07/18/2022, 05:50 AM

07/18/2022, 09:56 AM

It depends what eta means for you.

For me it is the point where a qualitative change happens:

When moving the base from below eta to above eta, \(b^x\) has two real fixed points (attracting and repelling), then one parabolic fixed point, and above eta two repelling complex conjugated fixed points.

When we look at 2sinh - or better \(b^x - b^{-x}\) the situation is a bit different. We have either 3 fixed points (attracting, repelling, attracting) or one fixed point (repelling) as this picture shows:

[attachment=1772]

So the goal would be to find out at which base there is the change, i.e. having one parabolic fixed point.

The parabolic fixed point of \(f(x)=b^x - b^{-x}\) would where \(f'(x) = 0\) and we know already that it would be at \(x=0\).

\begin{align}

f'(x) = \ln(b)b^x+\ln(b)b^{-x} &= 1\\

f'(0) = \ln(b)+\ln(b) &= 1\\

b &= e^{1/2}

\end{align}

Which then looks like this:

[attachment=1773]

So \(b=e^{1/2}\) would be our "eta" for 2sinh. However the Euler number itself has not much to do with our reasoning on iterating functions, so that should stay untouched. Also eta as we have it, should not be touched though, otherwise we come into languague confusion if someone talks about this eta and another one talks about that eta.

For me it is the point where a qualitative change happens:

When moving the base from below eta to above eta, \(b^x\) has two real fixed points (attracting and repelling), then one parabolic fixed point, and above eta two repelling complex conjugated fixed points.

When we look at 2sinh - or better \(b^x - b^{-x}\) the situation is a bit different. We have either 3 fixed points (attracting, repelling, attracting) or one fixed point (repelling) as this picture shows:

[attachment=1772]

So the goal would be to find out at which base there is the change, i.e. having one parabolic fixed point.

The parabolic fixed point of \(f(x)=b^x - b^{-x}\) would where \(f'(x) = 0\) and we know already that it would be at \(x=0\).

\begin{align}

f'(x) = \ln(b)b^x+\ln(b)b^{-x} &= 1\\

f'(0) = \ln(b)+\ln(b) &= 1\\

b &= e^{1/2}

\end{align}

Which then looks like this:

[attachment=1773]

So \(b=e^{1/2}\) would be our "eta" for 2sinh. However the Euler number itself has not much to do with our reasoning on iterating functions, so that should stay untouched. Also eta as we have it, should not be touched though, otherwise we come into languague confusion if someone talks about this eta and another one talks about that eta.

07/18/2022, 10:01 AM

Thank you for the reply, but what I was asking about is what is the largest real number x, such that if you used the 2sinh method to try calculate pentation base x, x pentated to the infinity would converge. And what would converge to.