Expand e^(-x³) in a power series in x, using known

Expand $e^{-x^3}$ in a power series in $x$, using known expansions, and indicate when it is valid.

Step 1 · Series for $e^t$

$$ e^t = \sum_{n=0}^{\infty}\frac{t^n}{n!}, \qquad t\in\mathbb R. $$

Step 2 · Substitute $t=-x^3$

$$ e^{-x^3} = \sum_{n=0}^{\infty}\frac{(-x^3)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^{3n} = 1-x^3+\frac{x^6}{2!}-\frac{x^9}{3!}+\cdots $$

Step 3 · Region of convergence

Since $e^t$ converges for every real $t$, and $t=-x^3$ ranges over all of $\mathbb R$ as $x$ does, the substituted series converges for all $x\in\mathbb R$.

Summary Table

Additional Notes

• Only exponents that are multiples of $3$ appear, mirroring how Problem 255's $1/(4-x^4)$ produces only multiples of $4$ — in both cases the sparsity comes directly from substituting a power of $x$ into a known series.
• Because $e^t$ is entire, every substitution $t=g(x)$ where $g$ is a polynomial (as here, or in Problem 249's shift) yields a series with infinite radius of convergence; only substitutions producing a non-polynomial or singular $g$ (like $t=1/x$) can restrict the domain.
• Differentiating termwise gives $\frac{d}{dx}e^{-x^3}=-3x^2e^{-x^3}=\sum(-1)^n\frac{3n}{n!}x^{3n-1}$ for $n\ge1$, a direct check consistent with the chain rule.