Hyperbolic functions

Beard · #1 $Old$ November 1st, 2009, 07:21 AM

Hi,

I'm having problem with this question:

If $y = \cosh(x)$ for x > 0, find x in terms of y and hence show that $\cosh^{-1}x = \ln (x + \sqrt{x^2-1})$ . State the domain and range of the function $\cosh^{-1}x$ .

Currently I have x in terms of y as $x = \frac{2}{e^y + e^{-y}}$ I am unsure if this is as far as it can go. This part: $\cosh^{-1}x = \ln (x + \sqrt{x^2-1})$ is where I am more confused.

Thankyou for any help. It is greatly appreciated

rain · #2 $Old$ November 1st, 2009, 09:15 AM

Hello there,

I did this pretty quickly but I think this will hopefully help you,

Given
$y=cosh(x)$ where $x\geq0$ [1]
then
$y=(e^{x}-e^{-x})/2$
so
$e^{x}-e^{-x}=2y$
then multiplying all by exp(x) gives
$e^{x}=\frac{2y\pm\sqrt{4y^2-4}}{2}$ (quadratic formula)
thus
$x = ln(y\pm\sqrt{y^2-1})$
and since y = cosh(x) = cosh( ln( y .... ) ) then
$cosh^{-1}(y) = ln(y\pm\sqrt{y^2 -1} )$

Now, ln(a) is a real valued function where a must be positive and real, so y +/- sqrt(y^2 - 1) must be positive and real; thus y^2 - 1 >= 0 => y >= 1 (which is true since by [1] that x >= 0 and so cosh(x) >= 1), now for y +/- sqrt(y^2 - 1) to be positive then we must take y + sqrt(y^2 - 1)

So
$cosh^{-1}(y) = ln(y+\sqrt{y^2 -1} )$
which (for letting y be x) is what was wanted.

The domain for arccosh(y) would then be such that y + sqrt(y^2 - 1) >= ie y > 1 (since ln(0) is also undefined) and the range would then be all positive real numbers (from 0 to +infinity) which is the range of ln for domain greater than 1.

Beard · #3 $Old$ November 1st, 2009, 09:43 AM

Quote:

Originally Posted by rain $View Post$

Hello there,

I did this pretty quickly but I think this will hopefully help you,

Unfortunately I am not brilliant at maths, hence why I come here for help. I prefer chemistry

rain · #4 $Old$ November 1st, 2009, 09:47 AM

Quote:

Originally Posted by Beard $View Post$

Unfortunately I am not brilliant at maths, hence why I come here for help. I prefer chemistry

I'd love to pick up on chemistry at some point - you're very lucky

Beard · #5 $Old$ November 4th, 2009, 09:55 AM

Quote:

Originally Posted by rain $View Post$

Hello there,

then multiplying all by exp(x) gives
$e^{x}=\frac{2y\pm\sqrt{4y^2-4}}{2}$ (quadratic formula)
thus
$x = ln(y\pm\sqrt{y^2-1})$
and since y = cosh(x) = cosh( ln( y .... ) ) then
$cosh^{-1}(y) = ln(y\pm\sqrt{y^2 -1} )$

Now, ln(a) is a real valued function where a must be positive and real, so y +/- sqrt(y^2 - 1) must be positive and real; thus y^2 - 1 >= 0 => y >= 1 (which is true since by [1] that x >= 0 and so cosh(x) >= 1), now for y +/- sqrt(y^2 - 1) to be positive then we must take y + sqrt(y^2 - 1)

So
$cosh^{-1}(y) = ln(y\pm\sqrt{y^2 -1} )$
which (for letting y be x) is what was wanted.

The domain for arccosh(y) would then be such that y + sqrt(y^2 - 1) >= ie y > 1 (since ln(0) is also undefined) and the range would then be all positive real numbers (from 0 to +infinity) which is the range of ln for domain greater than 1.

Sorry but I don't understand how the quadratic was formed when you multiply by exp(x) as you say

rain · #6 $Old$ November 4th, 2009, 02:16 PM

My fault, I should have made this clearer,

Given that you can get down to this equation,
$e^{x} -e^{-x} = 2y$
Then multiplying all the terms (both on the left hand side and the right hand side) by $e^{x}$ gives
$e^{x}(e^{x} -e^{-x}) = e^{x}2y$
Then, by expanding the bracket out, you get
$e^{x}e^{x} -e^{-x}e^{x} = e^{x}2y$
Now, by standard rules of indices, $a^ba^c=a^{b+c}$ and $a^0 = 1$ we get $e^{-x}e^{x} = e^{-x+x} = e^{0} = 1$
Then $e^{x}e^{x} -1 = e^{x}2y$ which can be written as
$(e^{x})^2 -e^{x}2y -1 = 0$ [1] by collecting the $e^{x}$ terms and bringing the 2y term to the left hand side
Now, say you let W be $e^{x}$ , just to show this is like a quadratic, you can rewrite [1] as $W^2 -W2y -1 = 0$ which is in the general formula for solving a quadratic equation, with roots at $W=\frac{(2y)\pm\sqrt{(2y)^2-4(-1)}}{2}=\frac{2y\pm\sqrt{4y^2+4}}{2}=\frac{2y\pm\sqrt{4(y^2+1)}}{2}$

$=\frac{2y\pm2\sqrt{y^2+1}}{2}$ (since $\sqrt{4} = 2$ )

$=y\pm\sqrt{y^2+1}$ (by dividing by 2)

thus $W=e^{x}=y\pm\sqrt{y^2+1}$

and taking the log of both sides gives

$x = log(y\pm\sqrt{y^2+1})$ and the rest carries on above

*The formula for solving a quadratic equation in general is as follows in case you can't remember,
for $ax^2 +bx +c = 0$
then $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Beard · #7 $Old$ November 5th, 2009, 06:03 AM

Quote:

Originally Posted by rain $View Post$

My fault, I should have made this clearer,

Given that you can get down to this equation,
$e^{x} -e^{-x} = 2y$
Then multiplying all the terms (both on the left hand side and the right hand side) by $e^{x}$ gives
$e^{x}(e^{x} -e^{-x}) = e^{x}2y$
Then, by expanding the bracket out, you get
$e^{x}e^{x} -e^{-x}e^{x} = e^{x}2y$
Now, by standard rules of indices, $a^ba^c=a^{b+c}$ and $a^0 = 1$ we get $e^{-x}e^{x} = e^{-x+x} = e^{0} = 1$
Then $e^{x}e^{x} -1 = e^{x}2y$ which can be written as
$(e^{x})^2 -e^{x}2y -1 = 0$ [1] by collecting the $e^{x}$ terms and bringing the 2y term to the left hand side
Now, say you let W be $e^{x}$ , just to show this is like a quadratic, you can rewrite [1] as $W^2 -W2y -1 = 0$ which is in the general formula for solving a quadratic equation, with roots at $W=\frac{(2y)\pm\sqrt{(2y)^2-4(-1)}}{2}=\frac{2y\pm\sqrt{4y^2+4}}{2}=\frac{2y\pm\sqrt{4(y^2+1)}}{2}$

$=\frac{2y\pm2\sqrt{y^2+1}}{2}$ (since $\sqrt{4} = 2$ )

$=y\pm\sqrt{y^2+1}$ (by dividing by 2)

thus $W=e^{x}=y\pm\sqrt{y^2+1}$

and taking the log of both sides gives

$x = log(y\pm\sqrt{y^2+1})$ and the rest carries on above

*The formula for solving a quadratic equation in general is as follows in case you can't remember,
for $ax^2 +bx +c = 0$
then $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

That made it a load clearer thanks again