To find the Maclaurin series of the function:
$f(x) = \frac{-2}{(x - 1)^2}$ we need to express it as a power series centered at $x = 0$.

Step 1: Rewrite the Function

Since the Maclaurin series is centered at $x = 0$, we substitute $x - 1 = 1 - (1 - x)$ and rewrite:

$f(x) = \frac{-2}{(1 - (1 - x))^2} = \frac{-2}{(1 - u)^2}, \quad \text{where } u = 1 - x.$

Using the binomial series expansion formula: $\frac{1}{(1 - u)^2} = \sum_{n=0}^{\infty} (n+1) u^n = \sum_{n=0}^{\infty} (n+1) (1 - x)^n.$

Multiplying both sides by $-2$:

$f(x) = -2 \sum_{n=0}^{\infty} (n+1) (1 - x)^n.$

Step 2: Expand in Terms of $x$

Now, expand $(1 - x)^n$ using the binomial series: $(1 - x)^n = \sum_{k=0}^{\infty} \binom{n}{k} (-x)^k.$

So, $f(x) = -2 \sum_{n=0}^{\infty} (n+1) \sum_{k=0}^{\infty} \binom{n}{k} (-x)^k.$

Rewriting this as a single sum:

$f(x) = -2 \sum_{m=0}^{\infty} \left( \sum_{n=m}^{\infty} (n+1) \binom{n}{m} (-1)^m \right) x^m.$

Computing the inner sum:

$\sum_{n=m}^{\infty} (n+1) \binom{n}{m} = \frac{(m+2)(m+1)}{2}.$

Thus, the final Maclaurin series expansion is:

$f(x) = -2 \sum_{m=0}^{\infty} \frac{(m+2)(m+1)}{2} (-1)^m x^m.$

Final Answer

$f(x) = \sum_{m=0}^{\infty} (-1)^{m+1} (m+2)(m+1) x^m.$

Would you like a step-by-step derivation of the sum evaluation?

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Related Questions

1. What is the radius of convergence for this Maclaurin series?
2. How can this series be used to approximate values of $f(x)$?
3. How does this compare to the Taylor series centered at $x = 1$?
4. Can we derive the Maclaurin series using derivatives instead of expansion techniques?
5. How does the convergence behavior change for different values of $x$?

Math Tip

For functions with singularities (like $f(x)$ at $x = 1$), power series expansions often converge only within a limited radius around the center. Always check the radius of convergence before using a series for approximations!