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

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:

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

Now, consider:

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

Let $\displaystyle u = e^y$ and you'll get a quadratic in $\displaystyle u$. Solve for $\displaystyle u$, substitute $\displaystyle u = e^y$ 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 $\displaystyle a = b - c$ and you take natural logs of both sides, you take the log of the *entire* side i.e. $\displaystyle \ln a = \ln (b - c)$ and as you know, in general $\displaystyle \ln (x - y) \neq \ln x - \ln y$.

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?

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 $\displaystyle u^2 - \tfrac{2}{x} u - 1 = 0$ as $\displaystyle x u^2 - 2u - x = 0$ by multiplying through by $\displaystyle x$ to make things easier but it shouldn't matter.

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

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