# When do the even part and odd part of a hypergeometric like function have common nonnegative real root?

I encountered the following problem in my research project:

Let $$f(x)=\sum_{k=0}^{\infty} a_{k} x^{k}$$

We can separate the even part $p(x^2)$ from the odd part $x q(x^2)$ and write $$f(x)=p(x^2)+x q(x^2)$$ $$p(x)=\sum_{k=0}^{\infty} a_{2k} x^{k}$$ $$q(x)=\sum_{k=0}^{\infty} a_{2k+1} x^{k}$$

Suppose $|a_0|>|a_1|>...>0$ and $a_{2k}=\left(-1\right)^k b_k$ and $b_k>0$ and $\left(b_{k}\right)^2>b_{k-1}b_{k+1}$ (ie. log concave) and $a_{2k+1}=\left(-1\right)^k c_k$ and $c_k>0$ and $\left(c_k\right)^2>c_{k-1}c_{k+1}$ (ie. log concave)

So we can rewrite them as $$p(x)=\sum_{k=0}^{\infty} \left(-1\right)^k b_{k} x^{k}$$ $$q(x)=\sum_{k=0}^{\infty} \left(-1\right)^k c_{k} x^{k}$$

Q: When do $p(x)$ and $q(x)$ have common nonnegative real root?

Thanks in advance for the help!

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One possible way to prove no common root is to calculate the resultant of $p\left( x \right)$ and $q \left( x \right)$. But the result is an infinite polynomial of the coefficients $a_k$ and $b_k$ and it seems too messy. A second way is to prove that $x \ge 0$: $$\left( p \left( x \right) \right)^2+\left( q \left( x \right) \right)^2>0$$ A third possible way is to prove that real roots of $p \left( x \right)$ and $q \left( x \right)$ interlace. But how? –  mike May 5 '13 at 5:57