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Taylor series of e²ˣ up to degree 4

The Taylor polynomial of degree 4 for e²ˣ about 0 (the Maclaurin polynomial) is 1 + 2x + 2x² + 4x³/3 + 2x⁴/3. The full series converges for every x.

e²ˣ and its degree 4 Taylor polynomial-3-2-1123100200300400e²ˣdegree 4
e²ˣ 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) = e²ˣ the first derivatives at 0 are:

    f(0)(0)=1,f(1)(0)=2,f(2)(0)=4,f(3)(0)=8,f(4)(0)=16f^{(0)}(0) = 1,\quad f^{(1)}(0) = 2,\quad f^{(2)}(0) = 4,\quad f^{(3)}(0) = 8,\quad f^{(4)}(0) = 16
  3. 3

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

    e2x≈1+2x+2x2+43x3+23x4e^{2 x} \approx 1 + 2x + 2x^{2} + \frac{4}{3}x^{3} + \frac{2}{3}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
\[ e^{2 x} \approx 1 + 2x + 2x^{2} + \frac{4}{3}x^{3} + \frac{2}{3}x^{4} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 1,\quad f^{(1)}(0) = 2,\quad f^{(2)}(0) = 4,\quad f^{(3)}(0) = 8,\quad f^{(4)}(0) = 16 \]

Degree 4 polynomial against the real e²ˣ

xPolynomialExactError
0.11.22140001.22140282.76e-6
0.52.70833332.71828189.95e-3
0.95.82940006.04964752.20e-1
234.33333354.5981502.03e+1

Questions

What is the Taylor series of e²ˣ?

1 + 2x + 2x² + 4x³/3 + 2x⁴/3 + 4x⁵/15 + 4x⁶/45 + 8x⁷/315 + 2x⁸/315 + …

What is the degree 4 Taylor polynomial of e²ˣ?

1 + 2x + 2x² + 4x³/3 + 2x⁴/3

What is the radius of convergence of the series of e²ˣ?

∞

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of e²ˣ

Degree 4 of other functions

    Taylor series of e²ˣ: degree 4 polynomial and error | WordGPT