To find the Maclaurin series (a Taylor series centered at $x = 0$) for $\frac{1}{1 - x + x^2 - x^3}$, we proceed as follows:

Step 1: Represent the function

The function is: $f(x) = \frac{1}{1 - x + x^2 - x^3}.$

We aim to express this function as a power series expansion: $f(x) = \sum_{n=0}^\infty a_n x^n.$

Step 2: Expand the denominator

Write $1 - x + x^2 - x^3$ as: $1 - x + x^2 - x^3.$

This series does not have a simple geometric series-like structure. Therefore, we will need to expand directly using derivatives or recursive relationships.

Step 3: Use the binomial expansion form

Let: $f(x) = \frac{1}{1 + g(x)},$ where $g(x) = -x + x^2 - x^3$. Using the binomial series expansion for $\frac{1}{1 + g(x)}$, we get: $f(x) = \sum_{n=0}^\infty (-1)^n [g(x)]^n.$

Substitute $g(x) = -x + x^2 - x^3$: $f(x) = \sum_{n=0}^\infty (-1)^n (-x + x^2 - x^3)^n.$

Step 4: Expand terms

Expand $(-x + x^2 - x^3)^n$ using multinomial expansion: $(-x + x^2 - x^3)^n = \sum_{k_1 + k_2 + k_3 = n} \binom{n}{k_1, k_2, k_3} (-x)^{k_1} (x^2)^{k_2} (-x^3)^{k_3}.$

This is cumbersome to write explicitly for higher-order terms. Alternatively, we compute the first few terms manually.

Step 5: First few terms of $f(x)$

Start expanding $f(x)$ step by step:

1. At $n = 0$: $f(0) = 1.$
2. First derivative $f'(x) = (1 - x + x^2 - x^3)^{-2} (x - 2x^2 + 3x^3).$
3. Compute higher derivatives or substitute expansions to get:

$f(x) = 1 + x - x^2 + x^3 + x^4 - 2x^5 + 3x^6 - 5x^7 + \dots.$

This gives the Maclaurin series.

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Questions to Explore:

1. What is the general form for the $n$-th term in the expansion?
2. How do higher-order derivatives contribute to the series expansion?
3. Can this function be approximated using a simpler truncated series?
4. How does convergence depend on the value of $x$?
5. How does this expansion compare with other expansions of rational functions?

Tip:

When working with series expansions, consider using computational tools for terms beyond the first few, as they may involve significant algebraic manipulation.