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

Question: Evaluate the function \(f(x) = \sum_{k=0}^{\infty} x^{k}\) at \(x = 0, 0.2, 0.5, 1,\) and \(1.5\) and determine the domain of \(f\). Answer: \(f(0) = 1, f(0.2) = 1.25, f(0.5) = 2\), and the function is not convergent at \(x=1\) and \(x=1.5\). The domain of \(f\) is \((-1, 1)\).

Step-by-step solution

Review the geometric series formula and convergence

For a geometric series $$\sum_{k=0}^{\infty} ar^k = a\frac{1}{1 - r},$$ where \(a\) is the first term of the series, \(r\) is the common ratio, and the series converges if \(|r| < 1\). In our case, \(a = 1\) and \(r = x\), so the formula for \(f(x)\) becomes $$f(x)=\sum_{k=0}^{\infty} x^{k} = \frac{1}{1-x}$$ as long as \(|x| < 1\).

Evaluate \(f(x)\) at given values

Evaluate the function \(f(x)\) at the given \(x\) values if the series converges: 1. \(f(0)\): In this case, \(x = 0\) and since \(|0| < 1\), the series converges, therefore $$f(0) = \frac{1}{1-0} = 1.$$ 2. \(f(0.2)\): Here, \(x = 0.2\) and since \(|0.2| < 1\), the series converges, thus $$f(0.2) = \frac{1}{1-0.2} = \frac{1}{0.8} = 1.25.$$ 3. \(f(0.5)\): In this case, \(x = 0.5\) and since \(|0.5| < 1\), the series converges, so $$f(0.5) = \frac{1}{1-0.5} = \frac{1}{0.5} = 2.$$ 4. \(f(1)\): When \(x = 1\), \(|1| = 1,\) which is not less than 1, so the series does not converge at this point, meaning we cannot evaluate \(f(1).\) 5. \(f(1.5)\): For this value, \(x = 1.5\) and since \(|1.5| > 1\), the series does not converge, and we cannot evaluate \(f(1.5).\)

Determine the domain of \(f(x)\)

From the convergence of the geometric series, we know that \(f(x)\) is defined and only converges for \(|x| < 1\). Therefore, the domain of the function \(f\) is the set of all real numbers in the interval \((-1, 1)\).

Key concepts

Series Convergence Criteria

When working with infinite series, a fundamental question to ask is whether the series converges or diverges. An infinite series converges if the sum of its terms approaches a finite value as the number of terms goes to infinity. In contrast, a series diverges if the sum does not settle on a finite value.

Geometric series have a specific convergence criterion that is easy to check. The convergence of a geometric series \( \sum_{k=0}^{\infty} ar^k \) depends entirely on the common ratio \(r\): if the absolute value of the common ratio, \( |r| \), is less than 1, the series will converge. This criterion is rooted in the idea that as terms with an \( |r| < 1 \) are repeatedly multiplied, they get smaller and smaller, ultimately allowing the sum to stabilize. However, if \( |r| \geq 1 \) the terms do not shrink sufficiently, which results in a sum that grows without bound—meaning the series diverges.

In the case of the function \( f(x) \) from our exercise, the series is of the geometric type with a first term \( a = 1 \) and the common ratio as \( x \). Thus, the series will converge if \( |x| < 1 \) and diverge otherwise.

Geometric Series Formula

The geometric series formula is a critical tool for evaluating the sum of a geometric series that converges. It's expressed as \( \sum_{k=0}^{\infty} ar^k = a\frac{1}{1 - r} \) when \( |r| < 1 \). In this formula \( a \) is the first term, and \( r \) is the common ratio. The formula gives us a straightforward way to find the exact sum of an infinite number of decreasing (or increasing) terms.

Considering our specific function \( f(x) \) which represents a geometric series with \( a = 1 \) and \( r = x \), applying the geometric series formula simplifies to \( f(x) = \frac{1}{1-x} \) given that \( |x| < 1 \) for convergence. This simplification makes the task of evaluating the function at different points straightforward, as long as those points fall within the interval where the series converges.

Function Evaluation

Function evaluation is the process of determining the output of a function for a particular input. For the function \( f(x) \) defined by the geometric series in our exercise, we evaluate the function by substituting the input value for \( x \) into the simplified series formula \( \frac{1}{1-x} \) when the value of \( x \) is within the domain for which the series converges.

As demonstrated in the provided solutions, evaluation is a matter of substitution and simplification. For instance, when evaluating \( f(0.2) \) we simply replace \( x \) with 0.2, yielding \( f(0.2) = \frac{1}{1-0.2} \) and then calculate the result. It's important to always first verify if the input value falls within the domain of convergence; otherwise, the function is not defined and cannot be evaluated at that input. Notably, \( f(1) \) and \( f(1.5) \) cannot be evaluated as \( 1 \) and \( 1.5 \) do not meet the convergence criteria, demonstrating the direct link between series convergence and function evaluation.

Domain of a Function

The domain of a function is the set of all possible input values (often \( x \) values) for which the function is defined. In other words, it's the collection of all real numbers that you can plug into the function without causing any mathematical issues such as division by zero or taking the square root of a negative number.

In terms of our function \( f(x) \) defined by a geometric series, the domain is dictated by our need for the series to converge. Since we've established that a geometric series converges only if \( |r| < 1 \) and for the function \( f(x) \), \( r = x \) this means the function will only be defined when \( |x| < 1 \). Therefore, the domain of \( f \) is the open interval \( (-1, 1) \), as values outside this interval will lead to divergence of the geometric series, leaving the function value undefined.