I've started going through Frank Bowman's book 'Introduction to Bessel Functions' and while trying to follow along there's a specific point that I'm not quite how how it's reached.
Image of the page I'm confused about
For point 5 in the middle of the page, it mentions taking $\frac{1}{xJ_o^2(x)}$ where $J_o(x)$ is the zeroth order bessel function of the first kind, and replacing $J_o(x)$ with it's power series.
This then results in the following first few terms:
$\frac{1}{xJ_o^2(x)} = \frac{1}{x}$ + $\frac{x}{2}$ + $\frac{5x^3}{32}$ + ...
I'm confused as to how this result is reached, because in my attempts to expand this using the power series I am getting:
$\frac{1}{xJ_o^2(x)} = \frac{1}{x} - \frac{2}{x^3} + \frac{32}{3x^5}$ + ...
I got this by essentially 'brute forcing' it and just expanding the equation using only the first three terms of the function, like so:
$J_o(x) = 1 - \frac{x^2}{4} + \frac{x^4}{64}$
I'm not sure of a different way to go about it to get the result from the book. Is there something I'm missing?
Start, as you did with $$J_0(x)=1-\frac{x^2}{4}+\frac{x^4}{64}+O\left(x^6\right)$$ Square it $$J_0(x){}^2=1-\frac{x^2}{2}+\frac{3 x^4}{32}+O\left(x^6\right)$$ $$x\,J_0(x){}^2=x-\frac{x^3}{2}+\frac{3 x^5}{32}+O\left(x^7\right)$$ Long division $$\frac 1 {x\,J_0(x){}^2}=\frac{1}{x}+\frac{x}{2}+\frac{5 x^3}{32}+O\left(x^5\right)$$