1

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:

  1. $a_{0}=1$;
  2. $a_{2n+1}=0$;
  3. $a_{2n}>a_{2(n+1)}$;
  4. $a_{2n}>0$.

My questions:

  1. Does this kind of power series allways have real zeros?
  2. 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.

Community
  • 1
Neves
  • 5,617

1 Answers1

2

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.

Community
  • 1
Jack D'Aurizio
  • 353,855
  • Note that $a_{2n}\in\left ]0,1\right] $ and that $a_0=1$. – Neves Jul 20 '16 at 19:35
  • @Neves: irrelevant, but fixed. – Jack D'Aurizio Jul 20 '16 at 19:40
  • How does the polynomial $p(z)$ relates to the power series $f(z)$? – Neves Jul 20 '16 at 20:50
  • A polynomial is a power series. We are taking $f(z)$ as the sum of a polynomial and $\varepsilon$ times an analytic function. If $\epsilon$ is small enough and $p(z)$ has a complex root, $f(z)$ has a complex root, too. – Jack D'Aurizio Jul 20 '16 at 20:52
  • But, $a_{2n+1}=0$ – Neves Jul 20 '16 at 23:50
  • @Neves: yes, in the first sequence giving the counter-example we have $a_{2n+1}=0$. I think it is easier this way - what is not clear in my answer, in your opinion? You do not seem to have grasped the central fact: if you set some constraints that allow a sequence to violate Newton's inequality, the associated analytic function may still have a complex root. Now, your constraints on ${a_{2n}}$ to be positive, decreasing, in $[0,1]$ etc... are kind of irrelevant, since they allow a violation of Newton's inequalities, so they cannot ensure that all the roots are real. – Jack D'Aurizio Jul 20 '16 at 23:56
  • The only relevant constraint is to start with a sequence that fulfills Newton's inequalities, but also that relevant constraint turns out to be inconclusive about the localization of zeroes, as shown in the second part of my answer. – Jack D'Aurizio Jul 20 '16 at 23:58