求幂级数∑x^(2n)/(2n)!的和函数

consider Taylor expansion of e^x
e^x = 1 + x + x^2/2!+ ...+ x^n/n!+.(1)
e^(-x) = 1-x+x^2/2!+..;...+ (2)
(1)+(2)
e^(x) +e^(-x) = 2(1+x^2/2!+x^4/4!+...) = 2∑x^(2n)/(2n)!
∑x^(2n)/(2n)!=(1/2)(e^(x) +e^(-x))