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Recently, I have become quite obsessed with the follow series:

$$f(t,x)=\sum_{m=-\infty}^{\infty} (-1)^m f(t+x m)$$

where $f$ is analytic. This series automatically produces a periodic function with period $2x$ (assuming it converges). Let use focus on four seemingly similar functions of the form $g(t)e^{-h(t)}$:

$$\begin{align} f1(t)&=t e^{-t^2} \\ f2(t)&=\tanh(t) e^{-\cosh(t)} \\ f3(t)&=\tanh(t/2) e^{-\cosh(t)} \\ f4(t)&=t e^{-\cosh(t)} \end{align}$$

There is an interplay between the increasing behavior of $g(t)$ and $h(t)$ that I don't understand. In particular, I would like to know when $f(t,x)$ is positive on the first quarter of it's periodicity, that is when $t\in (0,x/2)$.

Using mathematica, it appears that $f1$ and $f2$ are never negative. In fact, it appears:

$$f(t,x)>(f(x/2,x)/(x/2)) * t$$

for all $t\in(0,x/2)$. However, $f3$ and $f4$ appear to fail being positive. Indeed, focusing on $f4$ and choosing $x\approx 0.82$, it appears that $f(t,x)<0$ for all $t\in(0,x/2)$. The split appears to be at $c\approx 0.919583$. Zooming in very very close to zero, it does appear that at $c$, $f4$ has multiple simple zeros being both positive and negative. However, to the right and left of this value $f4$ appears to be either all positive or all negative.

Likewise, for $c\approx 0.61093879$ and function $f3$, it appears that $f(t,x)$ is positive for $x>c$ and negative for $x<c$, again for $t\in(0,x/2)$. This is very surprising considering the extreme similarity of $f2$ and $f3$. Also, zooming in very very close to zero at $x\approx 0.61093879$ it does not appear that $f3$ has any simple zeros, meaning that is appears there exists some constant $c\approx 0.61093879 $, such that

$$\sum_{m=-\infty}^\infty (-1)^m \tanh((t+cm)/2)e^{-\cosh(t+cm)}=0$$

for all $t\in\mathbb{R}$. That seems very interesting.

Question: Is it indeed the case for $f2$, that $f(t,x)>0$ for all $t\in(0,x/2)$? Moreover is $f(t,x)>(f(x/2,x)/(x/2)) * t$?

UPDATE

So, it appears that $\sin(xt)$ serves as a better lower bound instead of a linear equation, in fact the series appears to uniformly approach $\sin(xt)$ as $x\to 0$, yielding a ratio of $1$ in their limit. I then went ahead and flipped over the function so that is would be between 0 and 1 (assuming the series is positive), and I then multiplied $t$ by $x$ so as to map the periodicity between $(0,1)$. I then graphed through different values of $x$ to see what $t$ looks like. The below animation is

$$ g(t,x):=\frac{f(x/2,x)*\sin(\pi t)}{f(xt,x)}$$

where the series uses the functions $f_2(t)=\tanh(t)e^{-\cosh(t)}$ and $f_5(t)=\tanh(t/2)e^{-\cosh(t)}$. We see that $f_2$ appears to be strictly decreasing as $x$ increases. Likewise, we see that $f_5$ appears to oscillate around $1$ at the start, in particular since the series for $f_5$ appears to be negative for some values of $x$, we see here the ratio is $>1$ for those same values of $x$. How would we prove that for $t>0$ and $f(t)=\tanh(t)e^{-\cosh(t)}$, we have

$$\frac{d}{dx} g(t,x) = \frac{d}{dx}\frac{f(x/2,x)*\sin(\pi t)}{f(xt,x)}\le 0 $$

for any $x$? How do we prove $\lim g(t,x)=1$ when $x\to 0$ and $\lim g(t,x)=0$ ($t\not= 1/2$) when $x\to\infty$? Working with this series seems particularly difficult, even though we can see what it is doing.

Recently, I have become quite obsessed with the follow series:

$$f(t,x)=\sum_{m=-\infty}^{\infty} (-1)^m f(t+x m)$$

where $f$ is analytic. This series automatically produces a periodic function with period $2x$ (assuming it converges). Let use focus on four seemingly similar functions of the form $g(t)e^{-h(t)}$:

$$\begin{align} f1(t)&=t e^{-t^2} \\ f2(t)&=\tanh(t) e^{-\cosh(t)} \\ f3(t)&=\tanh(t/2) e^{-\cosh(t)} \\ f4(t)&=t e^{-\cosh(t)} \end{align}$$

There is an interplay between the increasing behavior of $g(t)$ and $h(t)$ that I don't understand. In particular, I would like to know when $f(t,x)$ is positive on the first quarter of it's periodicity, that is when $t\in (0,x/2)$.

