Iterative solution for algebraic equation with small parameter

Question

I am a bit confused. I don't understand what iterative solution is. In my solid state book it is written that from equation: $$\frac{2}{3}\mu^{3/2}+\frac{\pi^{2}}{12}\frac{\left(k_{B}T\right)^{2}}{\sqrt{\mu}}=\frac{2}{3}\varepsilon_{F}^{3/2}$$ having in mind that $k_{B}T/\varepsilon_F << 1$ by iterative methods you can show that: $$ \mu = \varepsilon_F\left[1 - \frac{\pi^{2}}{12}\frac{\left(k_{B}T\right)^{2}}{\varepsilon_F^2}\right]. $$ Can anyone explain how this iterative solution should be done here and in general case.

Answer 1

For more simplicity let us write the equation as $$x^{3/2}+\frac{a}{\sqrt{x}}=b^{3/2}$$ So, you want you find the zero of function $$f(x)=x^{3/2}+\frac{a}{\sqrt{x}}-b^{3/2}$$ $$f'(x)=\frac{3 \sqrt{x}}{2}-\frac{a}{2 x^{3/2}}$$ Let us apply Newton method with $x_0=b$; this will give for the fist iterate $$x_1=b\left(1+\frac{2 a }{a-3 b^2}\right)$$ Now, if $a$ is small, by Taylor, we have $$\frac{2 a }{a-3 b^2}=-\frac{2 a}{3 b^2}-\frac{2 a^2}{9 b^4}+O\left(a^3\right)$$ making $$x_1 \approx b\left(1-\frac{2 a}{3 b^2}\right)$$

Answer 2

For me, it's more like doing a power series expansion. We see that the solution can be written as $\mu = \varepsilon_F f(x)$ where $x = \Big(\dfrac{k_B T}{\varepsilon_F}\Big)^2$ and $f^{3/2}(x) + \dfrac{\pi^2}{8} x f^{-1/2}(x) = 1$, or $f^2(x) + \dfrac{\pi^2}{8}x = f^{1/2}(x)$.

Thus for $x \ll 1$ we have $f(x) = f(0) + f'(0)x + \ldots$ where $f(0) = 1$ and $$2 f(x)f'(x) + \frac{\pi^2}{8} = \frac{1}{2}f^{-1/2}(x)f'(x) \quad\Rightarrow\quad f'(0) = -\frac{\pi^2}{12},$$ arriving at your last equation.