Math Help Forum - View Single Post

Beard · #5 November 4th, 2009, 08:55 AM

Quote:

Originally Posted by rain

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