Question

How do you find the Maclaurin series of `f(x)=cosh(x)` ?

Answer 1

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

Let us look at some details.

We already know

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

and

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

so we have

`f(x)=coshx=1/2(e^x+e^{-x})`

`=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)!}`