How do I calculate the sum $\sum_{i=0}^n (-1)^{n-i}{{n}\choose{i}}f(i)$ if I know that $\sum_{n\geq0} f(n)\frac{x^n}{n!} = e^{x+\frac{x^2}{2}}$?

Question

I have a generating function $\sum_{n\geq0} f(n)\frac{x^n}{n!} = e^{x+\frac{x^2}{2}}$
And I need to compute the sum:
$$\sum_{i=0}^n (-1)^{n-i}{{n}\choose{i}}f(i) =\ ?$$

I understand that this sum is a coefficient $c_n$ from $\sum_{n\geq0}c_n\frac{x^n}{n!} = \Big( \sum_{n\geq0}a_n\frac{x^n}{n!} \Big)\Big( \sum_{i\geq0}b_i\frac{x^i}{i!} \Big) = \Big( \sum_{n\geq0}(-1)^n\frac{x^n}{n!} \Big)\Big( \sum_{i\geq0}f(i)\frac{x^i}{i!} \Big) = (*)$

Because $\sum_{n\geq0}c_nx^n = \Big( \sum_{n\geq0}a_nx^n \Big)\Big( \sum_{n\geq0}b_nx^n\Big)$, where $c_n=\sum_{i=0}^n a_ib_{n-i}$ and $(-1)^{-i}=(-1)^{i}$

But I don't understant:
1) Whether it's correct to move $(-1)^n$ in the $(-x)^n$ and get the exponent:
$$\sum_{n\geq0}(-1)^n\frac{x^n}{n!} = \sum_{n\geq0}\frac{(-x)^n}{n!} = e^{-x}$$

2) If it's correct - why?

3) If I use this formula in $(*)$ and get $\Big( \sum_{n\geq0}(-1)^n\frac{x^n}{n!} \Big)\Big( \sum_{i\geq0}f(i)\frac{x^i}{i!} \Big) = e^{-x+x+\frac{x^2}{2}} = e^{\frac{x^2}{2}}$
How do I get the $c_n$ component in the series: $\sum_{n\geq0}c_n\frac{x^n}{n!} = e^{\frac{x^2}{2}}$ ?

Answer 1

If $$ F(x) = \sum_{n\geq 0}f(n)\frac{x^n}{n!},\qquad e^{-x}=\sum_{n\geq 0}(-1)^n\frac{x^n}{n!} \tag{1}$$ then by Cauchy product $$ e^{-x} F(x) = \sum_{m\geq 0}\left(\sum_{m=0}^{n}\binom{n}{m}\,f(m)(-1)^{n-m}\right)\frac{x^n}{n!}\tag{2}$$ hence if we know in advance that $F(x)=e^{x+x^2/2}$ it follows that: $$ \sum_{m=0}^{n}\binom{n}{m}\,f(m)(-1)^{n-m} = \left[\frac{x^n}{n!}\right]e^{x^2/2} = \left\{\begin{array}{rcl}0&\text{if}&n\text{ is odd}\\\frac{n!}{2^{n/2}(n/2)!}&\text{if}&n\text{ is even}\end{array}\right.\tag{3}$$ since $e^{x^2/2}=\sum_{m\geq 0}\frac{x^{2m}}{2^m m!}.$