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The binomial transform states that if one has two real sequences $\{a_k \}$ and $\{b_k \}$ satisfying $b_n = \sum_{k = 0}^{n} \binom{n}{k} a_k$, then $a_n = \sum_{k = 0}^{n} (-1)^{n-k} \binom{n}{k} b_k$. Such a transform has many higher-order generalizations.

Is there a related transform for real, double sequences $\{ a_{k,l} \}$ and $\{ b_{k, l} \}$ satisfying \begin{align} b_{n,m} = \sum_{k = 0}^{n} \sum_{l = 0}^{m} \binom{n}{k} \binom{m}{l} B_{n-k} B_{m-l} a_{k,l}, \end{align} where $B_{n}$ is the $n^{\text{th}}$-Bernoulli number? Presumably, this can be derived from the single-sum specialization, \begin{align} b_n & = \sum_{k = 0}^{n} \binom{n}{k} B_{n -k} a_{k}. \end{align}

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1 Answer 1

This calls for a double application of a standard inversion formula involving Bernoulli numbers. I will use exponential generating functions to explain it.

Let $$\begin{align} f(x,y) &= \sum_{k=0}^\infty \sum_{l=0}^\infty a_{k,l}\frac{x^k}{k!}\frac{y^l}{l!} & g(x,y) &= \sum_{n=0}^\infty \sum_{m=0}^\infty b_{n,m}\frac{x^n}{n!}\frac{y^m}{m!} \end{align}$$ Using $$\begin{align} \frac{z}{\mathrm{e}^{z}-1} &= \sum_{n=0}^\infty \operatorname{B}_n\frac{z^n}{n!} & \frac{\mathrm{e}^{z}-1}{z} &= \sum_{k=0}^\infty \frac{1}{k+1}\frac{z^k}{k!} \end{align}$$ we find $$\begin{align} g(x,y) &= \frac{x}{\mathrm{e}^x-1}\frac{y}{\mathrm{e}^y-1}f(x,y) &\Leftrightarrow\qquad b_{n,m} &= \sum_{k = 0}^{n} \sum_{l = 0}^{m} \binom{n}{k} \binom{m}{l} \operatorname{B}_{n-k} \operatorname{B}_{m-l} a_{k,l} \\&\Updownarrow\\ f(x,y) &= \frac{\mathrm{e}^x-1}{x}\frac{\mathrm{e}^y-1}{y}g(x,y) &\Leftrightarrow\qquad a_{n,m} &= \sum_{k = 0}^{n} \sum_{l = 0}^{m} \binom{n}{k} \binom{m}{l} \frac{1}{n-k+1} \frac{1}{m-l+1} b_{k,l} \end{align}$$ And that's it.

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That's great. Presumably, infinitely many such transforms follow from two generating functions $E$ and $F$ such that $EF = 1$. –  user02138 Jan 6 at 21:50
    
Indeed. Here the only coupling of the sums is via $a_{k,l}$. Such cases can easily be generalized to any depth. –  ccorn Jan 6 at 22:07

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