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!
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