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

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

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1/(1 + x) and its degree 2 Taylor polynomial-1.5-1-0.50.511.5-10-55101/(1 + x)degree 2
1/(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) = 1/(1 + x) the first derivatives at 0 are:

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

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

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

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

xPolynomialExactError
0.10.910000000.909090919.09e-4
0.50.750000000.666666678.33e-2
0.90.910000000.526315793.84e-1
23.00000000.333333332.67e+0

Questions

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

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

What is the degree 2 Taylor polynomial of 1/(1 + x)?

1 - x + x²

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

1

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of 1/(1 + x)

Degree 2 of other functions

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