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 given power series is: ∑k=0∞(−1)kxk3k\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3^k}
This can be rewritten as: ∑k=0∞(−x3)k\sum_{k=0}^{\infty} \left(\frac{-x}{3}\right)^k
This is a geometric series of the form: ∑k=0∞rk=11−r,for ∣r∣<1\sum_{k=0}^{\infty} r^k = \frac{1}{1 – r}, \quad \text{for } |r| < 1
Step-by-step Identification:
We recognize the form: ∑k=0∞(−x3)k\sum_{k=0}^{\infty} \left(\frac{-x}{3}\right)^k
This means the common ratio r=−x3r = \frac{-x}{3}. Applying the geometric series formula: F(x)=11−(−x/3)=11+x/3F(x) = \frac{1}{1 – (-x/3)} = \frac{1}{1 + x/3}
Now simplify: F(x)=13+x3=33+xF(x) = \frac{1}{\frac{3 + x}{3}} = \frac{3}{3 + x}
✅ Final Answer:
F(x)=33+x\boxed{F(x) = \frac{3}{3 + x}}
✍️ Explanation
The given series is: ∑k=0∞(−1)kxk3k\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3^k}
To understand what function this series represents, we look for patterns that match known Taylor or power series. This series is in the form of a geometric series, which is one of the most common and easily recognizable types. A geometric series has the general form: ∑k=0∞rk=11−r\sum_{k=0}^{\infty} r^k = \frac{1}{1 – r}
provided that the absolute value of the common ratio, ∣r∣|r|, is less than 1 (ensuring convergence).
In our case, we can rewrite the series: ∑k=0∞(−x3)k\sum_{k=0}^{\infty} \left(\frac{-x}{3}\right)^k
Here, the common ratio r=−x/3r = -x/3. Because this is a geometric series, we can apply the formula for the sum of a geometric series: ∑k=0∞rk=11−r\sum_{k=0}^{\infty} r^k = \frac{1}{1 – r}
Plugging in r=−x/3r = -x/3: F(x)=11−(−x/3)=11+x/3F(x) = \frac{1}{1 – (-x/3)} = \frac{1}{1 + x/3}
Simplifying this expression to a single rational function: F(x)=33+xF(x) = \frac{3}{3 + x}
This is a well-defined function for all x≠−3x \neq -3, and the power series converges for ∣x∣<3|x| < 3. Hence, the function represented by the original power series is: F(x)=33+x\boxed{F(x) = \frac{3}{3 + x}}
