For fixed non-negative integers $n$ and $q \geq 2$, the $k$-th Krawtchouk (Kravchuk) polynomial is defined as $$K_k = \sum_{j=0}^k (-1)^j (q-1)^{k-j} \binom{X}{j} \binom{n-X}{k-j} \in \mathbb{Q}[X]$$ The definition is based on generalized binomial coefficients $$\binom{a}{b} = \frac{a \cdot (a-1) \cdot \ldots \cdot (a-b+1)}{b!}$$ where $b\in\mathbb{N}$ and $a$ comes from any ring ($\mathbb{Q}[X]$ in our case) containing the rationals.


The Wikipedia page lists two alternative expressions for $K_k$:

$$K_k = \sum_{j=0}^k (-q)^j (q-1)^{k-j} \binom{n-j}{k-j}\binom{X}{j}$$

$$K_k = \sum_{j=0}^k (-1)^j q^{k-j} \binom{n-k+j}{j}\binom{n-X}{k-j}$$

I'm looking for elegant proofs of these alternative forms. Preferably in an elementary way or using the generating series (see again Wikipedia)

$$\sum_{k=0}^\infty K_k Z^k = (1 + (q-1)Z)^{n-X} (1-Z)^X$$ where polynomial exponents are defined based on the binomial series $$(1 + Z)^X = \sum_{i=0}^\infty \binom{X}{i} Z^i.$$


The alternative expressions have also been mentioned (without proof) here as well as here (for $q=2$).


1 Answer 1


Starting from

$$K_k = \sum_{j=0}^k (-1)^j (q-1)^{k-j} {X\choose j} {n-X\choose k-j}$$

we first obtain

$$K_k = [z^k] (1+z)^{n-X} \sum_{j\ge 0} (-1)^j (q-1)^{k-j} z^j {X\choose j}.$$

Here we have extended to infinity due to the coefficient extractor. Continuing,

$$(q-1)^k [z^k] (1+z)^{n-X} \sum_{j\ge 0} (-1)^j (q-1)^{-j} z^j {X\choose j} \\ = (q-1)^k [z^k] (1+z)^{n-X} (1-z/(q-1))^X \\ = [z^k] (1+(q-1)z)^{n-X} (1-z)^X.$$

First identity

We seek to verify that

$$K_k = \sum_{j=0}^k (-q)^j (q-1)^{k-j} {n-j\choose k-j} {X\choose j}.$$

We get for the sum

$$[z^k] (1+z)^n \sum_{j\ge 0} (-q)^j (q-1)^{k-j} (1+z)^{-j} z^j {X\choose j}.$$

Here the coefficient extractor has once more enforced the range. Continuing,

$$(q-1)^k [z^k] (1+z)^n \sum_{j\ge 0} (-q)^j (q-1)^{-j} (1+z)^{-j} z^j {X\choose j} \\ = (q-1)^k [z^k] (1+z)^n \left[1-\frac{qz}{(q-1)(1+z)}\right]^X \\ = [z^k] (1+(q-1)z)^n \left[1-\frac{qz}{1+(q-1)z}\right]^X \\ = [z^k] (1+(q-1)z)^{n-X} [1+(q-1)z-qz]^X \\ = [z^k] (1+(q-1)z)^{n-X} (1-z)^X.$$

This is the claim.

Second identity

Here we set out to simplify

$$K_k = \sum_{j=0}^k (-1)^j q^{k-j} {n-k+j\choose j} {n-X\choose k-j}.$$

We get for the sum

$$[z^k] (1+z)^{n-X} \sum_{j\ge 0} (-1)^j q^{k-j} {n-k+j\choose j} z^j.$$

This is the third time with the extractor enforcing the range. Continuing,

$$q^k [z^k] (1+z)^{n-X} \sum_{j\ge 0} (-1)^j q^{-j} {n-k+j\choose j} z^j \\ = q^k [z^k] (1+z)^{n-X} \frac{1}{(1+z/q)^{n-k+1}} = [z^k] (1+qz)^{n-X} \frac{1}{(1+z)^{n-k+1}}.$$

This is

$$\;\underset{z}{\mathrm{res}}\; \frac{1}{z^{k+1}} (1+z)^{k+1} (1+qz)^{n-X} \frac{1}{(1+z)^{n+2}}.$$

Now we put $z/(1+z)=w$ so that $z=w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$ to get

$$\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{k+1}} (1+qw/(1-w))^{n-X} \frac{1}{(1+w/(1-w))^{n+2}} \frac{1}{(1-w)^2} \\ = \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{k+1}} (1+qw/(1-w))^{n-X} (1-w)^n \\ = \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{k+1}} (1-w+qw)^{n-X} (1-w)^X \\ = [w^k] (1+(q-1)w)^{n-X} (1-w)^X.$$

Once more we have the claim.

  • $\begingroup$ Many thanks for this answer. What is 'res'? $\endgroup$
    – azimut
    Commented Jul 7, 2023 at 8:29
  • $\begingroup$ This is the complex residue for use with Egorychev method. $\endgroup$ Commented Jul 7, 2023 at 17:48
  • 1
    $\begingroup$ @MarkoRiedel: Very nice! (+1) $\endgroup$ Commented Jul 7, 2023 at 18:19

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