Solving for a variable in an equation with fractional powers

[archaic]

Homework Statement

$$\epsilon^\frac32=\mu^\frac32+\frac{\left(\pi k_B T\right)^2}{8}\frac1{\mu^\frac12}$$
I am given that $\epsilon\approx\mu$ and $(1+x)^\alpha\approx1+\alpha x$ when $x\to0$.

Relevant Equations

$(1+x)^\alpha\approx1+\alpha x$ when $x\to0$

I have tried manipulating this to
$$1-\frac{8}{(\pi k_B T)^2}\mu^\frac12(\epsilon^\frac32-\mu^\frac32)=0\Leftrightarrow\left[1+\mu^\frac12(\epsilon^\frac32-\mu^\frac32)\right]^{-\frac{8}{(\pi k_B T)^2}}=0$$
but this doesn't seem to lead anywhere.
any hints please?

the solution is one of these $T_f=\epsilon/k_B$:

 

[fresh_42]

If you set $\mu^{\frac{1}{2}}=\nu$ for convenience, and multiply the equation with $\nu$ then you have a polynomial equation which may be solvable. The exponent in your formula is wrong.

 

[archaic]

If you set $\mu^{\frac{1}{2}}=\nu$ for convenience, and multiply the equation with $\nu$ then you have a polynomial equation which may be solvable. The exponent in your formula is wrong.

I don't think so. I tried that with wolfram, but the result was very bad.
how so?

 

[fresh_42]

I don't think so. I tried that with wolfram, but the result was very bad.
how so?

It's always best to make simplifications at the very last. So multiplying with $\mu^{\frac{1}{2}}$ results in
$$
0=\mu^{\frac{1}{2}}\cdot \left(\mu^{\frac{3}{2}}-\varepsilon^{\frac{3}{2}}\right) +\dfrac{\left(\pi k_B T\right)^2}{8}=\mu^{\frac{1}{2}}\cdot\left(\mu+\mu^{\frac{1}{2}}\varepsilon^{\frac{1}{2}}+\varepsilon \right)\cdot\left(\mu^{\frac{1}{2}}-\varepsilon^{\frac{1}{2}}\right)+\dfrac{\left(\pi k_B T\right)^2}{8}
$$
What if you now start to apply the hints? But I don't see how we can deal with $\mu^{\frac{1}{2}}-\varepsilon^{\frac{1}{2}}$.

 

[etotheipi]

Here's a quick-ish and dirty way, just making use of $\epsilon \approx \mu$. Since $\epsilon^\frac32 = \mu^\frac32+\frac{\left(\pi k_B T\right)^2}{8\mu^\frac12}$ you have $\epsilon^3 = \left( \mu^\frac32+\frac{\left(\pi k_B T\right)^2}{8\mu^\frac12} \right)^2 \approx \mu^3 \left( 1 + \frac{\left(\pi k_B T\right)^2}{4\mu^2} \right)$. This is a cubic $\mu^3 + \frac{\left(\pi k_B T\right)^2}{4} \mu - \epsilon^3 \approx 0$. But $\mu^3 - \epsilon^3 = (\mu - \epsilon)(\mu^2 + \mu \epsilon + \epsilon^2) \approx 3\mu^2 (\mu - \epsilon)$ and so $\mu^2 - \epsilon \mu + \frac{\left(\pi k_B T\right)^2}{12} \approx 0$ and \begin{align*}

\mu_{+} \approx \frac{\epsilon + \sqrt{\epsilon^2 - \frac{\left(\pi k_B T\right)^2}{3}} }{2} \approx \epsilon\left( 1 -\frac{\left(\pi k_B T\right)^2}{12 \epsilon^2} \right) = \epsilon\left( 1 - \frac{\pi^2}{12} \left( \frac{T}{T_f} \right)^2 \right)

\end{align*}

 

[archaic]

thank you! never mind my other message.

 

[haruspex]

I rewrote the equation as $x^3=y^3+c/y$ and substituted $y=x(1-\alpha)$. Discarding $\alpha^2$ etc. this gives $y=x(1-\frac c{3x^4-c})$. This does not seem to match any of the choices.

 

[archaic]

I rewrote the equation as $x^3=y^3+c/y$ and substituted $y=x(1-\alpha)$. Discarding $\alpha^2$ etc. this gives $y=x(1-\frac c{3x^4-c})$. This does not seem to match any of the choices.

a more formal way to find the result is to write$$\epsilon^\frac32\mu^\frac12-\mu^2-\frac{\left(\pi k_B T\right)^2}{8}=0$$$$f(x)=\epsilon^\frac32x^\frac12-x^2-\frac{\left(\pi k_B T\right)^2}{8}$$and compute the taylor expansion about $x=\epsilon$. since $\epsilon\approx\mu$, we ignore higher order terms.$$-\frac{\left(\pi k_B T\right)^2}{8}-\frac{3\epsilon}{2}\left(\mu-\epsilon\right)=0$$