I am trying to complete a simplification for a statistical mechanics problem.
Given $s = \frac{\phi\cdot (2\sigma+s)\cdot e^{\frac{s}{w}}}{\phi\cdot e^{\frac{s}{w}} + (1-\phi)\cdot e^{\frac{s}{\sigma}}}$
In the case where $w\geq \sigma \text{ and } \phi \to 0$, show that $s \approx 2\phi\sigma + O(\phi^2), \hspace{0.5cm} (\phi << 1) $.
I am not able to understand and use the hint provided by the question: Assume that $s$ can be expanded as $[c\phi+ c'(\phi) +o(\phi^2)]$, then plug this into the expression and proceed to identify the two sides of the equation order by order in $\phi$ to arrive at the result.
I am confused on how to approach the problem as trivially from the equation of $s$, with $\phi \to 0, s\to 0$.
Any help is extremely appreciated!
$s$ appears on both sides of this equation, so $s$ depends on $\phi$. This means your "trivially from the equation" isn't quite true.
Expanding the right-hand side in powers of $s$, we obtain $$ \frac{\phi (2\sigma+s) \mathrm{e}^{\frac{s}{w}}}{\phi \mathrm{e}^{\frac{s}{w}} + (1-\phi)\mathrm{e}^{\frac{s}{\sigma}}} = 2\sigma \phi - \frac{\phi}{w}(-2w\phi + 2\sigma(\phi-1)+w)s + \cdots \text{,}$$ so the proposed solution form is consistent with the right-hand side. (Whatever this power series in $s$ represents, we are requiring it equal $s$. Since $2\sigma\phi$ is not identically zero, the variables must be related in some complicated way to make the original equality hold. This complicated relation is what thwarts your "trivially from the equation".)
So let's take the hint. Set $$ s = c + c' \phi + O(\phi^2) $$ and obtain \begin{align*} s = c + c' \phi + O(\phi^2) &= \frac{\phi (2\sigma+(c + c' \phi + O(\phi^2))) \mathrm{e}^{\frac{(c + c' \phi + O(\phi^2))}{w}}}{\phi \mathrm{e}^{\frac{(c + c' \phi + O(\phi^2))}{w}} + (1-\phi)\mathrm{e}^{\frac{(c + c' \phi + O(\phi^2))}{\sigma}}} \end{align*} Expanding the right-hand side in powers of $\phi$, we have \begin{align*} s = c + c' \phi + O(\phi^2) &= (c+2\sigma)\mathrm{e}^{\frac{c}{w} - \frac{c}{\sigma}} \phi + O(\phi^2) \text{.} \end{align*} Since there is no constant term on the right, $c = 0$, reducing to $$ s = c' \phi + O(\phi^2) = 2\sigma \phi + O(\phi^2) \text{.} $$