Using mathematica, it appears that $f1$ and $f2$ are never negative. In fact, it appears:

$$f(t,x)>(f(x/2,x)/(x/2)) * t$$

for all $t\in(0,x/2)$. However, $f3$ and $f4$ appear to fail being positive. Indeed, focusing on $f4$ and choosing $x\approx 0.82$, it appears that $f(t,x)<0$ for all $t\in(0,x/2)$. The split appears to be at $c\approx 0.919583$. Zooming in very very close to zero, it does appear that at $c$, $f4$ has multiple simple zeros being both positive and negative. However, to the right and left of this value $f4$ appears to be either all positive or all negative.

Likewise, for $c\approx 0.61093879$ and function $f3$, it appears that $f(t,x)$ is positive for $x>c$ and negative for $x<c$, again for $t\in(0,x/2)$. This is very surprising considering the extreme similarity of $f2$ and $f3$. Also, zooming in very very close to zero at $x\approx 0.61093879$ it does not appear that $f3$ has any simple zeros, meaning that is appears there exists some constant $c\approx 0.61093879 $, such that

$$\sum_{m=-\infty}^\infty (-1)^m \tanh((t+cm)/2)e^{-\cosh(t+cm)}=0$$

for all $t\in\mathbb{R}$. That seems very interesting.

Question: Is it indeed the case for $f2$, that $f(t,x)>0$ for all $t\in(0,x/2)$? Moreover is $f(t,x)>(f(x/2,x)/(x/2)) * t$?

Recently, I have become quite obsessed with the follow series:

$$f(t,x)=\sum_{m=-\infty}^{\infty} (-1)^m f(t+x m)$$

where $f$ is analytic. This series automatically produces a periodic function with period $2x$ (assuming it converges). Let use focus on four seemingly similar functions of the form $g(t)e^{-h(t)}$:

$$\begin{align} f1(t)&=t e^{-t^2} \\ f2(t)&=\tanh(t) e^{-\cosh(t)} \\ f3(t)&=\tanh(t/2) e^{-\cosh(t)} \\ f4(t)&=t e^{-\cosh(t)} \end{align}$$

There is an interplay between the increasing behavior of $g(t)$ and $h(t)$ that I don't understand. In particular, I would like to know when $f(t,x)$ is positive on the first quarter of it's periodicity, that is when $t\in (0,x/2)$.

Using mathematica, it appears that $f1$ and $f2$ are never negative. In fact, it appears:

$$f(t,x)>(f(x/2,x)/(x/2)) * t$$

for all $t\in(0,x/2)$. However, $f3$ and $f4$ appear to fail being positive. Indeed, focusing on $f4$ and choosing $x\approx 0.82$, it appears that $f(t,x)<0$ for all $t\in(0,x/2)$. The split appears to be at $c\approx 0.919583$. Zooming in very very close to zero, it does appear that at $c$, $f4$ has multiple simple zeros being both positive and negative. However, to the right and left of this value $f4$ appears to be either all positive or all negative.

Likewise, for $c\approx 0.61093879$ and function $f3$, it appears that $f(t,x)$ is positive for $x>c$ and negative for $x<c$, again for $t\in(0,x/2)$. This is very surprising considering the extreme similarity of $f2$ and $f3$. Also, zooming in very very close to zero at $x\approx 0.61093879$ it does not appear that $f3$ has any simple zeros, meaning that is appears there exists some constant $c\approx 0.61093879 $, such that

$$\sum_{m=-\infty}^\infty (-1)^m \tanh((t+cm)/2)e^{-\cosh(t+cm)}=0$$

for all $t\in\mathbb{R}$. That seems very interesting.

Question: Is it indeed the case for $f2$, that $f(t,x)>0$ for all $t\in(0,x/2)$? Moreover is $f(t,x)>(f(x/2,x)/(x/2)) * t$?

