How can I invert the asymptotic form $x^{3/2}=y^{3/2}(1+a/y^2 + ... )$ to find $y=y(x)$?

Question

This might sound silly, but the fact there's a $a/y^2$ term in the expansion made me feel a little lost. Could anyone help?

Thanks

Answer 1

Write $y = x + u$ where $u = o(x)$ as $x \to \infty$. Then $y^{3/2} \sim x^{3/2} + \dfrac{3}{2} x^{1/2} u + \ldots$ and $1 + a/y^2 \sim 1 + a x^{-2} + \ldots$, making the right side $x^{3/2} + a x^{-1/2} + \dfrac{3}{2} x^{1/2} u + \ldots$. For this to work, we want $u \sim -\dfrac{2a}{3x}$. Thus $$y = x - \dfrac{2a}{3x} + \ldots$$ (which is as far as you can go without the next term).

Answer 2

One way could be to consider that we look for the zero of function $$f(y)=y^{3/2}(1+\frac a {y^2})-x^{3/2}$$ and use Newton method stating at $y_0=x$.$$f'(y)=\frac{3 y^2-a}{2 y^{3/2}}$$ So, the first iterate of Newton method is $$y_1=y_0-\frac{f(y_0)}{f'(y_0)}$$ which write $$y_1=x+\frac{2 a x}{a-3 x^2}$$ Considering that this is the solution of $f(y)=0$, use now Taylor series for the development of the rhs (for infinitely large values of $x$); this gives $$y\approx y_1=x-\frac{2 a}{3 x}-\frac{2 a^2}{9 x^3}+O\left(\frac{1}{x^4}\right)$$ Similarly, considering $$f(y)=y^{3/2}(1+\frac a {y^2}+\frac b {y^3})-x^{3/2}$$ using the same approach, we would have obtained $$y\approx y_1=x-\frac{2 a}{3 x}-\frac{2 b}{3 x^2}-\frac{2 a^2}{9 x^3}+O\left(\frac{1}{x^4}\right)$$ Similarly, considering $$f(y)=y^{3/2}(1+\frac a {y^2}+\frac b {y^3}+\frac c {y^4})-x^{3/2}$$ using the same approach, we would have obtained $$y\approx y_1=x-\frac{2 a}{3 x}-\frac{2 b}{3 x^2}-\frac{2 \left(a^2+3 c\right)}{9 x^3}+O\left(\frac{1}{x^4}\right)$$ If instead of Newton method, Halley method would be used (hoping it gives a better estimate of the solution of $f(y)=0$), we would have obtained $$y\approx y_1=x-\frac{2 a}{3 x}-\frac{2 b}{3 x^2}-\frac{ a^2+2 c}{3 x^3}+O\left(\frac{1}{x^4}\right)$$