f(x)=coshx=sum_{n=0}^infty{x^{2n}}/{(2n)!}f(x)=coshx=∞∑n=0x2n(2n)!
Let us look at some details.
We already know
e^x=sum_{n=0}^infty x^n/{n!}ex=∞∑n=0xnn!
and
e^{-x}=sum_{n=0}^infty {(-x)^n}/{n!}e−x=∞∑n=0(−x)nn!,
so we have
f(x)=coshx=1/2(e^x+e^{-x})f(x)=coshx=12(ex+e−x)
=1/2(sum_{n=0}^infty x^n/{n!}+sum_{n=0}^infty (-x)^n/{n!})=12(∞∑n=0xnn!+∞∑n=0(−x)nn!)
=1/2sum_{n=0}^infty( x^n/{n!}+(-x)^n/{n!})=12∞∑n=0(xnn!+(−x)nn!)
since terms are zero when nn is odd,
=1/2sum_{n=0}^infty{2x^{2n}}/{(2n)!}=12∞∑n=02x2n(2n)!
by cancelling out 22's,
=sum_{n=0}^infty{x^{2n}}/{(2n)!}=∞∑n=0x2n(2n)!
