Question

Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ f(x)=\frac{4}{5-x}, \quad c=-2 $$

Short answer

The power series representation for \( f(x) \) is \( \sum_{n=0}^{\infty} 4*(-1)^n \cdot \left( \frac{x+2}{7} \right)^n \). The interval of convergence is \( -9 < x < 5 \).

Step-by-step solution

Apply the geometric series representation

We'll start by writing \( \frac{4}{5-x} \) in the form of \( \frac{a}{1-r} \) to be able to express the function as a geometric series. We can write the function as \( \frac{4}{7 + (x+2)} \). Now, it is in the form that allows us to apply the geometric series representation. Therefore, the power series representation of \( f(x) \) is \( \sum_{n=0}^{\infty} 4*(-1)^n \cdot \left( \frac{x+2}{7} \right)^n \).

Find the interval of convergence

The interval of convergence for a geometric series \( \sum a \cdot r^n \) is \( |r| < 1 \). So in our series, our \( r \) is \( \left|\frac{x+2}{7}\right| \). Setting this less than 1 and solving gives \( -7 < x+2 < 7 \). Translating this for x, we get \( -9 < x < 5 \). Thus the interval of convergence of the series is \( -9 < x < 5 \).

Confirm its valid for endpoints

We need to verify the behaviour at the endpoints of the interval. We need to substitute \( x = -9 \) and \( x = 5 \) into the original function and the series to confirm they match. The function is undefined at \( x = 5 \), it does not need testing. At \( x = -9 \), the series becomes \( \sum_{n=0}^{\infty} 4*(-1)^n = 4 - 4 + 4 - 4 + ... \) which is undefined. Therefore the interval of convergence does not include its endpoints.

Write the final results

The power series representation for \( f(x) \) is \( \sum_{n=0}^{\infty} 4*(-1)^n \cdot \left( \frac{x+2}{7} \right)^n \) and its interval of convergence is \( -9 < x < 5 \).