Are all the zeros of $1-a_2x^2+a_4x^4-a_6x^6+\cdots$ real for $a_{2n}>a_{2(n+1)}$ with $a_{2n+1}=0$ and $a_{2n}>0$? This question is related to a previous question of mine. 
I was not pleased about the conditions I provided there. I had something different in mind but I failed in stating it. So here are the premises. Supose I have a power series $\sum_{k=0}^{\infty}a_{k}x^{k}$ and:


*

*$a_{0}=1$;

*$a_{2n+1}=0$;

*$a_{2n}>a_{2(n+1)}$;

*$a_{2n}>0$.


My questions:


*

*Does this kind of power series allways have real zeros?

*If not are there counterexamples?


I've made some quick checking for $\cos(x)$, shifting the value of $a_{2n}$, and it seems that all the roots remain real provided that we obey the above conditions nevertheless the number of roots changes from infinite to finite.
Thanks.

The same objection as before holds. If we consider
$$ f(z)=1-a_2 z+a_4 z^2-a_6 z^3 +\ldots $$
the fact that $\{a_{2n}\}_{n\geq 1}$ is a positive decreasing sequence do not give that Newton's inequalities are fulfilled. If Newton's inequalities are not fulfilled, $f(z)$ cannot have only real roots and the same applies to your original function. For instance, the disciminant of the third-degree polynomial
$$ p(z) = 1-\frac{z}{2}+\frac{z^2}{4}-\frac{z^3}{8} $$
is negative, hence $p(z)$ has some complex root and the sequence $a_2=\frac{1}{2},a_4=\frac{1}{4},a_6=\frac{1}{8},$ $a_8=\varepsilon,a_{10}=\frac{\varepsilon}{2},a_{12}=\frac{\varepsilon}{4},\ldots$ gives a counter-example for any sufficiently small $\varepsilon>0$. 

The only question that makes sense is the following:

If $\{a_n\}_{n\geq 0}$ is a real sequence fulfilling Newton's inequalities and 
  $$f(x)=\sum_{n\geq 0}a_n x^n $$ is an analytic/entire function, can we say that all the roots of $f$ are real?

Unluckily, the answer is negative also in that case, always by a perturbation argument: it is enough to consider $f(x)=\varepsilon+e^{-x}$ for some small $\varepsilon>0$. So Newton's inequalities can be used to prove the existence of some complex root, but not to prove that every root is real.
At some meta-level, theorems ensuring that every root of something is real have to be fairly complex (pun intended). Otherwise, RH would have been solved centuries ago.
