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Chapter 7: Problem 68

(i) Expand each of the following functions as a Taylor series about the given point, and (ii) find the values of \(x\) for which the series converges: \(\frac{1}{x}, 1\)

Short Answer

Expert verified
The Taylor series is \(\sum_{n=0}^{\infty} (-1)^n (x-1)^n\) and converges for \(0 < x < 2\).

Step by step solution

01

Identify the Function

We are given the function \(\frac{1}{x}\) and asked to expand it as a Taylor series about the point \(x = 1\).
02

Write the Taylor Series Formula

The Taylor series of a function \(f(x)\) about a point \(a\) is given by:\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f(x)\) evaluated at \(x=a\).
03

Calculate Derivatives of the Function

Let's find the derivatives of \(\frac{1}{x}\) at \(x = 1\):- First derivative: \(f'(x) = -\frac{1}{x^2}\)- Second derivative: \(f''(x) = \frac{2}{x^3}\)- Third derivative: \(f'''(x) = -\frac{6}{x^4}\)Note the pattern: \(f^{(n)}(x) = (-1)^n \cdot \frac{n!}{x^{n+1}}\).
04

Evaluate Derivatives at the Point

Evaluate these derivatives at \(x=1\):- \(f(1) = 1\)- \(f'(1) = -1\)- \(f''(1) = 2\)- \(f'''(1) = -6\)Generally: \(f^{(n)}(1) = (-1)^n \cdot n!\).
05

Construct the Taylor Series

Substitute the values of \(f^{(n)}(1)\) into the Taylor series formula:\[\frac{1}{x} = 1 - (x-1) + \frac{(x-1)^2}{1} - \frac{(x-1)^3}{1} + \ldots = \sum_{n=0}^{\infty} (-1)^n (x-1)^n\]
06

Determine the Convergence Interval

Using the Ratio Test, we find the interval of convergence for the series \(\sum_{n=0}^{\infty} (-1)^n (x-1)^n\). Calculate the limit:\[L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{(x-1)^{n+1}}{(x-1)^n}\right| = |x-1|\]The series converges when \(|x-1| < 1\), giving us the interval of convergence: \(0 < x < 2\).

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Key Concepts

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

Convergence Interval
The convergence interval is where a Taylor series is valid and accurately represents a function. It is crucial for determining whether the series sum matches the function within certain limits for the variable. In our example, the function is \(\frac{1}{x}\), expanded around the point \(x = 1\). The Ratio Test helps us find this interval. We calculated:
  • \[ L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = |x-1| \]
The series converges when \(|x-1| < 1\), translating to the interval \(0 < x < 2\). This means the series can accurately represent the function between these x-values.
Derivatives
Derivatives are a fundamental part of constructing a Taylor series. They represent the rate of change of the function at points close to the specified expansion point. We evaluated the function \(\frac{1}{x}\) at \(x = 1\), finding patterns in its derivatives:
  • First derivative: \(f'(x) = -\frac{1}{x^2}\)
  • Second derivative: \(f''(x) = \frac{2}{x^3}\)
  • Third derivative: \(f'''(x) = -\frac{6}{x^4}\)
Notice a pattern: \(f^{(n)}(x) = (-1)^n \cdot \frac{n!}{x^{n+1}}\). Evaluating them at \(x = 1\) gives values of \((-1)^n \cdot n!\). This systematic approach helps fill the Taylor series formula efficiently.
Ratio Test
The Ratio Test is a popular method to ensure the convergence of a series within a specific interval. By comparing terms in a series, it helps determine if the series converges or diverges:
  • Evaluate: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
  • If \(L < 1\), the series converges.
  • If \(L > 1\), it diverges.
In our case, the test showed \(|x-1| < 1\) for convergence. This solidifies our confidence in the Taylor series representing the function accurately between \(0\) and \(2\).
Function Expansion
Function expansion through a Taylor series enables us to express a function as an infinite sum of terms calculated from derivatives at a specified point. For \(\frac{1}{x}\) about \(x = 1\), the expansion is:
  • \[ \frac{1}{x} = 1 - (x-1) + (x-1)^2 - (x-1)^3 + \ldots \]
This series \(\sum_{n=0}^{\infty} (-1)^n (x-1)^n\) reveals how the function can be broken down into simpler components, providing an easier way to approximate or understand its behavior near the point of expansion. This technique is valuable in calculus for solving differential equations and analyzing functions.

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Most popular questions from this chapter

\(\sum_{n=2}^{\infty} \frac{1}{n \ln n}\)

Find the radius of convergence of each of the following series: \(\sum_{m=0}^{\infty} \frac{x^{m}}{4^{m}}\)

The energy density of black-body radiation at temperature \(T\) is given by the Planck formula $$ \rho(\lambda)=\frac{8 \pi h c}{\lambda^{5}}\left[e^{h c / \lambda k T}-1\right]^{-1} $$ where \(\lambda\) is the wavelength. Show that the formula reduces to the classical Rayleigh-Jeans law \(\rho=8 \pi k T / \lambda^{4}\) (i) for long wavelengths \((\lambda \rightarrow \infty)\), (ii) if Planck's constant is set to zero \((h \rightarrow 0)\).

Find \(\sum_{n=1}^{10} \frac{1}{n(n+1)}\).

\(u_{x}=\frac{1}{x(x+2)} ; \quad x=1,2,3, \ldots\)
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