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 $A_n \left({x, t}\right)$ be the polynomial of degree $n$ defined as:

$A_n \left({x, t}\right) := \dbinom {x - n t} n \dfrac x {x - n t}$

for $x \ne n t$.

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


Then:

$\displaystyle \sum_k A_k \left({r, t}\right) z^k = x^r$

for sufficiently small $z$.


Proof

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

$\displaystyle \sum_j \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$

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

Thus:

$\displaystyle \sum_{j, k} \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k = 1$


We have:

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


Now let:

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


Thus:

\(\displaystyle \sum_j A_j \left({r, t}\right) z^j \left({1 + w}\right)^r\) \(=\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_j A_j \left({r, t}\right) z^j\) \(=\) \(\displaystyle \left({1 + w}\right)^{- r}\)
\(\displaystyle \) \(=\) \(\displaystyle x^r\)



$\blacksquare$


Also see


Sources