Taylor Series Expansion Strategy

We need the coefficient of the $x^3$ term for $f(x) = e^{x^2 - x}$. We use the standard Maclaurin series for $e^u$: $ e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots $ Let $u = x^2 - x$. Substituting this gives:

$ f(x) = e^{x^2 - x} = 1 + (x^2 - x) + \frac{(x^2 - x)^2}{2!} + \frac{(x^2 - x)^3}{3!} + \dots $

Identifying $x^3$ Contributions

We examine the terms contributing to $x^3$:

• The term $\frac{(x^2 - x)^2}{2!}$ expands to $\frac{x^4 - 2x^3 + x^2}{2}$. The $x^3$ part is $-x^3$, giving a coefficient of $-1$.
• The term $\frac{(x^2 - x)^3}{3!}$ involves expanding $(x^2 - x)^3$. The term producing $x^3$ comes from $(-x)^3 = -x^3$. So, this part contributes $\frac{-x^3}{3!} = -\frac{x^3}{6}$. The coefficient is $-\frac{1}{6}$.

Higher terms in the expansion of $e^u$ yield powers of $x$ greater than 3.

Summing Coefficients

The total coefficient of $x^3$ is the sum of the coefficients identified:

$ \text{Coefficient of } x^3 = -1 + \left(-\frac{1}{6}\right) $

$ \text{Coefficient of } x^3 = -\frac{6}{6} - \frac{1}{6} = -\frac{7}{6} $