Find a power series representation for the function. (Give your power series representation centered at $x = 0$.) $f(x) = \frac{7 + x}{1 - x}$

$f(x) = 7 + \displaystyle\sum_{n=1}^{\infty} \boxed{(\ ? \ )}$

Determine the interval of convergence. (Enter your answer using interval notation.)

$\boxed{?}$

Neetesh Kumar | December 20, 2024

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This is the solution to Math 1c
Assignment: 11.9 Question Number 14
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Step-by-step solution:

Step 1: Split the Function into Simpler Terms:

The given function is:

$f(x) = \frac{7 + x}{1 - x}$.

Split this into two separate terms:

$f(x) = \frac{7}{1 - x} + \frac{x}{1 - x}$.

Step 2: Write Each Term as a Power Series:

1. First Term: $\frac{7}{1 - x}$

   The geometric series expansion for $\frac{1}{1 - x}$ is:

   $\frac{1}{1 - x} = \displaystyle\sum_{n=0}^{\infty} x^n, \ |x| < 1.$

   Multiply this by $7$:

   $\frac{7}{1 - x} = 7 \displaystyle\sum_{n=0}^{\infty} x^n = \displaystyle\sum_{n=0}^{\infty} 7x^n.$

2. Second Term: $\frac{x}{1 - x}$
   Factor out $x$:

   $\frac{x}{1 - x} = x \cdot \frac{1}{1 - x} = x \displaystyle\sum_{n=0}^{\infty} x^n.$

   Simplify:

   $\frac{x}{1 - x} = \displaystyle\sum_{n=0}^{\infty} x^{n+1}.$

   Rewrite the index of summation:

   $\frac{x}{1 - x} = \displaystyle\sum_{n=1}^{\infty} x^n.$

Step 3: Combine the Two Series:

Combine the results:

$f(x) = \frac{7}{1 - x} + \frac{x}{1 - x}.$

Substitute the series expansions:

$f(x) = \displaystyle\sum_{n=0}^{\infty} 7x^n + \displaystyle\sum_{n=1}^{\infty} x^n.$

Notice that the first term starts at $n=0$, while the second term starts at $n=1$.

Combine them:

$f(x) = 7 + \displaystyle\sum_{n=1}^{\infty} (7 + 1)x^n.$

Simplify:

$f(x) = 7 + \displaystyle\sum_{n=1}^{\infty} 8x^n.$

Step 4: Interval of Convergence

1. Convergence of the Geometric Series:

   The geometric series $\displaystyle\sum x^n$ converges when $|x| < 1$.

   Thus, the interval of convergence is:

   $(-1, 1).$

Final Answer:

The power series representation of $f(x)$ is:

$f(x) = 7 + \displaystyle\sum_{n=1}^{\infty} \boxed{8x^n}$

The interval of convergence is:

$\boxed{(-1, 1)}$

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