Taylor series of 1/(1 + x²) up to degree 10
The Taylor polynomial of degree 10 for 1/(1 + x²) about 0 (the Maclaurin polynomial) is 1 - x² + x⁴ - x⁶ + x⁸ - x¹⁰. 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 + x²) the first derivatives at 0 are:
- 3
Divide each by k! and keep the terms up to degree 10:
- 4
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
\[ \frac{1}{1 + x^{2}} \approx 1 - x^{2} + x^{4} - x^{6} + x^{8} - x^{10} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = 0,\quad f^{(2)}(0) = -2,\quad f^{(3)}(0) = 0,\quad f^{(4)}(0) = 24,\quad f^{(5)}(0) = 0,\quad f^{(6)}(0) = -720 \]Degree 10 polynomial against the real 1/(1 + x²)
| x | Polynomial | Exact | Error |
|---|---|---|---|
| 0.1 | 0.99009901 | 0.99009901 | 9.90e-13 |
| 0.5 | 0.79980469 | 0.80000000 | 1.95e-4 |
| 0.9 | 0.39644777 | 0.55248619 | 1.56e-1 |
| 2 | -819.00000 | 0.20000000 | 8.19e+2 |
Questions
What is the Taylor series of 1/(1 + x²)?
1 - x² + x⁴ - x⁶ + x⁸ + …
What is the degree 10 Taylor polynomial of 1/(1 + x²)?
1 - x² + x⁴ - x⁶ + x⁸ - x¹⁰
What is the radius of convergence of the series of 1/(1 + x²)?
1
What is the Maclaurin series?
A Taylor series centred at 0.