# Binomial transform

In combinatorics, the binomial transform is a sequence transformation (i.e., a transform of a sequence) that computes its forward differences. It is closely related to the Euler transform, which is the result of applying the binomial transform to the sequence associated with its ordinary generating function.

## Definition

The binomial transform, T, of a sequence, {an}, is the sequence {sn} defined by

$s_n = \sum_{k=0}^n (-1)^k {n\choose k} a_k.$

Formally, one may write (Ta)n = sn for the transformation, where T is an infinite-dimensional operator with matrix elements Tnk:

$s_n = (Ta)_n = \sum_{k=0}^\infty T_{nk} a_k.$

The transform is an involution, that is,

$TT = 1 \,$

or, using index notation,

$\sum_{k=0}^\infty T_{nk}T_{km} = \delta_{nm}$

where δ is the Kronecker delta function. The original series can be regained by

$a_n=\sum_{k=0}^n (-1)^k {n\choose k} s_k.$

The binomial transform of a sequence is just the nth forward differences of the sequence, with odd differences carrying a negative sign, namely:

$s_0 = a_0$
$s_1 = - (\triangle a)_0 = -a_1+a_0$
$s_2 = (\triangle^2 a)_0 = -(-a_2+a_1)+(-a_1+a_0) = a_2-2a_1+a_0$
$\vdots\,$
$s_n = (-1)^n (\triangle^n a)_0$

where Δ is the forward difference operator.

Some authors define the binomial transform with an extra sign, so that it is not self-inverse:

$t_n=\sum_{k=0}^n (-1)^{n-k} {n\choose k} a_k$

whose inverse is

$a_n=\sum_{k=0}^n {n\choose k} t_k.$

## Example

Binomial transforms can be seen in difference tables. Consider the following:

 0 1 10 63 324 1485 1 9 53 261 1161 8 44 208 900 36 164 692 128 528 400

The top line 0, 1, 10, 63, 324, 1485,... (a sequence defined by (2n2 + n)3n − 2) is the (noninvolutive version of the) binomial transform of the diagonal 0, 1, 8, 36, 128, 400,... (a sequence defined by n22n − 1).

## Shift states

The binomial transform is the shift operator for the Bell numbers. That is,

$B_{n+1}=\sum_{k=0}^n {n\choose k} B_k$

where the Bn are the Bell numbers.

## Ordinary generating function

The transform connects the generating functions associated with the series. For the ordinary generating function, let

$f(x)=\sum_{n=0}^\infty a_n x^n$

and

$g(x)=\sum_{n=0}^\infty s_n x^n$

then

$g(x) = (Tf)(x) = \frac{1}{1-x} f\left(\frac{x}{x-1}\right).$

## Euler transform

The relationship between the ordinary generating functions is sometimes called the Euler transform. It commonly makes its appearance in one of two different ways. In one form, it is used to accelerate the convergence of an alternating series. That is, one has the identity

$\sum_{n=0}^\infty (-1)^n a_n = \sum_{n=0}^\infty (-1)^n \frac {\Delta^n a_0} {2^{n+1}}$

which is obtained by substituting x=1/2 into the last formula above. The terms on the right hand side typically become much smaller, much more rapidly, thus allowing rapid numerical summation.

The Euler transform can be generalized (Borisov B. and Shkodrov V., 2007):

$\sum_{n=0}^\infty (-1)^n {n+p\choose n} a_n = \sum_{n=0}^\infty (-1)^n {n+p\choose n}\frac {\Delta^n a_0} {2^{n+p+1}}$,

where p = 0, 1, 2,...

The Euler transform is also frequently applied to the Euler hypergeometric integral $\,_2F_1$. Here, the Euler transform takes the form:

$\,_2F_1 (a,b;c;z) = (1-z)^{-b} \,_2F_1 \left(c-a, b; c;\frac{z}{z-1}\right).$

The binomial transform, and its variation as the Euler transform, is notable for its connection to the continued fraction representation of a number. Let $0 < x < 1$ have the continued fraction representation

$x=[0;a_1, a_2, a_3,\cdots]$

then

$\frac{x}{1-x}=[0;a_1-1, a_2, a_3,\cdots]$

and

$\frac{x}{1+x}=[0;a_1+1, a_2, a_3,\cdots].$

## Exponential generating function

For the exponential generating function, let

$\overline{f}(x)= \sum_{n=0}^\infty a_n \frac{x^n}{n!}$

and

$\overline{g}(x)= \sum_{n=0}^\infty s_n \frac{x^n}{n!}$

then

$\overline{g}(x) = (T\overline{f})(x) = e^x \overline{f}(-x).$

The Borel transform will convert the ordinary generating function to the exponential generating function.

## Integral representation

When the sequence can be interpolated by a complex analytic function, then the binomial transform of the sequence can be represented by means of a Nörlund–Rice integral on the interpolating function.

## Generalizations

Prodinger gives a related, modular-like transformation: letting

$u_n = \sum_{k=0}^n {n\choose k} a^k (-c)^{n-k} b_k$

gives

$U(x) = \frac{1}{cx+1} B\left(\frac{ax}{cx+1}\right)$

where U and B are the ordinary generating functions associated with the series $\{u_n\}$ and $\{b_n\}$, respectively.

The rising k-binomial transform is sometimes defined as

$\sum_{j=0}^n {n\choose j} j^k a_j.$

The falling k-binomial transform is

$\sum_{j=0}^n {n\choose j} j^{n-k} a_j$.

Both are homomorphisms of the kernel of the Hankel transform of a series.

In the case where the binomial transform is defined as

$\sum_{i=0}^n(-1)^{n-i}\binom{n}{i}a_i=b_n.$

Let this be equal to the function $\mathfrak J(a)_n=b_n.$

If a new forward difference table is made and the first elements from each row of this table are taken to form a new sequence $\{b_n\}$, then the second binomial transform of the original sequence is,

$\mathfrak J^2(a)_n=\sum_{i=0}^n(-2)^{n-i}\binom{n}{i}a_i.$

If the same process is repeated k times, then it follows that,

$\mathfrak J^k(a)_n=b_n=\sum_{i=0}^n(-k)^{n-i}\binom{n}{i}a_i.$

Its inverse is,

$\mathfrak J^{-k}(b)_n=a_n=\sum_{i=0}^nk^{n-i}\binom{n}{i}b_i.$

This can be generalized as,

$\mathfrak J^k(a)_n=b_n=(\mathbf E-k)^na_0$

where $\mathbf E$ is the shift operator.

Its inverse is

$\mathfrak J^{-k}(b)_n=a_n=(\mathbf E+k)^nb_0.$