Taylor series of ln(1 + x) up to degree 9
The Taylor polynomial of degree 9 for ln(1 + x) about 0 (the Maclaurin polynomial) is x - x²/2 + x³/3 - x⁴/4 + x⁵/5 - x⁶/6 + x⁷/7 - x⁸/8 + x⁹/9. 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) = ln(1 + x) the first derivatives at 0 are:
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Divide each by k! and keep the terms up to degree 9:
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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
\[ \ln\left(1 + x\right) \approx x - \frac{1}{2}x^{2} + \frac{1}{3}x^{3} - \frac{1}{4}x^{4} + \frac{1}{5}x^{5} - \frac{1}{6}x^{6} + \frac{1}{7}x^{7} - \frac{1}{8}x^{8} + \frac{1}{9}x^{9} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 0,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = -1,\quad f^{(3)}(0) = 2,\quad f^{(4)}(0) = -6,\quad f^{(5)}(0) = 24,\quad f^{(6)}(0) = -120 \]Degree 9 polynomial against the real ln(1 + x)
| x | Polynomial | Exact | Error |
|---|---|---|---|
| 0.1 | 0.095310180 | 0.095310180 | 9.17e-12 |
| 0.5 | 0.40553230 | 0.40546511 | 6.72e-5 |
| 0.9 | 0.66106595 | 0.64185389 | 1.92e-2 |
| 2 | 37.574603 | 1.0986123 | 3.65e+1 |
Questions
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 9 Taylor polynomial of ln(1 + x)?
x - x²/2 + x³/3 - x⁴/4 + x⁵/5 - x⁶/6 + x⁷/7 - x⁸/8 + x⁹/9
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.