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Chapter 9: Problem 52

Find the domain of the function $$ f(x)=\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{(2 k-2) !} x^{k} $$

Short Answer

Expert verified
The domain is \((-1, 1)\).

Step by step solution

01

Analyze the Structure of the Function

The given function is an infinite series, where each term is of the form \( \frac{(2k-1)!!}{(2k-2)!} x^k \). To understand the domain, we need to determine where the series converges. The domain of the function is the set of all \( x \) values for which this series converges.
02

Use the Ratio Test for Convergence

The Ratio Test is suitable for determining the convergence of series. Consider the \( k^{th} \) term as \( a_k = \frac{(2k-1)!!}{(2k-2)!} x^k \). To apply the Ratio Test, analyze the ratio \( \left| \frac{a_{k+1}}{a_k} \right| \) and simplify it.
03

Calculate \( \left| \frac{a_{k+1}}{a_k} \right| \)

Calculate \( \left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(2k+1)!! x^{k+1}}{(2k)!} \cdot \frac{(2k-2)!}{(2k-1)!! x^k} \right| = \left| \frac{(2k+1) x}{(2k)} \right| \). Simplifying, you get \( \left| x \right| \).
04

Determine Convergence from the Ratio Test

By the Ratio Test, a series \( \sum a_k \) converges if the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \). Hence, \( \lim_{k \to \infty} \left| x \right| < 1 \). This implies the series converges for all \( \left| x \right| < 1 \).
05

State the Domain of the Function

The function \( f(x) \) converges for \( \left| x \right| < 1 \). Therefore, the domain of the function is \( x \in (-1, 1) \).

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Key Concepts

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

Domain of a Function
The domain of a function refers to the set of input values (usually \(x\) values) for which the function is defined and provides meaningful outputs.
For many functions, this involves determining for which values the function does not lead to undefined behaviors, such as division by zero or taking the square root of a negative number.
  • In the case of an infinite series, the domain often relies on where the series converges.
  • Convergence ensures that as we add more terms, the series approaches a finite number rather than growing without bounds.
For the exercise given, the function is defined as an infinite series.
Our task is to find the \(x\) values that make the series sum converge to a finite value.
Through analysis, we determined that the domain of the function is \(x \in (-1, 1)\), meaning the series only converges for \(x\) values between \(-1\) and \(1\).
Understanding the domain helps us know where the function is valid, avoiding non-converging or undefined results.
Ratio Test
The Ratio Test is an essential tool in determining the convergence of an infinite series.
It examines the behavior of successive terms in a series to decide if the series converges or diverges.
  • To employ the Ratio Test, consider a series \(\sum a_k\) and evaluate the ratio \(\left| \frac{a_{k+1}}{a_k} \right| \).
  • Simplify this ratio and find its limit as \(k\) approaches infinity.
The result of this limit analysis determines the series' behavior:
  • If the limit \(L < 1\), the series converges absolutely.
  • If \(L > 1\) or the limit does not exist, the series diverges.
  • If \(L = 1\), the test is inconclusive, and a different method must be used.
In our case, applying the Ratio Test showed that \(\lim_{k \to \infty} \left| x \right| < 1\), indicating convergence for \(\left| x \right| < 1\).
This is a critical step in determining the domain of the function.
Infinite Series
An infinite series is a sum of infinitely many terms, typically expressed as \(\sum_{k=1}^{\infty} a_k\), where each term is a function of \(k\).
Infinite series can converge to a meaningful number or diverge, depending on the behavior of the terms.
  • A converging series adds up to a finite value as you include more terms.
  • A diverging series increases indefinitely or oscillates without settling on a number.
Understanding whether an infinite series converges or diverges is crucial for applications like calculus and analysis.
For our exercise, the series involves terms \(a_k = \frac{(2k-1)!!}{(2k-2)!} x^k\), where convergence depends on \(x\).
By applying the Ratio Test, we confirmed the series converges for \(|x| < 1\), a vital outcome for determining both the series' utility and the function's domain.
Knowledge of infinite series aids in evaluating complex functions and exploring advanced mathematical concepts.

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Most popular questions from this chapter

Find the Taylor polynomials of orders \(n=0,1,2,3\) and 4 about \(x=x_{0},\) and then find the \(n\) th Taylor polynomial for the function in sigma notation. $$\sin \pi x ; x_{0}=\frac{1}{2}$$

Determine whether the series converges. $$\sum_{k=1}^{\infty}\left(1+\frac{1}{k}\right)^{-k}$$

Determine whether the series converges. $$\sum_{k=1}^{\infty} \frac{1}{\sqrt{k^{2}+1}}$$

The limit of an indeterminate form as \(x \rightarrow x_{0}\) can sometimes be found by expanding the functions involved in Taylor series about \(x=x_{0}\) and taking the limit of the series term by term. Use this method to find the limits in these exercises. (a) \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) (b) \(\lim _{x \rightarrow 0} \frac{\tan ^{-1} x-x}{x^{3}}\)

Exercise will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisfied. This result will be needed in Exercises. It was stated in Exercise 35 that $$\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}$$ (a) Let \(s_{n}\) be the \(n\) th partial sum of the series above. Show that $$ s_{n}+\frac{1}{3(n+1)^{3}} < \frac{\pi^{4}}{90} < s_{n}+\frac{1}{3 n^{3}} $$ (b) We can use a partial sum of the series to approximate \(\pi^{4} / 90\) to three decimal-place accuracy by capturing the sum of the series in an interval of length 0.001 (or less). Find the smallest value of \(n\) such that the interval containing \(\pi^{4} / 90\) in part (a) has a length of 0.001 or less. (c) Approximate \(\pi^{4} / 90\) to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of \(\pi^{4} / 90\).
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