# Find the root of the given equation in terms of ##u##

• chwala
In summary, the conversation discusses finding the values of u in terms of x in a given equation. The conversation goes through the process of solving for x and simplifying the equation to two possibilities, x1 and x2. The conversation then shows how x1 can be written as tanh u and x2 can be written as -coth u. The final part of the conversation mentions the need to find u in terms of x, which is the main task at hand.
chwala
Gold Member
Homework Statement
##x^2\sinh 2u +2x - \sinh 2u=0##
Relevant Equations
hyperbolic equations
Hmmmm was a nice one... took me some time to figure out ...seeking alternative ways ...

My working;

##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+\sqrt{1+ \sinh^2 2u}}{\sinh 2u}## or ##x=\dfrac{-1-\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##

##x=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{\sinh u}{\cosh u }##

Firstly, we have the possibility ##x=0 \Longleftrightarrow u=0.## Next, we assume ##x,u\neq 0.## Then

\begin{align*}
0&=x^2+\dfrac{2}{\sinh 2u}x-1\\[6pt]
x&=-\dfrac{1}{\sinh 2u} \pm \sqrt{\dfrac{1}{\sinh^2 2u}+1}=-\dfrac{1}{\sinh 2u} \pm \dfrac{\sqrt{1+\sinh^22u}}{\sinh 2u}\\[6pt]
x&=\dfrac{1}{\sinh 2u}\left(-1\pm \cosh 2u\right)=\dfrac{1}{2\sinh u\cosh u}\left(-1\pm \left(1+2\sinh^2 u\right)\right)\\[6pt]
x_1&=\dfrac{\sinh u}{\cosh u}\, , \,x_2=-\dfrac{1+\sinh^2u}{\sinh u\cosh u}=-\dfrac{\cosh u}{\sinh u}\\[6pt]
x_1&=\tanh u=1-\dfrac{2}{e^{2u}+1}\, , \,x_2=-\coth u=-1-\dfrac{2}{e^{2u}-1}
\end{align*}

This is the first half with all you left out. Now, you have to find ##u_1=u_1(x_1)## and ##u_2=u_2(x_2).##

Last edited:
SammyS, topsquark and chwala
chwala said:
My working;
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u}##
##x=\dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##
Sort of a style issue: you don't need to start each equation with "x = ...". Instead, you can have a continued equality.
E.g.:
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u} = \dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##
chwala said:
##x=\dfrac{-1+\sqrt{1+ \sinh^2 2u}}{\sinh 2u}## or ##x=\dfrac{-1-\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##
No need for the line above. The last expression I wrote gives exactly the same values of x.
chwala said:
##x=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}##
You lost me here -- how did you get from the +/- values of x to this line?
chwala said:
##x=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}##

##x=\dfrac{\sinh u}{\cosh u }##
The expression on the right can be written as ##\tanh u##.

chwala
Mark44 said:
Sort of a style issue: you don't need to start each equation with "x = ...". Instead, you can have a continued equality.
E.g.:
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u} = \dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##
No need for the line above. The last expression I wrote gives exactly the same values of x.
You lost me here -- how did you get from the +/- values of x to this line?

The expression on the right can be written as ##\tanh u##.
Noted @Mark44 ... I will get back on lost part later in the day...cheers mate

Mark44 said:
Sort of a style issue: you don't need to start each equation with "x = ...". Instead, you can have a continued equality.
E.g.:
##x=\dfrac{-2±\sqrt{4+4 \sinh^2 2u}}{2 \sinh 2u} = \dfrac{-2±2\sqrt{1+ \sinh^2 2u}}{2 \sinh 2u}##
##=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}##
No need for the line above. The last expression I wrote gives exactly the same values of x.
You lost me here -- how did you get from the +/- values of x to this line?

The expression on the right can be written as ##\tanh u##.

by...

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}=\dfrac{-1±\sqrt{\cosh^2 2u}}{\sinh 2u}=\dfrac{-1±\cosh 2u}{\sinh 2u}=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}## ...

chwala said:
by...

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}=\dfrac{-1±\sqrt{\cosh^2 2u}}{\sinh 2u}=\dfrac{-1±\cosh 2u}{\sinh 2u}=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}## ...
If you write ##x=a\pm b## then this is an abbreviation of ##x\in \{a-b,a+b\}.##

If you split it, then you get ##x=a+b## or ##x=a-b.##

What you cannot do is ##x=a\pm b \neq a +b.##

chwala
fresh_42 said:
If you write ##x=a\pm b## then this is an abbreviation of ##x\in \{a-b,a+b\}.##

If you split it, then you get ##x=a+b## or ##x=a-b.##

What you cannot do is ##x=a\pm b \neq a +b.##

aaargh i get you @fresh_42 ...should have double checked... i will amend my response in post ##5## as follows ...just in following with your working,

##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}=\dfrac{-1±\sqrt{\cosh^2 2u}}{\sinh 2u}=\dfrac{-1±\cosh 2u}{\sinh 2u}=\dfrac{-1±(1+2\sinh^{2} u)}{2\sinh u \cosh u}##

Now, we have two possibilities;

##x_1=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{\sinh u}{\cosh u}##

i can also see different variations on this by you and also by @Mark44 to be specific;

##\dfrac{\sinh u}{\cosh u}=\tanh u##.

