I am trying to prove the equivalence of the following two statement:
- $(a_i)_{0}^{\infty}$ and $(b_i)_{0}^{\infty}$ are infinite sequences of numbers, such that their elements are related in the following manner: $b_n=\sum_{k=0}^{n}{{n}\choose{k}}a_k$.
- $(a_i)_{0}^{\infty}$ and $(b_i)_{0}^{\infty}$ are infinite sequences of numbers, such that their generating functions are related in the following manner: $B(x)=e^{x}A(x)$
The way I tried to prove it. The j-th derivative of function of given power series is:
$G(x)=\sum_{k=0}^{\infty}g_{i}x^i \iff \frac{d^j}{dx^j}G(x)=\sum_{k=0}^{\infty}g_{i}\frac{d^j}{dx^j}x^i=\sum_{k=0}^{\infty}g_{i}\frac{i!}{(i-j)!}x^{i-j}=\sum_{k=j}^{\infty}g_{i}\frac{i!}{(i-j)!}x^{i-j}$
Therefore:
$H(x)=\frac{d^j}{dx^j}G(x) \iff h_k=g_{k+j}\frac{(k+j)!}{k!}$
Now consider j-th derivative of $e^{x}A(x)$, by the product rule:
$\frac{d^j}{dx^j}(e^xA(x))=\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)(e^x)^{(j-l)}=\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)e^x=e^x\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)$
Therefore:
$\frac{d^j}{dx^j}B(x)=e^x\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)$
By pluggin in $x=0$:
$\frac{d^j}{dx^j}B(0)=e^0\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(0)$
$\frac{d^j}{dx^j}B(0)=\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(0)$
When zero is plugged in, only the coefficient of the zeroth power remains, thus:
$j!b_j=\sum_{l=0}^{j}{{j}\choose{l}}l!a_l$
$b_j={\frac{1}{j!}}\sum_{l=0}^{j}{\frac{j!}{(j-l)!l!}}l!a_l$
$b_j=\sum_{l=0}^{j}{\frac{a_l}{(j-l)!}}$
$b_n=\sum_{l=0}^{n}{\frac{a_k}{(n-k)!}}$
Fromula is obviously wrong, what was my mistake?
It appears to me that the problem is talking about exponential generating functions. In that case
$$A(x)=\sum_{n\ge 0}a_n\frac{x^n}{n!}\;,$$
and
$$e^xA(x)=\left(\sum_{n\ge 0}\frac{x^n}{n!}\right)\left(\sum_{n\ge 0}a_n\frac{x^n}{n!}\right)\;.$$
The coefficient of $x^n$ in this product is
$$\sum_{k=0}^n\left(\frac{a_k}{k!}\cdot\frac1{(n-k)!}\right)=\sum_{k=0}^n\frac{a_k}{k!(n-k)!}\;,$$
so
$$\begin{align*} e^xA(x)&=\sum_{n\ge 0}\sum_{k=0}^n\frac{a_k}{k!(n-k)!}x^n\\ &=\sum_{n\ge 0}\sum_{k=0}^n\frac{n!a_k}{k!(n-k)!}x^n\\ &=\sum_{n\ge 0}\sum_{k=0}^n\binom{n}ka_kx^n\\ &=\sum_{n\ge 0}b_nx^n\;. \end{align*}$$