To express the function $g(x) = \frac{1}{(2+x)^3}$ as a power series centered at $x = 0$, we can use the binomial series expansion.

Step 1: Rewrite the function in a form suitable for binomial expansion

We want to manipulate the function into the form $(1 + u)^n$, since the binomial series expansion works for that form.

Start with:

$g(x) = \frac{1}{(2 + x)^3}$

Factor out the 2 from the denominator:

$g(x) = \frac{1}{2^3 \left( 1 + \frac{x}{2} \right)^3} = \frac{1}{8} \cdot \frac{1}{\left( 1 + \frac{x}{2} \right)^3}$

Step 2: Apply the binomial series

The binomial series for $(1 + u)^n$ is given by:

$(1 + u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \dots$

In our case, $u = \frac{x}{2}$ and $n = -3$, so:

$\left( 1 + \frac{x}{2} \right)^{-3} = 1 + (-3) \cdot \frac{x}{2} + \frac{-3(-3-1)}{2!} \cdot \left( \frac{x}{2} \right)^2 + \frac{-3(-3-1)(-3-2)}{3!} \cdot \left( \frac{x}{2} \right)^3 + \dots$

Step 3: Compute the first few terms

Now, calculate each term:

1. First term: $1$
2. Second term: $-3 \cdot \frac{x}{2} = -\frac{3x}{2}$
3. Third term: $\frac{-3(-4)}{2!} \cdot \left( \frac{x}{2} \right)^2 = \frac{12}{2} \cdot \frac{x^2}{4} = \frac{3x^2}{4}$
4. Fourth term: $\frac{-3(-4)(-5)}{3!} \cdot \left( \frac{x}{2} \right)^3 = \frac{-60}{6} \cdot \frac{x^3}{8} = -\frac{5x^3}{8}$

Step 4: Combine the results

So, the power series expansion for $\left( 1 + \frac{x}{2} \right)^{-3}$ is:

$\left( 1 + \frac{x}{2} \right)^{-3} = 1 - \frac{3x}{2} + \frac{3x^2}{4} - \frac{5x^3}{8} + \dots$

Finally, multiply by the factor $\frac{1}{8}$ to get the full expansion for $g(x)$:

$g(x) = \frac{1}{8} \left( 1 - \frac{3x}{2} + \frac{3x^2}{4} - \frac{5x^3}{8} + \dots \right)$

So the power series expansion for $g(x)$ is:

$g(x) = \frac{1}{8} - \frac{3x}{16} + \frac{3x^2}{32} - \frac{5x^3}{64} + \dots$

Summary:

$g(x) = \frac{1}{8} - \frac{3x}{16} + \frac{3x^2}{32} - \frac{5x^3}{64} + \dots$

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Would you like further details or clarification on any part of the process?

Here are some related questions:

1. How do we derive a binomial series for negative exponents?
2. What happens if we try to expand a function with a different form, such as $\frac{1}{(1+x)^3}$?
3. How can we check the radius of convergence for a power series?
4. How do we obtain the nth term of a binomial series?
5. Can this method be applied to other functions, such as $\frac{1}{(a + bx)^3}$?

Tip: For binomial expansions, you can always expand $(1 + u)^n$ around 0 as long as $|u| < 1$ to ensure convergence.