Writing a formula in terms of $a_m$

Question

I have derived a formula, which is:

$$a_m=\sum^m_{i=0}\binom{m-i}{n-i}b_ik^{m-i}, m\geq n$$

The main problem is to write the formula in terms of $a_m$, or means that we can find $a_m$ using $b_i$, can anyone help?

Answer 1

We can write OPs expression somewhat more symmetrically as \begin{align*} \frac{a_m}{k^m}=\sum_{i=0}^m\binom{m-i}{n-i}\frac{b_i}{k^i}\qquad\qquad m\geq n\tag{1} \end{align*}

Substituting $A_m:=\frac{a_m}{k^m}$ and $B_m:=\frac{b_m}{k^m}$ the expression (1) can be written as \begin{align*} A_m=\sum_{i=0}^n\binom{m-i}{m-n}B_i\qquad\qquad m\geq n\tag{2} \end{align*} where we set the upper index of the sum to $n$ since we have from (1) that $\binom{m-i}{n-i}=0$ if $i>n$. We also use $\binom{p}{q}=\binom{p}{p-q}$.

We want to express the $B_m$ as sum of $A_k$, which means to find the binomial inverse pair. We show the validity of a strongly related expression of (2). Maybe there is just a typo in OP's expression and he is actually looking for the relation we are going to show.

We have the following binomial inverse pair for $m\geq n$: \begin{align*} \color{blue}{A_n=\sum_{i=0}^n\binom{m-i}{m-n}B_i\qquad\qquad B_n=\sum_{i=0}^n(-1)^{n-i}\binom{m-i}{m-n}A_i}\tag{3} \end{align*}

This pair is stated as formula (4a) in section 2.2 of Binomial Identities by J. Riordan. It is one representantive of a group of binomial identities with prototype: \begin{align*} \alpha_n=\sum_{i=n}^m\binom{i}{n}\beta_i\qquad\qquad\beta_n=\sum_{i=n}^m(-1)^{n-i}\binom{i}{n}\alpha_i\tag{4} \end{align*} We show the validity of the protoype (4) and transform it afterwards until we obtain the binomial inverse pair (3).

We consider generating functions : \begin{align*} A(z)=\sum_{n=0}^m\alpha_n z^n\qquad\qquad B(z)=\sum_{n=0}^m\beta_n z^n \end{align*} which are related by $A(z)=B(z+1)$. From this relation we have on one hand

\begin{align*} A(z)&=\sum_{n=0}^m\color{blue}{\alpha_n }z^n=\sum_{i=0}^m\beta_i(z+1)^i\tag{4.1}\\ &=\sum_{i=0}^m\beta_i\sum_{n=0}^i\binom{i}{n}z^n\tag{4.2}\\ &=\sum_{n=0}^m\left(\color{blue}{\sum_{i=n}^m\binom{i}{n}\beta_i}\right)z^n\tag{4.3} \end{align*} On the other hand we obtain from $B(z)=A(z-1)$ \begin{align*} B(z)&=\sum_{n=0}^m\color{blue}{\beta_n }z^n=\sum_{i=0}^m\alpha_i(z-1)^i\\ &=\sum_{i=0}^m\alpha_i\sum_{n=0}^i(-1)^{i-n}\binom{i}{n}z^n\\ &=\sum_{n=0}^m\left(\color{blue}{\sum_{i=n}^m(-1)^{i-n}\binom{i}{n}\alpha_i}\right)z^n \end{align*} showing the validity of (4).

Comment:

• In (4.1) we use the relation $A(z)=B(z+1)$.

• In (4.2) we expand the binomial.

• In (4.3) we change the order of summation. Note we have $\sum_{0\leq n\leq i \leq m}\binom{i}{n}\beta_iz^n$.

In order to prove (3) we substitute in (4) $\gamma_n:=\alpha_{m-n}$ and $\delta_n:=\beta_{m-n}$. We obtain \begin{align*} \alpha_n&=\sum_{i=n}^m\binom{i}{n}\beta_i\\ \\ \color{blue}{\gamma_n}&=\sum_{i=m-n}^m\binom{i}{m-n}\beta_i\tag{3.1}\\ &=\sum_{i=0}^{n}\binom{m-n+i}{m-n}\beta_{m-n+i}\tag{3.2}\\ &=\sum_{i=0}^{n}\binom{m-i}{m-n}\beta_{m-i}\tag{3.3}\\ &\,\,\color{blue}{=\sum_{i=0}^n\binom{m-i}{m-n}\delta_i}\tag{3.4} \end{align*} Similarly we obtain \begin{align*} \beta_n&=\sum_{i=n}^m(-1)^{n-i}\binom{i}{n}\alpha_i\\ \\ \color{blue}{\delta_n}&=\sum_{i=m-n}^m(-1)^{m-n-i}\binom{i}{m-n}\alpha_i\\ &=\sum_{i=0}^n(-1)^i\binom{m-n+i}{m-n}\alpha_{m-n+i}\\ &=\sum_{i=0}^n(-1)^{n-i}\binom{m-i}{m-n}\alpha_{m-i}\\ &\,\,\color{blue}{=\sum_{i=0}^n(-1)^{n-i}\binom{m-k}{m-n}\gamma_i} \end{align*} showing the validity of (3).

Comment:

• In (3.1) we substitute $n$ by $m-n$ and use $\gamma_n:=\alpha_{m-n}$.

• In (3.2) we shift the index $i$ to start with $i=0$.

• In (3.3) we change the order of summation $i\to n-i$.

• In (3.4) we use $\delta_n:=\beta_{m-n}$.