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Taylor series of sinh x up to degree 5

The Taylor polynomial of degree 5 for sinh x about 0 (the Maclaurin polynomial) is x + x³/6 + x⁵/120. The full series converges for every x.

sinh x and its degree 5 Taylor polynomial-3-2-1123-10-5510sinh xdegree 5
sinh x and its degree 5 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) = sinh x the first derivatives at 0 are:

    f(0)(0)=0,f(1)(0)=1,f(2)(0)=0,f(3)(0)=1,f(4)(0)=0,f(5)(0)=1f^{(0)}(0) = 0,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = 0,\quad f^{(3)}(0) = 1,\quad f^{(4)}(0) = 0,\quad f^{(5)}(0) = 1
  3. 3

    Divide each by k! and keep the terms up to degree 5:

    sinh⁡(x)≈x+16x3+1120x5\sinh\left(x\right) \approx x + \frac{1}{6}x^{3} + \frac{1}{120}x^{5}
  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
\[ \sinh\left(x\right) \approx x + \frac{1}{6}x^{3} + \frac{1}{120}x^{5} \]
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \]
\[ f^{(0)}(0) = 0,\quad f^{(1)}(0) = 1,\quad f^{(2)}(0) = 0,\quad f^{(3)}(0) = 1,\quad f^{(4)}(0) = 0,\quad f^{(5)}(0) = 1 \]

Degree 5 polynomial against the real sinh x

xPolynomialExactError
0.10.100166750.100166751.98e-11
0.50.521093750.521095311.56e-6
0.91.02642071.02651679.60e-5
23.60000003.62686042.69e-2

Questions

What is the Taylor series of sinh x?

x + x³/6 + x⁵/120 + x⁷/5040 + …

What is the degree 5 Taylor polynomial of sinh x?

x + x³/6 + x⁵/120

What is the radius of convergence of the series of sinh x?

∞

What is the Maclaurin series?

A Taylor series centred at 0.

Other degrees of sinh x

Degree 5 of other functions

    Taylor series of sinh x: degree 5 polynomial and error | WordGPT