The zeros of alternating sign, binomial coefficient polynomials

Question

I have a question regarding the zeros of the following polynomial, based on partial rows of pascal's triangle,

$$f(x)=\sum_{k=a}^n\binom{n}{k}(-1)^{k}x^k,$$

where $a,n\in \mathbb{Z}^+,n>a.$

Specifically, I'm wondering if there exists zeros for $f$ when $x \in (0,1)$. Obviously, when $a=0$, the zeros occur at $x=1$, but as we increase $a$, I haven't found any solutions for $x\in (0,1)$.

For any $a \geq 0,$ does there exist $x \in (0,1)$ such that $f(x)=0$?

Potential leads include the Nörlund–Rice integrals.

http://en.wikipedia.org/wiki/N%C3%B6rlund%E2%80%93Rice_integral

Answer 1

Let $$f_{a,n}=\sum_{k=a}^n\binom{n}{k}(-x)^{k}.$$ For $0 \lt a \lt n$ one has $f_{a,n}(0)=0.$ Small cases make it reasonable to conjecture that $f_{a,n}(x)$ increases (for even $a$) or decreases (for odd $a$) along $[0,1].$ If true, this means that there are no other zeros in that interval. Certainly this is true for $f_{1,n}=(1-x)^n-1.$

This suggests looking at the derivative. One can confirm the identity $f_{a,n}'=-nf_{a-1,n-1} .$ From this and the facts above, the conjecture follows by induction. The identity follows from $$k\binom{n}{k}=n\binom{n-1}{k-1}.$$

Answer 2

After you changed the sign to $(-1)^k)$, I want to point some literature related to the distribution of all zeros. It is MR2147385 and later papers which cite this.