Question

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{k}\) 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

Answer: The values of \(f(x)\) for the given values of \(x\) are: \(f(0) = 1\), \(f(0.2) = 1.25\), and \(f(0.5) = 2\). The function cannot be evaluated for \(x=1\) and \(x=1.5\) as the series does not converge for these values. The domain of the function \(f(x)\) is the open interval \((-1, 1)\).

Step-by-step solution

Sum of an infinite geometric series formula for f(x)

The sum of an infinite geometric series with common ratio r is given by the formula: \(S = \frac{a}{1-r}\) where \(a\) is the first term of the series. For our function \(f(x)\), the first term \(a=1\) and the common ratio is \(x\). So the sum formula for \(f(x)\) is: \(f(x) = \frac{1}{1-x}\)

Evaluate f(0), f(0.2), f(0.5), f(1), and f(1.5), if possible

Now, we'll use the formula found in Step 1 to evaluate f(x) for the given values of x: \(f(0) = \frac{1}{1-0} = 1\) \(f(0.2) = \frac{1}{1-0.2} = \frac{1}{0.8} = 1.25\) \(f(0.5) = \frac{1}{1-0.5} = \frac{1}{0.5} = 2\) For \(f(1)\) and \(f(1.5)\), the series does not converge, because the common ratio x is not between -1 and 1. Therefore, we cannot evaluate f(1) and f(1.5).

Find the domain of f(x)

The domain of \(f(x)\) is the set of all possible values of x for which the function f(x) is defined (i.e., the geometric series converges). As discussed earlier, the geometric series converges when the common ratio x is between -1 and 1: \(-1 < x < 1\) Therefore, the domain of \(f(x)\) is the open interval \((-1, 1)\).