5
$\begingroup$

I have

\begin{equation} G(x) = \sum_{i = 0}^{\infty} \sum_{r = 0}^{\infty} (-1)^i \binom{k}{2i} \binom{n-k}{r} x^{2i} x^{r} = \frac{1}{2} \left( (1 + x{\iota})^{k} + (1 - x \iota)^{k} \right)(1 + x)^{n-k} \end{equation}

where $\iota$ is the imaginary unit, i.e., $\iota^2 = -1$. The $l$th term of the above is the alternating sum of even coefficients in the Vandermonde convolution, with $l \le k$. I am interested in obtaining a form of the $l$th term that involves a single binomial coefficient (as in the Vandermonde convolution) as opposed to the convolution of two binomial coefficients. Is there a way to do so?

Edit: Sorry, I needed to explain better. By the $l$th term I mean the coefficient of $x^l$. If we fix such an $l$, then we see that for $i = 0, 1, \ldots $, we have $r = l - 0, l - 2, \ldots $, and therefore fixing $r = l - 2i$, we obtain the following as the coefficient of $x^l$:

$\sum_{i = 0}^{\infty} (-1)^{i} \binom{k}{2i} \binom{n-k}{l - 2i}$

So, for instance, if $l = 3$, we get

$\binom{k}{0} \binom{n-k}{3 - 0} - \binom{k}{2} \binom{n-k}{3 - 2}$, where I have used the convention that $\binom{a}{b} = 0$ if $a < b$.

The Vandermonde convolution is:

$\sum_{i = 0}^{\infty} \binom{k}{i} \binom{n-k}{l - i} = \binom{n}{l}$.

I was hoping to get something like the RHS of the above, i.e., $\binom{n}{l}$, for the $l$th term of the main equation. Note that the inner sum in the main equation has the closed form $(1 + x)^{n-k}$, and the outer sum has the closed form $\frac{1}{2}\left((1 + x\iota)^k + (1 - x\iota)^k\right)$. Please bear in mind the difference between $i$ and $\iota$.

$\endgroup$
2
  • $\begingroup$ The inner sum is just a binomial expansion. $\endgroup$ Aug 5, 2015 at 7:09
  • $\begingroup$ You also need to clarify what $l$ means. $\endgroup$ Aug 5, 2015 at 13:41

1 Answer 1

1
$\begingroup$

$$(-1)^i x^{2 i} \binom{k}{2 i} (x+1)^{n-k}$$

$\endgroup$
5
  • $\begingroup$ Can you explain? The answer is supposed to depend on some value $l$. $\endgroup$ Aug 5, 2015 at 13:59
  • 1
    $\begingroup$ @BrendanMcKay The sum was obtained by Mathematica (I am too lazy), as for $l,$ I am pretty sure the OP means $i$ (as you see, there is no $l$ in the question). $\endgroup$
    – Igor Rivin
    Aug 5, 2015 at 14:01
  • 1
    $\begingroup$ True, but there is an $x$. $\endgroup$ Aug 5, 2015 at 14:19
  • $\begingroup$ @BrendanMcKay Sorry, fixed. $\endgroup$
    – Igor Rivin
    Aug 5, 2015 at 14:31
  • $\begingroup$ The coefficient of $x^l$ can be expressed in terms of Krawtchouk polynomials. See Proposition 1.2 of arxiv.org/pdf/1101.1798.pdf. I doubt whether there is a "reasonable" formula involving only a single binomial coefficient. $\endgroup$ May 27, 2017 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.