Let $m,n\geq0$ are integers, show that
$$p_{m,n}(x)=\sum_{k=0}^{m} \binom{2x+2k}{2k+1} \binom{n+m-k-x-\frac{1}{2}}{m-k}$$
must be an odd polynomial(all coefficients of even power of $x$ is $0$) with positive coefficients.
Where the binomial coefficients is used as polynomials, which is defined by
$$
\binom{x}{n}=\frac{x(x-1)(x-2) \cdots(x-n+1)}{n(n-1)(n-2) \cdots 2 \cdot 1}
$$
Using the rising factorials, we can express $p_{m,n}(x)$ in the following way
$$
p_{m,n}(x)=\sum_{k=0}^{m} \frac{(2x)^{(2k+1)}}{(2k+1)!} \frac{(\frac12+n-x)^{(m-k)}}{(m-k)!},
$$
where the rising factorial is defined as follows
$$
\begin{aligned}
x^{(n)}& =\overbrace{x(x+1)(x+2) \cdots(x+n-1)}^{n \text { factors }} =\prod_{k=1}^n(x+k-1)=\prod_{k=0}^{n-1}(x+k) ,
\end{aligned}
$$
where the value of $x^{(n)}$ is taken to be 1 (an empty product) when $n = 0$.
I can only prove the special case of $n=0$ via complicated analytic bypass, and I failed to prove the general $n\geq1$. I wonder if there is a nature combinatorial proof…
Update: Is it possilble to calculate the asymptotic of $P_{m,n}(x)$ for $n$ goes to infinity? I can only get the asymptotic for special $x=-1/2$, is there some way to derive the asymptotics for general $x$?
Best Answer
We first try to simplify the sum
$$P_{m,n}(x) = \sum_{k=0}^m {2x+2k\choose 2k+1} {n+m-k-x-1/2\choose m-k}.$$
We get
$$- [z^m] (1+z)^{n+m-x-1/2} \sum_{k\ge 0} {-2x\choose 2k+1} z^k (1+z)^{-k}.$$
Here we have extended to infinity due to the coefficient extractor. Next,
$$- [z^m] (1+z)^{n+m-x-1/2} \sum_{k\ge 0} z^k (1+z)^{-k} [w^{2k+1}] (1+w)^{-2x} \\ = - [z^{2m}] (1+z^2)^{n+m-x-1/2} \sum_{k\ge 0} z^{2k} (1+z^2)^{-k} [w^{2k+1}] (1+w)^{-2x} \\ = - [z^{2m+1}] (1+z^2)^{n+m-x-1/2} \sqrt{1+z^2} \sum_{k\ge 0} z^{2k+1} (1+z^2)^{-k-1/2} [w^{2k+1}] (1+w)^{-2x} \\ = - [z^{2m+1}] (1+z^2)^{n+m-x} \frac{1}{2} ((1+z/\sqrt{1+z^2})^{-2x} -(1-z/\sqrt{1+z^2})^{-2x}) \\ = - [z^{2m+1}] (1+z^2)^{n+m} \frac{1}{2} ((z+\sqrt{1+z^2})^{-2x} -(-z+\sqrt{1+z^2})^{-2x}).$$
We know from the initial representation that this is a polynomial in $x.$ Extract the coefficient on $[x^q]$
$$[x^q] \exp(-2x \log(\pm z+\sqrt{1+z^2})) = \frac{(-1)^q 2^q}{q!} \log^q(\pm z+\sqrt{1+z^2}).$$
The logarithms are both valid formal power series because the arguments are formal power series that start with constant coefficient equal to one. Note however that
$$\log(-z+\sqrt{1+z^2}) = - \log\frac{1}{-z+\sqrt{1+z^2}} \\ = -\log\frac{z+\sqrt{1+z^2}}{1+z^2-z^2} = -\log(z+\sqrt{1+z^2}).$$
This yields
$$[x^q] P_{m,n}(x) = - [z^{2m+1}] (1+z^2)^{n+m} \frac{(-1)^q 2^q}{q!} \frac{1}{2} \log^q(z+\sqrt{1+z^2}) (1-(-1)^q).$$
This is zero when $q$ is even (inspect last term) and we have the claim. We get for $q$ odd the coefficient
$$\frac{2^q}{q!} [z^{2m+1}] (1+z^2)^{n+m} \log^q(z+\sqrt{1+z^2}).$$
The powered logarithm starts at $z^q$ so we get zero when $q\gt 2m+1.$ To see positivity note that an alternate representation is
$$(-1)^{m+1} [z^m] (1+z)^{x-1/2-n} \sum_{k\ge 0} {-2x\choose 2k+1} (-1)^k z^k \\ = (-1)^{m+1} [z^m] (1+z)^{x-1/2-n} \sum_{k\ge 0} (-1)^k z^k [w^{2k+1}] (1+w)^{-2x} \\ = (-1)^{m+1} [z^{2m+1}] (1+z^2)^{x-1/2-n} \sum_{k\ge 0} (-1)^k z^{2k+1} [w^{2k+1}] (1+w)^{-2x} \\ = i (-1)^m [z^{2m+1}] (1+z^2)^{x-1/2-n} \sum_{k\ge 0} i^{2k+1} z^{2k+1} [w^{2k+1}] (1+w)^{-2x} \\ = \frac{i}{2} (-1)^m [z^{2m+1}] (1+z^2)^{x-1/2-n} ((1+iz)^{-2x}-(1-iz)^{-2x}) \\ = \frac{i}{2} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n} \left[ \left( \frac{1-iz}{1+iz} \right)^x - \left( \frac{1+iz}{1-iz} \right)^x \right].$$
Extract the coefficient on $[x^q]$ to get
$$\frac{i}{2q!} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n} \left[ \log^q \frac{1-iz}{1+iz} - \log^q \frac{1+iz}{1-iz} \right] \\ = \frac{i}{2q!} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n} \log^q \frac{1-iz}{1+iz} (1-(-1)^q).$$
Zero again for $q$ even, and for $q$ odd,
$$\frac{i}{q!} (-1)^m [z^{2m+1}] (1+z^2)^{-1/2-n} \log^q \frac{1-iz}{1+iz} \\ = \frac{1}{q!} [z^{2m+1}] (1-z^2)^{-1/2-n} \log^q \frac{1+z}{1-z}.$$
The two terms in the product both have series in positive terms only. To see this, note that the first one is
$$\sum_{p\ge 0} {-1/2-n\choose p} (-1)^p z^{2p} = \sum_{p\ge 0} {p+n-1/2\choose p} z^{2p}.$$
The coefficients are positive because $(p+n-1/2)^{\underline{p}}$ is. Note also that
$$\log\frac{1+z}{1-z} = \log\left(1+\frac{2z}{1-z}\right) = \sum_{p\ge 1} (-1)^{p-1} \frac{2^p z^p}{p (1-z)^p}.$$
This means the coefficient on $[z^r]$ is
$$\sum_{p=1}^r (-1)^{p-1} 2^p [z^{r-p}] \frac{1}{p (1-z)^p} = \sum_{p=1}^r (-1)^{p-1} 2^p {r-1\choose p-1} \frac{1}{p} \\ = \frac{1}{r} \sum_{p=1}^r (-1)^{p-1} 2^p {r\choose p} = - \frac{1}{r} ((1-2)^r-1) = - \frac{1}{r} ((-1)^r-1).$$
This is zero when $r$ is even and $\frac{2}{r}$ when $r$ is odd. We have established the desired positivity of the coefficients, as we have the product of a series with positive coefficients times a power of a series that also has positive coefficients.