Taylor series of 1/(1 − x) up to degree 1

The Taylor polynomial of degree 1 for 1/(1 − x) about 0 (the Maclaurin polynomial) is 1 + x. The full series converges for |x| < 1.

\frac{1}{1 - x} \approx 1 + x

1/(1 − x) and its degree 1 Taylor polynomial. The polynomial hugs the curve inside the radius of convergence (1) and leaves it outside.

How to work it out

1
The Taylor series of f about 0 uses the derivatives of f at 0:
$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k$
2
For f(x) = 1/(1 − x) the first derivatives at 0 are:
$f^{(0)}(0) = 1,\quad f^{(1)}(0) = 1$
3
Divide each by k! and keep the terms up to degree 1:
$\frac{1}{1 - x} \approx 1 + x$
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

\[ \frac{1}{1 - x} \approx 1 + x \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = 1 \]

x	Polynomial	Exact	Error
0.1	1.1000000	1.1111111	1.11e-2
0.5	1.5000000	2.0000000	5.00e-1
0.9	1.9000000	10.000000	8.10e+0
2	3.0000000	-1.0000000	4.00e+0

What is the Taylor series of 1/(1 − x)?

1 + x + x² + x³ + x⁴ + x⁵ + x⁶ + x⁷ + x⁸ + …

What is the degree 1 Taylor polynomial of 1/(1 − x)?

1 + x

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.