Geometric progression with exponential function?

[decahal]

Homework Statement

A particle in magnetic field B with spin S has energy E = -μBm (m=-S,-S+1,...,S). Show that the partition function of a single particle is sinh(βμB(S+1/2))/sinh(βμB/2).

Relevant Equations

Sn= a(1-r^n)/(1-r)

I know that the z for one particle would be e^-βμB(-S) + e^-βμB(-S+1) +...+ e^-βμB(S)
but I have no idea how to equate this to the solution with two sinh in a quotient.

 

[pasmith]

How far can you get? Does <br /> 1 - e^{-x} = e^{-x/2}(e^{x/2} - e^{-x/2}) = 2e^{-x/2} \sinh(x/2) assist?

 

[decahal]

How far can you get? Does <br /> 1 - e^{-x} = e^{-x/2}(e^{x/2} - e^{-x/2}) = 2e^{-x/2} \sinh(x/2) assist?

I tried to do a geometric progression on the positive and negative separately, this seems like it could potentially help I'm working on it but can you please just give me the full solution.

This only helps to change the denominator on what I have but does not help at combining them much.
I don't think I am on the right track at all, a gentle nudge is just gonna send me free floating in deep space, I need the whole thing.

 

[decahal]

How far can you get? Does <br /> 1 - e^{-x} = e^{-x/2}(e^{x/2} - e^{-x/2}) = 2e^{-x/2} \sinh(x/2) assist?

This is what I have so far and I am not getting any further.

 

[WWGD]

Sorry, per pf policy we can't give you the full answer. Maybe you can ask ChatGpt and we can help you correct it if necessary, reverse -engineer it.

 

[kuruman]

You quoted the relevant equation $$S_n=a(1+r+r^2+\dots+r^n)=\frac{a(1-r^{n+1})}{1-r}.$$You have the summation $$Z=e^{-Sx}+e^{-(S-1)x}+e^{-(S-2)x}+\dots +e^{Sx}.~~~~(x\equiv \mu\beta).$$What if you rewrote this as $$Z=e^{-Sx}\left(1+e^{x}+e^{2x}+\dots +e^{2Sx}\right)~?$$

 

[renormalize]

The partition function for a single particle in a magnetic field is given by:

Your post is quite hard to read. Can you repost it using LaTeX? (There's a handy guide at the bottom left of this thread page.)

Mentor note: The post cited above was deleted for being overly helpful, in violation of forum policies.

 

[vela]

This is what I have so far and I am not getting any further.

While I recommend you try @kuruman's suggestion, you can get your attempt to work. First, you have to fix the sums. The numerators should be $1-e^{\pm \alpha(S+1)}$. After you fix that, you'll end up with
$$\frac{\sinh \frac{\alpha(S+1)}2}{\sinh \frac \alpha 2} [e^{\alpha(S-1)/2} + e^{-\alpha(S-1)/2}] - 1.$$ Rewrite the sum of exponentials in terms of $\cosh$, use a product-to-sum hyperbolic trig identity, and then simplify.