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Chapter 11: Problem 26

In Exercises \(21-28\) , find the Taylor series generated by \(f\) at \(x=a\) $$ f(x)=x /(1-x), \quad a=0 $$

Short Answer

Expert verified
The Taylor series for \(f(x) = \frac{x}{1-x}\) at \(x=0\) is \(x + x^2 + x^3 + \cdots = \sum_{n=1}^{\infty} x^n\).

Step by step solution

01

Understand the Function

The function given is \(f(x) = \frac{x}{1-x}\). We need to find its Taylor series expansion around \(x = 0\). A Taylor series is an infinite sum of terms calculated from the values of a function's derivatives at a single point.
02

Use Known Series Expansion

The function can be expressed as \(f(x) = x \cdot \frac{1}{1-x}\). We recognize \(\frac{1}{1-x}\) as the geometric series \(1 + x + x^2 + x^3 + \cdots\) for \(|x| < 1\). Thus, \(f(x)\) becomes \(x(1 + x + x^2 + x^3 + \cdots)\).
03

Multiply the Series by x

To obtain the Taylor series for \(f(x)\), multiply the geometric series by \(x\): \[ f(x) = x(1 + x + x^2 + x^3 + \cdots) = x + x^2 + x^3 + x^4 + \cdots \] This is the Taylor series of \(f(x)\) around \(x = 0\).
04

Write the General Term

Each term in the series is \(x^n\) for \(n \geq 1\). The general term of the Taylor series is \(x^n\) starting from \(n=1\). Thus, the series can be written as: \[ \sum_{n=1}^{\infty} x^n \]

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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 sum of terms in which each term is a constant multiple of the previous term. This concept is crucial to understanding many other types of series, including Taylor series.
In the case of the function \(f(x) = \frac{x}{1-x}\), the expression \(\frac{1}{1-x}\) is recognized as a geometric series.
The geometric series is expressed as \(1 + x + x^2 + x^3 + \cdots\) for \(|x| < 1\).
  • It starts from 1 and progresses by multiplying each term by a constant factor \(x\).
  • It converges when the absolute value of \(x\) is less than 1.
By substituting this series back into the function, the original function can be expanded into a series, helping us understand and solve problems related to the function's behavior near a specific point.
Series Expansion
Series expansion refers to expressing a function as the sum of a series of terms. This is widely used in calculus, particularly for functions that can be difficult to express otherwise.
In our example, we want to expand \(f(x) = \frac{x}{1-x}\) around \(x = 0\). By using the geometric series \(1 + x + x^2 + \cdots\), we convert \(f(x)\) into a more manageable form.
  • We first identify a known series form that can represent part of our function.
  • Next, we substitute and adjust it to fit the function's specific needs, as done by multiplying the series by \(x\) in this example.
  • This results in \(f(x) = x + x^2 + x^3 + \cdots\), an infinite series that represents the function near \(x = 0\).
Series expansion allows us to explore function behaviors and is the building block for approximations in mathematical analysis.
Derivatives
Derivatives are a fundamental tool in calculus, representing the rate at which a function changes. Understanding derivatives is key to constructing Taylor series.
In a Taylor series, each term is derived from the derivatives of the function at a single point. For the function \(f(x) = \frac{x}{1-x}\), we begin by observing its structure and recognizing patterns of derivation.
  • The process involves calculating successive derivatives, which become the coefficients of the Taylor series terms.
  • At \(x = 0\), each derivative informs the behavior of the function's expansion.
While the series created in our example is derived from a known algebraic manipulation, derivatives play a key role in more complex Taylor expansions, especially when such shorthand is unavailable.
Infinite Series
An infinite series is a series that continues indefinitely, allowing the summation of an endless number of terms.
This concept enables functions to be expressed in a form that can be analyzed and utilized efficiently for various mathematical purposes.
In the Taylor series for \(f(x) = \frac{x}{1-x}\), the infinite series \(x + x^2 + x^3 + \cdots\) conveys the function's behavior across a range of \(x\) values close to zero.
  • It highlights how the function appears as a sum of infinitely many terms, which, creatively, can be truncated for approximation purposes.
  • Infinite series are central in calculus for representing complex functions in an approachable manner.
    When \(n\) approaches infinity, the power of the representation becomes evident, allowing enhanced understanding and application.
This aspect of calculus enables scientists and engineers to make accurate predictions and calculations based on seemingly complicated phenomena.

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Most popular questions from this chapter

In Exercises \(97-100,\) determine if the sequence is nondecreasing and if it is bounded from above. $$ a_{n}=2-\frac{2}{n}-\frac{1}{2^{n}} $$

Sequences generated by Newton's method Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\) b. Starting with \(x_{0}=1\) and \(a=3\) , calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Show that the Taylor series for \(f(x)=\tan ^{-1} x\) diverges for \(|x|>1\)

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan \(^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3} ?\) Give reasons for your answer.

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ (1-x) y^{\prime}-y=0, \quad y(0)=2 $$
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