Taylor series of 1/(1 − x) up to degree 9
The Taylor polynomial of degree 9 for 1/(1 − x) about 0 (the Maclaurin polynomial) is 1 + x + x² + x³ + x⁴ + x⁵ + x⁶ + x⁷ + x⁸ + x⁹. The full series converges for |x| < 1.
How to work it out
- 1
The Taylor series of f about 0 uses the derivatives of f at 0:
- 2
For f(x) = 1/(1 − x) the first derivatives at 0 are:
- 3
Divide each by k! and keep the terms up to degree 9:
- 4
Radius of convergence: 1 (the geometric series). Inside it, more terms give a better approximation; outside, the series does not settle.
LaTeX source of the formulas on this page
\[ \frac{1}{1 - x} \approx 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} + x^{7} + x^{8} + x^{9} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = 2,\quad f^{(3)}(0) = 6,\quad f^{(4)}(0) = 24,\quad f^{(5)}(0) = 120,\quad f^{(6)}(0) = 720 \]Degree 9 polynomial against the real 1/(1 − x)
| x | Polynomial | Exact | Error |
|---|---|---|---|
| 0.1 | 1.1111111 | 1.1111111 | 1.11e-10 |
| 0.5 | 1.9980469 | 2.0000000 | 1.95e-3 |
| 0.9 | 6.5132156 | 10.000000 | 3.49e+0 |
| 2 | 1023.0000 | -1.0000000 | 1.02e+3 |
Questions
What is the Taylor series of 1/(1 − x)?
1 + x + x² + x³ + x⁴ + x⁵ + x⁶ + x⁷ + x⁸ + …
What is the degree 9 Taylor polynomial of 1/(1 − x)?
1 + x + x² + x³ + x⁴ + x⁵ + x⁶ + x⁷ + x⁸ + x⁹
What is the radius of convergence of the series of 1/(1 − x)?
1 (the geometric series)
What is the Maclaurin series?
A Taylor series centred at 0.