Expand (1+e^x)² in a power series in x
Expand $(1+e^x)^2$ in a power series in $x$.
Explanation of Method
Expanding the square algebraically first turns the problem into a sum of three entire pieces — a constant, $2e^x$, and $e^{2x}$ — each with its own known series; combining them requires care to keep the standalone constant $1$ separate from the two series' shared $n=0$ terms.
- Expand the square: $(1+e^x)^2=1+2e^x+e^{2x}$.
- Expand $e^x$ and $e^{2x}$ as Maclaurin series.
- Combine carefully, keeping the extra constant term separate.
Step 1 · Expand the square
$$ (1+e^x)^2 = 1+2e^x+e^{2x}. $$
Step 2 · Expand each exponential
$$ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}, \qquad e^{2x}=\sum_{n=0}^{\infty}\frac{2^nx^n}{n!}. $$
Step 3 · Combine, keeping the constant separate
$$ (1+e^x)^2 = 1 + \sum_{n=0}^{\infty}\frac{2+2^n}{n!}x^n. $$ The standalone $1$ cannot be absorbed into the summation formula $\frac{2+2^n}{n!}$: at $n=0$ that formula alone gives $\frac{2+1}{1}=3$, but the true constant term is $1+3=4$ (the extra $1$ from the original expansion). Written out: $$ (1+e^x)^2 = 4+4x+3x^2+\frac{5x^3}3+\frac{3x^4}4+\frac{17x^5}{60}+\cdots $$
Summary Table
| Property | Value |
|---|---|
| Series | $1+\sum_{n=0}^{\infty}\frac{2+2^n}{n!}x^n$ |
| Constant term | $4$ |
| Region of validity | $\mathbb R$ |
Additional Notes
- The general-term formula $\frac{2+2^n}{n!}$ correctly describes every coefficient for $n\ge1$; only the constant term needs the extra $+1$ correction from the standalone "$1$" in $1+2e^x+e^{2x}$, since both $2e^x$ and $e^{2x}$ already independently contribute at $n=0$.
- All coefficients are positive, and the series is entire since it is built entirely from $e^x$ and $e^{2x}$, both of which converge for every real $x$.
- As $x\to\infty$, the $e^{2x}$ term dominates, so $(1+e^x)^2$ grows like $e^{2x}$ — consistent with the coefficient ratio $2^n/n!$ eventually dominating the constant $2/n!$ piece for large $n$.