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Let $n$ be a positive integer. Let $z$ be a complex number. Let $a_1,a_2,...,a_n$ be a given sequence of distinct reals larger than $\ln(2)$.

How many distinct solutions are there to $\displaystyle \sum_{i=1}^n 2 \sinh(a_i z)=z$ ?

Or in other words : How many distinct fixpoints does $\displaystyle \sum_{i=1}^n 2 \sinh(a_i z)$ have ?

For small values of $n$ I can prove some things but for large $n$ I see trends , but no easy proofs.

Splitting the equation up into real and imaginary parts works for small $n$ but I do not know if that is a good method ( for either small or large $n$ ).

Probably related is the equation $exp(z) = z$ I guess ?

($exp(z)=z$ has 2 solutions.)

*edit : giving some background *

Lets say I want to solve $exp(z)=z$.

Method $1$ : Using the argument principle where we take the contour as a circle centered at the origin and with a radius going to infinity.

Method $2$ : Using the Lambert W function.

See : http://en.wikipedia.org/wiki/Lambert_W

http://en.wikipedia.org/wiki/Variation_of_argument

Now the problem with those is that I think they are too complicated for this problem and it hard to generalize them to solve the more general case

$$\displaystyle \sum_{i=1}^n 2 \sinh(a_i z)=z$$

The reason is that - in the case of growing $n$ - for method $1$ we get more and more complicated integrals and for method $2$ we need more and more special functions.

I could be wrong , but those are my ideas.

To solve $exp(z)=z$ I used a different method. Possible not efficient itself. Im not sure if this other method is good for small $n$ or large $n$.

Lets call this other method : method $4$.

I say method $4$ because I wonder about method$3$ which I will give first. Method$3$ still needs some work and might not be a good alternative to the first 2 methods.

method $3$ : going from one fixpoint to another ?

$exp(z) = z$

$exp(z+kz) = z + kz$

$exp(z)*exp(kz) = z + kz$

$z*exp(kz) = z + kz$

$z*exp(kz) = z(1+k)$

$exp(k)^z = 1+k$

That seems intresting , but I am stuck here. Solving the last with method$1$ or method$2$ could not make method$3$ any easier than those methods themselves I assume.

So finally method$4$ :

$exp(z)=z$

$exp(r+ti) = r+ti$ where $r$ and $t$ are real and $i$ is the imaginary unit.

We know $|exp(r+ti)|=|exp(r)|=exp(2r)$.

Now if $exp(r+ti)=r+ti$ it follows that $|exp(r+ti)|=|r+ti|$.

Then we get $exp(r)^2 = r^2 + t^2$. (yes this is Pythagoras in a way)

Now we can solve $t$ as function of $r$.

$t = \sqrt {exp(r)^2 - r^2}$

We get $exp(r+(\sqrt {exp(r)^2 - r^2}i) = r + (\sqrt {exp(r)^2 - r^2})i$.

Now you can probably see what I meant with splitting up into real and imaginary parts.

We now have the system of equations :

A) $exp(r) cos(\sqrt {exp(r)^2 - r^2}) = r$

B) $exp(r) sin(\sqrt {exp(r)^2 - r^2}) = \sqrt {exp(r)^2 - r^2}$

Now to avoid tautologies like $sin^2 + cos^2 = 1$ we probably better avoid considering A)^2 + B)^2.

So to combine the equations A) and B) I used f(r) = A)^4 + B)^4.

Thus : $f(r)=0$ or explicit :

$$(exp(r) cos(\sqrt {exp(r)^2 - r^2}) - r)^4 + (exp(r) sin(\sqrt {exp(r)^2 - r^2}) - \sqrt {exp(r)^2 - r^2})^4 = 0$$

It is now clear that this can only have a finite amount of solutions since $r$ is real. This reduces the case to real calculus.

We can consider $\frac{d f}{d r}$ and $\frac{d^2 f}{d^2 r}$ and some limit(s) to show that there are only 2 solutions to $exp(z)=z$.

The bigger picture is however

if method$4$ is the best ?

if method$4$ extends easily to the general case $\displaystyle \sum_{i=1}^n 2 \sinh(a_i z)=z$ ?

how we prove the amount of zero's a priori for the given sequence $a_1,a_2,...,a_n$ , where with a priori I mean without doing every $n$ seperately or even worse computing the actual zero's.

In other words something like e.g.

$\displaystyle \sum_{i=1}^n 2 \sinh(a_i z)=z$ has at least $2n+1$ solutions. Would be nice ...

I hope this edit clarifies the problem a bit.

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  • $\begingroup$ Why don't you share your observations with us? $\endgroup$ – Igor Rivin Dec 13 '13 at 0:15
  • $\begingroup$ @IgorRivin I edited. Hope this is what you wanted. $\endgroup$ – mick Dec 13 '13 at 22:16

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