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Function defined by a series 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: The values of the function are: - \(f(0) = 1\) - \(f(0.2) = \frac{25}{24}\) - \(f(0.5) = \frac{4}{3}\) - \(f(1)\) is undefined - \(f(1.5)\) is not possible due to the divergence of the geometric series. The domain of \(f(x)\) is \((-1,1)\).

Step by step solution

01

Identify the general form of the geometric series

The function \(f(x)\) is given by the geometric series \[f(x)=\sum_{k=0}^{\infty} x^{2k}.\] This series converges if and only if \(|x^{2}|<1\), which means \(|x|<1\). Thus, the domain of the function will be \((-1, 1)\).
02

Evaluate the function at \(f(0)\)

We can plug in \(0\) into the series: \[f(0) = \sum_{k=0}^{\infty} 0^{2k} = 1 + 0 + 0 + \cdots = 1.\] So, \(f(0) = 1\).
03

Evaluate the function at \(f(0.2)\), \(f(0.5)\), and \(f(1)\)

Since \(|0.2|<1\), \(|0.5|<1\), and \(|1|=1\), all these points lie within or on the border of the domain of the function. We can use the formula for the sum of a convergent geometric series to evaluate the function at these points: \[f(x) = \frac{a}{1-r} = \frac{1}{1-x^2}.\] Now, we can plug in \(x=0.2, 0.5, 1\) to find the corresponding function values: \[f(0.2) = \frac{1}{1-(0.2)^2} = \frac{1}{1-0.04} = \frac{1}{0.96} = \frac{25}{24},\] \[f(0.5) = \frac{1}{1-(0.5)^2} = \frac{1}{1-0.25} = \frac{1}{0.75} = \frac{4}{3},\] \[f(1) = \frac{1}{1-(1)^2} = \frac{1}{1-1}.\] The value \(f(1)\) is not defined since the denominator is \(0\). So, \(f(0.2) = \frac{25}{24}\), \(f(0.5) = \frac{4}{3}\), and \(f(1)\) is undefined.
04

Evaluate the function at \(f(1.5)\)

Since \(|1.5|>1\), the value \(1.5\) does not belong to the domain of the function, so evaluating the function at \(x=1.5\) is not possible due to divergence of the geometric series.
05

Determine the domain of the function

As mentioned in step 1, the domain of the function is given by \((-1, 1)\), where the geometric series converges. In conclusion, - \(f(0) = 1\) - \(f(0.2) = \frac{25}{24}\) - \(f(0.5) = \frac{4}{3}\) - \(f(1)\) is undefined - \(f(1.5)\) is not possible due to the divergence of the geometric series - The domain of \(f(x)\) is \((-1,1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of a Function
The domain of a function is the set of all possible input values (usually represented by 'x') for which the function is defined and provides a real number output. In simpler terms, it's a collection of values that you can safely plug into the function without running into mathematical issues such as division by zero or taking the square root of a negative number. For the function defined by a geometric series like \(f(x)=\sum_{k=0}^{\infty} x^{2k}\), the domain is determined by the convergence of the series.

Given the series \(x^{2k}\), convergence occurs when \(|x^{2}|<1\). This constraint implies that \(|x|<1\), meaning the only valid input values for this function are those within the interval \((-1, 1)\). If \(x\) falls outside this interval, the series does not converge, and the function is undefined. Therefore, the domain of the function \(f(x)\) described is all values of \(x\) between -1 and 1, exclusive.

Remember, identifying the domain is critical before evaluating a function, as it ensures that your calculations remain valid and produce meaningful results.
Series Convergence
When dealing with series, convergence is an essential concept. A series is said to "converge" if the sum of its terms approaches a specific, finite limit as the number of terms goes to infinity. For geometric series, convergence is determined by the value of the common ratio between successive terms.

In a geometric series, the series can be represented as \(a + ar + ar^2 + ar^3 + \ldots \). Here, \(r\) is the common ratio. The geometric series converges only if the absolute value of \(r\) is less than one (\(|r| < 1\)). If \(|r|\) is greater than or equal to one, the series diverges, meaning it does not have a finite sum.

For the function \(f(x)=\sum_{k=0}^{\infty} x^{2k}\), the series has a common ratio of \(x^2\). For convergence, we require \(|x^2| < 1\), leading to \(|x| < 1\). Thus, within this range, the series converges and enables us to calculate a finite sum for the function \(f(x)\). Convergence ensures the validity of the function over its specified domain.
Geometric Series Sum Formula
The geometric series sum formula is a handy tool for calculating the sum of a geometric series when it converges. This formula allows you to find the sum without manually adding an infinite number of terms. For a converging geometric series represented by \(a + ar + ar^2 + ar^3 + \ldots \), where \(|r| < 1\), the sum \(S\) of this series can be found using the formula: \[S = \frac{a}{1 - r}.\]For the function \(f(x)=\sum_{k=0}^{\infty} x^{2k}\), the first term \(a = 1\) and the common ratio \(r = x^2\). Thus, the sum of the function is given by the geometric series sum formula as \(\frac{1}{1-x^2}\), provided \(|x| < 1\).

Using this sum formula, you can swiftly calculate the function's value for any \(x\) in the domain. For example, \(f(0.2)\) finds its value as \(\frac{1}{1 - 0.04} = \frac{25}{24}\), and \(f(0.5)\) as \(\frac{1}{1 - 0.25} = \frac{4}{3}\). This formula is indispensable, saving time and ensuring accuracy when solving problems involving infinite geometric series.

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