Recently, I have become quite obsessed with the follow series:

$$f(t,x)=\sum_{m=-\infty}^{\infty} (-1)^m f(t+x m)$$

where $f$ is analytic. This series automatically produces a periodic function with period $2x$ (assuming it converges). Let use focus on four seemingly similar functions of the form $g(t)e^{-h(t)}$:

$$\begin{align} f1(t)&=t e^{-t^2} \\ f2(t)&=\tanh(t) e^{-\cosh(t)} \\ f3(t)&=\tanh(t/2) e^{-\cosh(t)} \\ f4(t)&=t e^{-\cosh(t)} \end{align}$$

There is an interplay between the increasing behavior of $g(t)$ and $h(t)$ that I don't understand. In particular, I would like to know when $f(t,x)$ is positive on the first quarter of it's periodicity, that is when $t\in (0,x/2)$.

Using mathematica, it appears that $f1$ and $f2$ are never negative. In fact, it appears:

$$f(t,x)>(f(x/2,x)/(x/2)) * t$$

for all $t\in(0,x/2)$. However, $f3$ and $f4$ appear to fail being positive. Indeed, focusing on $f4$ and choosing $x\approx 0.82$, it appears that $f(t,x)<0$ for all $t\in(0,x/2)$. The split appears to be at $c\approx 0.919583$. Zooming in very very close to zero, it does appear that at $c$, $f4$ has multiple simple zeros being both positive and negative. However, to the right and left of this value $f4$ appears to be either all positive or all negative.

Likewise, for $c\approx 0.61093879$ and function $f3$, it appears that $f(t,x)$ is positive for $x>c$ and negative for $x<c$, again for $t\in(0,x/2)$. This is very surprising considering the extreme similarity of $f2$ and $f3$. Also, zooming in very very close to zero at $x\approx 0.61093879$ it does not appear that $f3$ has any simple zeros, meaning that is appears there exists some constant $c\approx 0.61093879 $, such that

$$\sum_{m=-\infty}^\infty (-1)^m \tanh((t+cm)/2)e^{-\cosh(t+cm)}=0$$

for all $t\in\mathbb{R}$. That seems very interesting.

Question: Is it indeed the case for $f2$, that $f(t,x)>0$ for all $t\in(0,x/2)$? Moreover is $f(t,x)>(f(x/2,x)/(x/2)) * t$?

UPDATE

So, it appears that $\sin(xt)$ serves as a better lower bound instead of a linear equation, in fact the series appears to uniformly approach $\sin(xt)$ as $x\to 0$, yielding a ratio of $1$ in their limit. I then went ahead and flipped over the function so that is would be between 0 and 1 (assuming the series is positive), and I then multiplied $t$ by $x$ so as to map the periodicity between $(0,1)$. I then graphed through different values of $x$ to see what $t$ looks like. The below animation is

$$ g(t,x):=\frac{f(x/2,x)*\sin(\pi t)}{f(xt,x)}$$

where the series uses the functions $f_2(t)=\tanh(t)e^{-\cosh(t)}$ and $f_5(t)=\tanh(t/2)e^{-\cosh(t)}$. We see that $f_2$ appears to be strictly decreasing as $x$ increases. Likewise, we see that $f_5$ appears to oscillate around $1$ at the start, in particular since the series for $f_5$ appears to be negative for some values of $x$, we see here the ratio is $>1$ for those same values of $x$. How would we prove that for $t>0$ and $f(t)=\tanh(t)e^{-\cosh(t)}$, we have

$$\frac{d}{dx} g(t,x) = \frac{d}{dx}\frac{f(x/2,x)*\sin(\pi t)}{f(xt,x)}\le 0 $$

for any $x$? How do we prove $\lim g(t,x)=1$ when $x\to 0$ and $\lim g(t,x)=0$ ($t\not= 1/2$) when $x\to\infty$? Working with this series seems particularly difficult, even though we can see what it is doing.

I feel like this is a topic that must exist somewhere as a chapter of periodicity or something, but haven't found much. Recently, I have become quite obsessed with the follow series:

$$f(t,x)=\sum_{m=-\infty}^{\infty} (-1)^m f(t+x m)$$

where $f$ is an analytic. Let us attempt to make this as simple as possible and assume, $f$ is positive on positive reals, $f$ is odd, $f$ has a simple zero at zero (that is $f'(0)>0$), if fact, lets just assume $f(t)=h(t)e^{-g(t)}$ for odd function $h$ with simple zero at zero and $g$ even, and $h$ and $g$ are both increasing functions such that the series converges uniformly.

