Taylor series of e⁻ˣ up to degree 9
The Taylor polynomial of degree 9 for e⁻ˣ about 0 (the Maclaurin polynomial) is 1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + x⁶/720 - x⁷/5040 + x⁸/40320 - x⁹/362880. 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 9:
- 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
\[ e^{-x} \approx 1 - x + \frac{1}{2}x^{2} - \frac{1}{6}x^{3} + \frac{1}{24}x^{4} - \frac{1}{120}x^{5} + \frac{1}{720}x^{6} - \frac{1}{5040}x^{7} + \frac{1}{40320}x^{8} - \frac{1}{362880}x^{9} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = -1,\quad f^{(2)}(0) = 1,\quad f^{(3)}(0) = -1,\quad f^{(4)}(0) = 1,\quad f^{(5)}(0) = -1,\quad f^{(6)}(0) = 1 \]Degree 9 polynomial against the real e⁻ˣ
| x | Polynomial | Exact | Error |
|---|---|---|---|
| 0.1 | 0.90483742 | 0.90483742 | 1.11e-16 |
| 0.5 | 0.60653066 | 0.60653066 | 2.57e-10 |
| 0.9 | 0.40656957 | 0.40656966 | 8.88e-8 |
| 2 | 0.13509700 | 0.13533528 | 2.38e-4 |
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 9 Taylor polynomial of e⁻ˣ?
1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + x⁶/720 - x⁷/5040 + x⁸/40320 - x⁹/362880
What is the radius of convergence of the series of e⁻ˣ?
∞
What is the Maclaurin series?
A Taylor series centred at 0.