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I have recently encountered the Binomial transform:

The binomial transform, $T$, of a sequence, ${a_n}$, is the sequence ${s_n}$ defined by

$$ s_{n}=\sum _{k=0}^{n}(-1)^{k}{n \choose k}a_{k} $$

This transformation has the neat property that it's self inverse. I am now trying to find some simple applications or exercise on this particular tool.

Can it be used à la Fourier Transform applying it to both sides of an equation to simplify calculations and then inverting it back one finds the solution? Or really any application would be interesting to me.

Monolite
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  • Here it is an interesting application (it dates back to Newton): https://math.stackexchange.com/questions/2514512/why-does-sum-n-0-infty-fracn2n1-converge-to-frac-pi2/2514520#2514520 – Jack D'Aurizio Jan 06 '18 at 18:43
  • There also is https://math.stackexchange.com/questions/2384932/a-nice-but-somewhat-challenging-binomial-identity/2385879#2385879 but I wouldn't call it simple. – Jack D'Aurizio Jan 06 '18 at 19:21

2 Answers2

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The binomial transform is usually employed for the acceleration of series or, in the opposite direction, for simplifying the structure of the terms of a hypergeometric series (or twisted hypergeometric series, according to the terminology introduced here). Besides the well-known application I mentioned in the comments, it also leads to interesting identities related to Euler sums, like the following one: $$ \zeta(2) = \sum_{n\geq 1}\frac{1}{n^2} = \sum_{n\geq 1}\frac{H_n}{n 2^{n-1}}.$$ Further examples can be found in the first pages of my notes.

Jack D'Aurizio
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In the context of the OEIS the binomial transform is defined by $\;B(a_n):=\sum_{k=0}^n{n \choose k}\:a_k.$ This transform can be iterated. The signed transform defined by $S(a_n):=(-1)^n\:a_n\;$ can be composed with the binomial transform (either before or after) to produce a self-inverse transformation (or an involution) such as the one you proposed.

As an example, $T(L_n) = L_n$ where $L$ is the Lucas number sequence. Searching the OEIS yields many other examples of binomial transforms combined with or without the signed transform.

EDIT (Oct 6, 2022)

Given a sequence $\,\{a_n\},\,$ your transform is a special case of the two parameter transform $$ T_{m,b}(\{a_n\}) := \sum_{k=0}^n\, m^k\,b^{n-k}\,{n \choose k}\:a_k. $$ The binomial theorem is equivalent to $\, T_{m,b}(\{x^n\}) = \{(mx+b)^n\}.\,$ Your transform is $\,T_{-1,1}\,$ which corresponds to the involution $\,x\to 1-x.\,$ The OEIS binomial transform is $\,T_{1,1}\,$ which corresponds to $\,x\to 1+x\,$ (a shift) while the $\,S\,$ transform corresponds to $\,x\to-x.\,$

Somos
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