# Hyperbolic functions

1. Mar 12, 2008

### Brewer

[SOLVED] Hyperbolic functions

As part of a long winded "show that" question I've ended up at the point where I have $$xcothx$$ and I want to show that this is equal to $$cothx - \frac{1}{x}$$ only I have no ideas how to get there. I can't see any reason why this should be so, but I'm pretty confident that I'm correct so far (in fact I know I am!).

Obviously I could take the "magic step" when doing this kind of question (in that I could just write the final answer down, and hope that I'm close enough for this step to be intuitive) but I'd quite like to know the step to take.

2. Mar 12, 2008

### tiny-tim

erm … when x = 1, xcothx = coth1, but cothx - 1/x = coth1 - 1 … so they're not equal.

How did you get to that position?

3. Mar 12, 2008

### HallsofIvy

Staff Emeritus
Well, you aren't going to be able to prove that because it pretty obviously isn't true. In particular, you would be proving, taking x= 1, that coth(1)= coth(1)- 1 which can't be true.

4. Mar 12, 2008

### Brewer

Meh, so the guy I checked my working with up until now is wrong.

So what I did is
$$U=-\frac{d(lnZ)}{d\beta}$$
$$=-\frac{1}{z}\frac{dZ}{d\beta}$$
$$=\frac{-\beta \mu B}{sinh(\beta \mu B)}(\mu Bcosh(\beta \mu B)$$
$$=-\frac{\mu ^2 B^2 \beta cosh(\beta \mu B)}{sinh(\beta \mu B)$$
$$=-\mu B(\beta \mu Bcoth(\beta \mu B)$$

where $$Z = \frac{sinh(\beta \mu B)}{\beta \mu B}$$

Are there problems with what I've done then?

Last edited: Mar 12, 2008
5. Mar 12, 2008

### Brewer

I've tried to correct the latex, but if it hasnt come out properly then the line thats gone wrong is the unfactorised form of the final line

6. Mar 12, 2008

### Brewer

Solved. Helps when you remember your quotient rule