Sum over k of r-kt choose k by r over r-kt by z^k

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $\map {A_n} {x, t}$ be the polynomial of degree $n$ defined as:

$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$

for $x \ne n t$.

Let $z = x^{t + 1} - x^t$.


Then:

$\ds \sum_k \map {A_k} {r, t} z^k = x^r$

for sufficiently small $z$.


Proof

From Sum over $k$ of $\paren {-1}^k$ by $\dbinom n k$ by $\dbinom {r - k t} n$ by $\dfrac r {r - k t}$ and renaming variables:

$\ds \sum_j \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$

where $\delta_{k 0}$ is the Kronecker delta.

Thus:

$\ds \sum_{j, k} \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k = 1$


We have:

\(\ds \) \(\) \(\ds \sum_{j, k} \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k\)
\(\ds \) \(=\) \(\ds \sum_j \paren {-1}^j \dfrac r {r - j t} \sum_k \dbinom k j \dbinom {r - j t} k w^k\)
\(\ds \) \(=\) \(\ds \sum_j \paren {-1}^j \dfrac r {r - j t} \sum_k \dbinom {r - j t} j \dbinom {r - j t - j} {k - j} w^k\) Product of $\dbinom r m$ with $\dbinom m k$
\(\ds \) \(=\) \(\ds \sum_j \paren {-1}^j \dfrac r {r - j t} \dbinom {r - j t} j \sum_k \dbinom {r - j t - j} {k - j} w^k\)
\(\ds \) \(=\) \(\ds \sum_j \paren {-1}^j \map {A_j} {r, t} \sum_k \dbinom {r - j t - j} {k - j} w^k\) Definition of $\map {A_j} {r, t}$
\(\ds \) \(=\) \(\ds \sum_j \paren {-1}^j \map {A_j} {r, t} \paren {1 + w}^{r - j t - j} w^j\) Binomial Theorem


Now let:

$x = \dfrac 1 {1 + w}$
\(\ds x\) \(:=\) \(\ds \dfrac 1 {1 + w}\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds -\frac w {\paren {1 + w}^{1 + t} }\)
\(\ds \leadsto \ \ \) \(\ds \paren {1 + w}^{r - j k - j} w^j \paren {-1}^j\) \(=\) \(\ds z^r z^j\)


Thus:

\(\ds \sum_j \map {A_j} {r, t} z^j \paren {1 + w}^r\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \sum_j \map {A_j} {r, t} z^j\) \(=\) \(\ds \paren {1 + w}^{- r}\)
\(\ds \) \(=\) \(\ds x^r\)



$\blacksquare$


Also see


Sources