Question

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) 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 \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) at \(x=0\), \(0.2\), \(0.5\), \(1\), and \(1.5\), and determine the domain of the function. Answer: \(f(0)=1\), \(f(0.2)=\frac{5}{6}\), \(f(0.5)=\frac{2}{3}\), \(f(1)\) and \(f(1.5)\) are not possible to be evaluated. The domain of \(f(x)\) is the open interval \((-1,1)\).

Step-by-step solution

Evaluate \(f(0)\)

Plug in \(x=0\) into the geometric series: \(f(0)=\sum_{k=0}^{\infty}(-1)^{k} (0)^{k}\). Since \(0^k = 0\) for every \(k>0\), we have \(f(0)= 0^0=1\).

Evaluate \(f(0.2)\), \(f(0.5)\), \(f(1)\), and \(f(1.5)\) using the formula for the sum of a geometric series

Recall that \(\sum_{k=0}^{\infty}ar^{k} =\frac{a}{1-r}\) if \(\left| r \right|<1\). Here, \(a=(-1)^0=1\), and \(r=(-1)x\), so we can rewrite \(f(x)\) as follows: \(f(x)=\frac{1}{1-(-1)x}\). For \(x=0.2\), \(f(0.2)=\frac{1}{1-(-1)(0.2)}=\frac{1}{1+0.2}=\frac{1}{1.2}=\frac{5}{6}\). For \(x=0.5\), \(f(0.5)=\frac{1}{1-(-1)(0.5)}=\frac{1}{1+0.5}=\frac{1}{1.5}=\frac{2}{3}\). For \(x=1\), the denominator of the formula becomes 0, which implies that the function does not converge. Thus, \(f(1)\) is not possible to be evaluated. For \(x=1.5\), it is outside the interval of convergence \((-1,1)\), so \(f(1.5)\) is also not possible to be evaluated. Part b: Domain of \(f(x)\)

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

Recall that the geometric series converges when the ratio \(\left| r \right|<1\). In our case, \(\left| -1x \right|<1\), which simplifies to \(-1<x<1\) or \((-1,1)\). Therefore, the domain of \(f(x)\) is the open interval \((-1,1)\).

Key concepts

Domain of a Function

The domain of a function refers to all the possible input values (often represented by \(x\)) for which the function is defined. In simpler terms, it's the complete set of x-values that you can plug into the function without running into any problems. When looking at geometric series like \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\), the domain becomes crucial to identify where the series actually provides a valid result.

For this particular function, the series converges, or results in a meaningful sum, only when the absolute value of the ratio \(r\), which here is \((-1)x\), is less than 1. Mathematically, this is represented as \(\left| r \right|<1\).

Therefore, for \(f(x)\), this condition translates into \(-1 < x < 1\), indicating that the domain is the open interval \((-1, 1)\).x

Understanding and defining the domain is essential because it informs us of the x-values that will produce a meaningful output without leading to any undefined expressions or convergent issues.

Convergence

Convergence in the context of a series, especially a geometric series, refers to the tendency of the series to approach a specific value as the number of terms increases. For students studying calculus or sequences, understanding convergence is key to determining whether a series will sum to a finite number or not.

The given geometric series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) converges if, and only if, the sum of the series becomes constant as \(k\) goes towards infinity. The convergence relies on the condition \(\left| r \right|<1\). Here, \(r = (-1)x\), meaning that as long as \(x\) falls within the range of \(-1\) and \(1\) (i.e., the absolute value of \(r\) is less than 1), the series will converge.

If \(x = 1\), the convergence fails since the denominator in the geometric series formula becomes zero. Similarly, if \(x > 1\) or \(x < -1\), the series diverges, or in simpler terms, fails to converge, because \(r\) does not meet the required condition. Hence, understanding convergence helps in determining where the function is valid and where it spuriously diverges.

Interval of Convergence

The interval of convergence for a series specifies the range of values for which the series will converge. This concept is significant in understanding the behavior of functions defined with series, such as the given geometric series.

For the series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\), we see that the interval of convergence is determined by the condition that the absolute value of the common ratio \(r\) must be less than 1, i.e., \(\left| -1x \right| < 1\).

Solving this condition yields the interval \(-1 < x < 1\). This means that within these bounds, the series converges and the function \(f(x)\) gives valid results.

• If \(x < -1\) or \(x > 1\), the series doesn't converge, hence these are outside the interval.
• At \(x = -1\) or \(x = 1\), the series fails to converge because it doesn't meet the strict \(\left| r \right| < 1\) criterion, thus making the endpoints not part of the interval.

Grasping the interval of convergence assists in predicting where a series-derived function will properly operate without issues.