I am working with solution to near regular singular points.
I started with:
$$y_1(x)=x^\frac{1}{2}\left[1-\frac{3}{4}x+\frac{9}{64}x^2-\frac{3}{256}x^3+\cdots\right] $$
Then I squared it:
$$y_1^2(x) = x\left(1-\frac{3}{2}x+\frac{27}{32}x^2-\frac{15}{64}x^3+\cdots\right)$$
Why is the inverse:
$$\frac{1}{x}\left[1+\frac{3}{2}x+\frac{45}{32}x^2+\frac{69}{64}x^3+\cdots\right] \text{ ?}$$
I cannot seem to see how this works out. Any pointers
If $$y=\sqrt{x} \left(1-\frac{3 x}{4}+\frac{9 x^2}{64}-\frac{3 x^3}{256}+\cdots\right)$$ then effectively $$y^2=x \left(1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{64} +\cdots\right)$$ Now you want to compute $\frac{1}{y^2 }$. You can write $$\frac{x}{y^2 }=\frac{1}{1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{64} +\cdots}$$ and perform the long division. Another way is to consider the rhs as $\frac{1}{1-z}$ and expand it as $$\frac{1}{1-z}=1+z+z^2+z^3+\cdots$$ and reuse $$z=\frac{3 x}{2}-\frac{27 x^2}{32}+\frac{15 x^3}{64}+\cdots$$ to affectively arrive, after substitution and simplifications, to $$\frac{x}{y^2 }=1+\frac{3 x}{2}+\frac{45 x^2}{32}+\frac{69 x^3}{64}+\cdots$$ and finally $$\frac{1}{y^2 }=\frac{1}{x}\Big(1+\frac{3}{2}x+\frac{45}{32}x^2+\frac{69}{64}x^3+\cdots\Big)$$
From a semantic point of view, I shall not call that operation the inverse of a power series (which is something different).