- #1
Bill Foster
- 338
- 0
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 [tex]f(x^{*})=x^{*}[/tex]
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:
[tex]\tanh{ax}=\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}=x[/tex]
This reduces to [tex]e^{2ax}=\frac{1+x}{1-x}[/tex]
Then I can take the Taylor expansion...
[tex]e^{2ax}=1+2ax+\frac{(2ax)^2}{2!}+\frac{(2ax)^3}{3!}+\frac{(2ax)^4}{4!}+...=\frac{1+x}{1-x}[/tex]
But as you can see, that doesn't really help.
Then I tried this:
[tex]\tanh{(ax)}=x[/tex]
[tex]ax=\tanh^{-1}{(x)}[/tex]
Differentiate both sides:
[tex]a=\frac{1}{1-x^2}[/tex]
Solve for x:
[tex]x=\pm \sqrt{1-\frac{1}{a}}[/tex]
Problem is, it doesn't work out in the calculator.
Suppose I let a=2. That means [tex]x=\frac{1}{\sqrt{2}}[/tex]
So the following should hold true:
[tex]\tanh{(\frac{2}{\sqrt{2}})}=\frac{1}{\sqrt{2}}[/tex]
But it doesn't. The correct value is
[tex]\tanh{(\frac{2}{\sqrt{2}})}=0.88839[/tex]
Is there an analytical solution to tanh(ax)=x ?
Thanks.
Last edited: