Find the roots of the given hyperbolic equation

[chwala]

Homework Statement

Find the roots of the following equation in terms of $u$.

$x^2-2x \cosh u +1 = 0$

Relevant Equations

hyperbolic trigonometric equations

This is a textbook question and i have no solution. My attempt:

We know that $\cosh x = \dfrac{e^x + e^{-x}}{2}$

and $\cosh u = \dfrac{{x^2 + 1}}{2x}$ it therefore follows that;

$e^{2u} = x^2$

$⇒u = \dfrac {2\ln x}{2}$

$u=\ln x$

$x=e^u$

Your insight or any other approach welcome guys!

 

[PeroK]

Homework Statement:: Find the roots of the following equation in terms of $u$.

$x^2-2x \cosh u +1 = 0$
Relevant Equations:: hyperbolic trigonometric equations

This is a textbook question and i have no solution. My attempt:

We know that $\cosh x = \dfrac{e^x + e^{-x}}{2}$

and $\cosh u = \dfrac{{x^2 + 1}}{2x}$ it therefore follows that;

$e^{2u} = x^2$

$⇒u = \dfrac {2\ln x}{2}$

$u=\ln x$

$x=e^u$

Your insight or any other approach welcome guys!

What about $x = e^{-u}$?

 

[pasmith]

It's a quadratic; there should be two roots.

<br /> x^2 - 2x\cosh u + 1 = (x - \cosh u)^2 + 1 - \cosh^2 u = (x - \cosh u)^2 - \sinh^2 u. Hence <br /> x = \cosh u \pm \sinh u = e^{\pm u}.

 

[pasmith]

Simplest is perhaps <br /> x^2 - 2\cosh u + 1 = x^2 - (e^{u} + e^{-u})x + 1 = (x - e^u)(x - e^{-u}) since e^ue^{-u} = 1.

 

[robphy]

Simplest is perhaps <br /> x^2 - 2\cosh u + 1 = x^2 - (e^{u} + e^{-u})x + 1 = (x - e^u)(x - e^{-u}) since e^ue^{-u} = 1.

There's an x missing in the 2nd-term (compare with the previous post).

By the way,...
The equation looked familiar to me. It's related to special relativity.
It's the characteristic equation
0=\det\left( \begin{array}{cc} \cosh\theta -\lambda &amp; \sinh\theta \\ \sinh\theta &amp; \cosh\theta -\lambda \end{array}\right)
to find the eigenvalues of the Lorentz boost transformation
(where \theta is the rapidity, \tanh\theta =v/c is the dimensionless-velocity, and \cosh\theta =\gamma=\frac{1}{\sqrt{1-(v/c)^2}} is the time-dilation factor ).
The eigenvalues \lambda_{\pm}= e^{\pm \theta} are the Doppler factor (Bondi k-factor) and its reciprocal.
The eigenvectors are along the light-cone \left(\begin{array}{c}1\\\pm 1\end{array}\right).