About the binomial sum?

Question

I have two concerns regarding this: $$a_n=\sum_{k=0}^n \binom{n}{k}b_k$$

First, should I say that $a_n$ is the binomial transform of $b_n$.

Second, how could I write $b_n$ in terms of $a_n$, sort of inversing this transform, if it is called so?

Answer 1

It is also convenient to consider exponential generating functions : \begin{align*} A(z)=\sum_{n=0}^\infty a_n\frac{z^n}{n!}\qquad B(z)=\sum_{n=0}^\infty b_n\frac{z^n}{n!} \end{align*} Since \begin{align*} \color{blue}{B(z)e^z}&=\sum_{k=0}^{\infty}b_n\frac{z^k}{k!}\sum_{l=0}^\infty\frac{z^l}{l!}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\binom{n}{k}b_k\right)\frac{z^n}{n!}\\ &=\sum_{n=0}^\infty a_n\frac{z^n}{n!}\\ &\,\,\color{blue}{=A(z)} \end{align*}

we obtain from \begin{align*} B(z)&=\sum_{n=0}^{\infty}\color{blue}{b_n}\frac{z^n}{n!}\\ &=A(z)e^{-z}\\ &=\sum_{k=0}^{\infty}a_k\frac{z^k}{k!}\sum_{l=0}^{\infty}(-1)^l\frac{z^l}{l!}\\ &=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n\binom{n}{k}a_k(-1)^{n-k}}\right)\frac{z^n}{n!} \end{align*} by comparing coefficients the wanted relationship.

Answer 2

Let $M:=\left(\binom{j}{i}\right)_{0\leqslant i,j\leqslant n}$, $B:=(b_0,\ldots,b_n)\in\mathbb{R}^{n+1}$ and $A:=(a_0,\ldots,a_n)\in\mathbb{R}^{n+1}$ then $A=MB$. $M$ is the matrix of the linear map $P(X)\in\mathbb{R}_n[X]\mapsto P(X+1)\in\mathbb{R}_n[X]$ in the basis $(1,X,\ldots,X^n)$. The inverse of this map is $P(X)\mapsto P(X-1)$ whose matrix in the same base is $\left((-1)^{j-i}\binom{j}{i}\right)_{0\leqslant i,j\leqslant n}=M^{-1}$, therefore $B=M^{-1}A$ which means that $$ b_n=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}a_k $$

Answer 3

Another proof: $$b_k=\sum_{j=0}^{k} (-1)^{k-j} {k \choose j} a_j$$ Then $$S=\sum_{k=0}^{n} {n \choose k} b_k= \sum_{k=0}^n \sum_{j=0}^{k} {n \choose k}{k \choose j} (-1)^{k-j} a_j$$ Use $${n \choose k}{k \choose j}={n \choose j} {n-j \choose k-j}$$ to get $$S=\sum_{k=0}^{n} {n \choose k} b_k= \sum_{k=0}^n \sum_{j=0}^{k} {n \choose j}{n-j\choose k-j} (-1)^{k-j} a_j$$ Let $k-j=p$, $$S=\sum_{k=0}^{n} {n \choose k} b_k=\sum_{j=0}^{k} {n \choose j} a_j \sum_{p=-j}^{n-j} {n-j\choose p} (-1)^{p}=\sum_{j=0}^{k} {n \choose j} a_j ~(1-1)^{n-j}.$$ $$\implies S=\sum_{j=0}^{k} {n \choose j} a_j ~\delta_{nj}=a_n$$