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Find the function represented by the following series, and find the interval of convergence of the series. (Not all these series are power series.) $$\sum_{k=0}^{\infty}\left(\frac{x^{2}-1}{3}\right)^{k}$$

Short Answer

Expert verified
Write a short answer question for this step by step solution. Question: Given the geometric series $\sum_{k=0}^{\infty} \left(\frac{x^{2}-1}{3}\right)^{k}$, find the function it represents, and determine its interval of convergence. Answer: The function represented by the given series is $S(x) = \frac{3}{4 - x^2}$. The series converges only for $- \sqrt{2} < x < 2$, meaning the interval of convergence is (-√2, 2).

Step by step solution

01

Identify the first term and common ratio

First, let's identify the first term (a) and the common ratio (r) of the given geometric series. Since the sum starts from k = 0, this means we can plug 0 into the expression: $$\left(\frac{x^{2}-1}{3}\right)^{0} = 1$$ So, the first term (a) is 1. Next, the common ratio (r) can be identified as the base of the series, which is: $$r = \frac{x^{2}-1}{3}$$
02

Write down the sum formula for the series

Now that we know the first term and the common ratio, we can write down the formula for the sum of the given series: $$S = \frac{a}{1 - r} = \frac{1}{1 - \frac{x^{2}-1}{3}}$$
03

Simplify the function

Simplify the expression: $$S = \frac{1}{1 - \frac{x^{2}-1}{3}} = \frac{1}{\frac{3 - (x^{2}-1)}{3}} = \frac{3}{3 - (x^2-1)} = \frac{3}{4 - x^2}$$ So the function represented by the series is: $$S(x) = \frac{3}{4 - x^2}$$
04

Find the interval of convergence

We know a geometric series converges if and only if the absolute value of the common ratio is less than 1: $$|r| = \left|\frac{x^{2}-1}{3}\right| < 1$$ We can now solve the inequality for x: $$\left|\frac{x^{2}-1}{3}\right| < 1 \Rightarrow -(3) < x^2-1 < 3 \Rightarrow -2 < x^2 < 4$$ This inequality means the series converges for x values in the interval (-2, 4) in terms of x^2. By taking the square root of the x^2 terms of the inequality, we can find the interval for x itself: $$\sqrt{-2} < x < \sqrt{4}$$ This means the series converges only for: $$- \sqrt{2} < x < 2$$ The interval of convergence is (-√2, 2).

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Key Concepts

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

Interval of Convergence
Understanding the interval of convergence is vital when dealing with geometric series. The interval of convergence tells us for which values of the variable, in this case, \( x \), our series will actually sum to a meaningful number rather than approach infinity or oscillate. In many series problems, this is found by examining when the series converges. For geometric series, convergence occurs when the absolute value of the common ratio, \( r \), is less than 1. For our series, the common ratio is \( r = \frac{x^2 - 1}{3} \). To find the interval of convergence, we solve the inequality:
  • \( \left| \frac{x^2 - 1}{3} \right| < 1 \)
By solving the inequality, we arrived at:
  • \(-2 < x^2 < 4 \)
  • Taking square roots, for \( x \) itself: \( -\sqrt{2} < x < 2 \)
This can be tricky since taking square roots can be nuanced when extracting ranges or intervals. Ultimately, the series converges for the interval \((-\sqrt{2}, 2)\).
Common Ratio
In any geometric series, the common ratio \( r \) is what separates it from merely being a sequence. It is the factor by which we multiply each term to get the next one. For the geometric series given in the exercise, the common ratio is:
  • \( r = \frac{x^2 - 1}{3} \)
The importance of the common ratio cannot be overstated as it directly informs us about the behavior of the series, specifically its convergence properties. If \(|r| < 1\), the series converges; this criterion is crucial when exploring the interval of convergence.Checking the common ratio's condition ensures that when summed infinitely, the series approaches a finite number. For values of \( x \) where \(|r| \geq 1\), the series fails to converge. So it’s essential to apply this test to ensure proper function and reliability of the solution.
Series Representation
In the realm of infinite series, representation speaks to expressing the series as a function or in a simpler form. For geometric series, we can use the formula:
  • \( S = \frac{a}{1 - r} \)
where \( a \) is the first term, and \( r \) is the common ratio. These formulas boil down complex expressions into manageable forms. For our particular series, the function representation is:
  • \( S(x) = \frac{3}{4 - x^2} \)
Simplifying as such not only shows us a neat, continuous function but also makes understanding and working with the series easier in analytical contexts.By transforming a series to a function, you can evaluate and analyze over a domain, allowing better insights into the characteristics of the function itself and applicability to real-world problems and academic scenarios.

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Most popular questions from this chapter

a.Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b.Determine the radius of convergence of the series. $$f(x)=\sqrt{1-x^{2}}$$

Given the power series $$\frac{1}{\sqrt{1-x^{2}}}=1+\frac{1}{2} x^{2}+\frac{1 \cdot 3}{2 \cdot 4} x^{4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{6}+\cdots$$ for \(-1

Nonconvergence to \(f\) Consider the function $$ f(x)=\left\\{\begin{array}{ll} e^{-1 / x^{2}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ a. Use the definition of the derivative to show that \(f^{\prime}(0)=0\). b. Assume the fact that \(f^{(k)}(0)=0,\) for \(k=1,2,3, \ldots\) (You can write a proof using the definition of the derivative.) Write the Taylor series for \(f\) centered at 0 . c. Explain why the Taylor series for \(f\) does not converge to \(f\) for \(x \neq 0\)

Representing functions by power series Identify the functions represented by the following power series. $$\sum_{k=2}^{\infty} \frac{k(k-1) x^{k}}{3^{k}}$$

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