Taylor series of 1/(1 + x²) up to degree 2
The Taylor polynomial of degree 2 for 1/(1 + x²) about 0 (the Maclaurin polynomial) is 1 - x². 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 2:
- 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
\[ \frac{1}{1 + x^{2}} \approx 1 - x^{2} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = 0,\quad f^{(2)}(0) = -2 \]Degree 2 polynomial against the real 1/(1 + x²)
| x | Polynomial | Exact | Error |
|---|---|---|---|
| 0.1 | 0.99000000 | 0.99009901 | 9.90e-5 |
| 0.5 | 0.75000000 | 0.80000000 | 5.00e-2 |
| 0.9 | 0.19000000 | 0.55248619 | 3.62e-1 |
| 2 | -3.0000000 | 0.20000000 | 3.20e+0 |
Questions
What is the Taylor series of 1/(1 + x²)?
1 - x² + x⁴ - x⁶ + x⁸ + …
What is the degree 2 Taylor polynomial of 1/(1 + x²)?
1 - x²
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.