/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 96 Suppose a function \(f\) is defi... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 96

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{2 k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

Short Answer

Expert verified
Answer: For the given function \(f(x)\), we have \(f(0) = 1\), \(f(0.2) \approx 1.0417\), and \(f(0.5) \approx 1.3333\). The geometric series does not converge for \(f(1)\) and \(f(1.5)\), so we cannot compute their values. The domain of the function is \((-1,1)\).

Step by step solution

01

Find the formula for the sum of the geometric series

For a geometric series, the sum can be written as \(S = \frac{a(1-r^n)}{1-r}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. In our case, the first term, \(k=0\), is \(x^{2\cdot 0} = 1\), and the common ratio is \(x^2\). We need to find the sum up to infinity, so we'll use the limit formula for geometric series when the absolute value of the common ratio is less than 1: \(S = \frac{a}{1-r}.\)
02

Substitute the values into the formula

As \(a = 1\) and \(r = x^2\), we find that $$ f(x) = \frac{1}{1-x^2}. $$ Remember that this sum converges only when the common ratio's absolute value is less than 1, which in this case means when \(|x^2|<1\), or equivalently \(|x|<1\).
03

Evaluate the function for the given values

Using the formula above, we can compute \(f(0), f(0.2), f(0.5), f(1), f(1.5)\) as follows: $$ f(0) = \frac{1}{1-0^2} = 1, $$ $$ f(0.2) = \frac{1}{1-0.2^2} = \frac{1}{1-0.04} = \frac{1}{0.96} \approx 1.0417, $$ $$ f(0.5) = \frac{1}{1-0.5^2} = \frac{1}{1-0.25} =\frac{1}{0.75} \approx 1.3333, $$ For \(x=1\) and \(x=1.5\), the absolute values of their squares are not smaller than 1 (and, moreover, in the case of \(x=1\), the denominator of the fraction is zero), so the geometric series does not converge, and we cannot compute the value of the function for these two values.
04

Determine the domain of the function

The domain of \(f(x) = \frac{1}{1-x^2}\) is determined by the values of x for which the series converges, which is when \(|x| < 1\). Thus, the domain of \(f\) is \((-1,1)\).

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Most popular questions from this chapter

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