Solving Hyperbolic Functions: cothx - \frac{1}{x}

[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.

Thanks in advance guys.

 

[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?

 

[HallsofIvy]

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.

 

[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?

 

[Brewer]

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

 

[Brewer]

Solved. Helps when you remember your quotient rule