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In Problems 1-10, find the power series representation for \(f(x)\) and specify the radius of convergence. Each is somehow related to a geometric series (see Examples 1 and 2). $$ f(x)=\frac{x^{3}}{2-x^{3}} $$

Short Answer

Expert verified
The power series is \( \sum_{n=0}^{\infty} \frac{x^{3n}}{2^{n+1}} \) with radius \( \sqrt[3]{2} \).

Step by step solution

01

Identify the Geometric Series Formula

The formula for the geometric series is \( \frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n \), where \(|r| < 1\). In this case, we want to express \(f(x) = \frac{x^{3}}{2-x^{3}}\) in a form similar to \( \frac{1}{1 - r} \).
02

Rewrite the Function

Consider \( f(x) = \frac{x^3}{2-x^3} \). Rewrite it as \( \frac{1}{2-x^3} \) and factor out the 2 to fit the geometric series template, leading to: \( \frac{1}{2} \times \frac{1}{1 - \frac{x^3}{2}} \).
03

Apply the Geometric Series Formula

Using the geometric series formula, express \( \frac{1}{1 - \frac{x^3}{2}} \) as a power series: \( \sum_{n=0}^{\infty} \left(\frac{x^3}{2}\right)^n \). This becomes \( \sum_{n=0}^{\infty} \frac{x^{3n}}{2^n} \).
04

Multiply by Initial Factor

Multiply the entire series by \( \frac{1}{2} \) to adjust for the initial factor taken out, resulting in \( \frac{1}{2} \times \sum_{n=0}^{\infty} \frac{x^{3n}}{2^n} = \sum_{n=0}^{\infty} \frac{x^{3n}}{2^{n+1}} \).
05

Determine the Radius of Convergence

The geometric series converges if \(|r| < 1\). In this case, \( r = \frac{x^3}{2} \), so \( \left| \frac{x^3}{2} \right| < 1 \). Solving for \(x\), we have \( |x^3| < 2 \), or \( |x| < \sqrt[3]{2} \). Thus, the radius of convergence is \( \sqrt[3]{2} \).

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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 the terms in a sequence where each term is a multiple of the previous one by a fixed number, called the common ratio, denoted as \( r \). The infinite geometric series can be expressed by the formula \( \frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n \). This series converges when the absolute value of the common ratio \( |r| \) is less than 1.

Understanding this concept is crucial when finding power series representations. If you view the given function \( f(x) = \frac{x^3}{2-x^3} \), it can be manipulated to match the general geometric series formula. By rewriting it as \( \frac{1}{2} \times \frac{1}{1 - \frac{x^3}{2}} \), we see a clear resemblance to the geometric series expression. This allows us to apply the formula to express the function as a power series.
  • The common ratio is key when using a geometric series for power series representation.
  • Ensure that the expression aligns with \( \frac{1}{1 - r} \) for direct application.
Radius of Convergence
The concept of radius of convergence is essential when dealing with power series. It tells us the range of \( x \) values for which the series converges. For a geometric series of the form \( \sum_{n=0}^{\infty} r^n \), the requirement for convergence is \( |r| < 1 \).

In the context of the given function \( f(x) = \frac{x^3}{2-x^3} \), this translates to finding the range \( x \) values for which the series \( \sum_{n=0}^{\infty} \left(\frac{x^3}{2}\right)^n \) converges. Here, the ratio \( r = \frac{x^3}{2} \), and to satisfy the convergence condition \( \left|\frac{x^3}{2}\right| < 1 \), we solve \( |x^3| < 2 \).
  • The radius of convergence, therefore, is \( \sqrt[3]{2} \).
  • This means the series converges for \( |x| < \sqrt[3]{2} \).
Understanding the radius of convergence helps in determining how far we can extend our power series approximation around the center of expansion.
Convergence of Series
Convergence of a series is a critical concept in calculus and analysis, signifying whether a series approaches a finite limit as more terms are added. For a power series, convergence is generally concerned with the values of \( x \) that make the series sum to a finite number.

In particular, for the geometric series representation \( \sum_{n=0}^{\infty} \left(\frac{x^3}{2}\right)^n \), we determine convergence based on the radius calculated as \( |x| < \sqrt[3]{2} \). This specific condition means as long as \( x \) stays within \( -\sqrt[3]{2} \) and \( +\sqrt[3]{2} \), the series will converge to a finite value.
  • Convergence of series ensures a valid power series approximation only within a certain range of \( x \).
  • Understanding its convergence can help in practical applications such as estimating functions in engineering and physics.
By knowing the convergence of the given series, we ascertain where our power series approximation holds true and where it doesn't.

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Most popular questions from this chapter

Find the sum \(S(x)\) of \(\sum_{n=0}^{\infty}(x-3)^{n}\). What is the convergence set?

A famous sequence \(\left\\{f_{n}\right\\}\), called the Fibonacci Sequence after Leonardo Fibonacci, who introduced it around A.D. 1200 , is defined by the recursion formula $$ f_{1}=f_{2}=1, \quad f_{n+2}=f_{n+1}+f_{n} $$ (a) Find \(f_{3}\) through \(f_{10}\) - (b) Let \(\phi=\frac{1}{2}(1+\sqrt{5}) \approx 1.618034\). The Greeks called this number the golden ratio, claiming that a rectangle whose dimensions were in this ratio was "perfect." It can be shown that $$ \begin{aligned} f_{n} &=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] \\\ &=\frac{1}{\sqrt{5}}\left[\phi^{n}-(-1)^{n} \phi^{-n}\right] \end{aligned} $$ Check that this gives the right result for \(n=1\) and \(n=2\). The general result can be proved by induction (it is a nice challenge). More in line with this section, use this explicit formula to prove that \(\lim _{n \rightarrow \infty} f_{n+1} / f_{n}=\phi\). (c) Using the limit just proved, show that \(\phi\) satisfies the equation \(x^{2}-x-1=0\). Then, in another interesting twist, use the Quadratic Formula to show that the two roots of this equation are \(\phi\) and \(-1 / \phi\), two numbers that occur in the explicit formula for \(f_{n}\).

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9 !}-\cdots $$

Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems 43-48 to see that you get the same answers using the methods of Section 9.7. $$ \exp (\sin x) $$

In Problems 19-24, find the Taylor series in \(x-\) a through the term \((x-a)^{3}\). $$ e^{x}, a=1 $$

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