Also,

##x_2=\dfrac{-1-1-2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{-2-2\sinh^{2} u}{2\sinh u \cosh u}=-\dfrac{2(1+\sinh^{2} u)}{2\sinh u \cosh u}=-\dfrac{1+\sinh^{2} u}{\sinh u \cosh u}##

##=-\dfrac{\cosh^{2} u}{\sinh u \cosh u}=-\dfrac{\cosh u}{\sinh u }##

Thanks @fresh_42 , i actually did not go through earlier posts in detail because of other work related duties...regards,

Last edited:
chwala said:
Also,
##x_2= <snip>=-\dfrac{\cosh u}{\sinh u }##
##= -\coth u##.

chwala
chwala said:
aaargh i get you @fresh_42 ...should have double checked... i will amend my response in post ##5## as follows ...just in following with your working,
. . .

Thanks @fresh_42 , i actually did not go through earlier posts in detail because of other work related duties...regards,
The last line @fresh_42 wrote in Post #6, which you just responded to.
fresh_42 said:
@fresh_42 has some great ending lines. From Post #2 :
fresh_42 said:
This is the first half with all you left out. Now, you have to find ##u_1=u_1(x_1)## and ##u_2=u_2(x_2).##
That "first half" he refers to is your amendment (Post #7).

What you left out is to find ##u## in terms of ##x## .

chwala
SammyS said:
The last line @fresh_42 wrote in Post #6, which you just responded to.

@fresh_42 has some great ending lines. From Post #2 :

That "first half" he refers to is your amendment (Post #7).

What you left out is to find ##u## in terms of ##x## .
Noted...let me look at this later...cheers mate.

Just by quick working; i am having

From,

##x^2\sinh 2u+2x-\sinh2u=0##

i will get,

##2x=\sinh 2u - x^2\sinh2u=\sinh 2u(1-x^2)##

therefore,

##\dfrac{2x}{1-x^2}=\sinh 2u##

##2u=\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]####u=\dfrac{1}{2}\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]=\dfrac{1}{2}\sin^{-1} \left[\dfrac{1}{1-x}-\dfrac{1}{1+x}\right]##

Last edited:
chwala said:
Just by quick working; i am having

From,

##x^2\sinh 2u+2x-\sinh2u=0##

i will get,

##2x=\sinh 2u - x^2\sinh2u=\sinh 2u(1-x^2)##

therefore,

##\dfrac{2x}{1-x^2}=\sinh 2u##
The next line below doesn't follow from the line above. With as much LaTeX as you usually put in, it would be a good idea to proofread your work before posting it. The icon in the menu bar that looks like a sheet of paper with a magnifying glass is the Preview button.
chwala said:
##2u=\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]####u=\dfrac{1}{2}\sin^{-1} \left[\dfrac{2x}{1-x^2}\right]=\dfrac{1}{2}\sin^{-1} \left[\dfrac{1}{1-x}-\dfrac{1}{1+x}\right]##

Mark44 said:
The next line below doesn't follow from the line above. With as much LaTeX as you usually put in, it would be a good idea to proofread your work before posting it. The icon in the menu bar that looks like a sheet of paper with a magnifying glass is the Preview button.
Let me amend that now...

Amendment to post ##11##

From,

##x^2\sinh 2u+2x-\sinh2u=0##

i will get,

##2x=\sinh 2u - x^2\sinh2u=\sinh 2u(1-x^2)##

therefore,

##\dfrac{2x}{1-x^2}=\sinh 2u##

##2u=\sinh^{-1} \left[\dfrac{2x}{1-x^2}\right]####u=\dfrac{1}{2}\sinh^{-1} \left[\dfrac{2x}{1-x^2}\right]=\dfrac{1}{2}\sinh^{-1} \left[\dfrac{1}{1-x}-\dfrac{1}{1+x}\right]##

Mark44

## 1. What does it mean to "find the root" of an equation?

Finding the root of an equation means solving for the value(s) of the variable that make the equation true. In other words, it is finding the value(s) of the variable that satisfy the equation.

## 2. What does it mean to find the root "in terms of ##u##"?

Finding the root in terms of ##u## means expressing the value(s) of the variable in relation to the variable ##u##. This is usually done when the given equation involves multiple variables and we want to isolate the variable ##u##.

## 3. How do I find the root of a given equation in terms of ##u##?

To find the root of a given equation in terms of ##u##, you can use algebraic methods such as factoring, substitution, or the quadratic formula. The specific method used will depend on the type of equation and its complexity.

## 4. Why is it important to find the root of an equation?

Finding the root of an equation is important because it allows us to solve for unknown values and understand the relationship between different variables in the equation. It also helps in solving real-world problems and making predictions based on mathematical models.

## 5. Can there be more than one root for a given equation in terms of ##u##?

Yes, there can be more than one root for a given equation in terms of ##u##. This is especially true for equations with higher degrees or those with multiple variables. In such cases, there can be multiple solutions that satisfy the equation and each of these solutions can be considered a root in terms of ##u##.

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