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Chapter 10: Problem 32

Use the geometric series $$f(x)=\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k}, \quad \text { for }|x|<1$$ to find the power series representation for the following functions (centered at 0 ). Give the interval of convergence of the new series. $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$

Short Answer

Expert verified
Answer: The power series representation of the function $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$ is $$\sum_{k=0}^{\infty} x^{3k}$$. Its interval of convergence is \(|x|<1\).

Step by step solution

01

Substitute \(x^3\) into the geometric series representation

Instead of using \(x\) in the given geometric series representation, we will use \(x^3\). So, the function will become:$$f\left(x^{3}\right)=\frac{1}{1-x^{3}}=\sum_{k=0}^{\infty} (x^{3})^{k}$$
02

Simplify the power series representation

Now, we will simplify the exponent of \(x\) in the power series:$$\sum_{k=0}^{\infty} (x^{3})^{k} = \sum_{k=0}^{\infty} x^{3k}$$This is the power series representation of the function $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$.
03

Find the interval of convergence for the new series

To find the interval of convergence for the new power series, we will use the fact that the interval of convergence for the original geometric series is \(|x|<1\). Since we replaced \(x\) with \(x^3\), we need to find the interval for \(x^3\). The inequality will be:$$|x^3|<1$$Taking the cube root of both sides, we get:$$|x|<1^{\frac{1}{3}}$$Since \(1^{\frac{1}{3}}=1\), the interval of convergence for the new power series representation is still \(|x|<1\). So, the power series representation for the function $$f\left(x^{3}\right)=\frac{1}{1-x^{3}}$$ is $$\sum_{k=0}^{\infty} x^{3k}$$ with an interval of convergence of \(|x|<1\).

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Most popular questions from this chapter

Use properties of power series, substitution, and factoring of constants to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Use the Taylor series. $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1 < x < 1$$ $$\frac{1}{\left(1+4 x^{2}\right)^{2}}$$

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\frac{e^{x}+e^{-x}}{2}$$

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=e^{-x}, a=0$$

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. $$f(x)=e^{-3 x}$$

Suppose \(f\) and \(g\) have Taylor series about the point \(a.\) a. If \(f(a)=g(a)=0\) and \(g^{\prime}(a) \neq 0,\) evaluate \(\lim _{x \rightarrow a} f(x) / g(x).\) by expanding \(f\) and \(g\) in their Taylor series. Show that the result is consistent with l'H脙麓pital's Rule. b. If \(f(a)=g(a)=f^{\prime}(a)=g^{\prime}(a)=0\) and \(g^{\prime \prime}(a) \neq 0\) evaluate \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) by expanding \(f\) and \(g\) in their Taylor series. Show that the result is consistent with two applications of I'H脙麓pital's Rule.
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