Identify the function represented by the following power series Sigma (-1)kxk/3k Click the icon to view a table of Taylor series for common functions F(x) =
The Correct Answer and Explanation is:
The power series given is:
$$
\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3^k}
$$
This can be rewritten as:
$$
\sum_{k=0}^{\infty} \left(\frac{-x}{3}\right)^k
$$
This is a geometric series with the general form:
$$
\sum_{k=0}^{\infty} r^k = \frac{1}{1 – r}, \quad \text{for } |r| < 1
$$
In our case, the common ratio $r = \frac{-x}{3}$. Thus, the series converges when:
$$
\left|\frac{-x}{3}\right| < 1 \Rightarrow |x| < 3
$$
Using the geometric series formula:
$$
\sum_{k=0}^{\infty} \left(\frac{-x}{3}\right)^k = \frac{1}{1 – \left(\frac{-x}{3}\right)} = \frac{1}{1 + \frac{x}{3}} = \frac{1}{\frac{3 + x}{3}} = \frac{3}{3 + x}
$$
✅ Final Answer:
$$
\boxed{f(x) = \frac{3}{3 + x}}
$$
📚 300-Word Explanation:
The given power series is:
$$
\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3^k}
$$
To identify the function it represents, it’s helpful to recognize its structure. Notice that the numerator alternates in sign, due to $(-1)^k$, and the denominator has a growing power of 3, that is $3^k$. Thus, we can simplify the general term:
$$
\frac{(-1)^k x^k}{3^k} = \left(\frac{-x}{3}\right)^k
$$
This matches the general form of a geometric series:
$$
\sum_{k=0}^{\infty} r^k = \frac{1}{1 – r}, \quad \text{where } |r| < 1
$$
The value of $r$ in this case is $\frac{-x}{3}$. According to the geometric series rule, this infinite series converges to a function when the absolute value of the common ratio is less than 1, i.e., $|x| < 3$.
Now, we apply the formula:
$$
\sum_{k=0}^{\infty} \left(\frac{-x}{3}\right)^k = \frac{1}{1 – \left(-\frac{x}{3}\right)} = \frac{1}{1 + \frac{x}{3}}
$$
To simplify this expression, combine the terms in the denominator:
$$
\frac{1}{1 + \frac{x}{3}} = \frac{1}{\frac{3 + x}{3}} = \frac{3}{3 + x}
$$
Therefore, the original power series represents the function $\boxed{f(x) = \frac{3}{3 + x}}$, valid for $|x| < 3$.
This type of identification is useful in calculus, especially in understanding how power series can represent functions within an interval of convergence.
