problem regards hyperbolic function

• Sep 5th 2013, 03:51 AM
junkwisch
problem regards hyperbolic function
Hi I have been having a problem regards this question. I spent some time trying to sit through it but to no result.

The Question is

Show that

arccosech x = log((1+squareroot(1+x^2)) for x>0
log((1-squareroot(1+x^2)) for x<0

Justify the sign of any square roots.

So this is what I did

I assume that y=arccosech(x) thereby cosech(y)=x

1/(sinh(y)) = 1/(1/2*(e^y)-(e^-y))
2/x=e^y - e^-y
In(2/x)=In(e^y)-In(e^-y)
In(2/x)=2y
2arccosechx=In(2/x)

this is totally wrong but I do not know how to continue or to handle problem such as this one. Can any one give me a hint and direction about solving this problem?

Best Regards
Junks
• Sep 5th 2013, 04:13 AM
FelixFelicis28
Re: problem regards hyperbolic function
Quote:

Originally Posted by junkwisch
Hi I have been having a problem regards this question. I spent some time trying to sit through it but to no result.

The Question is

Show that

arccosech x = log((1+squareroot(1+x^2)) for x>0
log((1-squareroot(1+x^2)) for x<0

Justify the sign of any square roots.

So this is what I did

I assume that y=arccosech(x) thereby cosech(y)=x

1/(sinh(y)) = 1/(1/2*(e^y)-(e^-y))

I think there's a mistake and it should be:

$\displaystyle \text{csch}^{-1} x = \begin{cases} \ln \left(1 + \sqrt{1 + x^2}\right) - \ln x \quad \quad \text{if} \ x > 0 \\ \ln \left(1 - \sqrt{1 + x^2}\right) - \ln x \quad \quad \text{if} \ x < 0 \end{cases}$

Now, consider:

\displaystyle \begin{aligned} y = \text{csch}^{-1} x \implies x = \text{csch} y = \frac{1}{\sinh y} & = \frac{2}{e^y - e^{-y}} \\ \implies x\left(e^y - \frac{1}{e^y}\right) & = 2 \end{aligned}

Let $u = e^y$ and you'll get a quadratic in $u$. Solve for $u$, substitute $u = e^y$ again and take natural logs.

Oh, and I felt the need to point this out:

Quote:

2/x=e^y - e^-y
In(2/x)=In(e^y)-In(e^-y)
In(2/x)=2y
2arccosechx=In(2/x)

this is totally wrong but I do not know how to continue or to handle problem such as this one. Can any one give me a hint and direction about solving this problem?

Best Regards
Junks
If you have $a = b - c$ and you take natural logs of both sides, you take the log of the entire side i.e. $\ln a = \ln (b - c)$ and as you know, in general $\ln (x - y) \neq \ln x - \ln y$.
• Sep 7th 2013, 08:41 PM
junkwisch
Re: problem regards hyperbolic function
thank you for your reply Felix, yes I did made a mistake. I forgot to put -In X into the equation. Just one small question though, so I subbed in U=e^y as you suggested, I did some algebra and got the equation to become u^2 - 2u/x -1 =0 I use quadratic formula for u, (-b plus of minus squareroot of (b^2 -4ac))/2a. The problem is I got U = (2/x +1) and U=-1, which is wrong. Do you know what I did wrong?
• Sep 8th 2013, 04:31 AM
FelixFelicis28
Re: problem regards hyperbolic function
Quote:

Originally Posted by junkwisch
thank you for your reply Felix, yes I did made a mistake. I forgot to put -In X into the equation. Just one small question though, so I subbed in U=e^y as you suggested, I did some algebra and got the equation to become u^2 - 2u/x -1 =0 I use quadratic formula for u, (-b plus of minus squareroot of (b^2 -4ac))/2a. The problem is I got U = (2/x +1) and U=-1, which is wrong. Do you know what I did wrong?

Hmmm, you've done something incorrectly with the quadratic formula. I personally would've re-written $u^2 - \tfrac{2}{x} u - 1 = 0$ as $x u^2 - 2u - x = 0$ by multiplying through by $x$ to make things easier but it shouldn't matter.

\displaystyle \begin{aligned} u^2 - \tfrac{2}{x} u - 1 = 0 \implies u & = \frac{\tfrac{2}{x} \pm \sqrt{\left( \tfrac{2}{x} \right)^2 + 4}}{2} \\ & = \frac{1}{x} \pm \sqrt{\frac{1}{x^2} + 1} \\ & = \frac{1 + \sqrt{1 + x^2}}{x} \end{aligned}

And then replace $u$ with $e^y$ and bob's your uncle.