# The equation $2 \cosh(3.1786803659501505 z) = z$?

Let $a$ be a positive real number and $z$ a complex number.

I was wondering about the equation $2 \cosh(a z) = z$ where we solve for $z$.

Clearly if $z$ is a solution than so is its conjugate.

It appears that if $0<a<c$ (for some real $c$) then the equation $2\cosh(a z) = z$ has 2 solutions.

And also if $a>c$ then we have more than 2 solutions to $2 \cosh(a z) = z$. (if $a$ is slightly larger than $c$ the equation has 4 solutions)

So I wonder about $c$ and I have $4$ questions:

$1)$ $c$ is about $3.1786803659501505$ but can $c$ be expressed by an integral or a limit ?

$2)$ How many solutions does $2 \cosh(c z) = z$ have ? Is it $2$ or $4$ ?

$3)$ I also wonder if I can call $2 \cosh((c+\epsilon) z) = z$ a bifurcation near $c$ ? I think not because the 4 zero's are always quite far away from eachother whereas for the bifurcations of e.g. the logistic map we get zero's that are arbitrary close.

$4)$ Similar to question $1)$ : Consider the 4 zero's of $2 \cosh((c+\epsilon) z) = z$. Can those zero's be expressed by an integral or a limit ? How about the sum of some of the zero's ?

Btw is this considered chaos theory , algebra or complex analysis ? Or all of that ? Im a bit insecure about the title and tags I should use. Is it ok to use many decimals in a title ?

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What about $a = 0.3313717096745908$? Before this point there are two real solutions; afterwards there can only be complex solutions. – A Blumenthal Jan 18 '13 at 23:05
@ABlumenthal Well what about it ? – mick Jan 18 '13 at 23:25
I'm just confused as to why you're singling out this particular value $c$; is there some theoretical justification or did you find it exclusively numerically? – A Blumenthal Jan 18 '13 at 23:40
I also wanted to say something about dynamical systems. What differentiates DS from the local theory of (continuous, differentiable, measurable, etc) functions is that DS concentrates on the asymptotic behavior of systems, and not just the 'time one iterate' of a given map. You're asking questions about fixed points of the map $z \mapsto 2 \cosh (a z)$; such questions can usually be answered by applying the inverse function theorem in one of its guises. If you were interested in the asymptotics of the number of periodic points $P(n)$ of period $n$ for the map, then DS would apply. – A Blumenthal Jan 20 '13 at 1:05
I would say that there are too many decimal places in the title because more decimal places do not add much value. I think you could re-title it something like "Number of solutions to $2\cosh(cz) = z$ with $c \approx 3.1786...$. But it's up to you. – Brad Jun 11 '14 at 20:50

Generally speaking, zeroes of a complex function don't appear in isolation as a parameter is continuously varied. They can appear and disappear in sets (generically, pairs) via bifurcations, and can move around, but can't just come and go as they please.

In your case, the appearance of "additional" roots has more to do with the basins of attraction of your root finder than it does with your function. If you ask WolframAlpha to solve $2\cosh(az)=z$ with $a=3.17$, it finds two roots (at $z\approx0.08\pm 0.48 i$); if you ask it with $a=3.18$, it finds the same two roots, moved slightly, plus two more roots (at $z\approx -0.21 \pm 1.45 i$). But in fact these "new" roots were already present at $a=3.17$, and have simply moved into the range where the root finder detects them. To find them for smaller $a$, you can give WolframAlpha a hint, by first transforming $z\mapsto (-0.21+1.45i) + z;$ this moves one of the "new" roots near the origin, where it may be easier to find. And indeed, solving $2\cosh(a(z-0.21+1.45i))=z-0.21+1.45i$ with $a=3.17$ does yield a root at $z\approx -0.004 + 0.009 i$.

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How do you prove that zero's cant appear or disappear without bifurcations ? Note that even if we have an infinite amount of zero's , then removing one is still an infinite amount. – mick Jun 12 '14 at 21:55
Cauchy's argument principle says that the integral around a closed contour of $f'(z)/f(z)$ is the number of zeroes inside the contour, counting order, as long as no zeroes are on the contour. Now, if the function $f$ is changing continuously with a parameter $a$, this integral must also change continuously (unless there's a zero on the contour), but since it's always an integer, that means it must remain constant. So in fact if $f$ is analytic, its zeroes can only move around (or be created simultaneously with poles, or a zero of order $2$ can split into two zeroes of order $1$, or etc.). – mjqxxxx Jun 13 '14 at 3:21
Since $f(a,z)$ is differentiable with respect to $a$ in this case, you can even find the "velocity" of each zero with respect to $a$. – mjqxxxx Jun 13 '14 at 3:23

Too long for a comment.

Note that $f(z) = 2\cosh(az)-z$ is an entire function with an essential singularity at $\infty$ (for $a\neq 0$). Because of Picard's great theorem we would expect that there are infinitely many zeros of $f$ (Picard implies that the equation $f(z) = b$ has infinitely many solutions for every complex $b$ with at most two exceptions).

Of course, it's conceivable that $0$ is one of the exceptional values (I'll think about that some more), but even for $a$ close to $0$, I can find many solutions to your equation. For example, if $a = 0.1$, here are just a few solutions:

\begin{align} z &= 2.041835988 \\ z &= 35.7606533293 \\ z &= 44.49177815\pm73.07181538i \\ z &= -40.28136974\pm39.12782185i \\ z &= 68.66860456\pm957.4697985i \\ z &= -68.99027078\pm988.9051729i \end{align}

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That may be. I only considered " small complex numbers ". But it seems that locally 2 fixpoints became 4. Like a bifurcation or such. And then locally the questions remain. Not sure what locally means here , maybe a bound on the absolute value or such ... Btw : thanks. – mick Jun 11 '14 at 21:12
@mick What precision did you work to, and how did you rootfind? I would trust neither standard libraries nor e.g. Alpha's standard rootfinding package to the degree of precision necessary to find some of the additional solutions mentioned here, and I wonder if what you're seeing might in part be an artifact of precision issues. – Steven Stadnicki Jun 11 '14 at 21:59
I checked a few with 1000 digit working precision and they look as genuine solutions. – mrf Jun 11 '14 at 22:01
@mrf makes solid sense to me (assuming that cosh is being computed to that precision, of course); the question was aimed at OP, to make sure that their own computations were being done suitably precisely. – Steven Stadnicki Jun 11 '14 at 22:03
Perhaps OP means real solutions only? – vonbrand Jun 12 '14 at 0:46