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}
We can rewrite this series 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
Here, the common ratio r=−x3r = \frac{-x}{3}. Since this is a geometric series, we can express it as a function: F(x)=11−(−x/3)=11+x/3F(x) = \frac{1}{1 – (-x/3)} = \frac{1}{1 + x/3}
To simplify: F(x)=11+x3=13+x3=33+xF(x) = \frac{1}{1 + \frac{x}{3}} = \frac{1}{\frac{3 + x}{3}} = \frac{3}{3 + x}
Final Answer:
F(x)=33+xF(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 identify the function represented by this power series, we recognize it as a geometric series. A geometric series has the general form: ∑k=0∞rk=11−r,for ∣r∣<1\sum_{k=0}^{\infty} r^k = \frac{1}{1 – r}, \quad \text{for } |r| < 1
We manipulate the original expression: (−1)kxk3k=(−x3)k\frac{(-1)^k x^k}{3^k} = \left(\frac{-x}{3}\right)^k
Thus, the series becomes: ∑k=0∞(−x3)k\sum_{k=0}^{\infty} \left( \frac{-x}{3} \right)^k
This fits the form of a geometric series with ratio r=−x3r = \frac{-x}{3}. The series converges for ∣x∣<3|x| < 3, because ∣−x3∣<1\left|\frac{-x}{3}\right| < 1. Applying the geometric series sum formula, we get: ∑k=0∞(−x3)k=11−(−x3)=11+x3\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 further: 11+x3=33+x\frac{1}{1 + \frac{x}{3}} = \frac{3}{3 + x}
Hence, the power series represents the function F(x)=33+xF(x) = \frac{3}{3 + x}, valid within the radius of convergence ∣x∣<3|x| < 3. Recognizing and manipulating power series into geometric form is a powerful technique in calculus and mathematical analysis.
