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Taylor series of e⁻ˣ up to degree 4

The Taylor polynomial of degree 4 for e⁻ˣ about 0 (the Maclaurin polynomial) is 1 - x + x²/2 - x³/6 + x⁴/24. The full series converges for every x.

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e⁻ˣ and its degree 4 Taylor polynomial-3-2-11235101520e⁻ˣdegree 4
e⁻ˣ and its degree 4 Taylor polynomial. The polynomial follows the curve closely near 0 and drifts away further out; more terms stretch the good region.

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) = e⁻ˣ the first derivatives at 0 are:

    f(0)(0)=1,f(1)(0)=−1,f(2)(0)=1,f(3)(0)=−1,f(4)(0)=1f^{(0)}(0) = 1,\quad f^{(1)}(0) = -1,\quad f^{(2)}(0) = 1,\quad f^{(3)}(0) = -1,\quad f^{(4)}(0) = 1
  3. 3

    Divide each by k! and keep the terms up to degree 4:

    e−x≈1−x+12x2−16x3+124x4e^{-x} \approx 1 - x + \frac{1}{2}x^{2} - \frac{1}{6}x^{3} + \frac{1}{24}x^{4}
  4. 4

    Radius of convergence: ∞. Inside it, more terms give a better approximation; outside, the series does not settle.

LaTeX source of the formulas on this page
\[ e^{-x} \approx 1 - x + \frac{1}{2}x^{2} - \frac{1}{6}x^{3} + \frac{1}{24}x^{4} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = -1,\quad f^{(2)}(0) = 1,\quad f^{(3)}(0) = -1,\quad f^{(4)}(0) = 1 \]

Degree 4 polynomial against the real e⁻ˣ

xPolynomialExactError
0.10.904837500.904837428.20e-8
0.50.606770830.606530662.40e-4
0.90.410837500.406569664.27e-3
20.333333330.135335281.98e-1

Questions

What is the Taylor series of e⁻ˣ?

1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + x⁶/720 - x⁷/5040 + x⁸/40320 + …

What is the degree 4 Taylor polynomial of e⁻ˣ?

1 - x + x²/2 - x³/6 + x⁴/24

What is the radius of convergence of the series of e⁻ˣ?

∞

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of e⁻ˣ

Degree 4 of other functions

    Taylor series of e⁻ˣ: degree 4 polynomial and error | WordGPT