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.
 A: 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.
