Solve tanh(ax)=x: Fixed Points for Non-linear Dynamics

[Bill Foster]

Homework Statement

Obtain the fixed points for the following as a function of a, where 0<a<infinity:

f(x)=tanh(ax)

Homework Equations

The fixed point is given by f(x^{*})=x^{*}

The Attempt at a Solution

It's basically a math problem: tanh(ax)=x Find x in terms of a.

I tried it two ways:

\tanh{ax}=\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}=x

This reduces to e^{2ax}=\frac{1+x}{1-x}

Then I can take the Taylor expansion...

e^{2ax}=1+2ax+\frac{(2ax)^2}{2!}+\frac{(2ax)^3}{3!}+\frac{(2ax)^4}{4!}+...=\frac{1+x}{1-x}

But as you can see, that doesn't really help.

Then I tried this:

\tanh{(ax)}=x
ax=\tanh^{-1}{(x)}

Differentiate both sides:

a=\frac{1}{1-x^2}

Solve for x:

x=\pm \sqrt{1-\frac{1}{a}}

Problem is, it doesn't work out in the calculator.

Suppose I let a=2. That means x=\frac{1}{\sqrt{2}}

So the following should hold true:

\tanh{(\frac{2}{\sqrt{2}})}=\frac{1}{\sqrt{2}}

But it doesn't. The correct value is

\tanh{(\frac{2}{\sqrt{2}})}=0.88839

Is there an analytical solution to tanh(ax)=x ?

Thanks.

 

[nicksauce]

Is there an analytical solution to tanh(ax)=x ?

Answer: no.

 

[Bill Foster]

Is there an analytical solution to tanh(ax)=x ?

Answer: no.

So would using the Taylor series approximation be the best way?

Or do you know of a better way?

 

[salai]

As far as I know, the definition of fixed point you gave is wrong. The fixed point is defined as \frac{dx}{dt} = f(x*) = 0 (look at Nonlinear Dynamics and Chaos, Steven H. Strogatz) then the problem is to find x* satisfying: tanh(x*)= 0.. Additionally, although, tanh(x*) = x* seems not to be solved anaytically, it can be solved numerically, but there may be inappropriate results if you use nonliiinear fixed point anaysis.

 

[HallsofIvy]

As far as I know, the definition of fixed point you gave is wrong. The fixed point is defined as \frac{dx}{dt} = f(x*) = 0 (look at Nonlinear Dynamics and Chaos, Steven H. Strogatz) then the problem is to find x* satisfying: tanh(x*)= 0.. Additionally, although, tanh(x*) = x* seems not to be solved anaytically, it can be solved numerically, but there may be inappropriate results if you use nonliiinear fixed point anaysis.

No, his definition of "fixed point" is correct. What you are referring to is an "equilibrium" point of a differential equation.

As nicksause said, there is no "analytic" solution to that equation. I would use Newton's method to numeraically solve the equation.

 

[Bill Foster]

As far as I know, the definition of fixed point you gave is wrong. The fixed point is defined as \frac{dx}{dt} = f(x*) = 0 (look at Nonlinear Dynamics and Chaos, Steven H. Strogatz) then the problem is to find x* satisfying: tanh(x*)= 0.. Additionally, although, tanh(x*) = x* seems not to be solved anaytically, it can be solved numerically, but there may be inappropriate results if you use nonliiinear fixed point anaysis.

That is incorrect. A fixed point is where x=f(x).

If f(x)=ax(1-x) where a<1, then the only fixed point will be at x=0, which is a trivial fixed point. When a>1, then fixed point will be given by x*=1-1/a