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The energy is given by: $$ E(\sigma) = -\sum_{i=1}^{N}\sigma_i + a \sum_{i=1}^{N}\sigma_i^2 $$ where $a$ is constant.

I have to calculate the following: $$\left\langle \frac{1}{N}\sum_{i=1}^{N}\sigma_i \right\rangle$$

I did some calculations and I arrived at this: $$\frac{\sum_{\sigma_1}\sigma_1 \exp(\beta\sigma_1-aB\sigma_1^2)}{\sum_{\sigma_1} \exp(\beta\sigma_1-aB\sigma_1^2)}$$

I don't know how work out his anymore. In the solutions the next step was: $$\frac{e^{\beta-a\beta}-e^{-\beta-a\beta}}{1+e^{\beta-a\beta}-e^{-\beta-a\beta}}$$ Someone who could explain me this step? :) Thanks!

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closed as off-topic by Aaron Stevens, Feynmans Out for Grumpy Cat, John Rennie, M. Enns, Kyle Kanos May 21 at 11:34

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I don't think you've included all the necessary information in your question, so I'm going to make an educated guess: The variables $\sigma_i$ take on values $0,1,-1$. In this case, you have $\sum_{\sigma_1}$ is the sum of three terms, so just write them out!

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