Taylor series of ln(1 + x) up to degree 1

1 (converges for −1 < x ≤ 1)

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

\ln\left(1 + x\right) \approx x

ln(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) = ln(1 + x) the first derivatives at 0 are:
$f^{(0)}(0) = 0,\quad f^{(1)}(0) = 1$
3
Divide each by k! and keep the terms up to degree 1:
$\ln\left(1 + x\right) \approx x$
4
Radius of convergence: 1 (converges for −1 < x ≤ 1). Inside it, more terms give a better approximation; outside, the series does not settle.

LaTeX source of the formulas on this page

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

x	Polynomial	Exact	Error
0.1	0.10000000	0.095310180	4.69e-3
0.5	0.50000000	0.40546511	9.45e-2
0.9	0.90000000	0.64185389	2.58e-1
2	2.0000000	1.0986123	9.01e-1

What is the Taylor series of ln(1 + x)?

x - x²/2 + x³/3 - x⁴/4 + x⁵/5 - x⁶/6 + x⁷/7 - x⁸/8 + …

What is the degree 1 Taylor polynomial of ln(1 + x)?

What is the radius of convergence of the series of ln(1 + x)?

1 (converges for −1 < x ≤ 1)

What is the Maclaurin series?

A Taylor series centred at 0.