Taylor series of e⁻ˣ up to degree 2
The Taylor polynomial of degree 2 for e⁻ˣ about 0 (the Maclaurin polynomial) is 1 - x + x²/2. The full series converges for every x.
How to work it out
- 1
The Taylor series of f about 0 uses the derivatives of f at 0:
- 2
For f(x) = e⁻ˣ the first derivatives at 0 are:
- 3
Divide each by k! and keep the terms up to degree 2:
- 4
Radius of convergence: ∞. Inside it, more terms give a better approximation; outside, the series does not settle.
LaTeX source of the formulas on this page
\[ e^{-x} \approx 1 - x + \frac{1}{2}x^{2} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = -1,\quad f^{(2)}(0) = 1 \]Degree 2 polynomial against the real e⁻ˣ
| x | Polynomial | Exact | Error |
|---|---|---|---|
| 0.1 | 0.90500000 | 0.90483742 | 1.63e-4 |
| 0.5 | 0.62500000 | 0.60653066 | 1.85e-2 |
| 0.9 | 0.50500000 | 0.40656966 | 9.84e-2 |
| 2 | 1.0000000 | 0.13533528 | 8.65e-1 |
Questions
What is the Taylor series of e⁻ˣ?
1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + x⁶/720 - x⁷/5040 + x⁸/40320 + …
What is the degree 2 Taylor polynomial of e⁻ˣ?
1 - x + x²/2
What is the radius of convergence of the series of e⁻ˣ?
∞
What is the Maclaurin series?
A Taylor series centred at 0.