Find a power series representation for the function. (Give your power series representation centered at $x = 0$.) $f(x) = \frac{2}{17 + x}$

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

Neetesh Kumar | December 10, 2024

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This is the solution to Math 132
Assignment: 11.9 Question Number 3
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Step-by-step solution:

Step 1: Rewrite the function

The given function is: $f(x) = \frac{2}{17 + x}.$

Factor $17$ from the denominator: $f(x) = \frac{2}{17(1 + \frac{x}{17})}.$

This simplifies to: $f(x) = \frac{2}{17} \cdot \frac{1}{1 + \frac{x}{17}}.$

Step 2: Use the geometric series formula

We know the geometric series formula: $\frac{1}{1 - r} = \displaystyle\sum_{n=0}^{\infty} r^n, \quad |r| < 1.$

Substitute $r = -\frac{x}{17}$ into the formula: $\frac{1}{1 + \frac{x}{17}} = \displaystyle\sum_{n=0}^{\infty} \left(-\frac{x}{17}\right)^n.$

Multiply by $\frac{2}{17}$: $f(x) = \frac{2}{17} \cdot \displaystyle\sum_{n=0}^{\infty} \left(-\frac{x}{17}\right)^n.$

Simplify: $f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{2(-1)^n x^n}{17^{n+1}}.$

Thus, the power series representation is: $\boxed{f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{2(-1)^n x^n}{17^{n+1}}}.$

Step 3: Determine the interval of convergence

The geometric series converges when: $\left| -\frac{x}{17} \right| < 1.$

Simplify: $\frac{|x|}{17} < 1.$

Multiply through by $17$: $|x| < 17.$

Thus, the interval of convergence is: $\boxed{(-17, 17)}.$

Final Answer:

1. Power series representation: $\boxed{f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{2(-1)^n x^n}{17^{n+1}}}$

   
   

2. Interval of convergence: $\boxed{(-17, 17)}$

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