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Chapter 10: Problem 61

Obtain the Taylor series for 1\(/(1+x)^{2}\) from the series for \(-1 /(1+x) .\)

Short Answer

Expert verified
The Taylor series for \(\frac{1}{(1+x)^2}\) is \(-1 + 2x - 3x^2 + 4x^3 - \cdots\).

Step by step solution

01

Recognize the Given Function

The initial given function is \(-\frac{1}{1+x}\). This is the derivative of the geometric series \(\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots\).
02

Find the Derivative

The derivative of \(-\frac{1}{1+x}\) with respect to \(x\) is expressed as \(f'(x) = \frac{1}{(1+x)^2}\). This function's Taylor series representation is what we're trying to find.
03

Differentiate the Geometric Series

Take the derivative of the geometric series term by term: \((-1)' = 0\), \((-x)' = -1\), \((x^2)' = 2x\), \((-x^3)' = -3x^2\), and so on. Differentiating the series \(-1 + x - x^2 + x^3 - \cdots\), we get the series: \(0 - 1 + 2x - 3x^2 + 4x^3 - \cdots\).
04

Reinterpret the Differentiated Series

The resulting series \(-1 + 2x - 3x^2 + 4x^3 - \cdots\) is the Taylor series for the derivative \(\frac{1}{(1+x)^2}\), which is our desired result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

geometric series
A geometric series is a type of sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In our exercise, the series starts with
  • \(\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots\).
This series has a common ratio of
  • \(-x\).
Each term is obtained by multiplying the previous term by
  • \(-x\).
This power of the geometric series helps express functions as a sum of infinite terms which can be differentiated or integrated term by term.
This makes geometric series often serve as a foundation for more complex functions like the Taylor series, which we are examining here.
derivatives
Derivatives measure how a function changes as its input changes. For example, the derivative of a function \(f(x)\) gives us a function \(f'(x)\) that describes the rate of change or the slope of \(f(x)\) at any given point. In this exercise, we differentiated
  • \(-\frac{1}{1+x}\)
by finding its derivative, resulting in
  • \(\frac{1}{(1+x)^2}\).

When dealing with series, taking derivatives of each term in a series like
  • \(-1 + x - x^2 + x^3 - \cdots\)
can help us find a Taylor series for more complex functions. Term by term differentiation is a powerful technique that simplifies the manipulation of series using calculus rules, providing a bridge from simple geometric series to more generalized forms like power series.
power series expansion
A power series expansion is a way to express functions as an infinite sum of terms involving powers of the variable. This is crucial when constructing Taylor series, which is a specific type of power series centered at a particular point. The exercise provided
  • a power series expansion for the function \(\frac{1}{(1+x)^2}\)
by differentiating a geometric series. Taylor series expansions allow us to approximate functions using polynomials, making them useful for calculations and solving differential equations.
The mechanism showcased here involved expanding
  • a simpler geometric series
, then differentiating term by term to find the expansion for the derived function, illustrating the flexibility and broad applications of power series in mathematical problem-solving.

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Most popular questions from this chapter

In Exercises \(25-28\) , find a polynomial that will approximate \(F(x)\) throughout the given interval with an error of magnitude less than \(10^{-3} .\) \begin{equation} F(x)=\int_{0}^{x} \sin t^{2} d t, \quad[0,1] \end{equation}

The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}$$ to establish the following test. \begin{equation} \begin{array}{l}{\text { Let } f \text { have continuous first and second derivatives and }} \\ {\text { suppose that } f^{\prime}(a)=0 . \text { Then }} \\ {\text { a. } f \text { has a local maximum at } a \text { if } f^{\prime \prime} \leq 0 \text { throughout an interval }} \\ {\text { whose interior contains } a ;} \\ {\text { b. } f \text { has a local minimum at } a \text { if } f^{\prime \prime} \geq 0 \text { throughout an interval }} \\\ {\text { whose interior contains a. }}\end{array} \end{equation}

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{\sin 3 x^{2}}{1-\cos 2 x} \end{equation}

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}}\end{equation}

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \sin \frac{1}{n}\end{equation}
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