Defining the Arcsinh Function

Date: 08/09/2003 at 00:07:08
From: Simon
Subject: Solving hyperbolic equations

I just got an assignment on hyperbolic functions, which I've never 
seen before. The first part was proving several identities, similar to 
the trig identities, so i'm semi-familiar with cosh and sinh now.

The second part is to solve an exact value for x in: 4 sinh x = 3

I can work out that 

   e^x - e^-x = 3/2 and
 e^2x + e^-2x = 15/4

but with every identity I try to substitute, etc., I always end up
with one of these two equations.

Thanks.

Date: 08/09/2003 at 22:24:29
From: Doctor Warren
Subject: Re: Solving hyperbolic equations

Hi Simon,

The easiest way to deal with this equation is to use the definition of
the arcsinh function:

   arcsinh(z) = ln(z + sqrt(z^2 + 1))

Your equation's solution is, of course, x = arcsinh(3/4).

If you don't have the definition of the arcsinh function handy, you
can derive it with only a couple of identities:

   Identity 1) sinh(z) + cosh(z) = e^z
   Identity 2) cosh^2(z) - sinh^2(z) = 1

Start with

   x = sinh(y).                          (1)

Substitute x into Identity 1:

   x + cosh(y) = e^y                     (2)

Rearrange Identity 2 in terms of cosh(y):

   cosh(y) = sqrt( 1 + sinh^2(y) )       (3)

Substitute (3) into (2):

   x + sqrt( 1 + sinh^2(y) ) = e^y       (4)

Substitute (1) into (4):

   x + sqrt(x^2 + 1) = e^y               (5)

Take the natural logarithm of both sides:

           y = ln(x + sqrt(x^2 + 1))
  arcsinh(x) = ln(x + sqrt(x^2 + 1))     (done)

You can take it from here.  Let me know if you have any more 
questions.

- Doctor Warren, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 08/12/2003 at 06:19:34
From: Simon
Subject: Thank you (Solving hyperbolic equations)

Thanks heaps, you've made my life a whole lot easier!