Problem 61
Question
Obtain the Taylor series for 1\(/(1+x)^{2}\) from the series for \(-1 /(1+x) .\)
Step-by-Step Solution
Verified Answer
The series is \( \sum_{n=0}^{\infty} (-1)^n (n+1) x^n \).
1Step 1: Understand the given problem
We need to find the Taylor series for \( \frac{1}{(1+x)^2} \) using the Taylor series for \( \frac{-1}{(1+x)} \). Start by recalling the Maclaurin series for basic functions.
2Step 2: Recall the known series
The Maclaurin series for \( \frac{1}{1+x} \) is \( 1 - x + x^2 - x^3 + x^4 - \cdots \). Therefore, \( \frac{-1}{1+x} = -(1 - x + x^2 - x^3 + x^4 - \cdots) = -1 + x - x^2 + x^3 - x^4 + \cdots \).
3Step 3: Differentiate the known series
To transform \( \frac{-1}{1+x} \) into \( \frac{1}{(1+x)^2} \), we know that differentiating \( \frac{-1}{1+x} \) gives us \( \frac{1}{(1+x)^2} \). Differentiate each term of the series \( -1 + x - x^2 + x^3 - x^4 + \cdots \).
4Step 4: Apply term-wise differentiation
Differentiate the series term by term: the derivative of \(-1\) is 0, the derivative of \(x\) is 1, the derivative of \(-x^2\) is \(-2x\), the derivative of \(x^3\) is \(3x^2\), and so on. Hence, the series becomes \( 1 - 2x + 3x^2 - 4x^3 + \cdots \).
5Step 5: Write the Taylor series result
The Taylor series for \( \frac{1}{(1+x)^2} \) is obtained as \( 1 - 2x + 3x^2 - 4x^3 + \cdots \). This series can be written in the form \( \sum_{n=0}^{\infty} (-1)^n (n+1) x^n \).
Key Concepts
Maclaurin seriesTerm-wise differentiationPower series expansion
Maclaurin series
The Maclaurin series is a special type of Taylor series. It provides the approximation of functions as a sum of their derivatives at zero. This makes it extremely useful for simplifying complex functions into an infinite power series. The formula for a Maclaurin series of a function \( f(x) \) is:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
- Here, each term involves derivatives of \( f(x) \) evaluated at \( x = 0 \).- It's especially handy because it centers the approximation around \( x = 0 \), making calculations easier in many cases.
In our exercise, to find the series for \( \frac{1}{(1+x)^2} \), we start with the Maclaurin series of \( \frac{-1}{1+x} \). This helps us see the relationship between the basic series and derivatives that refine our approximation.
- Here, each term involves derivatives of \( f(x) \) evaluated at \( x = 0 \).- It's especially handy because it centers the approximation around \( x = 0 \), making calculations easier in many cases.
In our exercise, to find the series for \( \frac{1}{(1+x)^2} \), we start with the Maclaurin series of \( \frac{-1}{1+x} \). This helps us see the relationship between the basic series and derivatives that refine our approximation.
Term-wise differentiation
Term-wise differentiation involves taking derivatives of each term in a series individually. This technique is particularly useful in simplifying power series, as observed in transforming one function to another.
In our problem, we transform the function \( \frac{-1}{1+x} \) to \( \frac{1}{(1+x)^2} \) using differentiation of each term:
In our problem, we transform the function \( \frac{-1}{1+x} \) to \( \frac{1}{(1+x)^2} \) using differentiation of each term:
- Starting with \(-1 + x - x^2 + x^3 - \cdots\), differentiate term by term.
- Differentiate to get \(0, 1, -2x, 3x^2, \ldots\).
- This pattern gives us insights into the behavior of more complex derivatives.
Power series expansion
Power series expansion is a method of expressing a function as an infinite sum of terms, each term being a power of \( x \). This technique is not only used for approximating functions, but also for analyzing their properties in calculus.
In the expansion for \( \frac{1}{(1+x)^2} \), we recognize a pattern:
In the expansion for \( \frac{1}{(1+x)^2} \), we recognize a pattern:
- The series \( 1 - 2x + 3x^2 - 4x^3 + \cdots \) can be expressed as \( \sum_{n=0}^{\infty} (-1)^n (n+1) x^n \).
- This sum reveals how each term changes with \( n \), exhibiting both the negative sign and the increasing coefficient \( n+1 \).
- Such expansions allow complex functions to be manipulated similarly to simple polynomials within their radius of convergence.
Other exercises in this chapter
Problem 60
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