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This may be a strange question, but I've not found anything about this.

Well, anyone can observe that both $$ \cos(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n)!}z^{2n} $$ and $$ \sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}z^{2n+1} $$ are alternating power series and that they all the roots on the real axis.

My question:

  1. Does the alternating sign (of real coefficients), somehow, implies that all the zeros are real?
  2. If not, are there alternating power series with complex zeros, I mean counterexamples?

Thanks.

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    $\begingroup$ I don't think that alternation alone will ever imply the root being zero. Make any polynomial with alternating sum that has a complex root. This can be considered a power series with vanishing coefficients after some point. $\endgroup$ – Behnam Esmayli Jul 19 '16 at 16:32
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    $\begingroup$ note $\cos(\sqrt{z}) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} z^n$ is a power with alternated and fast decreasing coefficients. I'd propose $\frac{1}{2}+e^{-z} = \frac{3}{2}+\sum_{n=1}^\infty \frac{(-1)^n}{n!} z^n$ having some complex zeros at $z = -\log(-1/2) + 2 i \pi k$ $\endgroup$ – reuns Jul 19 '16 at 17:05
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    $\begingroup$ and you can search for 'zeros of entire functions' for seeing some of the criterion sufficient for showing an entire function $f(z)$ has all its zero on the real line, or more generally that the imaginary part of its zeros converge to $0$ $\endgroup$ – reuns Jul 19 '16 at 23:00
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    $\begingroup$ You should hit the "accept" tick more often, judging on your queue. $\endgroup$ – Jack D'Aurizio Jul 20 '16 at 19:42
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An analytic function may have coefficients with alternating signs and still violate Newton's inequalities, enforcing a complex root. For instance, $$ f(x) = x^2+\sum_{k\geq 0}\frac{(-1)^k}{k!}x^k = e^{-x}+x^2 $$ has complex roots at $2\cdot W\left(\pm\frac{i}{2}\right)\approx 0.325199\, \pm 0.785257\, i.$

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1. No. 2. Yes. Consider the function $f$ defined on $\mathbb{C}$ by the power series $$f(z) = \sum_{n=0}^\infty (-1)^n a_n z^n$$ where $a_0=a_2=1$, $a_1=0$, and $a_n=0$ for all $n\geq 2$ (so that, indeed, $a_n \geq 0$ for all $n\in\mathbb{N}$). We do have that this series is alternating... even though it is a bit trivial.

Now, what are the zeroes of $f(z) = 1+z^2$?

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Consider $$f(z) = \frac{1-z+z^2}{4+z}= \frac14 - \frac{5}{16}z + \sum_{n>1} (-1)^n\frac{21}{2^{2n+2}}z^n$$

The terms in its expansion about $z=0$ are of alternating signs. Yet it has two complex zeros, at $$ \frac{1\pm\sqrt{3}i}{2} $$

A simpler example is $$g(z) = \frac{1-z+z^2}{1+z}= 1 - 2z + \sum_{n>1} (-1)^n3z^n$$

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A more general method: Start with an alternating series $\sum_{k} (-1)^k a_k x^k$ (for simplicity of exposition, suppose $a_1 > 0$) for which all the zeroes lie on the real line. Now, add a multiple of $x$, deforming the locations of any surviving zeroes away from the real line (because a zero at $z$ now has a multiple of $z$ as its value). Or, start over and subtract a multiple of $x^2$, deforming the locations of any surviving zeroes away from the real line (because a zero at $z$ now has a multiple of $z^2$ as its value). Having zeroes on the real line is a delicate thing, which we can break easily.

Following the pattern here: increasing the first positive coefficient in the power series for sine, $x + \sin x$ only has one zero on the real line, at $0$. It has more, for instance, at $4.21239... + 2.25073... \mathrm{i}$.

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