In general, this series is periodic in $t$ with period $2x$. Likewise, since $f$ is odd, then this series is symmetric on $(0,x)$ with $f(t,x)=f(x-t,x)$. I would like to know what is happening inside the the interval $(0,x/2)$. Clearly, $f(0,x)=f(x,x)=f(2x,x)=0$ and we already know $(0,x)$ is a negative reflection of $(x,2x)$ and that $(0,x/2)$ is a positive reflection of $(x/2,x)$. Hence, the only interval of interest is $(0,x/2)$.

1. Can $f(t,x)$ achieve any simple zeros in the interval $(0,x)$? In all my experiments, $t e^{-t^2}$, $t e^{-\cosh(t)}$, $\tanh(t) e^{-\cosh(t)}$, this does not appear to be possible. I CAN achieve the entire interval being equal to zero, using $t e^{-\cosh(t)}$ and some positive $x$ value. But in general, the interval appears to be either all positive all negative or equal to zero. It appears to be impossible for $f(t,x)$ to be non-constant and have zeros in $(0,x)$.

2. For what functions can the interval be all zero? This appears to happen for some $x$ and for the function $t e^{-\cosh(t)}$. However, for $\tanh(t) e^{-\cosh(t)}$ and $t e^{-t^2}$, this appears to never happen. But why? What is different about those?

3. Can we find a lower bound? For example, it appears to be the case that $$f(t,x) > (f(x/2,x)/(x/2)) t$$ for $t\in (0,x/2)$, whenever $f(t,x)$ is actually positive on the interval $(0,x/2)$. More general, it appears $$|f(x,t)|\ge |f(x/2,x)/(x/2)| t$$ for $t$ in $(0,x/2)$ for any functions described above of the form $f(t)=h(t)e^{-g(t)}$.

4. Is it the case that if $f(x/2,x)=0$ then $f(x,t)=0$ for all $t$ in $(0,x)$?

(4) appears very surprising to me, if it is true. Any materials or references would be wonderful.

Recently, I have become quite obsessed with the follow series:

$$f(t,x)=\sum_{m=-\infty}^{\infty} (-1)^m f(t+x m)$$

where $f$ is analytic. This series automatically produces a periodic function with period $2x$ (assuming it converges). Let use focus on four seemingly similar functions of the form $g(t)e^{-h(t)}$:

$$\begin{align} f1(t)&=t e^{-t^2} \\ f2(t)&=\tanh(t) e^{-\cosh(t)} \\ f3(t)&=\tanh(t/2) e^{-\cosh(t)} \\ f4(t)&=t e^{-\cosh(t)} \end{align}$$

There is an interplay between the increasing behavior of $g(t)$ and $h(t)$ that I don't understand. In particular, I would like to know when $f(t,x)$ is positive on the first quarter of it's periodicity, that is when $t\in (0,x/2)$.

Using mathematica, it appears that $f1$ and $f2$ are never negative. In fact, it appears:

$$f(t,x)>(f(x/2,x)/(x/2)) * t$$

for all $t\in(0,x/2)$. However, $f3$ and $f4$ appear to fail being positive. Indeed, focusing on $f4$ and choosing $x\approx 0.82$, it appears that $f(t,x)<0$ for all $t\in(0,x/2)$. The split appears to be at $c\approx 0.919583$. Zooming in very very close to zero, it does appear that at $c$, $f4$ has multiple simple zeros being both positive and negative. However, to the right and left of this value $f4$ appears to be either all positive or all negative.

Likewise, for $c\approx 0.61093879$ and function $f3$, it appears that $f(t,x)$ is positive for $x>c$ and negative for $x<c$, again for $t\in(0,x/2)$. This is very surprising considering the extreme similarity of $f2$ and $f3$. Also, zooming in very very close to zero at $x\approx 0.61093879$ it does not appear that $f3$ has any simple zeros, meaning that is appears there exists some constant $c\approx 0.61093879 $, such that

$$\sum_{m=-\infty}^\infty (-1)^m \tanh((t+cm)/2)e^{-\cosh(t+cm)}=0$$

for all $t\in\mathbb{R}$. That seems very interesting.

Question: Is it indeed the case for $f2$, that $f(t,x)>0$ for all $t\in(0,x/2)$? Moreover is $f(t,x)>(f(x/2,x)/(x/2)) * t$?