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Taylor series of ln(1 + x) up to degree 2

The Taylor polynomial of degree 2 for ln(1 + x) about 0 (the Maclaurin polynomial) is x - x²/2. The full series converges for |x| < 1.

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ln(1 + x) and its degree 2 Taylor polynomial-1.5-1-0.50.511.5-2.5-2-1.5-1-0.50.51ln(1 + x)degree 2
ln(1 + x) and its degree 2 Taylor polynomial. The polynomial hugs the curve inside the radius of convergence (1) and leaves it outside.

How to work it out

  1. 1

    The Taylor series of f about 0 uses the derivatives of f at 0:

    f(x)=∑k=0∞f(k)(0)k! xkf(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k
  2. 2

    For f(x) = ln(1 + x) the first derivatives at 0 are:

    f(0)(0)=0,f(1)(0)=1,f(2)(0)=−1f^{(0)}(0) = 0,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = -1
  3. 3

    Divide each by k! and keep the terms up to degree 2:

    ln⁡(1+x)≈x−12x2\ln\left(1 + x\right) \approx x - \frac{1}{2}x^{2}
  4. 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
\[ \ln\left(1 + x\right) \approx x - \frac{1}{2}x^{2} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 0,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = -1 \]

Degree 2 polynomial against the real ln(1 + x)

xPolynomialExactError
0.10.0950000000.0953101803.10e-4
0.50.375000000.405465113.05e-2
0.90.495000000.641853891.47e-1
20.00000001.09861231.10e+0

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 2 Taylor polynomial of ln(1 + x)?

x - x²/2

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.

Other degrees of ln(1 + x)

Degree 2 of other functions

    Taylor series of ln(1 + x): degree 2 polynomial and error | WordGPT