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Taylor series of 1/(1 āˆ’ x) up to degree 3

The Taylor polynomial of degree 3 for 1/(1 āˆ’ x) about 0 (the Maclaurin polynomial) is 1 + x + xĀ² + xĀ³. The full series converges for |x| < 1.

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1/(1 āˆ’ x) and its degree 3 Taylor polynomial-1.5-1-0.50.511.5-10-55101/(1 āˆ’ x)degree 3
1/(1 āˆ’ x) and its degree 3 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)=āˆ‘k=0āˆžf(k)(0)k!ā€‰xkf(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k
  2. 2

    For f(x) = 1/(1 āˆ’ x) the first derivatives at 0 are:

    f(0)(0)=1,f(1)(0)=1,f(2)(0)=2,f(3)(0)=6f^{(0)}(0) = 1,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = 2,\quad f^{(3)}(0) = 6
  3. 3

    Divide each by k! and keep the terms up to degree 3:

    11āˆ’xā‰ˆ1+x+x2+x3\frac{1}{1 - x} \approx 1 + x + x^{2} + x^{3}
  4. 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
\[ \frac{1}{1 - x} \approx 1 + x + x^{2} + x^{3} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = 2,\quad f^{(3)}(0) = 6 \]

Degree 3 polynomial against the real 1/(1 āˆ’ x)

xPolynomialExactError
0.11.11100001.11111111.11e-4
0.51.87500002.00000001.25e-1
0.93.439000010.0000006.56e+0
215.000000-1.00000001.60e+1

Questions

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

1 + x + xĀ² + xĀ³ + xā“ + xāµ + xā¶ + xā· + xāø + ā€¦

What is the degree 3 Taylor polynomial of 1/(1 āˆ’ x)?

1 + x + xĀ² + xĀ³

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

1 (the geometric series)

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of 1/(1 āˆ’ x)

Degree 3 of other functions

    Taylor series of 1/(1 āˆ’ x): degree 3 polynomial and error | WordGPT