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Im looking for a real-analytic function $f(z)$ such that for any $z$

$1) $$f(z+p) =f(z)$

With $p$ a nonzero real number and where $z$ is close to , or onto the real line such that $z$ is in the domain of analyticity.

$2)$ $f(z)= 0 + a_1 z + a_2 z^2 + a_3 z^3 + ...$ where more than $50$ % of the nonzero (signs of the) $a_n$ are positive.

Thus let $f_n(z)$ be the truncated Taylor expansion of $f(z)$ of degree $n$. Let $T(n)$ be the amount of nonzero (signs in the) coefficients of the polynomial $f_n(z)$.

Let $v(n)$ be the amount of strict positive ($>0$) coefficients of $f_n(z)$.

Then $\lim_{n -> +\infty} v(n)/T(n) > 1/2$.

$3)$ $f(z)$ is nonconstant.

Also I prefer $f(z)$ to be entire if possible.

Is such a function $f(z)$ possible ?

Related :

Real-analytic $f(z)=f(\sqrt z) + f(-\sqrt z)$?

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mick
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    What do you mean by more than $50%$ in this context (i.e. what does it mean for a subset of an infinite set to consist of at least $50%$ of the elements)? – Michael Albanese Jun 19 '14 at 19:57
  • Lets say you take a truncated Taylor series of degree $n$ for some large $n$. Let $k$ be the amount of nonzero $a_n$ If then more than 60 % of $a_n$ is positive and by increasing $n$ this remains a fact , then more than 50% is positive in the limiting sense. Very intuitive. – mick Jun 19 '14 at 20:00
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    To explain why this is hard, nothing with a finite Fourier series has this property. If $f(z) = \sum_{k=1}^N a_k \sin(kz)$, then, for $n$ large, the main contribution to the coefficient of $z^{2n+1}$ comes from $\sin (N z)$, and has sign matching $a_N (-1)^{n+1}$. So we need to take an infinite Fourier series whose coefficients cancel enough to make more interesting sign patterns, and yet decay fast enough to make an analytic function in a neighborhood of the real axis. – David E Speyer Jun 25 '14 at 15:20
  • I will probably give the bounty to someone who can give an entire solution. – mick Jun 27 '14 at 11:07

1 Answers1

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Such a function is $f(z) = \cot(z-e^{2 \pi i/3}) + \cot(z-e^{-2 \pi i/3})$. If you'd like that written in terms of real functions, it is $$\frac{2 \sin(1+2z)}{\cosh(\sqrt{3}) - \cos(1+2z)}.$$

Proof: Notice that $$g(z) := f(z) - \frac{1}{z-e^{2 \pi i/3}} - \frac{1}{z - e^{- 2 \pi i/3}}$$ has no poles within the disc of radius $\sqrt{(\pi-1/2)^2 + (\sqrt{3}/2)^2} \approx 2.78$. Therefore, the Taylor coefficients of $g(z)$ decay exponentially fast. Meanwhile, $$\frac{1}{z-e^{2 \pi i/3}} + \frac{1}{z - e^{- 2 \pi i/3}} = 1+z-2z^2+z^3+z^4-2z^5+z^6+z^7-2 z^8 + \cdots.$$

So the coefficients of $f(z)$ are very close to $(1,1,-2,1,1,-2,\dots)$ and, in particular, have this sign pattern after possibly finitely many exceptions. In fact, the data suggests the convergence is quite rapid: Here are the coefficients of $z^n$ in $f(z)$ for $0 \leq n \leq 20$:

0.708823, 0.40783, -2.02933, 0.980937, 1.00110, -1.99877, 
1.000920, 1.00047, -1.99981, 1.000080, 1.00002, -1.99999, 
1., 1., -2., 1., 1., -2., 1., 1., -2.
David E Speyer
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  • How about an entire solution ? That could get the bounty. – mick Jun 27 '14 at 11:08
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    I've been thinking about it. It seems to me that the most natural approach is to find an entire function $f$ so that $f^{(n)}(z)$ has the desired sign pattern and decays rapidly as $z \to \pm \infty$. Then show that $g(z) := \sum_{k=-\infty}^{\infty} f(z+k)$ converges to a periodic function and has $g^{(n)}(0) \approx f^{(n)}(0)$. I haven't found a good choice of $f$ yet though. – David E Speyer Jun 27 '14 at 13:01
  • For example, taking $f(z) = e^{-z^2/2} \cos(\sqrt{3} z^2/2)$ has the nice $+--+--+-- \cdots$ pattern for $f^{(n)}(0)$ and decays fast enough for the sum to converge, but it looks like $g^{(n)}(0)$ isn't particularly close to $f^{(n)}(0)$. – David E Speyer Jun 27 '14 at 13:07
  • I would bet that the liminf of the ratio of positive derivatives cannot exceed $\frac{1}{2}$ for a periodic entire function, since the periodicity condition gives a strong structure to the zero set of $f$ and bounds the order of $f$, so we can try to study the signs of the derivatives through the Weierstrass product for $f$: the exponential factor must vanish and the zeroes tend to be almost symmetrically distributed in every big ball centered in zero, so I am expecting a limit ratio $\frac{1}{2}$. – Jack D'Aurizio Jun 29 '14 at 04:18
  • @JackD'Aurizio Why does periodicity bound the order? Doesn't $\exp(\exp(\exp(\cdots \exp(\sin(z)) \cdots ))$ have arbitrarily high order, as I nest in more parentheses? – David E Speyer Jun 29 '14 at 18:51
  • @DavidSpeyer: yes, you're right. I believed that $f(z)=f(z+T)$ implies order-$1$, but this is false. – Jack D'Aurizio Jun 29 '14 at 19:01
  • Anyway, Carlson's theorem (http://math.stackexchange.com/questions/14423/entire-periodic-function) and the Descartes sign rule seem to suggest that no entire function can satisfy the given properties. – Jack D'Aurizio Jun 30 '14 at 22:37
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    @JackD'Aurizio I'd certainly be interested to see you write up a solution. – David E Speyer Jul 01 '14 at 02:14