Lagrange Inversion of power series for fractional exponents?

Question

I understand how they obtained the inversion of sin(x) shown here, using the Lagrange Inversion Formula, and have even written a MATLAB script to solve the inversion when input and output exponents are integers.

However, I don't see how they computed the inversion of x*sin(x) (first link). The result uses fractional powers, which I assume relates to 1/GCD{input exponents}.

How do you solve for the inversion when either the input or output exponents can be fractional? Is there a worked out example somewhere I can use as reference? At first thought I could factor out the common x^m and go from there, but it doesn't seem to be correct method. I would like to eventually be able to generalize the code to handle those cases - but even understanding the theory behind it would be great in of itself.

Answer 1

To invert $y=x\sin(x):$

Take the square root of both sides of the equation and put Y = $\sqrt y \quad$ to give

$$ Y=\frac{x}{\phi(x)}\qquad \qquad (1)$$

where $$\phi(x) =\sqrt {\frac{x}{\sin(x)}}=1+\frac{x^2}{12}+\frac{x^4}{160}+\cdots.$$

Now apply the Lagrange inversion theorem to (1) to get $x$ as a series in $Y$: $$x=Y+\frac{1}{12}Y^{3}+\frac{29}{1440}Y^{5}+\cdots.$$which agrees with the result given in the Wolfram link.

Of course, if you are trying to program this you will need some script to calculate the series expansion of the square root of the function $\frac{x} {\sin(x)}.$ Alternatively, Lagrange's theorem gives the coefficients in the series reversion as

$$\lim_{x \to 0} \frac{1}{n!} \frac {d^{n-1}}{dx^{n-1}}\phi(x)$$

Perhaps this formula could be scripted with MATLAB.

Answer 2

$f(x)=x\sin\,x$ is an even function ($f(-x)=f(x)$), and since $f(0)=0$, its series expansion at the origin starts out with an $x^2$ term, viz.

$$x\sin\,x=x^2-\frac{x^4}{6}+\frac{x^6}{120}-\frac{x^8}{5040}+\cdots$$

Thus, to be able to directly apply Lagrange, the function you ought to be considering is $g(x)=\frac{\sin\sqrt{x}}{\sqrt{x}}$:

$$\frac{\sin\sqrt{x}}{\sqrt{x}}=1-\frac{x}{6}+\frac{x^2}{120}-\frac{x^3}{5040}+\cdots$$

where now you have a series for which the inversion should be easy to apply. Since $f(x)=x^2 g(x^2)$, you can transform the inverse series for $g$ into the inverse series you want...