x\(<1\).

Step 5: Write the final answer

The power series representation for g(x) is given by: $$g(x)=\sum_{n=1}^{\infty} n(n-1)x^{n-2}$$ And the interval of convergence is \(-1<\)x\(<1\).

Key Concepts

Interval of Convergence

Understanding the interval of convergence is crucial when dealing with power series. It tells us the range of values for which the series converges. In simple terms, it's the set of all real numbers, within which the terms of the series add up to form a function. For the geometric series \( f(x) = \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \), the interval of convergence is \(|x| < 1\). This means the series will converge if \( x \) is between \(-1\) and \(1\).

When differentiating or integrating power series, the interval of convergence usually stays the same, provided no endpoints are involved. In our example, since we start with a geometric series with a known convergence interval \(-1

Geometric Series

The geometric series is foundational to understanding many power series solutions. A geometric series is a series of terms that have a constant ratio between successive terms. The simplest form of a geometric series is \( \sum_{n=0}^{\infty} x^n \), which represents \( \frac{1}{1-x} \) for \(|x| < 1\).

The concept is vital because many complex functions can be represented as geometric series, which are much easier to work with. Recognizing the geometric series form allows us to apply familiar operations like differentiation directly to each term rather than struggling with the function as a whole.

• The ratio between terms in geometric series is constant.
• Convergence requires the ratio \(x\) to be less than 1 in absolute value.
• It's widely applicable in solving problems because it simplifies complex expressions.

Understanding these aspects of geometric series helps in appreciating why they are an optimal choice as the starting point for manipulating functions in power series form.

Differentiation of Power Series

Differentiating power series involves finding the derivative of each term in the series. It's a powerful technique because power series, like polynomial expansion, allows for term-by-term differentiation.

For instance, when we have \( f(x) = \sum_{n=0}^{\infty} x^n \), its derivative \( f'(x) \) becomes \( \sum_{n=1}^{\infty} nx^{n-1} \). Notice we start our summation from \(n = 1\) because the derivative of the constant term (when \(n = 0\)) is zero.

• The differentiation of each term results in multiplying the existing power of \( x \) by its exponent.
• The new exponent decreases by one.

We utilize differentiation of power series to find other related forms, like second derivatives directly, which was necessary to find \( g(x) \) in our exercise. When applied properly, it gives a direct path to transforming initial functions into desired forms.

Integration of Power Series

Integration of power series is the reverse process of differentiation. By integrating term-by-term, we essentially add back the variable power incrementally and divide each term by the new power.

Consider the integrated form of \( f(x) = \sum_{n=0}^{\infty} x^n \). Through integration, each term \( x^n \) becomes \( \frac{x^{n+1}}{n+1} \). This can be especially useful when seeking antiderivatives of such expressions.

• Integration adds one to the power of \(x\).
• It divides by this new power, effectively reversing derivation mechanics.

Integration of power series maintains the initial interval of convergence, allowing us to return to the original function's form or explore related forms. It is instrumental in solving differential equations or exploring new function representations.

Previous

Problem 41

Next

Problem 41

Other exercises in this chapter

Problem 40

a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation assuming the exact value is given b

View solution

Problem 41

a. Find the first four nonzero terms of the Taylor series centered at 0 for the given finction. b. Use the first four terms of the series to approximate the giv

View solution

Problem 41

a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation assuming the exact value is given b

View solution

Problem 42

Use a Taylor series to approximate the following definite integrals. Retain as many terms as needed to ensure the error is less than \(10^{-4}\). $$\int_{0}^{0.

View solution

Key Concepts

Interval of Convergence

Understanding the interval of convergence is crucial when dealing with power series. It tells us the range of values for which the series converges. In simple terms, it's the set of all real numbers, within which the terms of the series add up to form a function. For the geometric series \( f(x) = \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \), the interval of convergence is \(|x| < 1\). This means the series will converge if \( x \) is between \(-1\) and \(1\).

When differentiating or integrating power series, the interval of convergence usually stays the same, provided no endpoints are involved. In our example, since we start with a geometric series with a known convergence interval \(-1

Geometric Series

The geometric series is foundational to understanding many power series solutions. A geometric series is a series of terms that have a constant ratio between successive terms. The simplest form of a geometric series is \( \sum_{n=0}^{\infty} x^n \), which represents \( \frac{1}{1-x} \) for \(|x| < 1\).

The concept is vital because many complex functions can be represented as geometric series, which are much easier to work with. Recognizing the geometric series form allows us to apply familiar operations like differentiation directly to each term rather than struggling with the function as a whole.

• The ratio between terms in geometric series is constant.
• Convergence requires the ratio \(x\) to be less than 1 in absolute value.
• It's widely applicable in solving problems because it simplifies complex expressions.

Understanding these aspects of geometric series helps in appreciating why they are an optimal choice as the starting point for manipulating functions in power series form.

Differentiation of Power Series

Differentiating power series involves finding the derivative of each term in the series. It's a powerful technique because power series, like polynomial expansion, allows for term-by-term differentiation.

For instance, when we have \( f(x) = \sum_{n=0}^{\infty} x^n \), its derivative \( f'(x) \) becomes \( \sum_{n=1}^{\infty} nx^{n-1} \). Notice we start our summation from \(n = 1\) because the derivative of the constant term (when \(n = 0\)) is zero.

• The differentiation of each term results in multiplying the existing power of \( x \) by its exponent.
• The new exponent decreases by one.

We utilize differentiation of power series to find other related forms, like second derivatives directly, which was necessary to find \( g(x) \) in our exercise. When applied properly, it gives a direct path to transforming initial functions into desired forms.

Integration of Power Series

Integration of power series is the reverse process of differentiation. By integrating term-by-term, we essentially add back the variable power incrementally and divide each term by the new power.

Consider the integrated form of \( f(x) = \sum_{n=0}^{\infty} x^n \). Through integration, each term \( x^n \) becomes \( \frac{x^{n+1}}{n+1} \). This can be especially useful when seeking antiderivatives of such expressions.

• Integration adds one to the power of \(x\).
• It divides by this new power, effectively reversing derivation mechanics.

Integration of power series maintains the initial interval of convergence, allowing us to return to the original function's form or explore related forms. It is instrumental in solving differential equations or exploring new function representations.

Previous

Problem 41

Next

Problem 41

Other exercises in this chapter

Problem 40

a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation assuming the exact value is given b

View solution

Problem 41

a. Find the first four nonzero terms of the Taylor series centered at 0 for the given finction. b. Use the first four terms of the series to approximate the giv

View solution

Problem 41

a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation assuming the exact value is given b

View solution

Problem 42

Use a Taylor series to approximate the following definite integrals. Retain as many terms as needed to ensure the error is less than \(10^{-4}\). $$\int_{0}^{0.

View solution