# Homework Help: Transformation of expoential to hyperbolic

1. Jul 27, 2011

### mccoy1

1. The problem statement, all variables and given/known data

The book has it exp(-MgbH/KT) =(sinh(2S+1)x/2)/(sinh(x/2)) for M=2S+1, and x = gbH/(kt).

2. Relevant equations

3. The attempt at a solution

I'd have it as cosh(Mx)-sinh(Mx). How did they get the above result? Help please. Thanks.

Last edited: Jul 28, 2011
2. Jul 27, 2011

### Dick

If x=0 and y=1 then the book's formula gives exp(0)=0. 1=0 is nonsense. You don't need to ask where it comes from. Or is there a typo in there? Your formula is certainly correct.

Last edited: Jul 27, 2011
3. Jul 27, 2011

### mccoy1

Sorry I have made a mistake in writing that down. The book actually have it as exp(-MgbH/KT). M = 2S+1. The book let gbH/KT be x. And then it went ahead to say exp(-MgbH/KT) = sinh[(2S+1)x/2]/sinh(x/2). Which is sinh[Mx/2]/sinh(x/2) . Sorry for the inconvenience.

4. Jul 28, 2011

### uart

Well for $x=2$ and $M=1$ that equation yields $\exp(-2) = 1$?

Here I'm assuming k=K.

BTW. Does (sinh(2S+1)x/2) mean (x/2) sinh(2S+1). Or do you mean sinh((2S+1)x/2) ?

5. Jul 28, 2011

### mccoy1

K is a boltzmann constant. M is magnetisation. b = beta.
I meant the last one. The equation should read: exp(-MgbgH/KT) = sinh[(2S+1)x/2]/sinh(x/2) with M and x as defined above.
Cheers.

Last edited: Jul 29, 2011