Binomial Coefficient

If $b_n=\sum_{k=0}^{n}(-1)^k\binom{n}{k} a_k \forall n\in\mathbb{N_0},$ then prove that $a_n= \sum_{k=0}^{n}(-1)^k\binom{n}{k} b_k \forall n\in\mathbb{N_0}.$

My Working:

let $f(x)=a_0+a_1x+a_2x^2+\cdots;g(x)=b_0+b_1x+b_2x^2+\cdots$

$\implies g(x)=f(x)(x-1)^n$

$\implies f(x)=g(x)(x-1)^{-n}$

Solution 1:

The binomial coefficients in the sum representation of $a_n, b_n$ indicate that exponential generating functions could be employed.

We recall the product of two exponential generating functions $U(x)=\sum_{n=0}^{\infty}u_n\frac{x^n}{n!}, W(x)=\sum_{n=0}^\infty w_n\frac{x^n}{n!}$ is \begin{align*} U(x)W(x)&=\left(\sum_{k=0}^{\infty}u_k\frac{x^k}{l!}\right)\left(\sum_{l=0}^\infty w_l\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{u_k}{k!}\frac{w_l}{l!}\right)x^n\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{u_k}{k!}\frac{w_{n-k}}{(n-k)!}\right)x^n\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{n!}{k!(n-k)!}u_kw_{n-k}\right)\frac{x^n}{n!}\\ &=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n \binom{n}{k}u_kw_{n-k}}\right)\frac{x^n}{n!}\\ \end{align*}

Hint: Consider $A(x)=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}, B(x)=\sum_{n=0}^{\infty}b_n\frac{x^n}{n!}$ and check a functional identity of following kind: \begin{align*} \color{blue}{B(x) e^x=A(x)\qquad \longleftrightarrow \qquad B(x)=A(x)e^{-x}} \end{align*} Note $A(x)$ and $B(x)$ might be somewhat adapted in order to fit your needs.