Problem 249
Expand e^x in a series of powers of (x+2)
Expand $e^x$ in a series of powers of $(x+2)$.
Explanation of Method
Since every derivative of $e^x$ equals $e^x$ itself, all Taylor coefficients about any center $x=c$ share the common factor $e^c$; equivalently, $e^x=e^{-2}\cdot e^{x+2}$ directly, and expanding $e^{x+2}$ via the standard Maclaurin series gives the result immediately.
- Factor $e^x=e^{-2}e^{x+2}$.
- Expand $e^{x+2}$ using the Maclaurin series for $e^y$.
- Note the radius of convergence is infinite, since $e^x$ is entire.
Step 1 · Factor out the constant
$$ e^x = e^{(x+2)-2} = e^{-2}\cdot e^{x+2}. $$
Step 2 · Expand $e^{x+2}$
With $y=x+2$: $e^y=\sum_{n=0}^{\infty}y^n/n!$, so $$ e^x = e^{-2}\sum_{n=0}^{\infty}\frac{(x+2)^n}{n!}. $$
Step 3 · Region of convergence
Since $e^t$ is entire (its Maclaurin series has infinite radius of convergence), the same holds for this recentered series: it converges for every $x\in\mathbb R$.
Summary Table
| Property | Value |
|---|---|
| Series | $e^{-2}\sum(x+2)^n/n!$ |
| Radius of convergence | $R=\infty$ |
| Region of validity | $\mathbb R$ |
Additional Notes
- Since all derivatives of $e^x$ are equal to $e^x$ itself, the Taylor expansion about any center $c$ has the identical shape $e^c\sum(x-c)^n/n!$ — recentering an exponential function never changes anything about the series structure except an overall multiplicative constant.
- This is the simplest possible instance of the "entire function" behavior seen throughout Problems 252–254 (hyperbolic sine, hyperbolic cosine, general exponential $a^x$) — all of them, being built from $e^x$, inherit the infinite radius of convergence.
- Verifying at $x=-1$: $e^{-1}=e^{-2}\sum1^n/n!=e^{-2}\cdot e=e^{-1}$ — a direct confirmation that recentering and re-expanding recovers the original function exactly.
Result
$$
e^x = e^{-2}\sum_{n=0}^{\infty}\frac{(x+2)^n}{n!}, \quad \text{valid for all } x\in\mathbb R.
$$