For integers $n$ and $k$, let $P(n,k)(x) = \sum\_{i=0}^k \binom ni x^i$ be the truncated binomial polynomial. There has been work on whether $P(n,k)$ is irreducible, but this is a different question. According to some Mathematica computations, if $n \le 200$ and $1 \le r < n/2$, then $P(n,2r)(x) > 0$ for all $x$. In addition, $P(n,2)(x)$ is easily seen to be positive and $P(2r+1,2r)(x) = (1+x)^{2r+1} - x^{2r+1} > 0$, because $f(t) = t^{2r+1}$ is strictly increasing. This seems so simple that it has to be known!


I have an application of this result which is related to Hilbert's 17th Problem and expressing odd powers of psd forms in many variables as a sum of squares.



Are truncated even degree binomial polynomials psd?
We have $$P(n,k)'(x)=\sum\_{i=1}^k \binom ni ix^{i-1} \\
=n\sum\_{i=1}^k \binom{n-1}{i-1}x^{i-1}
=nP(n-1,k-1)(x)\tag{1}\label{1}$$
and
$$P(n,k)(x)-P(n-1,k)(x)
=\sum\_{i=1}^k \Big(\binom ni-\binom{n-1}i\Big)x^i \\
=\sum\_{i=1}^k \binom{n-1}{i-1} x^i 
=xP(n-1,k-1)(x). \tag{2}\label{2}$$
Also, for $r=1,2,\dots$ and $n\ge2r$ we have $P(n,2r)(x)\to\infty$ as $x\to\pm\infty$. So, the minimum of $P(n,2r)(x)$ in real $x$ is attained at a critical point $x\_0$ of $P(n,2r)$, where $P(n,2r)'(x\_0)=0$.
Also, as noted by the OP, $P(2r+1,2r)>0$ (for $r$ as above).
By \eqref{2}, \eqref{1}, and induction on $n$, for $n>2r+1$ we have
$$\min\_x P(n,2r)(x)=P(n,2r)(x\_0)=P(n-1,2r)(x\_0)>0.\quad\Box$$




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Moreover, it follows that $\min\_x P(n,2r)(x)$ is nondecreasing in $n\ge2r+1$.


