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It might be complicated to decide whether or not a given entire function has only real zeros. So, I'll focus on special examples. It is known that the following function given by the infinite series, $$\sum_{k\geq0}\frac{z^k}{k!^2}$$ has infinitely many real zeros only (from Bessel function theory).

QUESTION. Fix an integer $s\geq3$. Can we say the same about the following entire functions? $$\sum_{k\geq0}\frac{z^k}{k!^s}.$$

Semiclassical
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T. Amdeberhan
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  • what you claim in the post is not correct as $f(z)=\sum_{k\geq0}\frac{z^{2k}}{k!^2}$ has roots on the imaginary axis; $f(iz)=\sum_{k\geq0}(-1)^k\frac{z^{2k}}{k!^2}$ has real zeroes – Conrad Aug 09 '24 at 23:07
  • @Conrad: correct. I need to fix it. – T. Amdeberhan Aug 10 '24 at 04:16
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    Strictly speaking, the series you give initially is $\sum_{k=0}^\infty z^{k}/k!^2=I_0(2z^{1/2})$. It's a bit of a stretch to call that "a Bessel function" but it does have the advantage of localizing the zeros to the negative real axis. More generally, though, the series you're interested in is a generalized hypergeometric function: $$\sum_{k=0}^\infty \frac{z^k}{k!^s}={0}F{s-1}(;1,1,\ldots,1,z)$$ – Semiclassical Aug 10 '24 at 04:33
  • Check out the techniques and answers here : https://math.stackexchange.com/questions/4801306/0-leq-c-sum-n-1-infty-fracxnn1n-where-c-1-036055393

    https://math.stackexchange.com/questions/4800097/solve-for-mt-in-the-integral-transform-int-011-tn-mt-dt-fracn1

    https://math.stackexchange.com/questions/4440852/is-fx-v-sum-n-0-infty-xn-over-gammav-n-1-0-for-all-real

    https://math.stackexchange.com/questions/4802259/conjecture-about-fx-v-1-sum-n-1-infty-xn-over-tv-n-0-for

    Utilizing of those techniques I think the answer is yes. when have trouble with it I might give it a go when I have time

     – mick May 20 '25 at 21:45
  • No wait, I am mistaken. The answer is probably no. The series converges to fast so it is to well approximated by polynomials. You asked for $s>= 3$. Notice my links where for $0<s<1$. I think it already fails at $s=3$. The small error term of the taylor series implies that testing with small truncations can already disprove it. – mick May 20 '25 at 21:52
  • See also : https://math.stackexchange.com/questions/4965190/1-z-fracz24-fracz343-fracz446-fracz5410 – mick May 20 '25 at 22:02
  • I think it fails for $3<s<4$ – mick May 20 '25 at 22:07
  • Notice $1 + z + \frac{z^2}{2^3} + \frac{z^3}{6^3} = 0$ has nonreal zero's. (around $-12 +/- 4i$) – mick May 20 '25 at 22:13
  • You might like this google site : https://sites.google.com/site/tommy1729/special-polynomials – mick May 20 '25 at 22:17

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