# Find the roots of the given hyperbolic equation

• chwala
In summary, the roots of the equation ##x^2 - 2x \cosh u + 1 = 0## can be found by using the formula ##x = \cosh u \pm \sinh u = e^{\pm u}## or ##x = e^{-u}##. This equation is also related to the Lorentz boost transformation and has eigenvalues of the Doppler factor and its reciprocal.
chwala
Gold Member
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!

chwala said:
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}##?

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

$$x^2 - 2x\cosh u + 1 = (x - \cosh u)^2 + 1 - \cosh^2 u = (x - \cosh u)^2 - \sinh^2 u.$$ Hence $$x = \cosh u \pm \sinh u = e^{\pm u}.$$

chwala and robphy
Simplest is perhaps $$x^2 - 2\cosh u + 1 = x^2 - (e^{u} + e^{-u})x + 1 = (x - e^u)(x - e^{-u})$$ since $e^ue^{-u} = 1.$

Last edited:
malawi_glenn, robphy, chwala and 1 other person
pasmith said:
Simplest is perhaps $$x^2 - 2\cosh u + 1 = x^2 - (e^{u} + e^{-u})x + 1 = (x - e^u)(x - e^{-u})$$ 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
$0=\det\left( \begin{array}{cc} \cosh\theta -\lambda & \sinh\theta \\ \sinh\theta & \cosh\theta -\lambda \end{array}\right)$
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 $\left(\begin{array}{c}1\\\pm 1\end{array}\right)$.

Last edited:
PeroK

## 1. What is a hyperbolic equation?

A hyperbolic equation is a type of mathematical equation that contains at least one hyperbolic function, such as sinh, cosh, or tanh. These functions are similar to trigonometric functions, but are defined using exponential functions instead of circular functions.

## 2. How do I find the roots of a hyperbolic equation?

To find the roots of a hyperbolic equation, you can use a variety of methods such as substitution, factoring, or graphing. You can also use the quadratic formula for certain types of hyperbolic equations.

## 3. What is the importance of finding the roots of a hyperbolic equation?

Finding the roots of a hyperbolic equation can help you solve problems in various fields such as physics, engineering, and economics. It can also help you understand the behavior of hyperbolic functions and their applications.

## 4. Are there any special cases when finding the roots of a hyperbolic equation?

Yes, there are some special cases when finding the roots of a hyperbolic equation. For example, if the equation contains complex numbers, you may need to use techniques such as the complex conjugate to find the roots.

## 5. Can I use a calculator to find the roots of a hyperbolic equation?

Yes, you can use a scientific calculator or a graphing calculator to find the roots of a hyperbolic equation. However, it is important to understand the underlying concepts and methods to ensure accurate results.

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