Question

解释一下，求具体过程

Answer 1

e^x =1 +x/1! +x^2/2!+....
x= x/2
e^(x/2) =1 +(x/2)/1! +(x/2)^2/2!+....
(x/2)e^(x/2) =(x/2) +(x/2)^2/1! +(x/2)^3/2!+....
=∑(n: 1->∞) [1/(n-1)!] (x/2)^n

Answer 2

解析：依题意可得，

e^x =1 +x/1! +x^2/2!+....x

= x/2e^(x/2)

=1 +(x/2)/1! +(x/2)^2/2!+.... (x/2)e^(x/2)

=(x/2) +(x/2)^2/1! +(x/2)^3/2!+.... 

=∑(n: 1->∞) [1/(n-1)!] (x/2)^n