Question

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. \(\sum_{0}^{\infty} 8^{-n}\left(x^{2}-1\right)^{n}\) Method: Let \(y=x^{2}-1 .\) The power series \(\sum_{0}^{\infty} 8^{-n} y^{n}\) converges for \(|y|<8,\) so the original series converges for \(\left|x^{2}-1\right|<8,\) which means \(|x|<3\)

Short answer

The series converges for \( |x| < 3 \).

Step-by-step solution

Identify the given series

The given series is \(\sum_{0}^{\infty} 8^{-n} \left(x^{2}-1\right)^{n}\).

Set the change of variable

Let \( y = x^{2} - 1 \). This step will transform the given series.

Transform the series

Substitute \( y \) into the series: \(\sum_{0}^{\infty} 8^{-n} \left(x^{2}-1\right)^{n} = \sum_{0}^{\infty} 8^{-n} y^{n}\).

Determine the convergence of the power series

The series \(\sum_{0}^{\infty} 8^{-n} y^{n}\) is a geometric series with ratio \( y/8 \). It converges for \(|y| < 8\).

Find the convergence condition for the original variable

Given \(|y| < 8\), substitute back \( y = x^{2} - 1 \), so \(|x^{2} - 1| < 8\).

Solve for \( x \)

This inequality simplifies to \(|x^{2} - 1| < 8\), leading to \( -8 < x^{2} - 1 < 8 \). Rearrange this to get \( -7 < x^{2} < 9 \). Since \( x^{2} \) is always non-negative, it simplifies further to \( 0 \leq x^{2} < 9 \).

Determine the intervals for \( x \)

Since \( x^{2} < 9 \, \ -3 < x < 3 \). Therefore, the series converges for \( |x| < 3 \).

Key concepts

geometric series

A geometric series is one where each term is a constant multiple (the ratio) of the previous term. For a geometric series \(\sum_{n=0}^{\infty} ar^{n}\), it converges if the absolute value of the ratio \(r\) is less than 1. The sum of a convergent geometric series can be found using the formula \(sum = \frac{a}{1 - r}\), where \(a\) is the first term.

variable substitution

Variable substitution is a technique used to simplify a problem. In this exercise, letting \(y = x^{2} - 1\) simplifies the original series. It turns the complex series into a more recognizable form, making it easier to solve.

convergence interval

The convergence interval of a series is the range of variable values for which the series converges. For the geometric series \(sum_{0}^{\infty} 8^{-n} y^{n}\), the convergence condition is \(|y| < 8\). Substituting \(y = x^{2} - 1\), we find the interval for \(x\) by solving \(|x^{2} - 1| < 8\), which results in \( -3 < x < 3\). Thus, \(|x| < 3\) is the convergence interval for the original series.

absolute value inequality

An absolute value inequality involves terms within an absolute value expression and looks like \(|A| < B\). It describes all values of A that lie within B units of 0. For example, solving \(|x^{2} - 1| < 8\) means finding all values of \(x\) for which \(x^{2} - 1\) lies between \(-8\) and \(8\). This converts to \(-7 < x^{2} < 9\), handling positive and negative ranges of \(x\).

solving inequalities

Solving inequalities involves finding values that make the inequality true. For \(|x^{2} - 1| < 8\), we first remove the absolute value, resulting in \(-8 < x^{2} - 1 < 8\). Adding 1 to all parts, it becomes \(-7 < x^{2} < 9\). Since \(x^{2} < 9\) and \(x^{2} \geq 0 \)naturally,nx must lie in the range \(-3 < x < 3\), which can be summarized as \(|x| < 3\).