# How do you find the Taylor polynomial of degree 4 for f(x) = cosh(x) about x = 0  by using the given Taylor polynomial for e^x?

May 9, 2017

$\cosh x = 1 + {x}^{2} / 2 + {x}^{4} / 24 + o \left({x}^{4}\right)$

#### Explanation:

The MacLaurin series for ${e}^{x}$ is:

e^x = sum_(n=0)^oo x^n/(n!)

Truncating the series at $n = 4$ we obtain the Taylor polynomial of degree 4:

${e}^{x} = 1 + x + {x}^{2} / 2 + {x}^{3} / 6 + {x}^{4} / 24 + o \left({x}^{4}\right)$

and substituting $- x$ we have:

${e}^{- x} = 1 - x + {x}^{2} / 2 - {x}^{3} / 6 + {x}^{4} / 24 + o \left({x}^{4}\right)$

Using the exponential form of $\cosh x$ then we have:

$\cosh x = \frac{{e}^{x} + {e}^{- x}}{2} = \frac{1}{2} \left(1 + x + {x}^{2} / 2 + {x}^{3} / 6 + {x}^{4} / 24 + 1 - x + {x}^{2} / 2 - {x}^{3} / 6 + {x}^{4} / 24\right) + o \left({x}^{4}\right)$

and we can see that the terms of even order cancel each other, so that:

$\cosh x = 1 + {x}^{2} / 2 + {x}^{4} / 24 + o \left({x}^{4}\right)$