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Chapter 10: Problem 49

Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series. $$f(x)=\frac{3}{3+x}$$

Short Answer

Expert verified
$$ Answer: The power series representation of the function is $$f(x) = \sum_{n=0}^{\infty} (-1)^n\frac{x^n}{3^n}$$ and its interval of convergence is $$(-3, 3).$$

Step by step solution

01

Rewrite the function using known power series

First, rewrite the given function $$f(x) = \frac{3}{3+x}$$ as $$f(x) = 3 \cdot \frac{1}{(3+x)}.$$ We notice that this function can be written as a geometric series. To do so, we divide the numerator and denominator by 3, so $$f(x) = \frac{1}{1 - (-\frac{x}{3})}.$$ Now we can write this as a geometric series, remembering that $$\frac{1}{1 - r} = 1 + r + r^2 + r^3 + \cdots$$ for $$|r| < 1.$$ We have $$r = -\frac{x}{3}.$$
02

Find the power series representation

Using the geometric series formula, the power series representation for the function $$f(x)$$ is $$f(x) = 1 + \left(-\frac{x}{3}\right) + \left(-\frac{x}{3}\right)^2 + \left(-\frac{x}{3}\right)^3 + \cdots$$ We can simplify this expression: $$f(x) = 1 - \frac{x}{3} + \frac{x^2}{9} - \frac{x^3}{27} + \cdots$$ We can write this series in a more compact form as: $$f(x) = \sum_{n=0}^{\infty} (-1)^n\frac{x^n}{3^n}$$
03

Determine the interval of convergence

For a geometric series, the interval of convergence is given by the inequality $$|-(-\frac{x}{3})| < 1.$$ We can solve this inequality to find the interval of convergence for the function $$f(x)$$.\\ We have: $$\left|\frac{x}{3}\right| < 1$$ Divide both sides by 3: $$|x| < 3$$ This inequality gives the interval of convergence we are looking for, which can also be written as: $$-3 < x < 3$$ To sum it up, the power series representation of the given function is $$f(x) = \sum_{n=0}^{\infty} (-1)^n\frac{x^n}{3^n},$$ and its interval of convergence is $$(-3,3).$$

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Key Concepts

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

Interval of Convergence
The interval of convergence for a power series tells us over which values of \( x \) the series will converge, or in simpler terms, where the series will behave nicely and provide useful results. To find this interval, we look at the condition required for the series to converge. For a geometric series of the form \( \sum_{n=0}^{\infty} ar^n \), this interval is determined by \(|r| < 1\).

In our example with the function \( f(x)=\frac{3}{3+x} \), we shifted it into a geometric series format:
  • Recognizing \( r = -\frac{x}{3} \).
  • The interval becomes \(|-\frac{x}{3}| < 1 \).
  • Solving this problem leads to the inequality \(-3 < x < 3\).
Knowing this interval means you can plug in any \( x \) within \(-3\) to \(3\) into your power series and safely know it will converge.
Geometric Series
A geometric series is a special kind of infinite series where each term is a constant multiple of the previous one. It's expressed in the form \( a + ar + ar^2 + ar^3 + \ldots \). For these series to converge, the absolute value of \( r \) (the ratio between terms) must be less than 1.

For the function \( f(x) = \frac{3}{3+x} \), once rearranged as a geometric series, we found that our \( r \) was \( -\frac{x}{3} \). Applying the known formula for geometric series convergence \( \sum_{n=0}^{\infty} ar^n \) where \(|r| < 1\), led to:
  • A series that starts with 1 and continues as \( 1 - \frac{x}{3} + \frac{x^2}{9} - \frac{x^3}{27} + \ldots \).
  • This series depiction is convenient for functions that can be manipulated into \( \frac{1}{1 - r} \) format.
  • Each term's sign alternates due to \( (-1)^n \) in the series formula \( \sum_{n=0}^{\infty} (-1)^n\frac{x^n}{3^n} \).
These characteristics make geometric series a powerful tool in calculus, helping simplify otherwise complex expressions.
Power Series Representation
Power series provide an immensely useful way to express functions, especially when the function needs to be evaluated over an interval. It takes the form \( \sum_{n=0}^{\infty} c_n(x-a)^n \), where \( a \) is the center (or starting point) of the series.

For \( f(x)=\frac{3}{3+x} \), after rewriting it as a geometric series, we established its power series representation centered at 0:
  • The constant \( a \) is 0, and each term progressively involves higher powers of \( x \).
  • All coefficients \( c_n \) can be deduced from simplifying the geometric series term \((-1)^n\frac{x^n}{3^n}\).
  • This expression makes it simpler to evaluate or approximate \( f(x) \) across its interval of convergence.
Power series, like our example, are often easier to differentiate or integrate compared to the original function, extending their usefulness in various mathematical and practical applications.

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Most popular questions from this chapter

Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty} \frac{(-1)^{k} x^{k+1}}{4^{k}}$$

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\tan ^{-1}\left(\frac{1}{2}\right)$$

The period of a pendulum is given by $$T=4 \sqrt{\frac{\ell}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=4 \sqrt{\frac{\ell}{g}} F(k)$$ where \(\ell\) is the length of the pendulum, \(g \approx 9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, \(k=\sin \left(\theta_{0} / 2\right),\) and \(\theta_{0}\) is the initial angular displacement of the pendulum (in radians). The integral in this formula \(F(k)\) is called an elliptic integral and it cannot be evaluated analytically. a. Approximate \(F(0.1)\) by expanding the integrand in a Taylor (binomial) series and integrating term by term. b. How many terms of the Taylor series do you suggest using to obtain an approximation to \(F(0.1)\) with an error less than \(10^{-3} ?\) c. Would you expect to use fewer or more terms (than in part (b)) to approximate \(F(0.2)\) to the same accuracy? Explain.

Use composition of series to find the first three terms of the Maclaurin series for the following functions. a. \(e^{\sin x}\) b. \(e^{\tan x}\) c. \(\sqrt{1+\sin ^{2} x}\)

Exponential function In Section 3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty < x < \infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{2 x}$$
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