- #1
LAHLH
- 409
- 1
Hi,
I start with [tex] arctanh\left(\frac{A}{\sqrt{A^2-1}}\right)=\frac{1}{2}ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)[/tex]
The function [tex] \frac{A}{\sqrt{A^2-1}} [/tex] is real, and since A>1, it too is always greater than 1.
Is it true that it should really be the modulus around the argument of ln? therefore I can manipulate it as follows:
[tex] ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{-(1-\frac{A}{\sqrt{A^2-1}})}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{(1-\frac{A}{\sqrt{A^2-1}})}\right)[/tex]
[/tex]
thanks
I start with [tex] arctanh\left(\frac{A}{\sqrt{A^2-1}}\right)=\frac{1}{2}ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)[/tex]
The function [tex] \frac{A}{\sqrt{A^2-1}} [/tex] is real, and since A>1, it too is always greater than 1.
Is it true that it should really be the modulus around the argument of ln? therefore I can manipulate it as follows:
[tex] ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{-(1-\frac{A}{\sqrt{A^2-1}})}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{(1-\frac{A}{\sqrt{A^2-1}})}\right)[/tex]
[/tex]
thanks