# Constructing a Maclaurin Series

## Key Questions

• f(x)=coshx=sum_{n=0}^infty{x^{2n}}/{(2n)!}

Let us look at some details.

e^x=sum_{n=0}^infty x^n/{n!}

and

e^{-x}=sum_{n=0}^infty {(-x)^n}/{n!},

so we have

$f \left(x\right) = \cosh x = \frac{1}{2} \left({e}^{x} + {e}^{- x}\right)$

=1/2(sum_{n=0}^infty x^n/{n!}+sum_{n=0}^infty (-x)^n/{n!})

=1/2sum_{n=0}^infty( x^n/{n!}+(-x)^n/{n!})

since terms are zero when $n$ is odd,

=1/2sum_{n=0}^infty{2x^{2n}}/{(2n)!}

by cancelling out $2$'s,

=sum_{n=0}^infty{x^{2n}}/{(2n)!}

• The MacLaurin series of a function $f$ is a power series of the form:

${\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$

With the coefficients ${a}_{n}$ given by the relation

a_n=(f^((n))(0))/(n!),

where ${f}^{\left(n\right)} \left(0\right)$ is the $n$th derivative of $f \left(x\right)$ evaluated at $x = 0$.

Therefore,

f^((n))(0)=a_n n!

This reasoning can be extended to Taylor series around ${x}_{0}$, of the form:

${\sum}_{n = 0}^{\infty} {c}_{n} {\left(x - {x}_{0}\right)}^{n}$

With the relation

f^((n))(x_0)=c_n n!

It's important to emphasize that the function $n$th derivative of $f$ (that is, ${f}^{\left(n\right)} \left(x\right)$) cannot be obtained directly from the Taylor/MacLaurin series (only it's value on the point around wich the series is constructed).