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

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Answer 1 · 2014-09-25T19:42:41

#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)!}#