Transformation of expoential to hyperbolic

[mccoy1]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[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.

 

[mccoy1]

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?

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.

 

[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) ?

 

[mccoy1]

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) ?

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.