Question

Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ f(x)=\frac{4}{3 x+2}, \quad c=2 $$

Short answer

The power series for the given function is: \( f(x) = 2 \times \sum_{n=0}^\infty (-1)^n (\frac{3}{2})^n (x-2)^n\) and the interval of convergence is \( \frac{4}{3} < x < \frac{8}{3} \).

Step-by-step solution

Simplify the function to apply geometric series formula

First, we will rearrange the function to fit into the form \( f(x) = \frac{a}{1 - r} \), where \(a\) is the first term and \(r\) is the common ratio. We want to get it into this format because this will allow us to apply the geometric series formula. So, we can rewrite it as follows: \( f(x) = \frac{4}{(2-3x+6)} = 2 \times \frac{1}{1 + \frac{3}{2} (x-2)} \)

Apply the geometric series formula

Now we can use the formula for the sum of a geometric series: \( \sum ar^n = a + ar + ar^2 + \ldots \) , where \( |r| < 1 \) . In our case, \(a = 1\) and \(r = - \frac{3}{2} (x-2)\). This gives us the power series: \(f(x) = 2 \times \sum_{n=0}^\infty (-\frac{3}{2}(x-2))^n \)

Simplify the power series

After simplifying the power series we get: \( f(x) = 2 \times \sum_{n=0}^\infty (-1)^n (\frac{3}{2})^n (x-2)^n = 2 \times \sum_{n=0}^\infty (-1)^n (\frac{3}{2})^n (x-2)^n \)

Determine the interval of convergence

We use the ratio test to determine the interval of convergence. The series converges if the limit: \( lim_{n->\infty} |\frac{a_{n+1}}{a_n}| < 1 \). That condition gives \( |-\frac{3}{2}(x-2)| < 1 \) which simplifies to \( -\frac{2}{3} < x-2 < \frac{2}{3} \). Solving for \(x\), the interval of convergence is \( \frac{4}{3} < x < \frac{8}{3} \)