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:
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Radius of convergence: 1. 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 | 0.90909091 | 0.90909091 | 9.09e-11 |
| 0.5 | 0.66601563 | 0.66666667 | 6.51e-4 |
| 0.9 | 0.34280082 | 0.52631579 | 1.84e-1 |
| 2 | -341.00000 | 0.33333333 | 3.41e+2 |
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
What is the Maclaurin series?
A Taylor series centred at 0.